1. No, because E^°_cell = -2.733 V 2. Yes, because E^°_cell = +0.287 V 3. No, because E^°_cell = -0.287 V 4. Yes, because E^°_cell = +2.733 V

Topic: EMF, Nernst equation and applications· Redox Reactions and Electrochemistry

Given below are half cell reactions :MnO_4^- + 8H^+ + 5e^- → Mn^2+ + 4H_2O,E^°_Mn^2+/MnO_4^- = -1.510 V1/2O_2 + 2H^+ + 2e^- → H_2O,E^°_O_2/H_2O = +1.223 VWill the permanganate ion, MnO_4^- liberate O_2 from water in the presence of an acid ?

1. No, because \(E^\circ_{cell} = -2.733\) V

2. Yes, because \(E^\circ_{cell} = +0.287\) V

3. No, because \(E^\circ_{cell} = -0.287\) V

4. Yes, because \(E^\circ_{cell} = +2.733\) V

Doubtly Asked 30 Oct 2026 NEET UG 2022 Practice test neet Save WhatsApp Report

1 answer

Best answer

Answer: 2. Yes, because \(E^\circ_{cell} = +0.287\) V

Key idea: a reaction is spontaneous if \(E^\circ_{cell}>0\), using the reduction potentials of the two half reactions.

\(\mathrm{MnO_4^-}\) is reduced (the value -1.510 V is for the oxidation direction, so its reduction potential is +1.510 V). Water is oxidised to \(\mathrm{O_2}\), whose reduction potential is +1.223 V.

\(E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}=1.510-1.223=+0.287\) V.

It is positive, so permanganate does liberate \(\mathrm{O_2}\) from water in acid: option 2.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

Was this answer helpful?

Given below are half cell reactions :MnO_4^- + 8H^+ + 5e^- → Mn^2+ + 4H_2O,E^°_Mn^2+/MnO_4^- = -1.510 V1/2O_2 + 2H^+ + 2e^- → H_2O,E^°_O_2/H_2O = +1.223 VWill the permanganate ion, MnO_4^- liberate O_2 from water in the presence of an acid ?

1 answer

Your answer