Calculate E from the following Redox reaction

Question

Redox Reaction:

2MnO₄^-(aq) + 2H₂O (l) + 6H⁺ (aq) --> 2Mn²⁺ (aq) + 5H₂O₂ (l)

The standard reduction potentials for the species in the reaction are listed below:

H₂O₂ (l) + 2H⁺ (aq) + 2e^- --> 2H₂O (l) E (red H2O2) = 1.776 V

MnO₄^-(aq) + 8H⁺ (aq) +5e^- --> Mn²⁺ + 4H₂O (l) E(red, MnO4) = 1.507 V

Answer 1

You can easily calculate the Eºcell by using standard reduction potentials.

E_cathode - E_anode = E_cell

H₂O₂ (l) + 2H⁺ (aq) + 2e^- --> 2H₂O (l) E (red H2O2) = 1.776 V

MnO₄^-(aq) + 8H⁺ (aq) +5e^- --> Mn²⁺ + 4H₂O (l) E(red, MnO4) = 1.507 V

Since 1.776 V is greater than 1.507, this will be the reduction reaction (cathode), and the other (in reverse) will be the oxidation reaction.

Cathode (reduction): H₂O₂ (l) + 2H⁺ (aq) + 2e^- --> 2H₂O (l) E (red H₂O₂)

Anode (oxidation): Mn²⁺ + 4H₂O (l) --> MnO₄^-(aq) + 8H⁺ (aq) +5e^-

Eºcell = Eº_cathode - Eº_anode

Eºcell = 1.776 - 1.507 = 0.269 V