JEE Main · 2024 · Shift-IIeasyEC-002

Reduction potential of ions are given below: | Ion | ClO4- | IO4- | BrO4- | |---|---|---|---| | E° (V) | 1.19 | 1.65 |…

Electrochemistry · Class 12 · JEE Main Previous Year Question

Question

Reduction potential of ions are given below:

| Ion | ClO4−\mathrm{ClO_4^-} | IO4−\mathrm{IO_4^-} | BrO4−\mathrm{BrO_4^-} | |---|---|---|---| | E°E° (V) | 1.19 | 1.65 | 1.74 |

The correct order of their oxidising power is:

Options
  1. a

    ClO4−>IO4−>BrO4−\mathrm{ClO_4^- > IO_4^- > BrO_4^-}

  2. b

    BrO4−>IO4−>ClO4−\mathrm{BrO_4^- > IO_4^- > ClO_4^-}

    ✓
  3. c

    BrO4−>ClO4−>IO4−\mathrm{BrO_4^- > ClO_4^- > IO_4^-}

  4. d

    IO4−>BrO4−>ClO4−\mathrm{IO_4^- > BrO_4^- > ClO_4^-}

Correct Answerb

BrO4−>IO4−>ClO4−\mathrm{BrO_4^- > IO_4^- > ClO_4^-}

Detailed Solution

Strategy: The oxidising power of a species is directly related to its willingness to undergo reduction. The standard reduction potential (E°E°) measures this tendency: a higher E°E° value indicates a stronger oxidising agent.

Step 1: Tabulate and compare E°E° values From the data provided:

  1. \ceBrO4−\ce{BrO4^-}: E°=1.74 VE° = 1.74\ \mathrm{V} (Highest)
  2. \ceIO4−\ce{IO4^-}: E°=1.65 VE° = 1.65\ \mathrm{V} (Medium)
  3. \ceClO4−\ce{ClO4^-}: E°=1.19 VE° = 1.19\ \mathrm{V} (Lowest)

Step 2: Determine order of oxidising power Sin\ce oxidising power ∝E°reduction\propto E°_{\text{reduction}}: \ceBrO4−>IO4−>ClO4−\ce{BrO4^- > IO4^- > ClO4^-}

Answer: (b) \ceBrO4−>IO4−>ClO4−\boxed{\text{Answer: (b) } \ce{BrO4^- > IO4^- > ClO4^-}}

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