JEE Main · 2023 · 25 Jan 2023 · Shift 2 · Redox Reactions and Electrochemistry

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

The

E_{cell}

for the given cell is 0.1115 V at 298 K when

\frac{[ M ^{+} ( aq ) ]}{[ M ^{3 +} ( aq ) ]} = 1 0^{a}

The value of

a

is ____
Given :

E^{θ} M^{3 +} / M^{+} = 0.2 V

\frac{2.303 RT}{F} = 0.059 V

Answer: 3

Solution

The hydrogen electrode is the anode and the

M^{3 +} / M^{+}

couple the cathode, so the cell process is

H_{2} + 2 M^{3 +} \to 2 H^{+} + 2 M^{+}

with

n = 2

and

E_{ce l l}^{\circ} = 0.2 - 0 = 0.2 V

. The Nernst equation, with

[H^{+}] = 1 M

and

p_{H_{2}} = 1 bar

, reads

E_{ce l l} = 0.2 - \frac{0.059}{2} lo g \frac{[ M ^{+} ]}{[ M ^{3 +} ]} .

Putting

E_{ce l l} = 0.1115 V

gives

\frac{0.059}{2} lo g \frac{[ M ^{+} ]}{[ M ^{3 +} ]} = 0.0885

, so that

lo g \frac{[ M ^{+} ]}{[ M ^{3 +} ]} = 3

. The value of

a

is three.

JEE Main · 2023 · 25 Jan 2023 · Shift 2 · Integer answer