Q48.Given Fe3+(aq) + e−→Fe2+(aq); E0 = +0.77 V Al3+(aq) + 3e−→Al(s); E0 = −1.66 V Br2(aq) + 2e−→2Br−; E0 = +1.09 V Considering the electrode potentials, which of the following represents the correct order of reducing power? JEE Main 2014 (11 Apr Online) JEE Main Previous Year Paper (1) Fe2+ < Al < Br− (2) Br−< Fe2+ < Al (3) Al < Br−< Fe2+ (4) Al < Fe2+ < Br−
What This Question Tests
This question tests the relationship between standard electrode potentials and reducing power; a more negative reduction potential corresponds to a stronger reducing agent.
Concepts Tested
📚 NCERT Sections This Tests
2.2 — Given The Standard Electrode Potentials,
Chemistry Class 11 · Chapter 2
2.2 Given the standard electrode potentials, K+/K = –2.93V, Ag+/Ag = 0.80V, Hg2+/Hg = 0.79V Mg2+/Mg = –2.37 V, Cr3+/Cr = – 0.74V Arrange these metals in their increasing order of reducing power.
2.4 — Calculate The Standard Cell Potentials Of Galvanic Cell In Which The Following
Chemistry Class 11 · Chapter 2
2.4 Calculate the standard cell potentials of galvanic cell in which the following reactions take place: (i) 2Cr(s) + 3Cd2+(aq) ® 2Cr3+(aq) + 3Cd (ii) Fe2+(aq) + Ag+(aq) ® Fe3+(aq) + Ag(s) Calculate the DrGo and equilibrium constant of the reactions. 2.5 Write the Nernst equation and emf of the following cells at 298 K: (i) Mg(s)|Mg2+(0.001M)||Cu2+(0.0001 M)|Cu(s) (ii) Fe(s)|Fe2+(0.001M)||H+(1M)|H2(g)(1bar)| Pt(s) (iii) Sn(s)|Sn2+(0.050 M)||H+(0.020 M)|H2(g) (1 bar)|Pt(s) (iv) Pt(s)|Br–(0.010 M)|Br2(l )||H+(0.030 M)| H2(g) (1 bar)|Pt(s).
2.1 — Arrange The Following Metals In The Order In Which They Displace Each Other
Chemistry Class 11 · Chapter 2
2.1 Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn.
📋 Question Details
- Chapter
- Electrochemistry
- Topic
- Reducing power
- Year
- 2014
- Shift
- 11 Apr Online
- Q Number
- Q48
- Type
- MCQ
- NCERT Ref
- Class 12 Chemistry Ch 3: Electrochemistry
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