Question

$\textbf{Reduction potential of ions are given below:}$
\[ \begin{array}{ccc} \text{ClO}_4^- & \text{IO}_4^- & \text{BrO}_4^- \\ E^\circ = 1.19 \, \text{V} & E^\circ = 1.65 \, \text{V} & E^\circ = 1.74 \, \text{V} \\ \end{array} \]
The correct order of their oxidising power is:

• A. \(\text{ClO}_4^- > \text{IO}_4^- > \text{BrO}_4^-\)
• B. \(\text{BrO}_4^- > \text{IO}_4^- > \text{ClO}_4^-\)
• C. \(\text{BrO}_4^- > \text{ClO}_4^- > \text{IO}_4^-\)
• D. \(\text{IO}_4^- > \text{BrO}_4^- > \text{ClO}_4^-\)

Answer 1

The Correct Option is B

Solution: Standard reduction potentials (SRP) indicate the propensity of each ion to be reduced. A higher standard reduction potential signifies a greater tendency to gain electrons, meaning the ion acts as a more effective oxidizing agent.

Comparative Analysis: Based on the provided data, BrO₄⁻ exhibits the highest E^o value at 1.74 V, followed by IO₄⁻ at 1.65 V. ClO₄⁻ has the lowest value at 1.19 V. Consequently, the order of oxidizing strength determined by SRP is:

\(\text{BrO}_4^- > \text{IO}_4^- > \text{ClO}_4^-\).