Question

For the reaction $A_{(s)}+2B_{(aq)}^{+}\to A_{(aq)}^{2+}+2B_{(s)}$; the $E^o$ is $1.18V$. Then the equilibrium constant for the reaction is

• A. $10^{10}$
• B. $10^{20}$
• C. $10^{40}$
• D. $10^{60}$

Answer 1

Answer: (C)

$A(S)+2B^{+}(aq)\to A^{2+}(aq)+2B(s)$
Oxidation $A(s)\to A^{2+}(aq)+2e^{-}$
Reduction $2B^{+}(aq)+2e^{-}\to 2B(s)$
Here, $n=2E^o=1.18V{K}_{eq}=?$
$\because \log K_{eq}=\frac{n{E}^{o}F}{RT}$
$\left[\frac{F}{RT}=\frac{1}{0.059}\right]$
$\therefore \log K_{eq}=\frac{2\times 1.18\times 1}{0.059}$
$\therefore \log K_{eq}=40$
or $K_{eq}=10^{40}$