Question

The standard electrode potentials of the electrodes $C{u}^{2+}/Cu$ and $A{g}^{+}/Ag$ are 0.337 and 0.799 V respectively. What would be the concentration of $A{g}^{+}$ ion in solution containing 0.06 M $C{u}^{2+}$ 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

The correct option is B $0.37\times {10}^{-8}M$

For the simultaneous deposition of two metals present in a solution, the emf for their half-cell must be equal.
$E_{\left(C{u}^{2+}/Cu\right)}=E_{\left(A{g}^{+}/Ag\right)}$
$E_{\left(C{u}^{2+}/Cu\right)}=E_{\left(C{u}^{2+}/Cu\right)}^o-\frac{0.0591}{2}log\frac{1}{\left[C{u}^{2+}\right]}$

$E_{\left(A{g}^{+}/Ag\right)}=E_{\left(A{g}^{+}/Ag\right)}^o-0.0591log\frac{1}{\left[A{g}^{+}\right]}$

$E_{\left(C{u}^{2+}/Cu\right)}^o-\frac{0.0591}{2}log\frac{1}{\left[C{u}^{2+}\right]}=E_{\left(A{g}^{+}/Ag\right)}^o-0.0591log\frac{1}{\left[A{g}^{+}\right]}$
$0.337V-\frac{0.0591}{2}log\frac{1}{0.06}=0.799V-0.0591log\frac{1}{\left[A{g}^{+}\right]}$
$-0.0361=0.462V-0.0591log\frac{1}{\left[A{g}^{+}\right]}$
$0.0591log\frac{1}{\left[A{g}^{+}\right]}=0.4981$

$log\frac{1}{\left[A{g}^{+}\right]}=8.428$

$\frac{1}{\left[A{g}^{+}\right]}=267916832.5$

$\left[A{g}^{+}\right]=0.37\times {10}^{-8}M$