12.3 BEHAVIOUR OF GASES where M is the mass of the gas containing N Properties of gases are easier to understand than molecules, M0 is the molar mass and NA the those of solids and liquids. This is mainly Avogadro’s number. Using Eqs. (12.4) and (12.3) because in a gas, molecules are far from each can also be written as other and their mutual interactions are PV = kB NT or P = kB nT negligible except when two molecules collide. Gases at low pressures and high temperatures much above that at which they liquefy (or solidify) approximately satisfy a simple relation ) between their pressure, temperature and volume –1 given by (see Chapter 10) K –1 PV = KT (12.1) mol Jfor a given sample of the gas. Here T is the ( temperature in kelvin or (absolute) scale. K is a T pV µ constant for the given sample but varies with the volume of the gas. If we now bring in the idea of atoms or molecules, then K is proportional to the number of molecules, (say) N in the sample. We can write K = N k . Observation tells P (atm) us that this k is same for all gases. It is called Fig.12.1 Real gases approach ideal gas behaviour at Boltzmann constant and is denoted by k B. low pressures and high temperatures. P1V1 P2 V2 where n is the number density, i.e. number ofAs = = constant = kB (12.2) N 1 T1 N 2 T2 molecules per unit volume. kB is the Boltzmann constant introduced above. Its value in SI unitsif P, V and T are same, then N is also same for all is 1.38 × 10–23 J K–1.gases. This is Avogadro’s hypothesis, that the Another useful form of Eq. (12.3) isnumber of molecules per unit volume is ρRT the same for all gases at a fixed temperature and P = (12.5) pressure. The number in 22.4 litres of any gas M 0 Reprint 2025-26 KINETIC THEORY 247 where ρ is the mass density of the gas. etc. in a vessel of volume V at temperature T and A gas that satisfies Eq. (12.3) exactly at all pressure P. It is then found that the equation of pressures and temperatures is defined to be an state of the mixture is : ideal gas. An ideal gas is a simple theoretical PV = ( µ1 + µ2 +… ) RT (12.7)model of a gas. No real gas is truly ideal. Fig. 12.1 shows departures from ideal gas RT RT i.e. P = µ1 + µ2 + ... (12.8)behaviour for a real gas at three different V V temperatures. Notice that all curves approach = P1 + P2 + … (12.9)the ideal gas behaviour for low pressures and high temperatures. Clearly P1 = µ1 R T/V is the pressure that At low pressures or high temperatures the gas 1 would exert at the same conditions of molecules are far apart and molecular volume and temperature if no other gases were interactions are negligible. Without interactions present. This is called the partial pressure of the the gas behaves like an ideal one. gas. Thus, the total pressure of a mixture of ideal If we fix µ and T in Eq. (12.3), we get gases is the sum of partial pressures. This is Dalton’s law of partial pressures. PV = constant (12.6) i.e., keeping temperature constant, pressure of a given mass of gas varies inversely with volume. This is the famous Boyle’s law. Fig. 12.2 shows comparison between experimental P-V curves and the theoretical curves predicted by Boyle’s law. Once again you see that the agreement is good at high temperatures and low pressures. Next, if you fix P, Eq. (12.1) shows that V ∝ T i.e., for a fixed pressure, the volume of a gas is proportional to its absolute temperature T (Charles’ law). See Fig. 12.3. Fig. 12.3 Experimental T-V curves (solid lines) for CO2 at three pressures compared with Charles’ law (dotted lines). T is in units of 300 K and V in units of 0.13 litres. We next consider some examples which give us information about the volume occupied by the molecules and the volume of a single molecule. ⊳ Example 12.1 The density of water is 1000 kg m–3. The density of water vapour at 100 °C and 1 atm pressure is 0.6 kg m–3. The volume of a molecule multiplied by the total Fig.12.2 Experimental P-V curves (solid lines) for number gives ,what is called, molecular steam at three temperatures compared with volume. Estimate the ratio (or fraction) of Boyle’s law (dotted lines). P is in units of 22 the molecular volume to the total volume atm and V in units of 0.09 litres. occupied by the water vapour under the Finally, consider a mixture of non-interacting above conditions of temperature and pressure.ideal gases: µ moles of gas 1, µ moles of gas 2, 1 2 Reprint 2025-26 248 PHYSICS Answer For a given mass of water molecules, number of molecules and (ii) mass density the density is less if volume is large. So the of neon and oxygen in the vessel. Atomic volume of the vapour is 1000/0.6 = 1/(6 × 10 -4 ) mass of Ne = 20.2 u, molecular mass of O2times larger. If densities of bulk water and water = 32.0 u. molecules are same, then the fraction of molecular volume to the total volume in liquid Answer Partial pressure of a gas in a mixture is state is 1. As volume in vapour state has the pressure it would have for the same volume increased, the fractional volume is less by the and temperature if it alone occupied the vessel. same amount, i.e. 6×10-4. ⊳ (The total pressure of a mixture of non-reactive ⊳ gases is the sum of partial pressures due to its Example 12.2 Estimate the volume of a constituent gases.) Each gas (assumed ideal) water molecule using the data in Example obeys the gas law. Since V and T are common to 12.1. the two gases, we have P1V = µ 1 RT and P2V = Answer In the liquid (or solid) phase, the µ2 RT, i.e. (P1/P2) = (µ1 / µ2). Here 1 and 2 refer to neon and oxygen respectively. Since (P1/P2) =molecules of water are quite closely packed. The (3/2) (given), (µ1/ µ2) = 3/2.density of water molecule may therefore, be (i) By definition µ1 = (N1/NA ) and µ2 = (N2/NA)regarded as roughly equal to the density of bulk where N1 and N2 are the number of moleculeswater = 1000 kg m–3. To estimate the volume of of 1 and 2, and NA is the Avogadro’s number.a water molecule, we need to know the mass of Therefore, (N1/N2) = (µ1 / µ2) = 3/2.a single water molecule. We know that 1 mole (ii) We can also write µ1 = (m1/M1) and µ2 =of water has a mass approximately equal to (2 + 16)g = 18 g = 0.018 kg. (m2/M2) where m1 and m2 are the masses of 1 and 2; and M1 and M2 are their molecular Since 1 mole contains about 6 × 1023 masses. (Both m1 and M1; as well as m2 andmolecules (Avogadro’s number), the mass of a molecule of water is (0.018)/(6 × 1023) kg = M2 should be expressed in the same units). If ρ1 and ρ2 are the mass densities of 1 and3 × 10–26 kg. Therefore, a rough estimate of the 2 respectively, we havevolume of a water molecule is as follows : Volume of a water molecule ρ1 m 1 / V m 1 µ1  M 1  = (3 × 10–26 kg)/ (1000 kg m–3) = = = × ρ2 m 2 / V m 2 µ2  M 2  = 3 × 10–29 m3 = (4/3) π (Radius)3 3 20.2 Hence, Radius ≈ 2 ×10-10 m = 2 Å ⊳ = × = 0.947 2 32.0 ⊳ ⊳ Example 12.3 What is the average distance between atoms (interatomic distance) in water? Use the data given in 12.4 KINETIC THEORY OF AN IDEAL GAS Examples 12.1 and 12.2. Kinetic theory of gases is based on the molecular picture of matter. A given amount of gas is a Answer : A given mass of water in vapour state collection of a large number of molecules has 1.67×103 times the volume of the same mass (typically of the order of Avogadro’s number) that of water in liquid state (Ex. 12.1). This is also are in incessant random motion. At ordinary the increase in the amount of volume available pressure and temperature, the average distance for each molecule of water. When volume between molecules is a factor of 10 or more than increases by 103 times the radius increases by the typical size of a molecule (2 Å). Thus, V1/3 or 10 times, i.e., 10 × 2 Å = 20 Å. So the interaction between molecules is negligible and average distance is 2 × 20 = 40 Å. ⊳ we can assume that they move freely in straight lines according to Newton’s first law. However,⊳ Example 12.4 A vessel contains two non- occasionally, they come close to each other, reactive gases : neon (monatomic) and experience intermolecular forces and their oxygen (diatomic). The ratio of their partial velocities change. These interactions are called pressures is 3:2. Estimate the ratio of (i) collisions. The molecules collide incessantly against each other or with the walls and change Reprint 2025-26 KINETIC THEORY 249 their velocities. The collisions are considered to the wall. Thus, the number of molecules with be elastic. We can derive an expression for the velocity (vx, vy, vz ) hitting the wall in time ∆t is pressure of a gas based on the kinetic theory. ½A vx ∆t n, where n is the number of molecules We begin with the idea that molecules of a per unit volume. The total momentum gas are in incessant random motion, colliding transferred to the wall by these molecules in against one another and with the walls of the time ∆t is: container. All collisions between molecules Q = (2mvx) (½ n A vx ∆t ) (12.10) among themselves or between molecules and the The force on the wall is the rate of momentum walls are elastic. This implies that total kinetic transfer Q/∆t and pressure is force per unit energy is conserved. The total momentum is area : conserved as usual. P = Q /(A ∆t) = n m vx 2 (12.11) Actually, all molecules in a gas do not have 12.4.1 Pressure of an Ideal Gas the same velocity; there is a distribution in velocities. The above equation, therefore, standsConsider a gas enclosed in a cube of side l. Take for pressure due to the group of molecules withthe axes to be parallel to the sides of the cube, speed vx in the x-direction and n stands for theas shown in Fig. 12.4. A molecule with velocity number density of that group of molecules. The (vx, vy, vz ) hits the planar wall parallel to yz- total pressure is obtained by summing over theplane of area A (= l2). Since the collision is elastic, contribution due to all groups:the molecule rebounds with the same velocity; its y and z components of velocity do not change P = n m v x2 (12.12) in the collision but the x-component reverses where v 2x is the average of vx 2 . Now the gas sign. That is, the velocity after collision is is isotropic, i.e. there is no preferred direction (-vx, vy, vz ) . The change in momentum of the of velocity of the molecules in the vessel. molecule is: –mvx – (mvx) = – 2mvx . By the Therefore, by symmetry, principle of conservation of momentum, the momentum imparted to the wall in the collision v 2x = v y2 = v z2 = 2mvx . 2 2 2 2 = (1/3) [ v x + v y + v z ] = (1/3) v (12.13) where v is the speed and v 2 denotes the mean of the squared speed. Thus P = (1/3) n m v 2 (12.14) Some remarks on this derivation. First, though we choose the container to be a cube, the shape of the vessel really is immaterial. For a vessel of arbitrary shape, we can always choose a small infinitesimal (planar) area and carry through the steps above. Notice that both A and ∆t do not appear in the final result. By Pascal’s law, given in Ch. 9, pressure in one portion of Fig. 12.4 Elastic collision of a gas molecule with the the gas in equilibrium is the same as anywhere wall of the container. else. Second, we have ignored any collisions in To calculate the force (and pressure) on the the derivation. Though this assumption is wall, we need to calculate momentum imparted difficult to justify rigorously, we can qualitatively to the wall per unit time. In a small time interval see that it will not lead to erroneous results. The ∆t, a molecule with x-component of velocity vx number of molecules hitting the wall in time ∆t will hit the wall if it is within the distance vx ∆t was found to be ½ n Avx ∆t. Now the collisions from the wall. That is, all molecules within the are random and the gas is in a steady state. volume Avx ∆t only can hit the wall in time ∆t. Thus, if a molecule with velocity (vx, vy, vz ) But, on the average, half of these are moving acquires a different velocity due to collision with towards the wall and the other half away from some molecule, there will always be some other Reprint 2025-26 250 PHYSICS molecule with a different initial velocity which P = (1/3) [n1m1 v1 2 + n2 m2 v 22 +… ] (12.20) after a collision acquires the velocity (vx, vy, vz ). In equilibrium, the average kinetic energy of If this were not so, the distribution of velocities the molecules of different gases will be equal. would not remain steady. In any case we are That is, finding v x2 . Thus, on the whole, molecular ½ m1 v1 2 = ½ m2 v 22 = (3/2) kB Tcollisions (if they are not too frequent and the so thattime spent in a collision is negligible compared to time between collisions) will not affect the P = (n1 + n2 +… ) kB T (12.21) calculation above. which is Dalton’s law of partial pressures. From Eq. (12.19), we can get an idea of