Q23 / 75 · JEE Main 2020 (02 Sep Shift 2) · Chemistry

For the disproportionation reaction $2 {Cu}^{+} (aq) ⇌ Cu (s) + {Cu}^{2 +} (aq)$ at $298 K,$ $\ln K$ (where $K$ is the equilibrium constant) is _______ $\times 10^{- 1}$
Given $: (E °_{{Cu}^{2 +} / {Cu}^{+}} = 0.16 V E °_{{Cu}^{+} / Cu} = 0.52 V \frac{RT}{F} = 0.025)$

Correct answer: 144

Solution

$E_{cell}^{o} = E_{{Cu}^{+} / Cu}^{°} - E_{{Cu}^{2 +} / {Cu}^{+}}^{°}$

$= 0.52 - 0.16 = 0.36 V$

$E_{cell}^{°} = \frac{RT}{nF} {lnK}_{eq}$

$0.36 = \frac{0.025}{1} lnk$

$In k = 14.4 = 144 \times 10^{- 1}$

Hence, the correct answer is 144.