Question

\(Pt(s) ∣ H2​(g)(1atm) ∣ H+(aq, [H+]=1)\, ∥\, Fe3+(aq), Fe2+(aq) ∣ Pt(s)\)
Given\( E^∘_{Fe^{3+}Fe^{2+}}\)\(=0.771V\) and \(E^∘_{H^{+1/2}H_2}​=0\,V,T=298K\)
If the potential of the cell is 0.712V, the ratio of concentration of \(Fe2+\) to \(Fe3+\) is

Answer 1

Correct Answer: 10

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

Answer 2

1. Write the Nernst equation for the cell: \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q, \] where \(Q = \frac{[\text{Fe}^{3+}]}{[\text{Fe}^{2+}]}\). 
2. Determine \(E^\circ_{\text{cell}}\): \[ E^\circ_{\text{cell}} = E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} - E^\circ_{\text{H}^+/\text{H}_2} = 0.771 - 0 = 0.771 \, \text{V}. \] 
3. Rearrange the Nernst equation to find \(\log Q\): \[ 0.712 = 0.771 - \frac{0.0591}{1} \log Q. \] 
\[ \log Q = \frac{0.771 - 0.712}{0.0591} = \frac{0.059}{0.0591} \approx 1. \] 
4. Calculate \(Q\): \[ Q = 10^{\log Q} = 10^1 = 10. \] 
5. Determine the ratio of concentrations: \[ Q = \frac{[\text{Fe}^{3+}]}{[\text{Fe}^{2+}]} \implies \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]} = \frac{1}{Q} = \frac{1}{10}. \] 
Thus, the ratio is 10. 
The Nernst equation relates the cell potential to the ratio of reactant and product concentrations. Here, \([\text{Fe}^{2+}]/[\text{Fe}^{3+}]\) is calculated from the measured cell potential.