the12.4.2 Kinetic Interpretation of Temperature typical speed of molecules in a gas. At a Equation (13.14) can be written as temperature T = 300 K, the mean square speed PV = (1/3) nV m v 2 (12.15a) of a molecule in nitrogen gas is : PV = (2/3) N x ½ m v 2 (12.15b) M N 2 28 –26 where N (= nV) is the number of molecules in m = = 26 = 4.65 × 10 kg. N A 6.02 × 10the sample. The quantity in the bracket is the average v 2 = 3 kB T / m = (516)2 m2s-2 translational kinetic energy of the molecules in 2 The square root of v is known as root mean the gas. Since the internal energy E of an ideal square (rms) speed and is denoted by vrms,gas is purely kinetic*, 2 ( We can also write v 2 as < v2 >.) E = N × (1/2) m v (12.16) vrms = 516 m s-1 Equation (12.15) then gives : The speed is of the order of the speed of sound PV = (2/3) E (12.17) in air. It follows from Eq. (12.19) that at the same We are now ready for a kinetic interpretation temperature, lighter molecules have greater rms of temperature. Combining Eq. (12.17) with the speed. ⊳ideal gas Eq. (12.3), we get Example 12.5 A flask contains argon and E = (3/2) kB NT (12.18) chlorine in the ratio of 2:1 by mass. The or E/ N = ½ m v 2 = (3/2) kBT (12.19) temperature of the mixture is 27 °C. Obtain i.e., the average kinetic energy of a molecule is the ratio of (i) average kinetic energy per proportional to the absolute temperature of the molecule, and (ii) root mean square speed gas; it is independent of pressure, volume or vrms of the molecules of the two gases. the nature of the ideal gas. This is a fundamental Atomic mass of argon = 39.9 u; Molecular result relating temperature, a macroscopic mass of chlorine = 70.9 u. measurable parameter of a gas (a thermodynamic variable as it is called) to a Answer The important point to remember is thatmolecular quantity, namely the average kinetic the average kinetic energy (per molecule) of anyenergy of a molecule. The two domains are connected by the Boltzmann constant. We note (ideal) gas (be it monatomic like argon, diatomic in passing that Eq. (12.18) tells us that internal like chlorine or polyatomic) is always equal to energy of an ideal gas depends only on (3/2) kBT. It depends only on temperature, and temperature, not on pressure or volume. With is independent of the nature of the gas. this interpretation of temperature, kinetic theory (i) Since argon and chlorine both have the same of an ideal gas is completely consistent with the temperature in the flask, the ratio of average ideal gas equation and the various gas laws kinetic energy (per molecule) of the two gasesbased on it. is 1:1. For a mixture of non-reactive ideal gases, the total pressure gets contribution from each gas (ii) Now ½ m vrms2 = average kinetic energy per in the mixture. Equation (12.14) becomes molecule = (3/2) ) kBT where m is the mass * E denotes the translational part of the internal energy U that may include energies due to other degrees of freedom also. See section 12.5. Reprint 2025-26 KINETIC THEORY 251 of a molecule of the gas. Therefore, v 2 ( rms ) Ar (m )Cl ( M )Cl 70.9 = = 2 v rms (m ) Ar ( M ) Ar = 39.9 =1.77 ( )Cl where M denotes the molecular mass of the gas. (For argon, a molecule is just an atom of argon.) Taking square root of both sides, v ( rms ) Ar ( vrms )Cl = 1.33 You should note that the composition of the mixture by mass is quite irrelevant to the above calculation. Any other proportion by mass of argon and chlorine would give the same answers to (i) and (ii), provided the temperature remains Fig. 12.5 Molecules going through a porous wall. unaltered. ⊳ ⊳ ⊳ Example 12.7 (a) When a molecule (or an Example 12.6 Uranium has two isotopes elastic ball) hits a ( massive) wall, it of masses 235 and 238 units. If both are rebounds with the same speed. When a ball present in Uranium hexafluoride gas which hits a massive bat held firmly, the same would have the larger average speed ? If thing happens. However, when the bat is atomic mass of fluorine is 19 units, moving towards the ball, the ball rebounds estimate the percentage difference in with a different speed. Does the ball move speeds at any temperature. faster or slower? (Ch.5 will refresh your Answer At a fixed temperature the average memory on elastic collisions.) energy = ½ m <v2 > is constant. So smaller the (b) When gas in a cylinder is compressed mass of the molecule, faster will be the speed. by pushing in a piston, its temperature The ratio of speeds is inversely proportional to rises. Guess at an explanation of this in the square root of the ratio of the masses. The terms of kinetic theory using (a) above. masses are 349 and 352 units. So (c) What happens when a compressed gas v349 / v352 = ( 352/ 349)1/2 = 1.0044 . pushes a piston out and expands. What ∆ V would you observe ? Hence difference = 0.44 %. (d) Sachin Tendulkar used a heavy cricket V bat while playing. Did it help him in [235U is the isotope needed for nuclear fission. anyway ?To separate it from the more abundant isotope 238U, the mixture is surrounded by a porous cylinder. The porous cylinder must be thick and Answer (a) Let the speed of the ball be u relative narrow, so that the molecule wanders through to the wicket behind the bat. If the bat is moving individually, colliding with the walls of the long towards the ball with a speed V relative to the pore. The faster molecule will leak out more than wicket, then the relative speed of the ball to bat the slower one and so there is more of the lighter is V + u towards the bat. When the ball rebounds molecule (enrichment) outside the porous (after hitting the massive bat) its speed, relative cylinder (Fig. 12.5). The method is not very to bat, is V + u moving away from the bat. So efficient and has to be repeated several times relative to the wicket the speed of the rebounding for sufficient enrichment.]. ⊳ ball is V + (V + u) = 2V + u, moving away from the When gases diffuse, their rate of diffusion is wicket. So the ball speeds up after the collision inversely proportional to square root of the with the bat. The rebound speed will be less than masses (see Exercise 12.12 ). Can you guess the u if the bat is not massive. For a molecule this explanation from the above answer? would imply an increase in temperature. Reprint 2025-26 252 PHYSICS You should be able to answer (b) (c) and (d) to the axis joining the two oxygen atoms about based on the answer to (a). which the molecule can rotate*. The molecule (Hint: Note the correspondence, pistonà bat, thus has two rotational degrees of freedom, each of which contributes a term to the total energycylinder à wicket, molecule à ball.) ⊳ consisting of translational energy tε and rotational energy εr.12.5 LAW OF EQUIPARTITION OF ENERGY The kinetic energy of a single molecule is εt + εr = 1 mv x2 + 1 mv y2 + 1 mv z2 + 1 I 1ω12 + 1 I 2ω22 (12.25) 2 2 2 2 2 1 2 1 2 1 2 εt = mv x + mv y + mv z (12.22) 2 2 2 For a gas in thermal equilibrium at temperature T the average value of energy denoted by < tε > is 1 2 1 2 1 2 3 εt = mv x + mv y + mv z = k B T (12.23) 2 2 2 2 Since there is no preferred direction, Eq. (12.23) implies 1 2 1 1 2 1 mv x = k B T , mv y = k B T , 2 2 2 2 Fig. 12.6 The two independent axes of rotation of a diatomic molecule 1 2 1 mv z = k B T (12.24) 2 2 where ω1 and ω2 are the angular speeds about A molecule free to move in space needs three the axes 1 and 2 and I1, I2 are the corresponding coordinates to specify its location. If it is moments of inertia. Note that each rotational constrained to move in a plane it needs two; and degree of freedom contributes a term to the if constrained to move along a line, it needs just energy that contains square of a rotational one coordinate to locate it. This can also be variable of motion. expressed in another way. We say that it has We have assumed above that the O2 molecule one degree of freedom for motion in a line, two is a ‘rigid rotator’, i.e., the molecule does not for motion in a plane and three for motion in vibrate. This assumption, though found to be space. Motion of a body as a whole from one true (at moderate temperatures) for O2, is notpoint to another is called translation. Thus, a always valid. Molecules, like CO, even at molecule free to move in space has three moderate temperatures have a mode of translational degrees of freedom. Each vibration, i.e., its atoms oscillate along the translational degree of freedom contributes a interatomic axis like a one-dimensional term that contains square of some variable of 2 oscillator, and contribute a vibrational energymotion, e.g., ½ mvx and similar terms in term εv to the total energy:vy and vz. In, Eq. (12.24) we see that in thermal equilibrium, the average of each such term is 1  d y  2 1 2 εv = m + ky½ kBT . 2  d t  2 Molecules of a monatomic gas like argon have only translational degrees of freedom. But what ε = εt + εr + ε v (12.26) about a diatomic gas such as O2 or N2? A where k is the force constant of the oscillator molecule of O2 has three translational degrees and y the vibrational co-ordinate. of freedom. But in addition it can also rotate Once again the vibrational energy terms in about its centre of mass. Figure 12.6 shows the Eq. (12.26) contain squared terms of vibrational two independent axes of rotation 1 and 2, normal variables of motion y and dy/dt . * Rotation along the line joining the atoms has very small moment of inertia and does not come into play for quantum mechanical reasons. See end of section 12.6. Reprint 2025-26 KINETIC THEORY 253 At this point, notice an important feature in where Cp is the molar specific heat at constant Eq.(12.26). While each translational and pressure. Thus, rotational degree of freedom has contributed only 5 one ‘squared term’ in Eq.(12.26), one vibrational Cp = R (12.30) mode contributes two ‘squared terms’ : kinetic 2 and potential energies. C p 5 The ratio of specific heats γ = = (12.31) Each quadratic term occurring in the C v 3 expression for energy is a mode of absorption of energy by the molecule. We have seen that in 12.6.2 Diatomic Gases thermal equilibrium at absolute temperature T, As explained earlier, a diatomic molecule treated for each translational mode of motion, the as a rigid rotator, like a dumbbell, has 5 degrees average energy is ½ kBT. The most elegant of freedom: 3 translational and 2 rotational. principle of classical statistical mechanics (first Using the law of equipartition of energy, the total proved by Maxwell) states that this is so for each internal energy of a mole of such a gas is mode of energy: translational, rotational and 5 5 vibrational. That is, in equilibrium, the total U = k B T × N A = RT (12.32) 2 2 energy is equally distributed in all possible The molar specific heats are then given by energy modes, with each mode having an average energy equal to ½ kBT. This is known as the law 5 7 Cv (rigid diatomic) = R, Cp = R (12.33)of equipartition of energy. Accordingly, each 2 2 translational and rotational degree of freedom 7 of a molecule contributes ½ kBT to the energy, γ (rigid diatomic) = (12.34) while each vibrational frequency contributes 5 If the diatomic molecule is not rigid but has 2 × ½ kBT = kBT , since a vibrational mode has in addition a vibrational mode both kinetic and potential energy modes.  5  7 The proof of the law of equipartition of energy U =  k B T + k B T  N A = RT is beyond the scope of this book. Here, we shall  2  2 apply the law to predict the specific heats of gases 7 9 9 theoretically. Later, we shall also discuss briefly, C v = R , C p = R , γ = R (12.35) 2 2 7 the application to specific heat of solids. 