Question

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

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

Answer 1

Answer: (B)

At anode, $Mg⟶M{g}^{2+}+2e^{-}$

At cathode, $S{n}^{2+}+2e^{-}⟶Sn$

From Nernst equation,

$E_{\text{cell }}=E_{\text{cell }}^{\circ }-\frac{0.0591}{n}\log \frac{M{g}^{+2}}{S{n}^{+2}}$

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

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

$=2.2+0.0295\approx 2.23$