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:

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

Answer 2

Given the standard reduction potentials for the perhalate ions: \( \text{ClO}_4^- / \text{ClO}_3^- \) (\( E^\circ = 1.19 \, \text{V} \)), \( \text{IO}_4^- / \text{IO}_3^- \) (\( E^\circ = 1.65 \, \text{V} \)), and \( \text{BrO}_4^- / \text{BrO}_3^- \) (\( E^\circ = 1.74 \, \text{V} \)), we are to find the correct order of their oxidizing power.

Concept Used:

The oxidizing power of a species is directly related to its standard reduction potential (\( E^\circ \)). A higher (more positive) standard reduction potential indicates a greater tendency for the species to be reduced. Therefore, a species with a higher \( E^\circ \) is a stronger oxidizing agent. The order of oxidizing power will be the same as the order of decreasing \( E^\circ \) values.

Step-by-Step Solution:

Step 1: List the given standard reduction potentials.

\[ \begin{aligned} &\text{For } \text{ClO}_4^- / \text{ClO}_3^-: & E^\circ &= 1.19 \, \text{V} \\ &\text{For } \text{IO}_4^- / \text{IO}_3^-: & E^\circ &= 1.65 \, \text{V} \\ &\text{For } \text{BrO}_4^- / \text{BrO}_3^-: & E^\circ &= 1.74 \, \text{V} \end{aligned} \]

Step 2: Compare the reduction potentials to determine oxidizing strength. Since oxidizing power increases with increasing \( E^\circ \), we compare the numerical values.

\[ 1.74 \, \text{V} > 1.65 \, \text{V} > 1.19 \, \text{V} \]

Step 3: Arrange the corresponding ions in order of decreasing oxidizing power. The ion with the highest \( E^\circ \) (\( \text{BrO}_4^- \)) is the strongest oxidizing agent, followed by \( \text{IO}_4^- \), and then \( \text{ClO}_4^- \).

Thus, the correct order of oxidizing power is \( \text{BrO}_4^- > \text{IO}_4^- > \text{ClO}_4^- \).