12.6.3 Polyatomic Gases 12.6 SPECIFIC HEAT CAPACITY In general a polyatomic molecule has 3 12.6.1 Monatomic Gases translational, 3 rotational degrees of freedom and a certain number ( f ) of vibrational modes.The molecule of a monatomic gas has only three According to the law of equipartition of energy,translational degrees of freedom. Thus, the it is easily seen that one mole of such a gas hasaverage energy of a molecule at temperature 3T is (3/2)kBT . The total internal energy of a mole  3 U = kBT + kBT + f kBT  NA of such a gas is  2 2 3 3 U = k B T × N A = RT (12.27) i.e.,Cv = (3 + f ) R, Cp = (4 + f ) R, 2 2 ( 4 + f ) γ = (12.36) The molar specific heat at constant volume, ( 3 + f ) Cv, is Note that Cp – Cv = R is true for any ideal d U 3 gas, whether mono, di or polyatomic. Cv (monatomic gas) = = RT (12.28) d T 2 Table 12.1 summarises the theoretical For an ideal gas, predictions for specific heats of gases ignoring Cp – Cv = R (12.29) any vibrational modes of motion. The values are Reprint 2025-26 254 PHYSICS in good agreement with experimental values of Answer Using the gas law PV = µRT, you can specific heats of several gases given in Table 12.2. easily show that 1 mol of any (ideal) gas at Of course, there are discrepancies between standard temperature (273 K) and pressure predicted and actual values of specific heats of (1 atm = 1.01 × 105 Pa) occupies a volume of 22.4 several other gases (not shown in the table), such litres. This universal volume is called molar volume. Thus the cylinder in this exampleas Cl2, C2H6 and many other polyatomic gases. contains 2 mol of helium. Further, since heliumUsually, the experimental values for specific is monatomic, its predicted (and observed) molar heats of these gases are greater than the specific heat at constant volume, Cv = (3/2) R,predicted values as given in Table12.1 suggesting and molar specific heat at constant pressure, that the agreement can be improved by including Cp = (3/2) R + R = (5/2) R. Since the volume of vibrational modes of motion in the calculation. the cylinder is fixed, the heat required is The law of equipartition of energy is, thus, well determined by Cv. Therefore, verified experimentally at ordinary temperatures. Heat required = no. of moles × molar specific heat × rise in temperature Table 12.1 Predicted values of specific heat = 2 × 1.5 R × 15.0 = 45 R capacities of gases (ignoring vibrational modes) = 45 × 8.31 = 374 J. ⊳ 12.6.4 Specific Heat Capacity of Solids Nature of Cv Cp Cp - Cv g Gas (J mol-1 K-1) (J mol-1 K-1) (J mol-1 K-1) We can use the law of equipartition of energy to determine specific heats of solids. Consider a Monatomic 12.5 20.8 8.31 1.67 solid of N atoms, each vibrating about its mean Diatomic 20.8 29.1 8.31 1.40 position. An oscillation in one dimension has average energy of 2 × ½ kBT = kBT . In three Triatomic 24.93 33.24 8.31 1.33 dimensions, the average energy is 3 kBT. For a mole of solid, N = NA, and the total energy is Table12.2 Measured values of specific heat U = 3 kBT × NA = 3 RT capacities of some gases Now at constant pressure ∆Q = ∆U + P∆V = ∆U, since for a solid ∆V is negligible. Hence, ∆Q ∆ U C = = = 3 R (12.37) ∆T ∆T Table 12.3 Specific Heat Capacity of some solids at room temperature and atmospheric pressure As Table 12.3 shows the prediction generally ⊳ Example 12.8 A cylinder of fixed capacity agrees with experimental values at ordinary 44.8 litres contains helium gas at standard temperature (Carbon is an exception). temperature and pressure. What is the amount of heat needed to raise the 12.7 MEAN FREE PATH temperature of the gas in the cylinder by Molecules in a gas have rather large speeds of 15.0 °C ? (R = 8.31 J mo1–1 K–1). the order of the speed of sound. Yet a gas leaking Reprint 2025-26 KINETIC THEORY 255 from a cylinder in a kitchen takes considerable are moving and the collision rate is determined time to diffuse to the other corners of the room. by the average relative velocity of the molecules. The top of a cloud of smoke holds together for Thus we need to replace <v> by <v r> in Eq. hours. This happens because molecules in a gas (12.38). A more exact treatment gives have a finite though small size, so they are bound 2 2 nπd (12.40)to undergo collisions. As a result, they cannot l = 1/ ( ) move straight unhindered; their paths keep Let us estimate l and τ for air molecules with getting incessantly deflected. average speeds <v> = ( 485m/s). At STP 0.02 × 1023 ( ) n = –3 22.4 × 10 ( ) = 2.7 × 10 25 m -3. Taking, d = 2 × 10–10 m, τ = 6.1 × 10–10 s t and l = 2.9 × 10–7 m ≈ 1500 d (12.41) v As expected, the mean free path given by d Eq. (12.40) depends inversely on the number density and the size of the molecules. In a highly evacuated tube n is rather small and the mean d free path can be as large as the length of the tube. ⊳ Example 12.9 Estimate the mean free path for a water molecule in water vapour at 373 K. Use information from Exercises 12.1 and Eq. Fig. 12.7 The volume swept by a molecule in time ∆t (12.41) above. in which any molecule will collide with it. Answer The d for water vapour is same as that Suppose the molecules of a gas are spheres of of air. The number density is inverselydiameter d. Focus on a single molecule with the proportional to absolute temperature.average speed <v>. It will suffer collision with any molecule that comes within a distance d 25 273 25 –3 So n = 2.7 × 10 × = 2 × 10 mbetween the centres. In time ∆t, it sweeps a 373 volume πd2 <v> ∆t wherein any other molecule –7 Hence, mean free path l = 4 × 10 m ⊳will collide with it (see Fig. 12.7). If n is the number of molecules per unit volume, the Note that the mean free path is 100 times the molecule suffers nπd2 <v> ∆t collisions in time interatomic distance ~ 40 Å = 4 × 10-9 m calculated ∆t. Thus the rate of collisions is nπd2 <v> or the earlier. It is this large value of mean free path that time between two successive collisions is on the leads to the typical gaseous behaviour. Gases can average, not be confined without a container. τ = 1/(nπ <v> d2 ) (12.38) Using, the kinetic theory of gases, the bulk The average distance between two successive measurable properties like viscosity, heat collisions, called the mean free path l, is : conductivity and diffusion can be related to the l = <v> τ = 1/(nπd2) (12.39) microscopic parameters like molecular size. It is In this derivation, we imagined the other through such relations that the molecular sizes molecules to be at rest. But actually all molecules were first estimated. Reprint 2025-26 256 PHYSICS SUMMARY 1. The ideal gas equation connecting pressure (P), volume (V) and absolute temperature (T ) is PV = µ RT = kB NT where µ is the number of moles and N is the number of molecules. R and kB are universal constants. R R = 8.314 J mol–1 K–1, kB = N A = 1.38 × 10–23 J K–1 Real gases satisfy the ideal gas equation only approximately, more so at low pressures and high temperatures. 2. Kinetic theory of an ideal gas gives the relation 1 2 P = n m v 3 where n is number density of molecules, m the mass of the molecule and v 2 is the mean of squared speed. Combined with the ideal gas equation it yields a kinetic interpretation of temperature. 1 2 3 2 1/2 3k B T m v = k B T , v rms = v = 2 2 ( ) m This tells us that the temperature of a gas is a measure of the average kinetic energy of a molecule, independent of the nature of the gas or molecule. In a mixture of gases at a fixed temperature the heavier molecule has the lower average speed. 3. The translational kinetic energy 3 E = kB NT. 2 This leads to a relation 2 PV = E 3 4. The law of equipartition of energy states that if a system is in equilibrium at absolute temperature T, the total energy is distributed equally in different energy modes of absorption, the energy in each mode being equal to ½ kB T. Each translational and rotational degree of freedom corresponds to one energy mode of absorption and has energy ½ kB T. Each vibrational frequency has two modes of energy (kinetic and potential) with corresponding energy equal to 2 × ½ kB T = kB T. 5. Using the law of equipartition of energy, the molar specific heats of gases can be determined and the values are in agreement with the experimental values of specific heats of several gases. The agreement can be improved by including vibrational modes of motion. 6. The mean free path l is the average distance covered by a molecule between two successive collisions : 1 l = 2 2 n πd where n is the number density and d the diameter of the molecule. Reprint 2025-26 KINETIC THEORY 257 POINTS TO PONDER 1. Pressure of a fluid is not only exerted on the wall. Pressure exists everywhere in a fluid. Any layer of gas inside the volume of a container is in equilibrium because the pressure is the same on both sides of the layer. 2. We should not have an exaggerated idea of the intermolecular distance in a gas. At ordinary pressures and temperatures, this is only 10 times or so the interatomic distance in solids and liquids. What is different is the mean free path which in a gas is 100 times the interatomic distance and 1000 times the size of the molecule. 3. The law of equipartition of energy is stated thus: the energy for each degree of freedom in thermal equilibrium is ½ k T. Each quadratic term in the total energy expression of a molecule is to be counted asB a degree of freedom. Thus, each vibrational mode gives 2 (not 1) degrees of freedom (kinetic and potential energy modes), corresponding to the energy 2 × ½ k T = k T. B B 4. Molecules of air in a room do not all fall and settle on the ground (due to gravity) because of their high speeds and incessant collisions. In equilibrium, there is a very slight increase in density at lower heights (like in the atmosphere). The effect is small since the potential energy (mgh) for ordinary heights is much less than the average kinetic energy ½ mv2 of the molecules. 5. < v2 > is not always equal to ( < v >)2. The average of a squared quantity is not necessarily the square of the average. Can you find examples for this statement. EXERCISESEXERCISESEXERCISESEXERCISESEXERCISES 12.112.112.112.112.1 Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 Å. 12.212.212.212.212.2 Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, 0 °C). Show that it is 22.4 litres. 12.312.312.312.312.3 Figure 12.8 shows plot of PV/T versus P for 1.00×10–3 kg of oxygen gas at two different temperatures. y T1 PV –1 T2 (J K ) T x P Fig.Fig.Fig.Fig.Fig. 12.812.812.812.812.8 (a) What does the dotted plot signify? (b) Which is true: T1 > T2 or T1 < T2? (c) What is the value of PV/T where the curves meet on the y-axis? Reprint 2025-26 258 PHYSICS (d) If we obtained similar plots for 1.00×10–3 kg of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of PV/T (for low pressure high temperature region of the plot) ? (Molecular mass of H2 = 2.02 u, of O2 = 32.0 u, R = 8.31 J mo1–1 K–1.) 12.412.412.412.412.4 An oxygen cylinder of volume 30 litre has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol–1 K–1, molecular mass of O2 = 32 u). 12.512.512.512.512.5 An air bubble of volume 1.0 cm3 rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C ? 12.612.612.612.612.6 Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m3 at a temperature of 27 °C and 1 atm pressure. 12.712.712.712.712.7 Estimate the average thermal energy of a helium atom at (i) room temperature (27 °C), (ii) the temperature on the surface of the Sun (6000 K), (iii) the temperature of 10 million kelvin (the typical core temperature in the case of a star). 12.812.812.812.812.8 Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules ? Is the root mean square speed of molecules the same in the three cases? If not, in which case is vrms the largest ? 12.912.912.912.912.9 At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20 °C ? (atomic mass of Ar = 39.9 u, of He = 4.0 u). 12.1012.1012.1012.1012.10 Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 0C. Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N2 = 28.0 u). Reprint 2025-26 CHAPTER THIRTEEN OSCILLATIONS 13.1 INTRODUCTION In our daily life we come across various kinds of motions. You have already learnt about some of them, e.g., rectilinear 13.1 Introduction motion and motion of a projectile. Both these motions are non-repetitive. We have also learnt about uniform circular13.2 Periodic and oscillatory motions motion and orbital motion of planets in the solar system. In 13.3 Simple harmonic motion these cases, the motion is repeated after a certain interval of 13.4 Simple harmonic motion time, that is, it is periodic. In your childhood, you must have and uniform circular enjoyed rocking in a cradle or swinging on a swing. Both motion these motions are repetitive in nature but different from the 13.5 Velocity and acceleration periodic motion of a planet. Here, the object moves to and fro in simple harmonic motion about a mean position. The pendulum of a wall clock executes 13.6 Force law for simple a similar motion. Examples of such periodic to and fro harmonic motion motion abound: a boat tossing up and down in a river, the

1.36 100 g of liquid A (molar mass 140 g mol–1) was dissolved in 1000 g of liquid B (molar mass 180 g mol–1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr. 29 Solutions Reprint 2025-26

