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: The standard reduction potentials (SRP) provide insight into the tendency of each ion to undergo reduction:
Understanding Standard Reduction Potential: The higher the value of the standard reduction potential, the greater the tendency to gain electrons (undergo reduction). Hence, ions with higher reduction potentials act as better oxidizing agents.

Comparative Analysis: From the given data: BrO₄⁻ has the highest E^o value (1.74 V). IO₄⁻ follows with 1.65 V. ClO₄⁻ has the lowest at 1.19 V. Thus, the correct order of oxidizing power based on SRP is:

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