JEE Main · 2019 · Shift-ImediumRDX-034

Given: Co3+ + e- Co2+; E = +1.81\ V Pb4+ + 2e- Pb2+; E = +1.67\ V Ce4+ + e- Ce3+; E = +1.61\ V Bi3+ + 3e- Bi; E =…

Redox Reactions · Class 11 · JEE Main Previous Year Question

Question

Given:

Co3++e−→Co2+;E∘=+1.81 V\mathrm{Co^{3+} + e^{-} \rightarrow Co^{2+}};\quad E^\circ = +1.81\ \mathrm{V} Pb4++2e−→Pb2+;E∘=+1.67 V\mathrm{Pb^{4+} + 2e^{-} \rightarrow Pb^{2+}};\quad E^\circ = +1.67\ \mathrm{V} Ce4++e−→Ce3+;E∘=+1.61 V\mathrm{Ce^{4+} + e^{-} \rightarrow Ce^{3+}};\quad E^\circ = +1.61\ \mathrm{V} Bi3++3e−→Bi;E∘=+0.20 V\mathrm{Bi^{3+} + 3e^{-} \rightarrow Bi};\quad E^\circ = +0.20\ \mathrm{V}

Oxidizing power of the species will increase in the order:

Options
  1. a

    Co3+<Ce4+<Bi3+<Pb4+\mathrm{Co^{3+} < Ce^{4+} < Bi^{3+} < Pb^{4+}}

  2. b

    Bi3+<Ce4+<Pb4+<Co3+\mathrm{Bi^{3+} < Ce^{4+} < Pb^{4+} < Co^{3+}}

    ✓
  3. c

    Co3+<Pb4+<Ce3+<Bi4+\mathrm{Co^{3+} < Pb^{4+} < Ce^{3+} < Bi^{4+}}

  4. d

    Ce4+<Pb4+<Bi3+<Co3+\mathrm{Ce^{4+} < Pb^{4+} < Bi^{3+} < Co^{3+}}

Correct Answerb

Bi3+<Ce4+<Pb4+<Co3+\mathrm{Bi^{3+} < Ce^{4+} < Pb^{4+} < Co^{3+}}

Detailed Solution

Principle: Higher E∘E^\circ (reduction potential) → stronger oxidising agent.

Step 1 — Arrange in increasing order of E∘E^\circ:

| Species | E∘E^\circ (V) | Oxidising power | |---|---|---| | Bi3+\mathrm{Bi^{3+}} | +0.20+0.20 | Weakest | | Ce4+\mathrm{Ce^{4+}} | +1.61+1.61 | ↑ | | Pb4+\mathrm{Pb^{4+}} | +1.67+1.67 | ↑ | | Co3+\mathrm{Co^{3+}} | +1.81+1.81 | Strongest |

Step 2 — Increasing order of oxidising power: Bi3+<Ce4+<Pb4+<Co3+\mathrm{Bi^{3+} < Ce^{4+} < Pb^{4+} < Co^{3+}}

Answer: Option (2)

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