Given the standard half-cell potentials $\left(E^{\circ}\right)$ of the following as $$ \begin{array}{l} \text { Zn } \longrightarrow \mathrm{Zn}^{2+}+2 e^{-} ; \quad E^{\circ}=+0.76 \mathrm{V} \\ \mathrm{Fe} \longrightarrow \mathrm{Fe}^{2+}+2 e^{-} ; \quad E^{\circ}=0.41 \mathrm{V} \end{array} $$ Then the standard e.m.f. of the cell with the reaction $\mathrm{Fe}^{2+}+\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{Fe}$ is

• A

  $-0.35 V$
• Correct answer
  B

  $+0.35 \vee$
• C

  $+117 V$
• D

  $-117 \mathrm{V}$

Solution

Given, $\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+2 \mathrm{e}^{-} ; E^{\circ}=+0.76 \mathrm{V}$ $$ \mathrm{Fe} \longrightarrow \mathrm{Fe}^{2+}+2 e^{-} ; E^{\circ}=+0.41 \mathrm{V} $$ On reversing the above equation (i) and (ii), we get $\mathrm{Zn}^{2+}+2 e^{-} \longrightarrow \mathrm{Zn} ; E^{\circ}=-0.76 \mathrm{V}$ $\mathrm{Fe}^{2+}+2 e^{-} \longrightarrow \mathrm{Fe} ; E^{\circ}=-0.41 \mathrm{V}$ [where, $E^{\circ}=$ standard reduction potential] To find, the standard emf of the cell with the reaction. So, $E_{\text {cell }}^{\circ}=E_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\circ}-E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}$ $=-0.41 \mathrm{V}+0.76 \mathrm{V}=+0.35 \mathrm{V}$

Hence, the correct answer is option: 'B'.