Question

A 1.0 M solution of Cd²⁺ is added to excess iron and the system is allowed to reach equilibrium. What is the concentration in mol of Cd²⁺?

$\mathrm{Cd}^{2+}(\mathrm{aq})+\mathrm{Fe}^{(s)} \rightarrow \mathrm{Cd}(s)+\mathrm{Fe}^{2+}(\mathrm{aq}) ; \mathrm{E}^o=0.037 \mathrm{~V}$

Given: log 18 = 1.25

Report your answer upto two decimal places.

Answer 1

At equilibrium, $\mathrm{E}^0=\frac{0.0591}{2} \log \left(\frac{\mathrm{Fe}^{2+}}{\mathrm{Cd}^{2+}}\right)$

$\text{0.037} = \frac{\text{0.0591}}{2} log \left(\frac{\text{x}}{1 - \text{x}}\right)$

$\text{x} = \left[\text{Fe}^{2 +}\right] \Rightarrow \text{0.947 M}$

$\therefore \left[\text{Cd}^{2 +}\right] = \text{0.053 M}$