1.37 Vapour pressures of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot ptotal, pchloroform, and pacetone as a function of xacetone. The experimental data observed for different compositions of mixture is: 100 x xacetone 0 11.8 23.4 36.0 50.8 58.2 64.5 72.1 pacetone /mm Hg 0 54.9 110.1 202.4 322.7 405.9 454.1 521.1 pchloroform /mm Hg 632.8 548.1 469.4 359.7 257.7 193.6 161.2 120.7 Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal solution. 1.38 Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene. 1.39 The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if the Henry’s law constants for oxygen and nitrogen at 298 K are 3.30 × 107 mm and 6.51 × 107 mm respectively, calculate the composition of these gases in water. 1.40 Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27° C. 1.41 Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 litre of water at 25° C, assuming that it is completely dissociated. Answers to Some Intext Questions 1.1 C6H6 = 15.28%, CCl4 = 84.72% 1.2 0.459, 0.541 1.3 0.024 M, 0.03 M 1.4 36.946 g 1.5 1.5 mol kg–1 , 1.45 mol L–1 0.0263 1.9 23.4 mm Hg 1.10 121.67 g 1.11 5.077 g 1.12 30.96 Pa Chemistry 30 Reprint 2025-26 UnitUnitUnitUnit Unit22 Objectives ElectrochemistryElectrochemistry After studying this Unit, you will be able to · describe an electrochemical cell Chemical reactions can be used to produce electrical energy, and differentiate between galvanic conversely, electrical energy can be used to carry out chemical and electrolytic cells; reactions that do not proceed spontaneously.· apply Nernst equation for calculating the emf of galvanic cell and define standard potential of Electrochemistry is the study of production of the cell; · derive relation between standard electricity from energy released during spontaneous potential of the cell, Gibbs energy chemical reactions and the use of electrical energy of cell reaction and its equilibrium to bring about non-spontaneous chemical constant; transformations. The subject is of importance both · define resistivity (r), conductivity for theoretical and practical considerations. A large (k) and molar conductivity (✆m) of number of metals, sodium hydroxide, chlorine, ionic solutions; fluorine and many other chemicals are produced by · differentiate between ionic electrochemical methods. Batteries and fuel cells (electrolytic) and electronic convert chemical energy into electrical energy and are conductivity; · describe the method for used on a large scale in various instruments and measurement of conductivity of devices. The reactions carried out electrochemically electrolytic solutions and can be energy efficient and less polluting. Therefore, calculation of their molar study of electrochemistry is important for creating new conductivity; technologies that are ecofriendly. The transmission of · justify the variation of sensory signals through cells to brain and vice versa conductivity and molar and communication between the cells are known to conductivity of solutions with have electrochemical origin. Electrochemistry, is change in their concentration and therefore, a very vast and interdisciplinary subject. In define m (molar conductivity at this Unit, we will cover only some of its important zero concentration or infinite elementary aspects. dilution); · enunciate Kohlrausch law and learn its applications; · understand quantitative aspects of electrolysis; · describe the construction of some primary and secondary batteries and fuel cells; · explain corrosion as an electrochemical process. Reprint 2025-26 2.12.12.12.12.1 ElectrochemicalElectrochemicalElectrochemicalElectrochemicalElectrochemical We had studied the construction and functioning of Daniell cell CellsCellsCellsCellsCells (Fig. 2.1). This cell converts the chemical energy liberated during the redox reaction Zn(s) + Cu2+(aq) ® Zn2+(aq) + Cu(s) (2.1) to electrical energy and has an electrical potential equal to 1.1 V when concentration of Zn2+ and Cu2+ ions is unity (1 mol dm–3)*. Such a device is called a galvanic or a voltaic cell. If an external opposite potential is applied in the galvanic cell [Fig. 2.2(a)] and increased slowly, we find that the reaction continues to take place till the opposing voltage reaches the value 1.1 V [Fig. 2.2(b)] when, the reaction stops altogether and no current flows through the cell. Any further increase in the external potential again starts the reaction but in the opposite direction [Fig. 2.2(c)]. It now functions as an electrolytic cell, a device for using electrical energy to carry non-spontaneous chemical reactions. Both types of cells are Fig. 2.1: Daniell cell having electrodes of zinc and quite important and we shall study some of copper dipping in the solutions of their their salient features in the following pages. respective salts. Eext < 1.1V Eext = 1.1V (a) (b) e current cathodeanode I=0 Zn salt Cu Zn Cu -ve bridge +ve When Eext = 1.1 V (i) No flow of electrons or current. (ii) No chemical ZnSO4 CuSO4 ZnSO4 CuSO4 reaction. When Eext < 1.1 V Eext >1.1 (i) Electrons flow from Zn rod to (c) Cu rod hence current flows from Cu to Zn. – When Eext > 1.1 V (ii) Zn dissolves at anode and e (i) Electrons flow copper deposits at cathode. Cathode Current Anode from Cu to Zn +ve –ve and current flows Zn Cu from Zn to Cu. Fig. 2.2 (ii) Zinc is deposited Functioning of Daniell at the zinc cell when external electrode and voltage Eext opposing the copper dissolves at cell potential is applied. copper electrode. *Strictly speaking activity should be used instead of concentration. It is directly proportional to concentration. In dilute solutions, it is equal to concentration. You will study more about it in higher classes. Chemistry 32 Reprint 2025-26 2.22.22.22.22.2 GalvanicGalvanicGalvanicGalvanicGalvanic CellsCellsCellsCellsCells As mentioned earlier a galvanic cell is an electrochemical cell that converts the chemical energy of a spontaneous redox reaction into electrical energy. In this device the Gibbs energy of the spontaneous redox reaction is converted into electrical work which may be used for running a motor or other electrical gadgets like heater, fan, geyser, etc. Daniell cell discussed earlier is one such cell in which the following redox reaction occurs. Zn(s) + Cu2+(aq) ® Zn2+ (aq) + Cu(s) This reaction is a combination of two half reactions whose addition gives the overall cell reaction: (i) Cu2+ + 2e– ® Cu(s) (reduction half reaction) (2.2) (ii) Zn(s) ® Zn2+ + 2e– (oxidation half reaction) (2.3) These reactions occur in two different portions of the Daniell cell. The reduction half reaction occurs on the copper electrode while the oxidation half reaction occurs on the zinc electrode. These two portions of the cell are also called half-cells or redox couples. The copper electrode may be called the reduction half cell and the zinc electrode, the oxidation half-cell. We can construct innumerable number of galvanic cells on the pattern of Daniell cell by taking combinations of different half-cells. Each half- cell consists of a metallic electrode dipped into an electrolyte. The two half-cells are connected by a metallic wire through a voltmeter and a switch externally. The electrolytes of the two half-cells are connected internally through a salt bridge as shown in Fig. 2.1. Sometimes, both the electrodes dip in the same electrolyte solution and in such cases we do not require a salt bridge. At each electrode-electrolyte interface there is a tendency of metal ions from the solution to deposit on the metal electrode trying to make it positively charged. At the same time, metal atoms of the electrode have a tendency to go into the solution as ions and leave behind the electrons at the electrode trying to make it negatively charged. At equilibrium, there is a separation of charges and depending on the tendencies of the two opposing reactions, the electrode may be positively or negatively charged with respect to the solution. A potential difference develops between the electrode and the electrolyte which is called electrode potential. When the concentrations of all the species involved in a half-cell is unity then the electrode potential is known as standard electrode potential. According to IUPAC convention, standard reduction potentials are now called standard electrode potentials. In a galvanic cell, the half-cell in which oxidation takes place is called anode and it has a negative potential with respect to the solution. The other half-cell in which reduction takes place is called cathode and it has a positive potential with respect to the solution. Thus, there exists a potential difference between the two electrodes and as soon as the switch is in the on position the electrons flow from negative electrode to positive electrode. The direction of current flow is opposite to that of electron flow. 33 Electrochemistry Reprint 2025-26 The potential difference between the two electrodes of a galvanic cell is called the cell potential and is measured in volts. The cell potential is the difference between the electrode potentials (reduction potentials) of the cathode and anode. It is called the cell electromotive force (emf) of the cell when no current is drawn through the cell. It is now an accepted convention that we keep the anode on the left and the cathode on the right while representing the galvanic cell. A galvanic cell is generally represented by putting a vertical line between metal and electrolyte solution and putting a double vertical line between the two electrolytes connected by a salt bridge. Under this convention the emf of the cell is positive and is given by the potential of the half- cell on the right hand side minus the potential of the half-cell on the left hand side i.e., Ecell = Eright – Eleft This is illustrated by the following example: Cell reaction: Cu(s) + 2Ag+(aq) ¾® Cu2+(aq) + 2 Ag(s) (2.4) Half-cell reactions: Cathode (reduction): 2Ag+(aq) + 2e– ® 2Ag(s) (2.5) Anode (oxidation): Cu(s) ® Cu2+(aq) + 2e– (2.6) It can be seen that the sum of (3.5) and (3.6) leads to overall reaction (2.4) in the cell and that silver electrode acts as a cathode and copper electrode acts as an anode. The cell can be represented as: Cu(s)|Cu2+(aq)||Ag+(aq)|Ag(s) and we have Ecell = Eright – Eleft = EAg+úAg – ECu2+úCu (2.7) 2.2.1 The potential of individual half-cell cannot be measured. We can Measurement measure only the difference between the two half-cell potentials that of Electrode gives the emf of the cell. If we arbitrarily choose the potential of one Potential electrode (half-cell) then that of the other can be determined with respect to this. According to convention, a half-cell called standard hydrogen electrode (Fig.3.3) represented by Pt(s)ú H2(g)ú H+(aq), is assigned a zero potential at all temperatures corresponding to the reaction 1 H+ (aq) + e– ® H2(g) 2 The standard hydrogen electrode consists of a platinum electrode coated with platinum black. The electrode is dipped in an acidic solution and pure hydrogen gas is bubbled through it. The concentration of both the reduced and oxidised forms of hydrogen is maintained at unity (Fig. 2.3). This implies that the pressure of hydrogen gas is one bar and the concentration of hydrogen ion in the Fig. 2.3: Standard Hydrogen Electrode (SHE). solution is one molar. Chemistry 34 Reprint 2025-26 At 298 K the emf of the cell, standard hydrogen electrode ççsecond half-cell constructed by taking standard hydrogen electrode as anode (reference half-cell) and the other half-cell as cathode, gives the reduction potential of the other half-cell. If the concentrations of the oxidised and the reduced forms of the species in the right hand half-cell are unity, then the cell potential is equal to standard electrode potential, Eo R of the given half-cell. Eo = EoR – Eo L As Eo L for standard hydrogen electrode is zero. Eo = Eo R – 0 = EoR The measured emf of the cell: Pt(s) ç H2(g, 1 bar) ç H + (aq, 1 M) çç Cu 2+ (aq, 1 M) ú Cu is 0.34 V and it is also the value for the standard electrode potential of the half-cell corresponding to the reaction: Cu2+ (aq, 1M) + 2 e – ® Cu(s) Similarly, the measured emf of the cell: Pt(s) ç H2(g, 1 bar) ç H+ (aq, 1 M) çç Zn2+ (aq, 1M) ç Zn is -0.76 V corresponding to the standard electrode potential of the half-cell reaction: Zn2+ (aq, 1 M) + 2e– ® Zn(s) The positive value of the standard electrode potential in the first case indicates that Cu2+ ions get reduced more easily than H+ ions. The reverse process cannot occur, that is, hydrogen ions cannot oxidise Cu (or alternatively we can say that hydrogen gas can reduce copper ion) under the standard conditions described above. Thus, Cu does not dissolve in HCl. In nitric acid it is oxidised by nitrate ion and not by hydrogen ion. The negative value of the standard electrode potential in the second case indicates that hydrogen ions can oxidise zinc (or zinc can reduce hydrogen ions). In view of this convention, the half reaction for the Daniell cell in Fig. 2.1 can be written as: Left electrode: Zn(s) ® Zn 2+ (aq, 1 M) + 2 e – Right electrode: Cu 2+ (aq, 1 M) + 2 e – ® Cu(s) The overall reaction of the cell is the sum of above two reactions and we obtain the equation: Zn(s) + Cu 