KCET · Chemistry · Electrochemistry
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 \(+1.20 \mathrm{~V}\)
- D \(+0.32 \mathrm{~V}\)
Answer & Solution
Correct Answer
(D) \(+0.32 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn} ; \mathrm{E}^{\circ}=-0.76 \mathrm{~V}\) \(\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe} ; \mathrm{E}^{\circ}=-0.44 \mathrm{~V}\)
Cell reaction is
\(\begin{aligned} \mathrm{Fe}^{2+}+\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{Fe} \\ \mathrm{E}_{\text {cell }} &=\mathrm{E}_{\text {cathode }}-\mathrm{E}_{\text {anode }} \\ &=-0.44-(-0.76) \\ &=-0.44+0.76 \\ &=0.32 \mathrm{~V} \end{aligned}\)
Cell reaction is
\(\begin{aligned} \mathrm{Fe}^{2+}+\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{Fe} \\ \mathrm{E}_{\text {cell }} &=\mathrm{E}_{\text {cathode }}-\mathrm{E}_{\text {anode }} \\ &=-0.44-(-0.76) \\ &=-0.44+0.76 \\ &=0.32 \mathrm{~V} \end{aligned}\)
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