JEE PYQ: Electrochemistry Question 10
Question 10 - 2020 (02 Sep Shift 1)
The Gibbs energy change (in J) for the given reaction at $[\text{Cu}^{2+}] = [\text{Sn}^{2+}] = 1$ M and 298 K is: $\text{Cu}(s) + \text{Sn}^{2+}(\text{aq.}) \rightarrow \text{Cu}^{2+}(\text{aq.}) + \text{Sn}(s)$ $(E^0_{\text{Sn}^{2+}|\text{Sn}} = -0.16$ V, $E^0_{\text{Cu}^{2+}|\text{Cu}} = 0.34$ V, Take $F = 96500$ C mol$^{-1}$)
Show Answer
Answer: 96500
Solution
$E^0_{\text{cell}} = -0.16 - 0.34 = -0.50$ V. $\Delta G^0 = -2 \times 96500 \times (-0.5) = 96500$ J.
BETA
