The magnitude of the change in oxidising power of the $\mathrm{MnO}_{4}^{-} / \mathrm{Mn}^{2+}$ couple is $\mathrm{x} \times 10^{-4} \mathrm{V}$, if the $\mathrm{H}^{+}$ concentration is decreased from 1 M to $10^{-4} \mathrm{M}$ at $25^{\circ} \mathrm{C}$ (Assume concentration of $\mathrm{MnO}_{4}^{-}$ and $\mathrm{Mn}^{2+}$ to be same on change in $\mathrm{H}^{+}$ concentration). The value of x is $\_\_\_\_$ (Rounded off to the nearest integer) [Given: $\frac{2.303 RT}{F}=0.059$]

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Redox Reactions and Electrochemistry

The magnitude of the change in oxidising power of the

MnO_{4}^{-} / Mn^{2 +}

couple is

x \times 1 0^{- 4} V

, if the

H^{+}

concentration is decreased from 1 M to

1 0^{- 4} M

2 5^{\circ} C

(Assume concentration of

MnO_{4}^{-}

and

Mn^{2 +}

to be same on change in

H^{+}

concentration). The value of x is

____

(Rounded off to the nearest integer)
[Given:

\frac{2.303 R T}{F} = 0.059

]

Answer: 3776

The reduction half-reaction of the couple is

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

so the Nernst equation for five electrons reads

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

Both manganese species are held at the same concentrations, so only the hydrogen ion term changes, and lowering

[H^{+}]

from

1

M to

1 0^{- 4}

M changes

lo g [H^{+}]

- 4

∣Δ E ∣ = \frac{8 \times 0.059}{5} \times 4 = 0.3776 V

The oxidising power falls by

0.3776 V = 3776 \times 1 0^{- 4} V

, so

x = 3776

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Integer answer