Question

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

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

Answer 1

Answer: (A)

$10^{-37}$

Answer 2

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