Consider the cell $$\operatorname{Pt}(\mathrm{s})\left|\mathrm{H}_{2}(\mathrm{g})(1 \mathrm{~atm})\right| \mathrm{H}^{+}\left(\mathrm{aq},[\mathrm{H}^{+}]=1\right)|| \mathrm{Fe}^{3+}(\mathrm{aq}), \mathrm{Fe}^{2+}(\mathrm{aq}) | \mathrm{Pt}(\mathrm{s})$$ Given: $\mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\circ}=0.771 \mathrm{~V}$ and $\mathrm{E}_{\mathrm{H}^{+} / \frac{1}{2} \mathrm{H}_{2}}^{\circ}=0 \mathrm{~V}$, $\mathrm{T}=298 \mathrm{~K}$ If the potential of the cell is 0.712 V the ratio of concentration of $\mathrm{Fe}^{2+}$ to $\mathrm{Fe}^{3+}$ is ______ (Nearest integer)

JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Redox Reactions and Electrochemistry

Consider the cell

Pt (s) ∣ H_{2} (g) (1 atm) ∣ H^{+} (aq, [H^{+}] = 1) ∣∣ Fe^{3 +} (aq), Fe^{2 +} (aq) ∣ Pt (s)

Given:

E_{Fe^{3 +} / Fe^{2 +}}^{\circ} = 0.771 V

and

E_{H^{+} / \frac{1}{2} H_{2}}^{\circ} = 0 V

T = 298 K

If the potential of the cell is 0.712 V the ratio of concentration of

Fe^{2 +}

Fe^{3 +}

is ______ (Nearest integer)

Answer: 10

Hydrogen is oxidised at the left electrode and iron(III) reduced at the right, giving the cell change

\frac{1}{2} H_{2} + F e^{3 +} \to H^{+} + F e^{2 +}

with one electron transferred and a standard potential of

0.771

V. The Nernst equation reads

E = 0.771 - 0.059 lo g \frac{[ F e ^{2 +} ]}{[ F e ^{3 +} ]}

. Putting

E = 0.712

lo g \frac{[ F e ^{2 +} ]}{[ F e ^{3 +} ]} = \frac{0.771 - 0.712}{0.059} = 1

, so the concentration of iron(II) exceeds that of iron(III) by a factor of ten.

JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Integer answer