A-Level Redox, electrode potentials and electrolysis: worked solution

Question

Acidified KMnO4 and acidified K2Cr2O7 both serve as oxidants in organic chemistry. By referring to appropriate E° values, explain why KMnO4 is the stronger oxidising agent.

Relevant reduction potentials: MnO4^- + 8H^+ + 5e^- -> Mn²+ + 4H2O, E° = +1.52 V Cr2O7²- + 14H^+ + 6e^- -> 2Cr³+ + 7H2O, E° = +1.33 V E°(MnO4^-/Mn²+) = +1.52 V is more positive than E°(Cr2O7²-/Cr³+) = +1.33 V. A more positive (more negative reduction... more positive) electrode potential means a greater tendency to be reduced, i.e. to gain electrons. Therefore MnO4^- is the more readily reduced species and so acidified KMnO4 is the stronger oxidising agent.