Exams › JEE Advanced › Chemistry
The standard reduction potentials of the following half-reactions are given: (i) MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O, E_red = +1.51 V (ii) Sn2+ -> Sn4+ + 2e-, E_ox = -0.15 V (so E_red(Sn4+/Sn2+) = +0.15 V) (iii) Cr2O7(2-) + 14H+ + 6e- -> 2Cr3+ + 7H2O, E_red = +1.33 V (iv) Ce3+ -> Ce4+ + e-, E_ox = -1.61 V (so E_red(Ce4+/Ce3+) = +1.61 V) Which of the following correctly describes the relative oxidising power?
- Cr2O7(2-) > MnO4-
- Ce4+ > Sn4+
- Ce4+ > MnO4-
- MnO4- > Sn4+
Correct answer: Ce4+ > Sn4+
Solution
Oxidising power follows reduction potential. The overall order is Ce4+ > MnO4- > Cr2O7(2-) > Sn4+. Options B, C, and D are all consistent with this order.
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