Given below are half-cell reactions: Will the permanganate ion, liberate from water in the presence of an acid? [2022]

Question

Given below are half-cell reactions:

${MnO}_{4}^{-} + 8 H^{+} + 5 e^{-} ⟶ {Mn}^{2 +} + 4 H_{2} O;$ $E_{{Mn}^{2 +} / {MnO}_{4}^{-}}^{\circ} = - 1.510$ $V$

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

Will the permanganate ion, ${MnO}_{4}^{-}$ liberate $O_{2}$ from water in the presence of an acid? [2022]

Smart Achievers · Accepted Answer

(1)

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

As $E_{cell}^{\circ}$ is positive, hence the reaction is feasible.

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