2+ (aq) ® Zn2+ (aq) + Cu(s) emf of the cell = Eocell = Eo R – Eo L = 0.34V – (– 0.76)V = 1.10 V Sometimes metals like platinum or gold are used as inert electrodes. They do not participate in the reaction but provide their surface for oxidation or reduction reactions and for the conduction of electrons. For example, Pt is used in the following half-cells: Hydrogen electrode: Pt(s)|H2(g)| H+(aq) With half-cell reaction: H+ (aq)+ e– ® ½ H2(g) Bromine electrode: Pt(s)|Br2(aq)| Br–(aq) 35 Electrochemistry Reprint 2025-26 With half-cell reaction: ½ Br2(aq) + e– ® Br–(aq) The standard electrode potentials are very important and we can extract a lot of useful information from them. The values of standard electrode potentials for some selected half-cell reduction reactions are given in Table 2.1. If the standard electrode potential of an electrode is greater than zero then its reduced form is more stable compared to hydrogen gas. Similarly, if the standard electrode potential is negative then hydrogen gas is more stable than the reduced form of the species. It can be seen that the standard electrode potential for fluorine is the highest in the Table indicating that fluorine gas (F2) has the maximum tendency to get reduced to fluoride ions (F–) and therefore fluorine gas is the strongest oxidising agent and fluoride ion is the weakest reducing agent. Lithium has the lowest electrode potential indicating that lithium ion is the weakest oxidising agent while lithium metal is the most powerful reducing agent in an aqueous solution. It may be seen that as we go from top to bottom in Table 2.1 the standard electrode potential decreases and with this, decreases the oxidising power of the species on the left and increases the reducing power of the species on the right hand side of the reaction. Electrochemical cells are extensively used for determining the pH of solutions, solubility product, equilibrium constant and other thermodynamic properties and for potentiometric titrations. IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 2.1 How would you determine the standard electrode potential of the system Mg2+|Mg? 2.2 Can you store copper sulphate solutions in a zinc pot? 2.3 Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions. 2.32.32.32.32.3 NernstNernstNernstNernstNernst We have assumed in the previous section that the concentration of all EquationEquationEquationEquationEquation the species involved in the electrode reaction is unity. This need not be always true. Nernst showed that for the electrode reaction: Mn+(aq) + ne–® M(s) the electrode potential at any concentration measured with respect to standard hydrogen electrode can be represented by: RT o [M] E = E ln ( M n + / M ) ( M n + / M ) – nF [M n+ ] but concentration of solid M is taken as unity and we have o RT 1 E = E (2.8) ( M n + / M ) ( M n + /M ) – nF ln [M n+ ] o E ( M n + / M ) has already been defined, R is gas constant (8.314 JK–1 mol–1), F is Faraday constant (96487 C mol–1), T is temperature in kelvin and [Mn+] is the concentration of the species, Mn+. Chemistry 36 Reprint 2025-26 Table 2.1: Standard Electrode Potentials at 298 K Ions are present as aqueous species and H2O as liquid; gases and solids are shown by g and s. Reaction (Oxidised form + ne– ® Reduced form) E o/V ® 2F– 2.87 F2(g) + 2e– Co3+ + e– ® Co2+ 1.81 H2O2 + 2H+ + 2e– ® 2H2O 1.78 MnO4– + 8H+ + 5e– ® Mn2+ + 4H2O 1.51 Au3+ + 3e– ® Au(s) 1.40 Cl2(g) + 2e– ® 2Cl– 1.36 Cr2O72– + 14H+ + 6e– ® 2Cr3+ + 7H2O 1.33 O2(g) + 4H+ + 4e– ® 2H2O 1.23 MnO2(s) + 4H+ + 2e– ® Mn2+ + 2H2O 1.23 Br2 + 2e– ® 2Br– 1.09 NO3– + 4H+ + 3e– ® NO(g) + 2H2O 0.97 2Hg2+ + 2e– ® Hg22+ 0.92 Ag+ + e– ® Ag(s) 0.80 agent agent Fe3+ + e– ® Fe2+ 0.77 O2(g) + 2H+ + 2e– ® H2O2 0.68 I2 + 2e– ® 2I– 0.54 oxidising reducing 0.52 of Cu+ + e– ® Cu(s) of Cu2+ + 2e– ® Cu(s) 0.34 AgCl(s) + e– ® Ag(s) + Cl– 0.22 strength AgBr(s) + e– ® Ag(s) + Br– strength 0.10 2H+ + 2e– ® H2(g) 0.00 Pb2+ + 2e– ® Pb(s) –0.13 Sn2+ + 2e– ® Sn(s) –0.14 Increasing Increasing Ni2+ + 2e– ® Ni(s) –0.25 Fe2+ + 2e– ® Fe(s) –0.44 Cr3+ + 3e– ® Cr(s) –0.74 Zn2+ + 2e– ® Zn(s) –0.76 2H2O + 2e– ® H2(g) + 2OH–(aq) –0.83 Al3+ + 3e– ® Al(s) –1.66 Mg2+ + 2e– ® Mg(s) –2.36 Na+ + e– ® Na(s) –2.71 Ca2+ + 2e– ® Ca(s) –2.87 K+ + e– ® K(s) –2.93 Li+ + e– ® Li(s) –3.05 1. A negative Eo means that the redox couple is a stronger reducing agent than the H+/H2 couple. 2. A positive Eo means that the redox couple is a weaker reducing agent than the H+/H2 couple. 37 Electrochemistry Reprint 2025-26 In Daniell cell, the electrode potential for any given concentration of Cu2+ and Zn2+ ions, we write For Cathode: E E o RT 1 (2.9)  Cu 2  /Cu  = (Cu 2 + /Cu ) – 2F ln   Cu 2   aq   For Anode: E E o RT 1 (2.10)  Zn 2  /Zn  = ( Zn 2 + / Zn ) – 2F ln   Zn 2   aq   E E 2  2  /Zn  The cell potential, E(cell) =  Cu /Cu  –  Zn o RT 1 E o RT 1 E = (Cu – ( Zn 2 + / Cu ) – 2 F ln 2 + / Zn ) + 2 F ln    Zn 2+ (aq)   Cu 2+ (aq) E o E o RT 1 1 – ln = (Cu 2 + / Cu ) – ( Zn 2 + / Zn ) – 2F ln    Cu 2+  aq    Zn 2+  aq   2  ] RT [ Zn o E(cell) = E ( cell ) – 2 F ln 2 + (2.11) [Cu ] It can be seen that E(cell) depends on the concentration of both Cu2+ and Zn2+ ions. It increases with increase in the concentration of Cu2+ ions and decrease in the concentration of Zn2+ ions. By converting the natural logarithm in Eq. (2.11) to the base 10 and substituting the values of R, F and T = 298 K, it reduces to 2 + ] 0 .059 [ Zn (2.12) 2 + ] E(cell) = E (ocell ) – 2 log [Cu We should use the same number of electrons (n) for both the electrodes and thus for the following cell Ni(s)ú Ni2+(aq) úú Ag+(aq)ú Ag The cell reaction is Ni(s) + 2Ag+(aq) ® Ni2+(aq) + 2Ag(s) The Nernst equation can be written as RT [Ni 2+ ] o + E(cell) = E ( cell ) – 2F ln [Ag ]2 and for a general electrochemical reaction of the type: a A + bB ne– cC + dD Nernst equation can be written as: RT E(cell) = E (ocell ) – nF 1nQ RT [C]c [D]d o (2.13) = E ( cell ) – nF ln [A] a [B]b Chemistry 38 Reprint 2025-26 Represent the cell in which the following reaction takes place ExampleExampleExampleExampleExample 2.12.12.12.12.1 Mg(s) + 2Ag+(0.0001M) ® Mg2+(0.130M) + 2Ag(s) Calculate its E(cell) if E (ocell ) = 3.17 V. The cell can be written as Mgú Mg2+(0.130M)úú Ag+(0.0001M)ú Ag SolutionSolutionSolutionSolutionSolution 2 + Mg RT o E = E ln (  cell  cell ) – 2F + 2 Ag 0 .059V 0.130 = 3.17 V – log 2 = 3.17 V – 0.21V = 2.96 V. 2 ( 0 . 0001) 2.3.1 Equilibrium If the circuit in Daniell cell (Fig. 2.1) is closed then we note that the reaction Constant Zn(s) + Cu2+(aq) ® Zn2+(aq) + Cu(s) (2.1) from Nernst takes place and as time passes, the concentration of Zn2+ keeps Equation on increasing while the concentration of Cu2+ keeps on decreasing. At the same time voltage of the cell as read on the voltmeter keeps on decreasing. After some time, we shall note that there is no change in the concentration of Cu2+ and Zn2+ ions and at the same time, voltmeter gives zero reading. This indicates that equilibrium has been attained. In this situation the Nernst equation may be written as: o 2.303 RT [Zn 2 + ] 2 + E(cell) = 0 = E ( cell ) – 2 F log [Cu ] o 2.303 RT [Zn 2  ] or E ( cell ) = log 2  2 F [Cu ] But at equilibrium, [ Zn 2 + ] = Kc for the reaction 2.1 [Cu2 + ] and at T = 298K the above equation can be written as o 0. 059 V o E ( cell ) = log KC = 1.1 V ( E ( cell ) = 1.1V) 2 (1.1V × 2) log KC =  37.288 0.059 V KC = 2 × 1037 at 298K. In general, o 2.303RT E ( cell ) = log KC (2.14) nF Thus, Eq. (2.14) gives a relationship between equilibrium constant of the reaction and standard potential of the cell in which that reaction takes place. Thus, equilibrium constants of the reaction, difficult to measure otherwise, can be calculated from the corresponding Eo value of the cell. 39 Electrochemistry Reprint 2025-26 ExampleExampleExampleExampleExample 2.22.22.22.22.2 Calculate the equilibrium constant of the reaction: Cu(s) + 2Ag+(aq) ® Cu2+(aq) + 2Ag(s) Eo( cell ) = 0.46 V o 0. 059 V SolutionSolutionSolutionSolutionSolution E ( cell ) = log KC = 0.46 V or 2 0 .46 V × 2 = 15.6 log KC = 0 .059 V KC = 3.92 × 1015 2.3.2 Electro- Electrical work done in one second is equal to electrical potential chemical multiplied by total charge passed. If we want to obtain maximum work Cell and from a galvanic cell then charge has to be passed reversibly. The Gibbs reversible work done by a galvanic cell is equal to decrease in its Gibbs Energy of energy and therefore, if the emf of the cell is E and nF is the amount the Reaction of charge passed and DrG is the Gibbs energy of the reaction, then DrG = – nFE(cell) (2.15) It may be remembered that E(cell) is an intensive parameter but DrG is an extensive thermodynamic property and the value depends on n. Thus, if we write the reaction Zn(s) + Cu2+(aq) ¾® Zn2+(aq) + Cu(s) (2.1) DrG = – 2FE(cell) but when we write the reaction 2 Zn (s) + 2 Cu2+(aq) ¾®2 Zn2+(aq) + 2Cu(s) DrG = – 4FE(cell) If the concentration of all the reacting species is unity, then E(cell) = E (ocell ) and we have DrGo = – nF E(cell)o (2.16) Thus, from the measurement of E (ocell ) we can obtain an important thermodynamic quantity, DrGo, standard Gibbs energy of the reaction. From the latter we can calculate equilibrium constant by the equation: DrGo = –RT ln K. ExampleExampleExampleExampleExample 2.32.32.32.32.3 The standard electrode potential for Daniell cell is 1.1V. Calculate the standard Gibbs energy for the reaction: Zn(s) + Cu2+(aq) ¾® Zn2+(aq) + Cu(s) SolutionSolutionSolutionSolutionSolution DrGo = – nF E(cell)o n in the above equation is 2, F = 96487 C mol–1 and E o( cell ) = 1.1 V Therefore, DrGo = – 2 × 1.1V × 96487 C mol–1 = – 21227 J mol–1 = – 212.27 kJ mol–1 Chemistry 40 Reprint 2025-26 IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 2.4 Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10. 2.5 Calculate the emf of the cell in which the following reaction takes place: Ni(s) + 2Ag+ (0.002 M) ® Ni2+ (0.160 M) + 2Ag(s) Given that Ecello = 1.05 V 2.6 The cell in which the following reaction occurs: E o = 0.236 V at 298 K. 2Fe 3 + ( aq ) + 2I − ( aq ) → 2Fe 2 + ( aq ) + I 2 ( s ) has cell Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction. 2.42.42.42.42.4 ConductanceConductanceConductanceConductanceConductance It is necessary to define a few terms before we consider the subject of ofofofofof ElectrolyticElectrolyticElectrolyticElectrolyticElectrolytic conductance of electricity through electrolytic solutions. The electrical resistance is represented by the symbol ‘R’ and it is measured in ohm (W) SolutionsSolutionsSolutionsSolutionsSolutions which in terms of SI base units is equal to (kg m2)/(S3 A2). It can be measured with the help of a Wheatstone bridge with which you are familiar from your study of physics. The electrical resistance of any object is directly proportional to its length, l, and inversely proportional to its area of cross section, A. That is, l l R µ or R = r (2.17) A A The constant of proportionality, r (Greek, rho), is called resistivity (specific resistance). Its SI units are ohm metre (W m) and quite often its submultiple, ohm centimetre (W cm) is also used. IUPAC recommends the use of the term resistivity over specific resistance and hence in the rest of the book we shall use the term resistivity. Physically, the resistivity for a substance is its resistance when it is one metre long and its area of cross section is one m2. It can be seen that: 1 W m = 100 W cm or 1 W cm = 0.01 W m The inverse of resistance, R, is called conductance, G, and we have the relation: 1 A A G = = = κ (2.18) R ρ l l The SI unit of conductance is siemens, represented by the symbol ‘S’ and is equal to ohm–1 (also known as mho) or W–1. The inverse of resistivity, called conductivity (specific conductance) is represented by the symbol, k (Greek, kappa). IUPAC has recommended the use of term conductivity over specific conductance and hence we shall use the term conductivity in the rest of the book. The SI units of conductivity are S m–1 but quite often, k is expressed in S cm–1. Conductivity of a material in S m–1 is its conductance when it is 1 m long and its area of cross section is 1 m2. It may be noted that 1 S cm–1 = 100 S m–1. 