Electrochemistry 1 Question 18
The standard reduction potentials $E^{\circ}$ for the half reactions are as
$ \begin{aligned} Zn & =Zn^{2+}+2 e^{-}, E^{\circ}=-0.76 \mathrm{V} \\ \mathrm{Fe}^{2+} & +2 e^{-} \rightarrow \mathrm{Fe}, 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. (b)
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} $
$\Rightarrow \mathrm{Fe}^{2+}+\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{Fe} ; \quad E^{\circ}=+0.35 \mathrm{V}$
BETA
