Electrochemical Cell — Galvanic, notation, cell reaction
Electrochemistry
9
JEE Qs
8%
Hard
75
min
Mastering cell notation and identifying oxidation/reduction at respective electrodes is fundamental for all subsequent electrochemistry calculations and problem-solving.
🧮 Key Formulas
✅ Key Points for JEE
- 1In a galvanic (voltaic) cell, the anode is the negative electrode (site of oxidation) and the cathode is the positive electrode (site of reduction).
- 2Electrons flow from the anode to the cathode through the external circuit, while conventional current flows from cathode to anode.
- 3The salt bridge connects the two half-cells, allowing the migration of ions to maintain electrical neutrality and complete the circuit, preventing charge build-up.
- 4Cell notation always represents the anode half-cell on the left and the cathode half-cell on the right, separated by a double vertical line (salt bridge), with phase boundaries denoted by a single vertical line.
- 5The overall cell reaction is the sum of the oxidation half-reaction at the anode and the reduction half-reaction at the cathode, ensuring electron balance and conservation of mass.
⚠️ Common Mistakes
- ✕Confusing the polarity of anode/cathode between galvanic cells (anode negative, cathode positive) and electrolytic cells (anode positive, cathode negative).
- ✕Incorrectly writing cell notation, such as reversing the order of anode/cathode, omitting phase boundaries, or misrepresenting the states of matter.
- ✕Errors in balancing the overall cell reaction, either by not cancelling electrons correctly or not balancing atoms and charges in the individual half-reactions.
📝 Practice Questions
See allQ54.The element that does not belong to the same period of the remaining elements (modern periodic table) is: 2025 (23 Jan Shift 1) JEE Main Previous Year Paper (1) Iridium (2) Platinum (3) Osmium (4) Palladium
Q67.For the given cell Fe2+ (aq) + Ag+ (aq) →Fe3+(aq) + Ag(s) The standard cell potential of the above reaction is Ag+ + e−→Ag Eθ = xV Given: Fe2+ + 2e−→Fe Eθ = yV Fe3+ + 3e−→Fe Eθ = zV (1) x + y −z (2) x + 2y (3) x + 2y −3z (4) y −2x
Q58.Which of the following electrolyte can be used to obtain H2 S2O8 by the process of electrolysis? (1) Dilute solution of sodium sulphate. (2) Acidified dilute solution of sodium sulphate. (3) Dilute solution of sulphuric acid (4) Concentrated solution of sulphuric acid
Q66.A solution of aluminium chloride is electrolysed for 30 minutes using a current of 2 A . The amount of the aluminium deposited at the cathode is [Given : molar mass of aluminium and chlorine are 27 g mol−1 and 35.5 g mol−1 respectively. Faraday constant = 96500Cmol−1] (1) 1.660 g (2) 0.336 g (3) 0.441 g (4) 1.007 g 2025 (22 Jan Shift 1) JEE Main Previous Year Paper
Q58.Standard electrode potentials for a few half cells are mentioned below : E∘ = 0.34 V, E∘ = −0.76 V Cu2+/Cu Zn2+/Zn Which one of the following cells gives the most negative value of E∘Ag+/Ag = 0.80 V, E∘Mg2+/Mg = −2.37 V ΔG∘ ? (1) Zn Zn2+(1M) Ag+(1M) Ag (2) Zn Zn2+(1M) Mg2+(1M) Mg (3) Ag Ag+(1M) Mg2+(1M) Mg (4) Cu Cu2+(1M)∥Ag+(1M) Ag
Q57. E∘Cr2O2−7 /Cr3+ = 1.33 V E∘Cl2/Cl(−) = 1.36 V Based on the data given below : the strongest reducing agent is : E∘ = 1.51 V E∘ = −0.74 V MnO−4 /Mn2+ Cr3+/Cr (1) Cr (2) Cl− (3) MnO−4 (4) Mn2+
NCERT Chapters
- Class 12 Chemistry Ch 3: Electrochemistry