Question

The standard reduction potential for $C{u}^{2+}|Cu$ is $+0.34V$. Calculate the reduction potential at pH $=14$ for the above couple if $K_{sp}$ of $Cu(OH{)}_2$ is $1.0\times {10}^{-19}$.

• A. $-0.2205V$
• B. $+0.2205V$
• C. $-0.11V$
• D. $+0.11V$

Answer 1

The correct option is A $-0.2205V$

pH = 14 means pOH = 0 or $\lceil {OH}^{-}\rceil =1M.$

$Cu(OH{)}_2⟶C{u}^{2+}+2{OH}^{-}$

Also, $K_{sp}=\lceil {Cu}^{2+}\rceil {\lceil {OH}^{-}\rceil }^2.$

Substituting values in the above equation.
$1\times {10}^{-19}=\lceil {Cu}^{2+}\rceil {\lceil 1\rceil }^2\lceil {Cu}^{2+}\rceil =1\times {10}^{-19}$

The expression for the reduction potential of the electrode is:

$E_{cell}=E_{cell}^0+\frac{0.0592}{2}log\lceil {Cu}^{2+}\rceil$

$=0.34V+\frac{0.0592}{2}log1\times {10}^{-19}$

$=-0.2205V.$

Hence, option $A$ is correct.