Question

The standard electrode potential of $C{u}^{2+}/Cu$ is $+0.34\text{V}$. At what concentration of $C{u}^{2+}$ ions will this electrode potential be zero?

Take:
$Antilog(-11.52)=3\times {10}^{-12}$

• A. $\left[C{u}^{2+}\right]=3.01\times {10}^{-12}\text{M}$
• B. $\left[C{u}^{2+}\right]=0.5\times {10}^{-12}\text{M}$
• C. $\left[C{u}^{2+}\right]=3.01\times {10}^{-5}\text{M}$
• D. $\left[C{u}^{2+}\right]=1.5\times {10}^{-5}\text{M}$

Answer 1

The correct option is A $\left[C{u}^{2+}\right]=3.01\times {10}^{-12}\text{M}$

For the electrode reaction (reduction)
$C{u}^{2+}+2e^{-}\rightleftharpoons Cu;E^{\circ }=+0.34\text{V}$

By Nernst equation,

$E_{C{u}^{2+}/Cu}=E_{C{u}^{2+}/Cu}^{\circ }-\frac{0.0591}{2}log\frac{1}{\left[C{u}^{2+}\right]}$

$E_{C{u}^{2+}/Cu}=E_{C{u}^{2+}/Cu}^{\circ }+\frac{0.0591}{2}log\left[C{u}^{2+}\right]$

For, $E_{C{u}^{2+}/Cu}=0$

$\Rightarrow 0=0.34+\frac{0.0591}{2}log\left[C{u}^{2+}\right]$

$\Rightarrow log\left[C{u}^{2+}\right]=\frac{-0.34\times 2}{0.059}$

$\Rightarrow log\left[C{u}^{2+}\right]\approx -11.52$

$\Rightarrow \left[C{u}^{2+}\right]=Antilog(-11.52)$

$\text{Hence,}\left[C{u}^{2+}\right]=3.01\times {10}^{-12}\text{M}$