Question

$E^{\circ }$ for the cell, $Zn|Z{n}^{2+}(aq)||C{u}^{2+}(aq)|Cu$ is 1.10 V at $25^{\circ }C.$ The equilibrium constant for the reaction, $Zn(s)+C{u}^{2+}(aq)\rightleftharpoons Cu(s)+Z{n}^{2+}(aq)$ is of the order

• A. $10^{37}$
• B. $10^{-28}$
• C. $10^{18}$
• D. $10^{17}$

Answer 1

Answer: (A)

$Zn(s)+C{u}^{2+}(aq)\rightleftharpoons Cu(s)+Z{n}^{2+}(aq)$.
$\therefore E^{\circ }=\frac{0.0591}{n}{\log }_{10}{K}_{\text{eq }}$
$E^{\circ }=+1.10V$
because at equilibrium, $E_{\text{cell }}=0$
$(n=$ number of electrons exchanged $=2$ )
$1.10=\frac{0.0591}{2}{\log }_{10}{K}_{eq}$
$\frac{2.20}{0.0591}={\log }_{10}{K}_{eq}$
$K_{eq}=\text{ antilog }37.225$
$K_{eq}=1.66\times 10^{-37}$