Question

The emf of the following cell $Mg \left| Mg ^{+2}(0.01\, M ) \| Sn ^{+2}(0.1 \,M )\right| Sn at$ $298\, K$ in ${ }^{'} V ^{'}$ is $\left(\right.$ Given, $\left.E_{ Mg ^{+2} \mid Mg }^{\circ}=-2.34\, V , E_{ Sn ^{+2} \mid Sn }^{\circ}=-0.14\, V \right)$

• A. 2.17
• B. 2.23
• C. 2.51
• D. 2.45

Answer 1

Answer: (B)

At anode, $Mg \longrightarrow Mg ^{2+}+2 e^{-}$

At cathode, $Sn ^{2+}+2 e^{-} \longrightarrow Sn$

From Nernst equation,

$E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.0591}{n} \log \frac{ Mg ^{+2}}{ Sn ^{+2}}$

$E_{\text {cell }} =-0.14-(-2.34)-\frac{0.0591}{2} \log 10^{-1} \,\,\,\left(\because n=2 e^{-}\right)$

$ =2.2+\frac{0.0591}{2} $

$=2.2+0.0295 \approx 2.23$