JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Redox Reactions and Electrochemistry

At what pH , given half cell

MnO_{4}^{-} (0.1 M) ∣ Mn^{2 +} (0.001 M)

will have electrode potential of 1.282 V ?
____ (Nearest Integer)

Given

E_{MnO_{4}^{-} / Mn^{2 +}}^{o} = 1.54 V, \frac{2.303 R T}{F} = 0.059 V

Answer: 3

Solution

For the couple

Mn O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O

the Nernst equation is

E = E^{\circ} - \frac{0.059}{5} lo g \frac{[ M n ^{2 +} ]}{[ Mn O _{4}^{-} ] [ H ^{+} ] ^{8}}

. Substituting

E = 1.282 V

E^{\circ} = 1.54 V

[M n^{2 +}] = 1 0^{- 3}

and

[Mn O_{4}^{-}] = 1 0^{- 1}

gives

1.282 - 1.54 = - 0.0118 lo g \frac{1 0 ^{- 2}}{[ H ^{+} ] ^{8}}

, so

lo g \frac{1 0 ^{- 2}}{[ H ^{+} ] ^{8}} = 21.86

. Writing this as

- 2 + 8 pH = 21.86

gives

pH = \frac{23.86}{8}

. To the nearest integer the pH is 3.

JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Integer answer