Question

If $E_{\text {cell }}^{\circ}$ is $1.05 V$, the emf of the cell for the following cell reaction
$Ni (s)+2 Ag ^{+}(0.004 M ) \longrightarrow Ni ^{2+}(0.16 M )+2 Ag (s)$ at $298 K$ in $V$ is

• A. 0.932
• B. 1.227
• C. 0.732
• D. 1.397

Answer 1

Answer: (A)

Cell reaction

$\underset{( s)}{Ni} +\underset{(0.004 M)}{2 Ag ^{+}} \longrightarrow \underset{(0.16 M)}{Ni ^{2+}}+\underset{(s)}{2 Ag}$

Given,

$\left[ Ag ^{+}\right] =0.004 \,M $

$\left[ Ni ^{2+}\right] =0.16 \,M $

$n =2 $

$E_{ cell }^{\circ} =1.05 \,V$

Apply in the Nernst equation,

$E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.0591}{n} \log \frac{\left[ Ni ^{2+}\right]}{\left[ Ag ^{+}\right]^{2}}$

Put the values in equation

$E_{\text {cell }}=1.05-\frac{0.0591}{2} \log \frac{[0.16]}{[0.004]^{2}} $

$E_{\text {cell }}=1.05-\frac{0.0591}{2} \log \frac{16}{16} \times 10^{4} $

$E_{\text {cell }}=1.05-\frac{0.059}{2} \log 10^{4} $

$E_{\text {cell }}=1.05-\frac{0.059}{2} 4 \log 10 \,\,\,[\because \log 10=1] $

$E_{\text {cell }}=1.05-\frac{0.059}{2} \times 4 $

$ E_{\text {cell }}=1.05-0.118 \Rightarrow E_{\text {cell }}=0.932\, V$