Chemistry · PYQ NEET UG 2022
हिन्दीGiven below are half cell reactions: \( \text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\,\,, E^\circ_{\text{Mn}^{2+}/\text{MnO}_4^-} = -1.510\text{ V} \) \( \frac{1}{2}\text{O}_2 + 2\text{H}^+ + 2e^- \rightarrow \text{H}_2\text{O}\,\,, E^\circ_{\text{O}_2/\text{H}_2\text{O}} = +1.223\text{ V} \) Will the permanganate ion, \( \text{MnO}_4^- \) liberate \( \text{O}_2 \) from water in the presence of an acid?
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Correct answer · Option D
No, because \( E^\circ_{\text{cell}} = -0.287\text{ V} \)
Explanation
\( E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 1.223 - 1.510 = -0.287\text{ V} \). Since \( E^\circ_{\text{cell}} \) is negative, the reaction is non-spontaneous.