Exams › JEE Main › Chemistry

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.

Related JEE Main Chemistry questions

• What volume of 20-volume hydrogen peroxide is needed to produce 5 L of oxygen gas at STP?
• Which one of the following reactions shows sulfuric acid acting as an oxidizing agent?
• Which of the following statements about redox behavior are true? (i) A metal M whose standard reduction potential for the half-reaction Mⁿ⁺ + ne⁻ ⇌ M is highly negative acts as a strong reducing agent. (ii) The oxidizing strength of the halogens falls in the order chlorine to iodine. (iii) The reducing strength of hydrogen halides rises from hydrogen chloride to hydrogen iodide.
• For the balanced redox equation X MnO4⁻ + Y C2O4²⁻ + Z H⁺ → X Mn²⁺ + 2Y CO2 + Z/2 H2O, what are the values of X, Y, and Z, respectively?
• Balance the redox equation below and identify the coefficients X, Y and Z in order: XNa2HAsO3 + YNaBrO3 + ZHCl → NaBr + H3AsO4 + NaCl
• Arrange the following sulphur-containing anions in order of increasing oxidation state of sulphur: SO3²⁻, S2O4²⁻ and S2O6²⁻.

⚔️ Practice JEE Main Chemistry free + battle 1v1 →