41 Electrochemistry Reprint 2025-26 Table 2.2: The values of Conductivity of some Selected Materials at 298.15 K Material Conductivity/ Material Conductivity/ S m–1 S m–1 Conductors Aqueous Solutions Sodium 2.1×103 Pure water 3.5×10–5 Copper 5.9×103 0.1 M HCl 3.91 Silver 6.2×103 0.01M KCl 0.14 Gold 4.5×103 0.01M NaCl 0.12 Iron 1.0×103 0.1 M HAc 0.047 Graphite 1.2×10 0.01M HAc 0.016 Insulators Semiconductors Glass 1.0×10–16 CuO 1×10–7 Teflon 1.0×10–18 Si 1.5×10–2 Ge 2.0 It can be seen from Table 2.2 that the magnitude of conductivity varies a great deal and depends on the nature of the material. It also depends on the temperature and pressure at which the measurements are made. Materials are classified into conductors, insulators and semiconductors depending on the magnitude of their conductivity. Metals and their alloys have very large conductivity and are known as conductors. Certain non-metals like carbon-black, graphite and some organic polymers* are also electronically conducting. Substances like glass, ceramics, etc., having very low conductivity are known as insulators. Substances like silicon, doped silicon and gallium arsenide having conductivity between conductors and insulators are called semiconductors and are important electronic materials. Certain materials called superconductors by definition have zero resistivity or infinite conductivity. Earlier, only metals and their alloys at very low temperatures (0 to 15 K) were known to behave as superconductors, but nowadays a number of ceramic materials and mixed oxides are also known to show superconductivity at temperatures as high as 150 K. Electrical conductance through metals is called metallic or electronic conductance and is due to the movement of electrons. The electronic conductance depends on (i) the nature and structure of the metal (ii) the number of valence electrons per atom (iii) temperature (it decreases with increase of temperature). * Electronically conducting polymers – In 1977 MacDiarmid, Heeger and Shirakawa discovered that acetylene gas can be polymerised to produce a polymer, polyacetylene when exposed to vapours of iodine acquires metallic lustre and conductivity. Since then several organic conducting polymers have been made such as polyaniline, polypyrrole and polythiophene. These organic polymers which have properties like metals, being composed wholly of elements like carbon, hydrogen and occasionally nitrogen, oxygen or sulphur, are much lighter than normal metals and can be used for making light-weight batteries. Besides, they have the mechanical properties of polymers such as flexibility so that one can make electronic devices such as transistors that can bend like a sheet of plastic. For the discovery of conducting polymers, MacDiarmid, Heeger and Shirakawa were awarded the Nobel Prize in Chemistry for the year 2000. Chemistry 42 Reprint 2025-26 As the electrons enter at one end and go out through the other end, the composition of the metallic conductor remains unchanged. The mechanism of conductance through semiconductors is more complex. We already know that even very pure water has small amounts of hydrogen and hydroxyl ions (~10–7M) which lend it very low conductivity (3.5 × 10–5 S m–1). When electrolytes are dissolved in water, they furnish their own ions in the solution hence its conductivity also increases. The conductance of electricity by ions present in the solutions is called electrolytic or ionic conductance. The conductivity of electrolytic (ionic) solutions depends on: (i) the nature of the electrolyte added (ii) size of the ions produced and their solvation (iii) the nature of the solvent and its viscosity (iv) concentration of the electrolyte (v) temperature (it increases with the increase of temperature). Passage of direct current through ionic solution over a prolonged period can lead to change in its composition due to electrochemical reactions (Section 2.4.1). 2.4.1 Measurement We know that accurate measurement of an unknown resistance can be of the performed on a Wheatstone bridge. However, for measuring the resistance Conductivity of an ionic solution we face two problems. Firstly, passing direct current of Ionic (DC) changes the composition of the solution. Secondly, a solution cannot Solutions be connected to the bridge like a metallic wire or other solid conductor. The first difficulty is resolved by using an alternating current (AC) source of power. The second problem is solved by using a specially designed vessel called conductivity cell. It is available in several designs and two simple ones are shown in Fig. 2.4. Connecting Connecting wires wires Platinized Pt Fig. 2.4 electrodes Two different types of conductivity cells. Platinized Pt electrode Platinized Pt electrode Basically it consists of two platinum electrodes coated with platinum black (finely divided metallic Pt is deposited on the electrodes electrochemically). These have area of cross section equal to ‘A’ and are separated by distance ‘l’. Therefore, solution confined between these electrodes is a column of length l and area of cross section A. The resistance of such a column of solution is then given by the equation: l l R = r = (2.17) A A 43 Electrochemistry Reprint 2025-26 The quantity l/A is called cell constant denoted by the symbol, G*. It depends on the distance between the electrodes and their area of cross-section and has the dimension of length–1 and can be calculated if we know l and A. Measurement of l and A is not only inconvenient but also unreliable. The cell constant is usually determined by measuring the resistance of the cell containing a solution whose conductivity is already known. For this purpose, we generally use KCl solutions whose conductivity is known accurately at various concentrations (Table 2.3) and at different temperatures. The cell constant, G*, is then given by the equation: l G* = = R k (2.18) A Table 2.3: Conductivity and Molar conductivity of KCl solutions at 298.15K Concentration/Molarity Conductivity Molar Conductivity mol L–1 mol m–3 S cm–1 S m–1 S cm2mol–1 S m2 mol–1 1.000 1000 0.1113 11.13 111.3 111.3×10–4 0.100 100.0 0.0129 1.29 129.0 129.0×10–4 0.010 10.00 0.00141 0.141 141.0 141.0×10–4 Once the cell constant is determined, we can use it for measuring the resistance or conductivity of any solution. The set up for the measurement of the resistance is shown in Fig. 2.5. It consists of two resistances R3 and R4, a variable resistance R1 and the conductivity cell having the unknown resistance R2. The Wheatstone bridge is fed by an oscillator O (a source of a.c. power in the audio frequency range 550 to 5000 cycles per second). P is a suitable detector (a headphone or other electronic device) and the bridge is balanced when no current passes through the detector. Under these conditions: Fig. 2.5: Arrangement for measurement of R 1 R 4 resistance of a solution of an Unknown resistance R2 = (2.19) R 3 electrolyte. These days, inexpensive conductivity meters are available which can directly read the conductance or resistance of the solution in the conductivity cell. Once the cell constant and the resistance of the solution in the cell is determined, the conductivity of the solution is given by the equation: cell constant G*   (2.20) R R The conductivity of solutions of different electrolytes in the same solvent and at a given temperature differs due to charge and size of the Chemistry 44 Reprint 2025-26 ions in which they dissociate, the concentration of ions or ease with which the ions move under a potential gradient. It, therefore, becomes necessary to define a physically more meaningful quantity called molar conductivity denoted by the symbol Lm (Greek, lambda). It is related to the conductivity of the solution by the equation:  Molar conductivity = Lm = (2.21) c In the above equation, if k is expressed in S m–1 and the concentration, c in mol m–3 then the units of Lm are in S m2 mol–1. It may be noted that: 1 mol m–3 = 1000(L/m3) × molarity (mol/L), and hence  (S cm  1 ) Lm(S cm2 mol–1) =  3 1 1000 L m × molarity (mol L ) If we use S cm–1 as the units for k and mol cm–3, the units of concentration, then the units for Lm are S cm2 mol–1. It can be calculated by using the equation:  (S cm 1 ) × 1000 (cm 3 /L) Lm (S cm2 mol–1) = molarity (mol/L) Both type of units are used in literature and are related to each other by the equations: 1 S m2mol–1 = 104 S cm2mol–1 or 1 S cm2mol–1 = 10–4 S m2mol–1. Resistance of a conductivity cell filled with 0.1 mol L–1 KCl solution is ExampleExampleExampleExampleExample 2.42.42.42.42.4 100 W . If the resistance of the same cell when filled with 0.02 mol L–1 KCl solution is 520 W , calculate the conductivity and molar conductivity of 0.02 mol L–1 KCl solution. The conductivity of 0.1 mol L–1 KCl solution is 1.29 S/m. SolutionSolutionSolutionSolutionSolution The cell constant is given by the equation: Cell constant = G* = conductivity × resistance = 1.29 S/m × 100 W = 129 m–1 = 1.29 cm–1 Conductivity of 0.02 mol L–1 KCl solution = cell constant / resistance G * 129 m –1 = = = 0.248 S m–1 R 520  Concentration = 0.02 mol L–1 = 1000 × 0.02 mol m–3 = 20 mol m–3  Molar conductivity = m  c 248 × 10 –3 S m –1 = –3 = 124 × 10–4 S m2mol–1 20 mol m 1.29 cm –1 Alternatively, k = = 0.248 × 10–2 S cm–1 520  45 Electrochemistry Reprint 2025-26 and Lm = k × 1000 cm3 L–1 molarity–1 0.248×10 –2 S cm –1 ×1000 cm 3 L–1 = –1 0.02 mol L = 124 S cm2 mol–1 ExampleExampleExampleExampleExample 2.52.52.52.52.5 The electrical resistance of a column of 0.05 mol L–1 NaOH solution of diameter 1 cm and length 50 cm is 5.55 × 103 ohm. Calculate its resistivity, conductivity and molar conductivity. SolutionSolutionSolutionSolutionSolution A = p r2 = 3.14 × 0.52 cm2 = 0.785 cm2 = 0.785 × 10–4 m2 l = 50 cm = 0.5 m  l RA 5.55  10 3  0.785cm 2 R = or    = 87.135 W cm A l 50cm 1  1  Conductivity =  = =   S cm–1   87.135  = 0.01148 S cm–1  × 1000 Molar conductivity, m = cm3 L–1 c 0.01148 S cm –1 ×1000 cm 3 L–1 = –1 0.05 mol L = 229.6 S cm2 mol–1 If we want to calculate the values of different quantities in terms of ‘m’ instead of ‘cm’, RA = l 5.55 × 10 3  × 0.785×10 –4 m 2 = = 87.135 ×10–2 W m 0.5 m 1 100  =  m = 1.148 S m–1 = 87.135  1.148 S m –1 and m = = –3 = 229.6 × 10–4 S m2 mol–1. c 50 mol m 2.4.2 Variation of Both conductivity and molar conductivity change with the Conductivity concentration of the electrolyte. Conductivity always decreases with and Molar decrease in concentration both, for weak and strong electrolytes. Conductivity This can be explained by the fact that the number of ions per unit with volume that carry the current in a solution decreases on dilution. Concentration The conductivity of a solution at any given concentration is the conductance of one unit volume of solution kept between two Chemistry 46 Reprint 2025-26 platinum electrodes with unit area of cross section and at a distance of unit length. This is clear from the equation: A G = =  (both A and l are unity in their appropriate units in l m or cm) Molar conductivity of a solution at a given concentration is the conductance of the volume V of solution containing one mole of electrolyte kept between two electrodes with area of cross section A and distance of unit length. Therefore, κA Λm = =κ l Since l = 1 and A = V ( volume containing 1 mole of electrolyte) Lm = k V (2.22) Molar conductivity increases with decrease in concentration. This is because the total volume, V, of solution containing one mole of electrolyte also increases. It has been found that decrease in k on dilution of a solution is more than compensated by increase in its volume. Physically, it means that at a given concentration, Lm can be defined as the conductance of the electrolytic solution kept between the electrodes of a conductivity cell at unit distance but having area of cross section large enough to accommodate sufficient volume of solution that contains one mole of the electrolyte. When concentration approaches zero, the molar conductivity is known as limiting molar conductivity and is represented by theFig. 2.6: Molar conductivity versus c½ for acetic acid (weak electrolyte) and potassium symbol L°m . The variation in Lm with chloride (strong electrolyte) in aqueous concentration is different (Fig. 2.6) for solutions. strong and weak electrolytes. Strong Electrolytes For strong electrolytes, Lm increases slowly with dilution and can be represented by the equation: Lm = L°m – A c ½ (2.23) It can be seen that if we plot (Fig. 2.6) Lm against c1/2, we obtain a straight line with intercept equal to L°m and slope equal to ‘–A’. The value of the constant ‘A’ for a given solvent and temperature depends on the type of electrolyte i.e., the charges on the cation and anion produced on the dissociation of the electrolyte in the solution. Thus, NaCl, CaCl2, MgSO4 are known as 1-1, 2-1 and 2-2 electrolytes respectively. All electrolytes of a particular type have the same value for ‘A’. 