Question

Consider the cell
$Pt(s)\left|H_2(g)(1atm)\right|H^{+}\text{(aq, }\left[H^{+}\right]=1)\Vert F{e}^{3+}(aq),F{e}^{2+}(aq)\mid Pt(s)$
Given $E^{\circ }F{e}^{3+}F{e}^{2+}=0.771V^{2+}$ and $E^{\circ }{H}^{+}1/2H_2=0V,T=298K$
If the potential of the cell is $0.712V$, the ratio of concentration of $F{e}^{2+}$ to $F{e}^{3+}$ is

Answer 1

Answer: 10

$\frac{1}{2}{H}_2(g)+F{e}^{3+}(aq.)⟶H^{+}(aq)+F{e}^{2+}(aq.)$
$E=E^o-\frac{0.059}{1}\log \frac{\left[F{e}^{2+}\right]}{\left[F{e}^{3+}\right]}$
$\Rightarrow 0.712=(0.771-0)-\frac{0.059}{1}\log \frac{\left[F{e}^{2+}\right]}{\left[F{e}^{3+}\right]}$
$\Rightarrow \log \frac{\left[F{e}^{2+}\right]}{\left[F{e}^{3+}\right]}=\frac{(0.771-0712)}{0.059}=1$
$\Rightarrow \frac{\left[F{e}^{2+}\right]}{\left[F{e}^{3+}\right]}=10$