Question

The Gibbs change (in J) for the given reaction at [Cu²⁺] = [Sn²⁺] = 1 M and 298 K is :
Cu(s) + Sn²⁺(aq.) \(\to\) Cu²⁺(aq.) + Sn(s);
(\(E^0_{\mathrm{Sn^{2+}|Sn}} = -0.16\) V, \(E^0_{\mathrm{Cu^{2+}|Cu}} = 0.34\) V)
Take F = 96500 C mol^-1

Hint:

\(\Delta G = -nFE_{cell}\), with \(E_{cell} = E^0_{red}(\text{cathode}) - E^0_{red}(\text{anode})\); here Sn²⁺ is reduced and Cu is oxidised.

Answer 1

Correct Answer: 96500

In the reaction, Sn²⁺ is reduced to Sn (cathode) and Cu is oxidised to Cu²⁺ (anode); n = 2.
\(E^0_{cell} = E^0_{cathode} - E^0_{anode} = (-0.16) - (0.34) = -0.50\) V.
Since both concentrations are 1 M, \(E = E^0 = -0.50\) V.
\(\Delta G = -nFE = -2 \times 96500 \times (-0.50) = +96500\) J.
Answer: 96500