Question

Consider the cell
$P{t}_{(s)}\left|H_2(g,1atm)\right|H^{+}(aq,1M)||F{e}^{3+}(aq),F{e}^{2+}(aq)Pt(s)$
When the potential of the cell is $0.712V$ at $298K$, the ratio $\left[F{e}^{2+}\right]/\left[F{e}^{3+}\right]$ is ______(Nearest integer)
Given : $F{e}^{3+}+e^{-}-F{e}^{2+},E^{\circ }F{e}^{3+},F{e}^{2+}Pt=0.771$
$\frac{2.303RT}{F}=0.06V$

Answer 1

Answer: 10

$P{t}_{(s)}\left|H_2(g,1atm)\right|H^{+}(aq,1M)\Vert F{e}^{3+}(aq),F{e}^{2+}(aq)|Pt|(s)$
$\text{ at anode }{H}_2⟶2H^{+}+2e^{-}$
$\text{At cathode }F{e}_{aq}^{3+}+e^{-}⟶F{e}_{aq}^{2+}$
$E^{\circ }=E_{H_2\mid H^{+}}^{\circ }+E_{F{e}^{3+}\mid F{e}^{2+}}^{\circ }=0\cdot 771V$
$E=E^{\circ }-\frac{0\cdot 06}{1}\log \frac{F{e}^{2+}}{F{e}^{3+}}$
$0\cdot 712=(0+0\cdot 771)-\frac{0\cdot 06}{1}\log \frac{F{e}^{2+}}{F{e}^{3+}}$
$\log \frac{F{e}^{2+}}{F{e}^{3+}}=\frac{0\cdot 059}{0\cdot 06}\approx 1$
$\frac{F{e}^{2+}}{F{e}^{3+}}=10$