NEET 2022 — Chemistry solution

NEET2022Chemistry

Given below are half cell reactions :

MnO $_{4}^{-}$ + 8H $^{+}$ + 5e $^{-}$ $\to$ Mn $^{2 +}$ + 4H $_{2}$ O, $E_{Mn^{2 +} / MnO_{4}^{-}}^{\circ} = - 1.510$ V

$\frac{1}{2}$ O $_{2}$ + 2H $^{+}$ + 2e $^{-}$ $\to$ H $_{2}$ O, $E_{O_{2} / H_{2} O}^{\circ} = + 1.223$ V

Will the permanganate ion, MnO $_{4}^{-}$ liberate O $_{2}$ from water in the presence of an acid ?

A
No, because $E_{cell}^{\circ} = - 0.287$ V
B
Yes, because $E_{cell}^{\circ} = + 2.733$ V
C
No, because $E_{cell}^{\circ} = - 2.733$ V
D
Yes, because $E_{cell}^{\circ} = + 0.287$ V

Solution

In the proposed reaction, MnO $_{4}^{-}$ is reduced (cathode) and water is oxidised to O $_{2}$ (anode).

Use the standard reduction potentials of the two half-reactions: $E_{cathode}^{\circ} = + 1.510$ V (reduction of MnO $_{4}^{-}$ to Mn $^{2 +}$ ) and $E_{anode (as reduction)}^{\circ} = + 1.223$ V (reduction of O $_{2}$ to H $_{2}$ O).

$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = 1.510 - 1.223 = + 0.287$ V.

Since $E_{cell}^{\circ}$ is positive, the reaction is spontaneous and MnO $_{4}^{-}$ does liberate O $_{2}$ from water in acidic medium.

← Back to full paper

Practice more Chemistry

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →