Question

Reduction potentials of some ions are given below. Arrange
them in decreasing order of oxidising power.

Ion ClO₄⁻ IO₄⁻ BrO₄⁻
Reduction potential E° = 1.19V E° = 1.65V E° = 1.74V
E° / V
(a) ClO₄⁻ > IO₄⁻ < BrO₄⁻ (b) IO₄⁻ > BrO₄⁻ < ClO₄⁻
(c) BrO₄⁻ > IO₄⁻ < ClO₄⁻ (d) BrO₄⁻ > ClO₄⁻ < IO₄⁻

Answer 1

Answer:

Reduction potentials of some ions are given below. Arrange

them in decreasing order of oxidising power.

Ion ClO₄⁻ IO₄⁻ BrO₄⁻

Reduction potential E° = 1.19V E° = 1.65V E° = 1.74V

E° / V

(a) ClO₄⁻ > IO₄⁻ < BrO₄⁻

(b) IO₄⁻ > BrO₄⁻ < ClO₄⁻

(c) BrO₄⁻ > IO₄⁻ < ClO₄⁻

(d) BrO₄⁻ > ClO₄⁻ < IO₄⁻

Answer 2

Option (d), BrO₄⁻ > ClO₄⁻ < IO₄⁻ is the correct way of arranging the given ions in the decreasing order of their oxidising power.

• Reduction potential of an ion denotes its ability to get reduced. Higher the reduction potential of an ion, greater is its ability to get reduced.

• We know that an ion which gets reduced itself, oxidises another ion or element in a compound. This implies that an ion with a high reduction potential will also have a high oxidising power.

• The respective values of reduction potential for ClO₄⁻,  IO₄⁻, and BrO₄⁻ are given as E° = 1.19V, E° = 1.65V, and E° = 1.74V.

• Looking at the values of reduction potential for the three ions, it can be concluded that BrO₄⁻ (E° = 1.74V) has the highest oxidising power as it has the highest reduction potential, followed by IO₄⁻ (E° = E° = 1.65V), having the second highest oxidising power, followed by ClO₄⁻ (E° = 1.19V), having the lowest oxidising power among the three.

• Therefore, the arrangement of the given ions in decreasing order of their oxidising power is BrO₄⁻ > IO₄⁻ > ClO₄⁻,  rearranged as BrO₄⁻ > ClO₄⁻ < IO₄⁻. Hence, option (d) is the correct answer.