Electrochemistry Question 9
Question 9 - 2024 (30 Jan Shift 2)
Reduction potential of ions are given below:
$$ \begin{array}{ccc} \mathrm{ClO}{4}^{-} & \mathrm{IO}{4}^{-} & \mathrm{BrO}_{4}^{-} \ \mathrm{E}^{\circ}=1.19 \mathrm{~V} & \mathrm{E}^{\circ}=1.65 \mathrm{~V} & \mathrm{E}^{\circ}=1.74 \mathrm{~V} \end{array} $$
The correct order of their oxidising power is:
(1) $\mathrm{ClO}{4}^{-}>\mathrm{IO}{4}^{-}>\mathrm{BrO}_{4}^{-}$
(2) $\mathrm{BrO}{4}^{-}>\mathrm{IO}{4}^{-}>\mathrm{ClO}_{4}^{-}$
(3) $\mathrm{BrO}{4}^{-}>\mathrm{ClO}{4}^{-}>\mathrm{IO}_{4}^{-}$
(4) $\mathrm{IO}{4}^{-}>\mathrm{BrO}{4}^{-}>\mathrm{ClO}_{4}^{-}$
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Answer (2)
Solution
Higher the value of $\oplus$ ve SRP (Std. reduction potential) more is tendency to undergo reduction, so better is oxidising power of reactant.
Hence, ox. Power:- $\mathrm{BrO}{4}^{-}>\mathrm{IO}{4}^{-}>\mathrm{ClO}_{4}^{-}$
