Electrochemistry 1 Question 18
18. The standard reduction potentials $E^{\circ}$, for the half reactions are as
$$ \begin{aligned} \mathrm{Zn} & =\mathrm{Zn}^{2+}+2 e^{-}, E^{\circ}=+0.76 \mathrm{~V} \ \mathrm{Fe} & =\mathrm{Fe}^{2+}+2 e^{-}, E^{\circ}=0.41 \mathrm{~V} \end{aligned} $$
The emf for the cell reaction,
$$ \mathrm{Fe}^{2+}+\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+}+\mathrm{Fe} \text { is } $$
$(1989,1 \mathrm{M})$
(a) $-0.35 \mathrm{~V}$
(b) $+0.35 \mathrm{~V}$
(c) $+1.17 \mathrm{~V}$
(d) $-1.17 \mathrm{~V}$
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Answer:
Correct Answer: 18. (d)
Solution:
- $\mathrm{Fe}^{2+}+2 e^{-} \longrightarrow \mathrm{Fe} ; \quad E^{\circ}=-0.41 \mathrm{~V}$
$$ \mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+2 e^{-} ; E^{\circ}=+0.76 \mathrm{~V} $$
$\Rightarrow \mathrm{Fe}^{2+}+\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{Fe} ; \quad E^{\circ}=+0.35 \mathrm{~V}$
