ExamBro
ExamBro
JEE Mains · Chemistry · STD 12 - 2. Electrochemistry

Reduction potential of ions are given below:
\(ClO_4^{-}\)\(IO_4^{-}\)\(BrO_4^{-}\)
\(E^{\circ}=1.19 V\)\(E^{\circ}=1.65 V\)\(E^{\circ}=1.74 V\)
The correct order of their oxidising power is _______.

  1. A \(\mathrm{ClO}_4^{-}>\mathrm{IO}_4^{-}>\mathrm{BrO}_4^{-}\)
  2. B \(\mathrm{BrO}_4^{-}>\mathrm{IO}_4^{-}>\mathrm{ClO}_4^{-}\)
  3. C \(\mathrm{BrO}_4^{-}>\mathrm{ClO}_4^{-}>\mathrm{IO}_4^{-}\)
  4. D \(\mathrm{IO}_4^{-}>\mathrm{BrO}_4^{-}>\mathrm{ClO}_4^{-}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\mathrm{BrO}_4^{-}>\mathrm{IO}_4^{-}>\mathrm{ClO}_4^{-}\)

Step-by-step Solution

Detailed explanation

Higher the value of \(\oplus\) ve \(SRP\) (Std. reduction potential) more is tendency to undergo reduction, so better is oxidising power of reactant. \(Hence, ox. Power:- \mathrm{BrO}_4^{-}>\mathrm{IO}_4^{-}>\mathrm{ClO}_4^{-}\)
From JEE Mains
Explore more questions on app