Question

\(Cu(S)+Sn^{2+} (0.001M) Cu^{2+} (0.01M) + Sn(s)\)
The Gibbs free energy change for the above reaction at 298 K is x × 10^-1 × kJ mol^-1. The value of x is ________ .[nearest integer]
[Given \(E^{-}_{cu^{2+} / cu} = 0.34V; E^{-}_{Sn^{2+} / Sn} = - 0.14V; F = 96500 C ∼ mol^{-1}\)]

Answer 1

Correct Answer: 983

The correct answer is 983
\(Cu + Sn^{2+} → Cu^{2+} + Sn(s)\)
\(E°_{cell} = E°_{Ox} + E°_{Red }\)
= - 0.34 - 0.14
= 0.48 V
\(E = E° - \frac{0.0591}{2} log\frac{[ Cu^{2+}]}{[ Sn^{2+}]}\)
= - 0.48 - 0.0295 log 10
= - 0.5095 V
\(ΔG = -nFE\)
= -2 × 96500 × 0.5095 J / mol
= 98333.5 × 10^-3 kJ / mol
= 983.3 × 10^-1 kJ / mol
= 983 × 10^-1 kJ / mol