47 Electrochemistry Reprint 2025-26 ExampleExampleExampleExampleExample 2.62.62.62.62.6 The molar conductivity of KCl solutions at different concentrations at 298 K are given below: c/mol L–1 Lm/S cm2 mol–1 0.000198 148.61 0.000309 148.29 0.000521 147.81 0.000989 147.09 Show that a plot between Lm and c1/2 is a straight line. Determine the values of L°m and A for KCl. SolutionSolutionSolutionSolutionSolution Taking the square root of concentration we obtain: c1/2/(mol L–1 )1/2 Lm/S cm2mol–1 0.01407 148.61 0.01758 148.29 0.02283 147.81 0.03145 147.09 A plot of Lm ( y-axis) and c1/2 (x-axis) is shown in (Fig. 3.7). It can be seen that it is nearly a straight line. From the intercept (c1/2 = 0), we find that L°m = 150.0 S cm2 mol–1 and A = – slope = 87.46 S cm2 mol–1/(mol/L–1)1/2. Fig. 2.7: Variation of Lm against c½. Chemistry 48 Reprint 2025-26 Kohlrausch examined L°m values for a number of strong electrolytes and observed certain regularities. He noted that the difference in L°m of the electrolytes NaX and KX for any X is nearly constant. For example at 298 K: m L°m (KCl) – L°m (NaCl) = L°m (KBr) – L° (NaBr) = L°m (KI) – L°m (NaI) ≃ 23.4 S cm2 mol–1 and similarly it was found that L°m (NaBr)– L°m (NaCl) = L°m (KBr) – L°m (KCl) ≃ 1.8 S cm2 mol–1 On the basis of the above observations he enunciated Kohlrausch law of independent migration of ions. The law states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte. Thus, – are limiting molar conductivity of the sodium and chlorideif l°Na+ and l°Cl ions respectively, then the limiting molar conductivity for sodium chloride is given by the equation: l° l° L°m – (2.24) (NaCl) = Na+ + Cl In general, if an electrolyte on dissociation gives n+ cations and n– anions then its limiting molar conductivity is given by: L°m = n+ l°+ + n– l°– (2.25) Here, l°+ and l°– are the limiting molar conductivities of the cation and anion respectively. The values of l° for some cations and anions at 298 K are given in Table 2.4. Table 2.4: Limiting Molar Conductivity for some Ions in Water at 298 K Ion l0/(S cm2mol–1) Ion l 0/(S cm2 mol–1) H+ 349.6 OH– 199.1 Na+ 50.1 Cl– 76.3 K+ 73.5 Br– 78.1 Ca2+ 119.0 CH3COO– 40.9 2 Mg2+ 106.0 SO4 160.0 Weak Electrolytes Weak electrolytes like acetic acid have lower degree of dissociation at higher concentrations and hence for such electrolytes, the change in Lm with dilution is due to increase in the degree of dissociation and consequently the number of ions in total volume of solution that contains 1 mol of electrolyte. In such cases Lm increases steeply (Fig. 2.6) on dilution, especially near lower concentrations. Therefore, L°m cannot be obtained by extrapolation of Lm to zero concentration. At infinite dilution (i.e., concentration c ® zero) electrolyte dissociates completely (a =1), but at such low concentration the conductivity of the solution is so low that it cannot be measured accurately. Therefore, L°m for weak electrolytes is obtained by using Kohlrausch law of independent migration of ions (Example 2.8). At any concentration c, if a is the degree of dissociation 49 Electrochemistry Reprint 2025-26 then it can be approximated to the ratio of molar conductivity Lm at the concentration c to limiting molar conductivity, L0m . Thus we have: m  = ° (2.26) m But we know that for a weak electrolyte like acetic acid (Class XI, Unit 7), c  2 cm2 c m2 K = = =   a 1   m m  m   m  (2.27) m 2 1     m  Applications of Kohlrausch law Using Kohlrausch law of independent migration of ions, it is possible to calculate L0m for any electrolyte from the lo of individual ions. Moreover, for weak electrolytes like acetic acid it is possible to determine the value of its dissociation constant once we know the L0m and Lm at a given concentration c. ExampleExampleExampleExampleExample 2.72.72.72.72.7 Calculate L0m for CaCl2 and MgSO4 from the data given in Table 3.4. SolutionSolutionSolutionSolutionSolution We know from Kohlrausch law that – = 119.0 S cm2 mol–1 + 2(76.3) S cm2 mol–1 m  CaCl 2  = Ca 2+  2 Cl = (119.0 + 152.6) S cm2 mol–1 = 271.6 S cm2 mol–1 2+  m  MgSO 4  = Mg  SO 2–4 = 106.0 S cm2 mol–1 + 160.0 S cm2 mol–1 = 266 S cm2 mol–1 . ExampleExampleExampleExampleExample 2.82.82.82.82.8 L0m for NaCl, HCl and NaAc are 126.4, 425.9 and 91.0 S cm2 mol–1 respectively. Calculate L0 for HAc.       +  Ac – H + Cl – Ac – Na + Cl – Na + SolutionSolutionSolutionSolutionSolution m  HAc  = H = m  HCl   m  NaAc   m  NaCl  = (425.9 + 91.0 – 126.4 ) S cm2 mol –1 = 390.5 S cm2 mol–1 . ExampleExampleExampleExampleExample 2.92.92.92.92.9 The conductivity of 0.001028 mol L–1 acetic acid is 4.95 × 10–5 S cm–1. Calculate its dissociation constant if L0m for acetic acid is 390.5 S cm2 mol–1.  4 . 95 10  5 Scm  1 1000cm 3 SolutionSolutionSolutionSolutionSolution m =  1  = 48.15 S cm3 mol–1 c 0 . 001028 mol L L m 48.15 Scm 2 mol 1 a =   2  1 = 0.1233 m 390.5 Scm mol c2 0 .001028molL–1  (0 .1233) 2 k = = 1.78 × 10–5 mol L–1  1   1  0 .1233 Chemistry 50 Reprint 2025-26 IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 2.7 Why does the conductivity of a solution decrease with dilution? 2.8 Suggest a way to determine the L°m value of water. 2.9 The molar conductivity of 0.025 mol L–1 methanoic acid is 46.1 S cm2 mol–1. Calculate its degree of dissociation and dissociation constant. Given l0(H+) = 349.6 S cm2 mol–1 and l0 (HCOO–) = 54.6 S cm2 mol–1. 2.52.52.52.52.5 ElectrolyticElectrolyticElectrolyticElectrolyticElectrolytic In an electrolytic cell external source of voltage is used to bring about a chemical reaction. The electrochemical processes are of great importance CellsCellsCellsCellsCells andandandandand in the laboratory and the chemical industry. One of the simplest electrolytic ElectrolysisElectrolysisElectrolysisElectrolysisElectrolysis cell consists of two copper strips dipping in an aqueous solution of copper sulphate. If a DC voltage is applied to the two electrodes, then Cu 2+ ions discharge at the cathode (negatively charged) and the following reaction takes place: Cu2+(aq) + 2e– ® Cu (s) (2.28) Copper metal is deposited on the cathode. At the anode, copper is converted into Cu2+ ions by the reaction: Cu(s) ® Cu2+(s) + 2e– (2.29) Thus copper is dissolved (oxidised) at anode and deposited (reduced) at cathode. This is the basis for an industrial process in which impure copper is converted into copper of high purity. The impure copper is made an anode that dissolves on passing current and pure copper is deposited at the cathode. Many metals like Na, Mg, Al, etc. are produced on large scale by electrochemical reduction of their respective cations where no suitable chemical reducing agents are available for this purpose. Sodium and magnesium metals are produced by the electrolysis of their fused chlorides and aluminium is produced by electrolysis of aluminium oxide in presence of cryolite. Quantitative Aspects of Electrolysis Michael Faraday was the first scientist who described the quantitative aspects of electrolysis. Now Faraday’s laws also flow from what has been discussed earlier. Faraday’s Laws of Electrolysis After his extensive investigations on electrolysis of solutions and melts of electrolytes, Faraday published his results during 1833-34 in the form of the following well known Faraday’s two laws of electrolysis: (i) First Law: The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte (solution or melt). (ii) Second Law: The amounts of different substances liberated by the same quantity of electricity passing through the electrolytic solution are proportional to their chemical equivalent weights (Atomic Mass of Metal ÷ Number of electrons required to reduce the cation). 51 Electrochemistry Reprint 2025-26 There were no constant current sources available during Faraday’s times. The general practice was to put a coulometer (a standard electrolytic cell) for determining the quantity of electricity passed from the amount of metal (generally silver or copper) deposited or consumed. However, coulometers are now obsolete and we now have constant current (I) sources available and the quantity of electricity Q, passed is given by Q = It Q is in coloumbs when I is in ampere and t is in second. The amount of electricity (or charge) required for oxidation or reduction depends on the stoichiometry of the electrode reaction. For example, in the reaction: Ag +(aq) + e– ® Ag(s) (2.30) One mole of the electron is required for the reduction of one mole of silver ions. We know that charge on one electron is equal to 1.6021 × 10–19C. Therefore, the charge on one mole of electrons is equal to: NA × 1.6021 × 10–19 C = 6.02 × 1023 mol–1 × 1.6021 × 10–19 C = 96487 C mol–1 This quantity of electricity is called Faraday and is represented by the symbol F. For approximate calculations we use 1F ≃ 96500 C mol–1. For the electrode reactions: Mg2+(l) + 2e– ¾® Mg(s) (2.31) Al3+(l) + 3e– ¾® Al(s) (2.32) It is obvious that one mole of Mg2+ and Al3+ require 2 mol of electrons (2F) and 3 mol of electrons (3F) respectively. The charge passed through the electrolytic cell during electrolysis is equal to the product of current in amperes and time in seconds. In commercial production of metals, current as high as 50,000 amperes are used that amounts to about 0.518 F per second. ExampleExampleExampleExampleExample 2.102.102.102.102.10 A solution of CuSO4 is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode? SolutionSolutionSolutionSolutionSolution t = 600 s charge = current × time = 1.5 A × 600 s = 900 C According to the reaction: Cu2+(aq) + 2e– = Cu(s) We require 2F or 2 × 96487 C to deposit 1 mol or 63 g of Cu. For 900 C, the mass of Cu deposited = (63 g mol–1 × 900 C)/(2 × 96487 C mol–1) = 0.2938 g. 2.5.1 Products of Products of electrolysis depend on the nature of material being Electrolysis electrolysed and the type of electrodes being used. If the electrode is inert (e.g., platinum or gold), it does not participate in the chemical reaction and acts only as source or sink for electrons. On the other hand, if the electrode is reactive, it participates in the electrode reaction. Thus, the products of electrolysis may be different for reactive and inert Chemistry 52 Reprint 2025-26 electrodes.The products of electrolysis depend on the different oxidising and reducing species present in the electrolytic cell and their standard electrode potentials. Moreover, some of the electrochemical processes although feasible, are so slow kinetically that at lower voltages these do not seem to take place and extra potential (called overpotential) has to be applied, which makes such process more difficult to occur. For example, if we use molten NaCl, the products of electrolysis are sodium metal and Cl2 gas. Here we have only one cation (Na+) which is reduced at the cathode (Na+ + e– ® Na) and one anion (Cl–) which is oxidised at the anode (Cl– ® ½Cl2 + e– ). During the electrolysis of aqueous sodium chloride solution, the products are NaOH, Cl2 and H2. In this case besides Na+ and Cl– ions we also have H+ and OH– ions along with the solvent molecules, H2O. At the cathode there is competition between the following reduction reactions: Na+ (aq) + e– ® Na (s) E (ocell ) = – 2.71 V H+ (aq) + e– ® ½ H2 (g) E (ocell ) = 0.00 V The reaction with higher value of Eo is preferred and therefore, the reaction at the cathode during electrolysis is: H+ (aq) + e– ® ½ H2 (g) (2.33) but H+ (aq) is produced by the dissociation of H2O, i.e., H2O (l ) ® H+ (aq) + OH– (aq) (2.34) Therefore, the net reaction at the cathode may be written as the sum of (2.33) and (2.34) and we have H2O (l ) + e– ® ½H2(g) + OH– (2.35) At the anode the following oxidation reactions are possible: Cl– (aq) ® ½ Cl2 (g) + e– E (ocell ) = 1.36 V (2.36) 2H2O (l ) ® O2 (g) + 4H+(aq) + 4e– E (ocell ) = 1.23 V (2.37) The reaction at anode with lower value of E o is preferred and therefore, water should get oxidised in preference to Cl– (aq). However, on account of overpotential of oxygen, reaction (2.36) is preferred. Thus, the net reactions may be summarised as: NaCl (aq) H 2 O → Na+ (aq) + Cl– (aq) Cathode: H2O(l ) + e– ® ½ H2(g) + OH– (aq) Anode: Cl– (aq) ® ½ Cl2(g) + e– Net reaction: NaCl(aq) + H2O(l) ® Na+(aq) + OH–(aq) + ½H2(g) + ½Cl2(g) The standard electrode potentials are replaced by electrode potentials given by Nernst equation (Eq. 2.8) to take into account the concentration effects. During the electrolysis of sulphuric acid, the following processes are possible at the anode: 2H2O(l) ® O2(g) + 4H+(aq) + 4e– E (ocell ) = +1.23 V (2.38) 53 Electrochemistry Reprint 2025-26 2SO42– (aq) ® S2O8 2– (aq) + 2e– E (ocell ) = 1.96 V (2.39) For dilute sulphuric acid, reaction (2.38) is preferred but at higher concentrations of H2SO4, reaction (2.39) is preferred. IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 2.10 If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire? 2.11 Suggest a list of metals that are extracted electrolytically. 2.12 Consider the reaction: Cr2O7 2– + 14H+ + 6e– ® 2Cr3+ + 7H2O What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr2O7 2–? 