Question

The standard electrode potentials of the electrodes $C{u}^{2+}/Cu$ and $A{g}^{+}/Ag$ are $0.337$ and $0.799V$ respectively. What would be the concentration of $A{g}^{+}$ ion in solution containing $0.06MC{u}^{2+}$ ion such that both metals can be deposited together?

• A. $0.37\times 10^{-6}M$
• B. $0.37\times 10^{-8}M$
• C. $0.42\times 10^{-6}M$
• D. $0.42\times 10^{-8}M$

Answer 1

Answer: (B)

For the simultaneous deposition of two metals present in a solution, the emf for their half-cell must be equal.
$E_{\left(C{u}^{2+}/Cu\right)}=E_{\left(A{g}^{+}/Ag\right)}$
$E_{\left(C{u}^{2+}/Cu\right)}=E_{\left(C{u}^{2+}/Cu\right)}^{\circ }-\frac{0.0591}{2}\log \frac{1}{\left[C{u}^{2+}\right]}$
$E_{\left(A{g}^{+}/Ag\right)}=E_{\left(A{g}^{+}/Ag\right)}^{\circ }-0.0591\log \frac{1}{\left[A{g}^{+}\right]}$
$E_{\left(C{u}^{2+}-Cu\right)}^{\circ }-\frac{0.0591}{2}\log \frac{1}{\left[C{u}^{2+}\right]}=E_{\left(A{g}^{+}/Ag\right)}^{\circ }-0.0591\log \frac{1}{\left[A{g}^{+}\right]}$
$0.337V-\frac{0.0591}{2}\log \frac{1}{0.06}=0.799V-0.0591\log \frac{1}{\left[A{g}^{+}\right]}$
$-0.0361=0.462V-0.0591\log \frac{1}{\left[A{g}^{+}\right]}$
$0.05911\log \frac{1}{\left[A{g}^{+}\right]}=0.4981$
$\log \frac{1}{\left[A{g}^{+}\right]}=8.428$
$\frac{1}{\left[A{g}^{+}\right]}=267916832.5$
$\left[A{g}^{+}\right]=0.37\times 10^{-8}M$