Question

The standard reduction potential for $Cu^{2+}/Cu$ is $+0.34 \,V$. What will be the reduction potential at $pH = 14$ ? [Given: $K_{sp}$ of $Cu(OH)_2$ is $1.0 \times 10^{-19}$]

• A. $2.2\,V$
• B. $3.4\,V$
• C. $-0.22\,V$
• D. $-2.2\,V$

Answer 1

Answer: (C)

$pH=-log\left[H^{+}\right]=14$
$\Rightarrow \left[H^{+}\right]=10^{-14}$
Hence, $\left[OH^{-}\right]=1\,M\,\,\left[H^{+}\right]\left[OH^{-}\right]=10^{-4}$
$Cu\left(OH\right)_{2} \to Cu^{2+}+2OH^{-}$
$K_{sp}=\left[Cu^{2+}\right]\left[OH^{-}\right]^{2}$
For the reaction, $Cu^{2+}+2e^{-} \to Cu$, $n=2$
$E_{cell}=E^{\circ}_{cell}=\frac{0.0591}{n} log \frac{1}{\left[Cu^{2+}\right]}$
$E_{cell}=0.34-\frac{0.0591}{2} log \frac{1}{1\times10^{-19}}$
$=-0.22\,V$