Question

The emf of the cell Zn $\left.\left| Zn ^{2+}{ }_{(0.01 M )}\right| \mid Fe ^{2+}{ }_{(0} .001 M \right) \mid Fe$ at $298 \quad K$ is $0.2905 V$. 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)

$Zn+Fe^{2 +} \rightarrow Zn^{2 +}+Fe\left(\right.n=2\left.\right)$
$\text{E} = \text{E}^{\text{o}} - \frac{0.0591}{\text{n}} \text{log} \, \text{Q}$
$0.02905=E^{o}-\frac{0.0591}{2}log\frac{0.01}{0.001}$
$\text{E}^{\text{o}} = 0.2905 + 0.0295 = 0.32 \, \text{volt}$
$\text{E}^{\text{o}} = \frac{0.0591}{\text{n}} \text{log} \, \text{K}_{\text{eq}}$
$0.32 = \frac{0.0591}{2} \text{log} \, \text{K}_{\text{eq}} = 0.02945 \, \text{log} \, \text{K}_{\text{eq}}$
$\text{K}_{\text{eq}} = 10^{0.32 / 0.0295}$