JEE PYQ: Electrochemistry Question 11
Question 11 - 2020 (02 Sep Shift 2)
For the disproportionation reaction $2\text{Cu}^+(\text{aq}) \rightleftharpoons \text{Cu}(s) + \text{Cu}^{2+}(\text{aq})$ at 298 K, ln K (where K is the equilibrium constant) is ____ $\times 10^{-1}$.
Given: $(E^0_{\text{Cu}^{2+}/\text{Cu}^+} = 0.16$ V; $E^0_{\text{Cu}^+/\text{Cu}} = 0.52$ V; $\frac{RT}{F} = 0.025)$
Show Answer
Answer: 144
Solution
$E^0_{\text{cell}} = 0.52 - 0.16 = 0.36$ V. $0.36 = 0.025 \ln K$. $\ln K = 14.4 = 144 \times 10^{-1}$.
BETA
