JEE MAIN - Chemistry (2024 - 30th January Evening Shift - No. 19)
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 :
$$\mathrm{IO}_4^{-}>\mathrm{BrO}_4^{-}>\mathrm{ClO}_4^{-}$$
$$\mathrm{BrO}_4^{-}>\mathrm{ClO}_4^{-}>\mathrm{IO}_4^{-}$$
$$\mathrm{ClO}_4^{-}>\mathrm{IO}_4^{-}>\mathrm{BrO}_4^{-}$$
$$\mathrm{BrO}_4^{-}>\mathrm{IO}_4^{-}>\mathrm{ClO}_4^{-}$$
Explanation
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:- $$\mathrm{BrO}_4^{-}>\mathrm{IO}_4^{-}>\mathrm{ClO}_4^{-}$$
Comments (0)
