Electrochemistry - Result Question 70

####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:

  1. $Fe^{2+}+2 e^{-} \longrightarrow Fe ; \quad E^{\circ}=-0.41 V$

$$ Zn \longrightarrow Zn^{2+}+2 e^{-} ; E^{\circ}=+0.76 V $$

$\Rightarrow Fe^{2+}+Zn \longrightarrow Zn^{2+}+Fe ; \quad E^{\circ}=+0.35 V$



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