Question

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

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

Answer 1

Answer: (C)

$A(S)+2 B^{+}(a q) \rightarrow A^{2+}(a q)+2 B(s)$
Oxidation $A(s) \rightarrow A^{2+}(a q)+2 e^{-}$
Reduction $2 B^{+}(a q)+2 e^{-} \rightarrow 2 B(s)$
Here, $n=2 E^{o}=1.18\, V\, K_{e q}=?$
$\because \log K_{e q}=\frac{n E^{o} F}{R T}$
$\left[\frac{F}{R T}=\frac{1}{0.059}\right]$
$\therefore \log K_{e q}=\frac{2 \times 1.18 \times 1}{0.059}$
$\therefore \log K_{e q}=40$
or $K_{e q}=10^{40}$