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⁺?

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

Given: log 18 = 1.25

Report your answer upto two decimal places.

Answer 1

Answer: 0.05

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

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

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

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