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Using the standard reduction potentials given below (acidic medium, E in volts vs NHE), identify the redox change that is thermodynamically feasible: I2 + 2e- -> 2I- E = 0.54 Cl2 + 2e- -> 2Cl- E = 1.36 Mn3+ + e- -> Mn2+ E = 1.50 Fe3+ + e- -> Fe2+ E = 0.77 O2 + 4H+ + 4e- -> 2H2O E = 1.23

1. Chloride ion is oxidised by O2
2. Iodide ion is oxidised by chlorine
3. Mn2+ is oxidised by chlorine
4. Fe2+ cannot be oxidised by O2

Correct answer: Iodide ion is oxidised by chlorine

Solution

Cl2 (E = 1.36 V) has a higher reduction potential than the I2/I- couple (0.54 V), so Cl2 can oxidise I- to I2 (Ecell = 1.36 - 0.54 = +0.82 V). For chloride being oxidised by O2: O2 (1.23) < Cl2 (1.36), so O2 cannot oxidise Cl-. For Mn2+ oxidised by chlorine: Mn3+/Mn2+ is 1.50 > 1.36, so Cl2 cannot oxidise Mn2+. O2 (1.23) > Fe3+/Fe2+ (0.77), so O2 can oxidise Fe2+, making that statement false.

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