Question

Given below are half cell reactions:

\[\ce{MnO^-_4 + 8H+ + 5e- -> Mn^{+2} + 4H2O}\]

\[\ce{E^{\circ}_{Mn^{+2}/MnO^-_4}}\] = −1.510 V

\[\ce{\frac{1}{2} O2 + 2H+ + 2e- -> H2O}\]

\[\ce{E^{\circ}_{O_2/H_2O}}\] = +1.223 V

Will the permanganate ion \[\ce{MnO^-_4}\] liberate O₂ from water in presence of an acid?

Options

• No, because \[\ce{E^{\circ}_{cell}}\] = −0.287 V

• Yes, because \[\ce{E^{\circ}_{cell}}\] = +2.733 V

• No, because \[\ce{E^{\circ}_{cell}}\] = −2.733 V

• Yes, because \[\ce{E^{\circ}_{cell}}\] = +0.287 V

Answer 1

Yes, because \[\ce{\mathbf{E^{\circ}_{cell}}}\] = +0.287 V

Explanation:

\[\ce{MnO^-_4}\] is reduced to Mn²⁺; E° = +1.510 V

H₂O is oxidised to O₂; E° = −1.223 V

So, \[\ce{E^{\circ}_{cell}}\] = 1.510 − 1.223

= +0.287 V