Chemistry Question on Galvanic Cells

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

Accepted Answer

The correct answer is 10.
21H2(g)+Fe3+(aq.)⟶H+(aq)+Fe2+(aq.)
E=Eo−10.059log[Fe3+][Fe2+]
⇒0.712=(0.771−0)−10.059log[Fe3+][Fe2+]
⇒log[Fe3+][Fe2+]=0.059(0.771−0712)=1
⇒[Fe3+][Fe2+]=10