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