Question

The standard electrode potentials of the electrodes $ Cu^{2+}/Cu $ and $ Ag^{+}/Ag $ are $0.337$ and $0.799\, V$ respectively. What would be the concentration of $ Ag^{+} $ ion in solution containing $0.06\, M \, Cu^{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.
$E_{\left( Cu ^{2 +} / Cu \right)}=E_{\left( Ag ^{+} / Ag \right)}$
$E_{\left( Cu ^{2+} / Cu \right)}=E_{\left( Cu ^{2+} / Cu \right)}^{\circ}-\frac{0.0591}{2} \log \frac{1}{\left[ Cu ^{2+}\right]}$
$E_{\left( Ag ^{+} / Ag \right)}=E_{\left( Ag ^{+} / Ag \right)}^{\circ}-0.0591 \log \frac{1}{\left[ Ag ^{+}\right]}$
$E _{\left( Cu ^{2+}- Cu \right)}^{\circ}-\frac{0.0591}{2} \log \frac{1}{\left[ Cu ^{2+}\right]}= E _{\left( Ag ^{+} / Ag \right)}^{\circ}-0.0591 \log \frac{1}{\left[ Ag ^{+}\right]}$
$0.337\, V -\frac{0.0591}{2} \log \frac{1}{0.06}=0.799 \,V -0.0591 \log \frac{1}{\left[ Ag ^{+}\right]}$
$-0.0361=0.462 \,V -0.0591 \log \frac{1}{\left[ Ag ^{+}\right]}$
$0.05911 \log \frac{1}{\left[ Ag ^{+}\right]}=0.4981$
$\log \frac{1}{\left[ Ag ^{+}\right]}=8.428$
$\frac{1}{\left[ Ag ^{+}\right]}=267916832.5$
$\left[ Ag ^{+}\right]=0.37 \times 10^{-8} M$