Q9 / 180 · KCET 2009 · Chemistry
The standard electrode potential for the half-cell reactions are
$$
\begin{aligned}
&\mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn} ; \mathrm{E}^{\circ}=-0.76 \mathrm{~V} \\
&\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe} ; \mathrm{E}^{\circ}=-0.44 \mathrm{~V}
\end{aligned}
$$
The emf of the cell reaction,
$\mathrm{Fe}^{2+}+\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{Fe}$ is
- A$-0.32 \mathrm{~V}$
- B$-1.20 \mathrm{~V}$
- C$+1.20 \mathrm{~V}$
- Correct answerD$+0.32 \mathrm{~V}$
Solution
$\mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn} ; \mathrm{E}^{\circ}=-0.76 \mathrm{~V}$ $\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe} ; \mathrm{E}^{\circ}=-0.44 \mathrm{~V}$
Cell reaction is
$\begin{aligned} \mathrm{Fe}^{2+}+\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{Fe} \\ \mathrm{E}_{\text {cell }} &=\mathrm{E}_{\text {cathode }}-\mathrm{E}_{\text {anode }} \\ &=-0.44-(-0.76) \\ &=-0.44+0.76 \\ &=0.32 \mathrm{~V} \end{aligned}$
Hence, the correct answer is option: 'D'.