Electrochemistry - Result Question 53
####55. $\quad E^{\circ}$ for the cell, $Zn \mid Zn^{2+}$ (aq) || $Cu^{2+}$ (aq) $\mid Cu$ is $1.10 V$ at $25^{\circ} C$. The equilibrium constant for the cell reaction:
$Zn+Cu^{2+}(aq) \rightarrow Cu+Zn^{2+}(aq)$, is of the order of
[1997]
(a) $10^{-18}$
(b) $10^{-37}$
(c) $10^{18}$
(d) $10^{37}$
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Solution:
- (d) Given for the reaction
$Zn(s)+Cu^{2+}(aq) \rightarrow Cu(s)+Zn^{2+}(aq)$,
$E^{\circ}=+1.10 V$.
At equilibrium
$\therefore E _{\text{cell }}^{\circ}=\frac{0.0591}{n} \log _{10} K _{\text{eq }}$
here ( $n \to$ number of exchange of electrons)
or $1.10=\frac{0.0591}{2} \log _{10} K _{\text{eq }} \frac{2.20}{0.059}=\log _{10} K _{\text{eq }}$
$=37.22$ or $K _{eq}=1.66 \times 10^{37}$
