The reduction potential of the two half cell reactions (occuring in an electrochemical…

Chemistry · JEE Advanced · NTA ExamsElectrochemistry

The reduction potential of the two half cell reactions (occuring in an electrochemical cell) are
PbSO4 + 2e → Pb + \(\mathrm{SO}_{4}^{2-}\)(Eº = – 0.31 V)
Ag+ (aq) + e → Ag (s) (Eº = 0.80 V)
The feasible reaction will be
  1. Pb + \(\mathrm{SO}_{4}^{2-}\) + 2 Ag+ (aq) 2 Ag (s) + PbSO4
  2. PbSO4 + 2 Ag+ (aq) Pb + \(\mathrm{SO}_{4}^{2-}\)+ 2 Ag (s)
  3. Pb + \(\mathrm{SO}_{4}^{2-}\)+ Ag (s) Ag+ (aq) + PbSO4
  4. PbSO4 + Ag (s) Ag+ (aq) + Pb + \(\mathrm{SO}_{4}^{2-}\).

Answer

(A) Pb + SO _ 4 ^ 2- + 2 Ag + (aq) → 2 Ag (s) + PbSO 4

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