Q37 / 125 · BITSAT 2016 · Chemistry
The equilibrium constant for the disproportionation reaction $2 \mathrm{Cu}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{ag})$ at $25^{\circ} \mathrm{C}$
$\left(\mathrm{E}^{\circ} \mathrm{Cu}^{+} / \mathrm{Cu}=0.52 \mathrm{~V}, \mathrm{E}^{\circ} \mathrm{Cu}^{2+} / \mathrm{Cu}=0.16 \mathrm{~V}\right)$ is
- A$6 \times 10^{4}$
- B$6 \times 10^{6}$
- Correct answerC$1.2 \times 10^{6}$
- D$1.2 \times 10^{-6}$
Solution
The reaction
$$
\begin{array}{l}
2 \mathrm{Cu}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \\
\mathrm{E}_{\mathrm{cell}}=\mathrm{E}_{\mathrm{cell}}^{\circ}-\frac{0.0592}{1} \log \frac{\left[\mathrm{Cu}^{2+}\right]}{\left[\mathrm{Cu}^{+}\right]^{2}}
\end{array}
$$
At equilibrium $\mathrm{E}_{\mathrm{cell}}=0$
$$
\therefore \quad \mathrm{E}_{\text {cell }}^{\circ}=0.0592 \log \mathrm{K}_{\mathrm{c}}
$$
or, $\quad \log \mathrm{K}_{\mathrm{c}}=\frac{0.52-0.16}{0.0592}$
$\therefore \mathrm{K}_{\mathrm{c}}=1.2 \times 10^{6}$
Hence, the correct answer is option: 'C'.