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Electrochemistry question

2024 · 30 Jan · Shift 2 · Q19
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Electrochemistry question

2024 · 30 Jan · Shift 2 · Q19

JEE MainChemistryElectrochemistryMCQ+4 / −1
Reduction potential of ions are given below: ClO4−IO4−BrO4−E∘=1.19 VE∘=1.65 VE∘=1.74 V\begin{array}{ccc} \mathrm{ClO}_4^{-} & \mathrm{IO}_4^{-} & \mathrm{BrO}_4^{-} \\ \mathrm{E}^{\circ}=1.19 \mathrm{~V} & \mathrm{E}^{\circ}=1.65 \mathrm{~V} & \mathrm{E}^{\circ}=1.74 \mathrm{~V} \end{array}ClO4−​E∘=1.19 V​IO4−​E∘=1.65 V​BrO4−​E∘=1.74 V​ The correct order of their oxidising power is :
  1. A
    IO4−>BrO4−>ClO4−\mathrm{IO}_4^{-}\gt \mathrm{BrO}_4^{-}\gt \mathrm{ClO}_4^{-}IO4−​>BrO4−​>ClO4−​
  2. B
    BrO4−>ClO4−>IO4−\mathrm{BrO}_4^{-}\gt \mathrm{ClO}_4^{-}\gt \mathrm{IO}_4^{-}BrO4−​>ClO4−​>IO4−​
  3. C
    ClO4−>IO4−>BrO4−\mathrm{ClO}_4^{-}\gt \mathrm{IO}_4^{-}\gt \mathrm{BrO}_4^{-}ClO4−​>IO4−​>BrO4−​
  4. D
    BrO4−>IO4−>ClO4−\mathrm{BrO}_4^{-}\gt \mathrm{IO}_4^{-}\gt \mathrm{ClO}_4^{-}BrO4−​>IO4−​>ClO4−​
View written solutionFree

Correct answer: D

  1. Concept used

    Oxidising power of a species depends on its tendency to get reduced.

    The greater the standard reduction potential E^, the greater is the tendency to undergo reduction, and hence the stronger is the oxidising agent.

  2. Given standard reduction potentials

    E(ClO4−)=1.19 VE^(\mathrm{ClO}_4^-) = 1.19\,\text{V}E(ClO4−​)=1.19V E(IO4−)=1.65 VE^(\mathrm{IO}_4^-) = 1.65\,\text{V}E(IO4−​)=1.65V E(BrO4−)=1.74 VE^(\mathrm{BrO}_4^-) = 1.74\,\text{V}E(BrO4−​)=1.74V

  3. Arrange in decreasing order of oxidising power

    Since higher E^ means stronger oxidising agent:

    BrO4−>IO4−>ClO4−\mathrm{BrO}_4^- > \mathrm{IO}_4^- > \mathrm{ClO}_4^-BrO4−​>IO4−​>ClO4−​

  4. Match with options

    This corresponds to Option D.

  5. Verification with stored answer

    Stored correct answer: D

    Our derived answer: D

    So, the answer agrees with the stored answer.

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