2.62.62.62.62.6 BatteriesBatteriesBatteriesBatteriesBatteries Any battery (actually it may have one or more than one cell connected in series) or cell that we use as a source of electrical energy is basically a galvanic cell where the chemical energy of the redox reaction is converted into electrical energy. However, for a battery to be of practical use it should be reasonably light, compact and its voltage should not vary appreciably during its use. There are mainly two types of batteries. 2.6.1 Primary In the primary batteries, the reaction occurs only once and after use Batteries over a period of time battery becomes dead and cannot be reused again. The most familiar example of this type is the dry cell (known as Leclanche cell after its discoverer) which is used commonly in our transistors and clocks. The cell consists of a zinc container that also acts as anode and the cathode is a carbon (graphite) rod surrounded by powdered manganese dioxide and carbon (Fig.2.8). The space between the electrodes is filled by a moist paste of ammonium chloride (NH4Cl) and zinc chloride (ZnCl2). The electrode reactions are complex, but they can be written approximately as follows : Anode: Zn(s) ¾® Zn2+ + 2e– Cathode: MnO2+ NH4 ++ e–¾® MnO(OH) + NH3 In the reaction at cathode, manganese is reduced from the + 4 oxidation state to the +3 state. Ammonia produced in the reaction forms a complex with Zn2+ to give [Zn (NH3)4]2+. The cell has a potential of nearly 1.5 V. Mercury cell, (Fig. 2.9) suitable for low current devices like hearing aids, watches, etc. consists of zinc – mercury amalgam as anode and a paste of HgO and carbon as the Fig. 2.8: A commercial dry cell cathode. The electrolyte is a paste of KOH and ZnO. The consists of a graphite electrode reactions for the cell are given below: (carbon) cathode in a Anode: Zn(Hg) + 2OH– ¾® ZnO(s) + H2O + 2e– zinc container; the latter Cathode: HgO + H2O + 2e– ¾® Hg(l) + 2OH– acts as the anode. Chemistry 54 Reprint 2025-26 The overall reaction is represented by Zn(Hg) + HgO(s) ¾® ZnO(s) + Hg(l) The cell potential is approximately 1.35 V and remains constant during its Fig. 2.9 life as the overall reaction does not Commonly used involve any ion in solution whose mercury cell. The concentration can change during its life reducing agent is time. zinc and the oxidising agent is mercury (II) oxide. 2.6.2 Secondary A secondary cell after use can be recharged by passing current Batteries through it in the opposite direction so that it can be used again. A good secondary cell can undergo a large number of discharging and charging cycles. The most important secondary cell is the lead storage battery (Fig. 2.10) commonly used in automobiles and invertors. It consists of a lead anode and a grid of lead packed with lead dioxide (PbO2 ) as cathode. A 38% solution of sulphuric acid is used as an electrolyte. The cell reactions when the battery is in use are given below: Anode: Pb(s) + SO42–(aq) ® PbSO4(s) + 2e– Cathode: PbO2(s) + SO42–(aq) + 4H+(aq) + 2e– ® PbSO4 (s) + 2H2O (l) i.e., overall cell reaction consisting of cathode and anode reactions is: Pb(s) + PbO2(s) + 2H2SO4(aq) ® 2PbSO4(s) + 2H2O(l) On charging the battery the reaction is reversed and PbSO4(s) on anode and cathode is converted into Pb and PbO2, respectively. Fig. 2.10: The Lead storage battery. 55 Electrochemistry Reprint 2025-26 Another important secondary cell is the nickel-cadmium cell (Fig. 2.11) which has longer life than the lead storage cell but Fig. 2.11 more expensive to manufacture. A rechargeable We shall not go into details of nickel-cadmium cell working of the cell and the Positive plate in a jelly roll electrode reactions during arrangement and Separator charging and discharging. separated by a layer Negative plate The overall reaction during soaked in moist discharge is: sodium or potassium hydroxide. Cd (s) + 2Ni(OH)3 (s) ® CdO (s) + 2Ni(OH)2 (s) + H2O (l ) 2.72.72.72.72.7 FuelFuelFuelFuelFuel CellsCellsCellsCellsCells Production of electricity by thermal plants is not a very efficient method and is a major source of pollution. In such plants, the chemical energy (heat of combustion) of fossil fuels (coal, gas or oil) is first used for converting water into high pressure steam. This is then used to run a turbine to produce electricity. We know that a galvanic cell directly converts chemical energy into electricity and is highly efficient. It is now possible to make such cells in which reactants are fed continuously to the electrodes and products are removed continuously from the electrolyte compartment. Galvanic cells that are designed to convert the energy of combustion of fuels like hydrogen, methane, methanol, etc. directly into electrical energy are called fuel cells. One of the most successful fuel cells uses the reaction of hydrogen with oxygen to form water (Fig. 2.12). The cell was used for providing electrical power in the Apollo space programme. The water vapours produced during the reaction were condensed and added to the drinking water supply for the astronauts. In the cell, hydrogen and oxygen are bubbled through porous carbon electrodes into concentrated aqueous sodium hydroxide solution. Catalysts like finely divided platinum or palladium metal are incorporated into the electrodes for increasing the rate of electrode Fig. 2.12: Fuel cell using H2 and O2 produces electricity. reactions. The electrode reactions are given below: Cathode: O2(g) + 2H2O(l) + 4e–¾® 4OH–(aq) Anode: 2H2 (g) + 4OH–(aq) ¾® 4H2O(l) + 4e– Overall reaction being: 2H2(g) + O2(g) ¾® 2H2O(l ) The cell runs continuously as long as the reactants are supplied. Fuel cells produce electricity with an efficiency of about 70 % compared Chemistry 56 Reprint 2025-26 to thermal plants whose efficiency is about 40%. There has been tremendous progress in the development of new electrode materials, better catalysts and electrolytes for increasing the efficiency of fuel cells. These have been used in automobiles on an experimental basis. Fuel cells are pollution free and in view of their future importance, a variety of fuel cells have been fabricated and tried. 2.82.82.82.82.8 CorrosionCorrosionCorrosionCorrosionCorrosion Corrosion slowly coats the surfaces of metallic objects with oxides or other salts of the metal. The rusting of iron, tarnishing of silver, development of green coating on copper and bronze are some of the examples of corrosion. It causes enormous damage to buildings, bridges, ships and to all objects made of metals especially that of iron. We lose crores of rupees every year on account of corrosion. In corrosion, a metal is oxidised by loss of electrons to oxygen and formation of oxides. Corrosion of iron (commonly known as rusting) occurs in presence of water and air. The chemistry of corrosion is quite complex but it may be considered Oxidation: Fe (s)® Fe2+ (aq) +2e– essentially as an electrochemical Reduction: O2 (g) + 4H+(aq) +4e– ® 2H2O(l) phenomenon. At a particular spot Atomospheric (Fig. 2.13) of an object made of iron,oxidation: 2Fe2+(aq) + 2H2O(l) + ½O2(g) ® Fe2O3(s) + 4H+(aq) oxidation takes place and that spot Fig. 2.13: Corrosion of iron in atmosphere behaves as anode and we can write the reaction E o Anode: 2 Fe (s) ¾® 2 Fe2+ + 4 e– (Fe 2+ /Fe) = – 0.44 V Electrons released at anodic spot move through the metal and go to another spot on the metal and reduce oxygen in the presence of H+ (which is believed to be available from H2CO3 formed due to dissolution of carbon dioxide from air into water. Hydrogen ion in water may also be available due to dissolution of other acidic oxides from the atmosphere). This spot behaves as cathode with the reaction E o =1.23 V Cathode: O2(g) + 4 H+(aq) + 4 e– ¾® 2 H2O (l) H + | O 2 | H 2 O The overall reaction being: 2Fe(s) + O2(g) + 4H+(aq) ¾® 2Fe2 +(aq) + 2 H2O (l) E o(cell) =1.67 V The ferrous ions are further oxidised by atmospheric oxygen to ferric ions which come out as rust in the form of hydrated ferric oxide (Fe2O3. x H2O) and with further production of hydrogen ions. Prevention of corrosion is of prime importance. It not only saves money but also helps in preventing accidents such as a bridge collapse or failure of a key component due to corrosion. One of the simplest methods of preventing corrosion is to prevent the surface of the metallic object to come in contact with atmosphere. This can be done by covering the surface with paint or by some chemicals (e.g. bisphenol). Another simple method is to cover the surface by other metals (Sn, Zn, etc.) that are inert or react to save the object. An electrochemical method is to provide a sacrificial electrode of another metal (like Mg, Zn, etc.) which corrodes itself but saves the object. 57 Electrochemistry Reprint 2025-26 IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 2.13 Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging. 2.14 Suggest two materials other than hydrogen that can be used as fuels in fuel cells. 2.15 Explain how rusting of iron is envisaged as setting up of an electrochemical cell. TheTheTheTheThe HydrogenHydrogenHydrogenHydrogenHydrogen EconomyEconomyEconomyEconomyEconomy At present the main source of energy that is driving our economy is fossil fuels such as coal, oil and gas. As more people on the planet aspire to improve their standard of living, their energy requirement will increase. In fact, the per capita consumption of energy used is a measure of development. Of course, it is assumed that energy is used for productive purpose and not merely wasted. We are already aware that carbon dioxide produced by the combustion of fossil fuels is resulting in the ‘Greenhouse Effect’. This is leading to a rise in the temperature of the Earth’s surface, causing polar ice to melt and ocean levels to rise. This will flood low-lying areas along the coast and some island nations such as Maldives face total submergence. In order to avoid such a catastrope, we need to limit our use of carbonaceous fuels. Hydrogen provides an ideal alternative as its combustion results in water only. Hydrogen production must come from splitting water using solar energy. Therefore, hydrogen can be used as a renewable and non polluting source of energy. This is the vision of the Hydrogen Economy. Both the production of hydrogen by electrolysis of water and hydrogen combustion in a fuel cell will be important in the future. And both these technologies are based on electrochemical principles. SummarySummarySummarySummarySummary An electrochemical cell consists of two metallic electrodes dipping in electrolytic solution(s). Thus an important component of the electrochemical cell is the ionic conductor or electrolyte. Electrochemical cells are of two types. In galvanic cell, the chemical energy of a spontaneous redox reaction is converted into electrical work, whereas in an electrolytic cell, electrical energy is used to carry out a non- spontaneous redox reaction. The standard electrode potential for any electrode dipping in an appropriate solution is defined with respect to standard electrode potential of hydrogen electrode taken as zero. The standard potential of the cell can be obtained by taking the difference of the standard potentials of cathode and anode ( E (ocell ) = Eocathode – Eoanode). The standard potential of the cells are related to standard Gibbs energy (DrGo = –nF E (ocell ) ) and equilibrium constant (DrGo = – RT ln K) of the reaction taking place in the cell. Concentration dependence of the potentials of the electrodes and the cells are given by Nernst equation. The conductivity, k, of an electrolytic solution depends on the concentration of the electrolyte, nature of solvent and temperature. Molar conductivity, Lm, is defined by = k/c where c is the concentration. Conductivity decreases but molar conductivity increases with decrease in concentration. It increases slowly with decrease in concentration for strong electrolytes while the increase is very steep for weak electrolytes in very dilute solutions. Kohlrausch found that molar conductivity at infinite dilution, for an electrolyte is sum of the contribution of the Chemistry 58 Reprint 2025-26 molar conductivity of the ions in which it dissociates. It is known as law of independent migration of ions and has many applications. Ions conduct electricity through the solution but oxidation and reduction of the ions take place at the electrodes in an electrochemical cell. Batteries and fuel cells are very useful forms of galvanic cell. Corrosion of metals is essentially an electrochemical phenomenon. Electrochemical principles are relevant to the Hydrogen Economy. ExercisesExercisesExercisesExercisesExercises