UPSC Combined Geo-Scientist, Mains · 2026 · Mains Paper I

Q6(c). Nernst equation for the silver–copper cell

Question

Derive Nernst equation for the cell potential of silver-copper cell.

For Cu | Cu²⁺(a₁) || Ag+(a₂) | Ag: E=E^°-(RT)/(2F)lna_(Cu²⁺)a_(Ag⁺)²; at 298 K E=0.46-(0.0591)/(2)log([Cu²⁺])/([Ag⁺]²) V.

anode (oxidation) Cu → Cu²⁺ + 2e⁻; cathode (reduction) 2Ag⁺ + 2e⁻ → 2Ag. Cell reaction (n=2): Cu + 2Ag⁺ → Cu²⁺ + 2Ag.
the van't Hoff isotherm: Δ G=Δ G^°+RTln Q, with Q=a_(Cu²⁺) a_(Ag)²a_(Cu) a_(Ag⁺)²=a_(Cu²⁺)a_(Ag⁺)² (activity of a pure solid is 1).
electrical work for n electrons: Δ G=-nFE and Δ G^°=-nFE^°. Substituting, -nFE=-nFE^°+RTln Q ⇒ E=E^°-(RT)/(nF)ln Q
E=E^°-(RT)/(2F)lna_(Cu²⁺)a_(Ag⁺)² 298 K, log E=E^°-(0.0591)/(2)log[Cu²⁺][Ag⁺]²

The full model answer

Worked out step by step, as it should go in the answer booklet. The solution sheet follows it.

Half-cells: anode (oxidation) Cu → Cu²⁺ + 2e⁻; cathode (reduction) 2Ag⁺ + 2e⁻ → 2Ag. Cell reaction (n=2): Cu + 2Ag⁺ → Cu²⁺ + 2Ag.
Free energy: the van't Hoff isotherm: Δ G=Δ G^°+RTln Q, with Q=a_(Cu²⁺) a_(Ag)²a_(Cu) a_(Ag⁺)²=a_(Cu²⁺)a_(Ag⁺)² (activity of a pure solid is 1).
Link with e.m.f.: electrical work for n electrons: Δ G=-nFE and Δ G^°=-nFE^°. Substituting, -nFE=-nFE^°+RTln Q ⇒ E=E^°-(RT)/(nF)ln Q
For the silver–copper cell: E=E^°-(RT)/(2F)lna_(Cu²⁺)a_(Ag⁺)² 298 K, log E=E^°-(0.0591)/(2)log[Cu²⁺][Ag⁺]²
Standard e.m.f.: E^°=E^°_(Ag+/Ag)-E^°_(Cu²⁺/Cu)=0.80-0.34=0.46 V, Δ G^°=-2(96485)(0.46)=-88.8 kJ mol⁻¹.
At equilibrium E=0: log K=(nE^°)/(0.0591)=(2(0.46))/(0.0591)=15.6, K=4×10¹⁵.

each ten-fold rise in Ag⁺ concentration raises E by 0.0591 V (two electrons, power 2); this is the principle of a silver-ion electrode (a pAg meter).

Coverscell e.m.f.Nernst equation for the silver–copper cell