JEE Main 2023 — Chemistry Physical Chemistry

2023integermedium

Consider the cell

${\mathrm{Pt}}_{(s)}|{H}_{2}(g,1\mathrm{atm})|{H}^{+}(\mathrm{aq},1M)||{\mathrm{Fe}}^{3+}(\mathrm{aq}),{\mathrm{Fe}}^{2+}(\mathrm{aq})|Pt(s)$

When the potential of the cell is $0.712V$ at $298K$ , the ratio $[{\mathrm{Fe}}^{2+}]/[{\mathrm{Fe}}^{3+}]$ is

(Nearest integer)

Given: ${\mathrm{Fe}}^{3+}+{e}^{-}={\mathrm{Fe}}^{2+},E^{\circ}{\mathrm{Fe}}^{3+},{\mathrm{Fe}}^{2+}\mathrm{Pt}=0.771\frac{2.303\mathrm{RT}}{F}=0.06V$

Official previous-year question

Held on 30 Jan 2023 · Verified 6 Jul 2026.

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Solution

Given cell reaction:

${\mathrm{Pt}}_{(s)}|{H}_{2}(g,1\mathrm{atm})|{H}^{+}(\mathrm{aq},1M)||{\mathrm{Fe}}^{3+}(\mathrm{aq}),{\mathrm{Fe}}^{2+}(\mathrm{aq})Pt(s)$

at anode(oxidation) ${H}_{2}\longrightarrow 2{H}^{+}+2{e}^{-}$

At cathode(reduction) ${\mathrm{Fe}}_{\mathrm{aq}}^{3+}+{e}^{-}\longrightarrow {\mathrm{Fe}}_{\mathrm{aq}}^{2+}$

$E{^{\circ}}_{\mathrm{cell}}={E}_{\mathrm{cathode}}^{o}-{E}_{\mathrm{anode}}^{o}={{E}^{o}}_{{H}_{2}|{H}^{+}}+{{E}^{o}}_{{\mathrm{Fe}}^{3+}\mid {\mathrm{Fe}}^{2+}}=0.771V$

Thus, using Nernst equation,

$\Rightarrow E=E^{\circ}-\frac{0\cdot 06}{1}\mathrm{log}\frac{{\mathrm{Fe}}^{2+}}{{\mathrm{Fe}}^{3+}}$

$\Rightarrow 0.712=(0+0.771)-\frac{0.06}{1}\mathrm{log}\frac{{\mathrm{Fe}}^{2+}}{{\mathrm{Fe}}^{3+}}$

$\Rightarrow \mathrm{log}\frac{{\mathrm{Fe}}^{2+}}{{\mathrm{Fe}}^{3+}}=\frac{0.059}{0.06}\approx 1$

$\frac{{\mathrm{Fe}}^{2+}}{{\mathrm{Fe}}^{3+}}=10$

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Solution

Given cell reaction:

${\mathrm{Pt}}_{(s)}|{H}_{2}(g,1\mathrm{atm})|{H}^{+}(\mathrm{aq},1M)||{\mathrm{Fe}}^{3+}(\mathrm{aq}),{\mathrm{Fe}}^{2+}(\mathrm{aq})Pt(s)$

at anode(oxidation) ${H}_{2}\longrightarrow 2{H}^{+}+2{e}^{-}$

At cathode(reduction) ${\mathrm{Fe}}_{\mathrm{aq}}^{3+}+{e}^{-}\longrightarrow {\mathrm{Fe}}_{\mathrm{aq}}^{2+}$

$E{^{\circ}}_{\mathrm{cell}}={E}_{\mathrm{cathode}}^{o}-{E}_{\mathrm{anode}}^{o}={{E}^{o}}_{{H}_{2}|{H}^{+}}+{{E}^{o}}_{{\mathrm{Fe}}^{3+}\mid {\mathrm{Fe}}^{2+}}=0.771V$

Thus, using Nernst equation,

$\Rightarrow E=E^{\circ}-\frac{0\cdot 06}{1}\mathrm{log}\frac{{\mathrm{Fe}}^{2+}}{{\mathrm{Fe}}^{3+}}$

$\Rightarrow 0.712=(0+0.771)-\frac{0.06}{1}\mathrm{log}\frac{{\mathrm{Fe}}^{2+}}{{\mathrm{Fe}}^{3+}}$

$\Rightarrow \mathrm{log}\frac{{\mathrm{Fe}}^{2+}}{{\mathrm{Fe}}^{3+}}=\frac{0.059}{0.06}\approx 1$

$\frac{{\mathrm{Fe}}^{2+}}{{\mathrm{Fe}}^{3+}}=10$

JEE Main 2023 — Chemistry Physical Chemistry

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JEE Main 2023 — Chemistry Physical Chemistry

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry