Question

The emf of the cell Zn $\left|Z{n}^{2+}{}_{(0.01M)}\right|\mid F{e}^{2+}{}_{(0}.001M)\mid Fe$ at $298K$ is $0.2905V$. Then the value of equilibrium constant for the cell reaction is

• A. $e^{0.32/0.0295}$
• B. $10^{0.32/0.0295}$
• C. $10^{0.26/0.0295}$
• D. $10^{0.32/0.0591}$

Answer 1

Answer: (B)

$Zn+F{e}^{2+}\to Z{n}^{2+}+Fe(n=2)$
$\text{E}={\text{E}}^{\text{o}}-\frac{0.0591}{\text{n}}\text{log}\text{Q}$
$0.02905=E^o-\frac{0.0591}{2}log\frac{0.01}{0.001}$
${\text{E}}^{\text{o}}=0.2905+0.0295=0.32\text{volt}$
${\text{E}}^{\text{o}}=\frac{0.0591}{\text{n}}\text{log}{\text{K}}_{\text{eq}}$
$0.32=\frac{0.0591}{2}\text{log}{\text{K}}_{\text{eq}}=0.02945\text{log}{\text{K}}_{\text{eq}}$
${\text{K}}_{\text{eq}}=10^{0.32/0.0295}$