Q45 / 200 · WBJEE 2014 · Chemistry
$\begin{array}{lllll}\text {The two } & \text { half-cell } & \text { reactions } & \text { of } & \text { an }\end{array}$ electrochemical cell is given as $\mathrm{Ag}^{+}+e^{-} \longrightarrow \mathrm{Ag} ; \quad E_{\mathrm{Ax}^{+} / \mathrm{Ag}}^{\circ}=-0.3995 \mathrm{V}$ $\mathrm{Fe}^{3+} \longrightarrow \mathrm{Fe}^{3 *}+e^{-} ; E_{\mathrm{Fe}^{3+} \mathrm{Fv}^{2+}}=-0.7120 \mathrm{V}$. The value of cell EMF will be
- A$-0.3125 \mathrm{V}$
- Correct answerB$0.3125 \mathrm{V}$
- C$1.114 \mathrm{V}$
- D$-1.114 \mathrm{V}$
Solution
Species with more negative $E^{\circ}$ (standard reduction potential) generally acts as reducing agent while with less negative $E^{\circ}$ acts as oxidising agent. Thus, the overall reaction is
$\begin{aligned} \mathrm{Ag}^{+}+\mathrm{Fe}^{2+} & \longrightarrow \mathrm{Fe}^{3+}+\mathrm{Ag} \\ \Delta E^{\circ} &=E_{\mathrm{OA}}^{\circ}-E_{\mathrm{RA}}^{\circ} \\ &=-0.3995-(-0.7120) \mathrm{V} \\ &=-0.3995+0.7120 \mathrm{V} \\ &=+0.3125 \mathrm{V} \end{aligned}$
Hence, the correct answer is option: 'B'.