Consider the cell \(\mathrm{Pt}_{(\mathrm{s})}\left|\mathrm{H}_{2}(\mathrm{~g}, 1 \mathrm{~atm})\right| \mathrm{H}^{+}(\mathrm{aq}, 1 \mathrm{M})| | \mathrm{Fe}^{3+}(\mathrm{aq}), \mathrm{Fe}^{2+}(\mathrm{aq}) \mid \mathrm{Pt}(\mathrm{s})\) When the potential of the cell is 0.712 V at 298 K, the ratio \(\left[\mathrm{Fe}^{2+}\right] /\left[\mathrm{Fe}^{3+}\right]\) is ______. (Nearest integer) Given: \(\mathrm{Fe}^{3+}+\mathrm{e}^{-}=\mathrm{Fe}^{2+}, \mathrm{E}^{0} \mathrm{Fe}^{3+}, \mathrm{Fe}^{2+} \mid \mathrm{Pt}=0.771\) \(\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V}\)

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Redox Reactions and Electrochemistry

Consider the cell

Pt_{(s)} ∣ H_{2} (g, 1 atm) ∣ H^{+} (aq, 1 M) ∣∣ Fe^{3 +} (aq), Fe^{2 +} (aq) ∣ Pt (s)

When the potential of the cell is 0.712 V at 298 K, the ratio

[Fe^{2 +}] / [Fe^{3 +}]

is ______.
(Nearest integer)
Given:

Fe^{3 +} + e^{-} = Fe^{2 +}, E^{0} Fe^{3 +}, Fe^{2 +} ∣ Pt = 0.771

\frac{2.303 RT}{F} = 0.06 V

Answer: 10

The right-hand electrode is the reduction

F e^{3 +} + e^{-} \to F e^{2 +}

and the left is the standard hydrogen electrode, so the standard cell potential is

0.771 - 0 = 0.771 V

for one electron transferred. The Nernst equation gives

E = E^{\circ} - 0.06 lo g \frac{[ F e ^{2 +} ]}{[ F e ^{3 +} ]}

. Substituting,

0.712 = 0.771 - 0.06 lo g \frac{[ F e ^{2 +} ]}{[ F e ^{3 +} ]}

, so

lo g \frac{[ F e ^{2 +} ]}{[ F e ^{3 +} ]} = \frac{0.059}{0.06} \approx 1

. The ratio

[F e^{2 +}] / [F e^{3 +}]

10

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Integer answer