Which of the following reaction involves neither oxidation
nor reduction
(a) CrO₄²⁻ → Cr₂O₇²⁻ (b) Cr → CrCl₃
(c) Na → Na⁺ (d) 2S₂O₃²⁻ → S₄O₆²⁻
Answers
Answer:
option a correct there is no change in oxidation number of Cr before after the reaction. ie +6.so that reaction involves neither oxidation nor reduction
The reaction (a) CrO₄²⁻ → Cr₂O₇²⁻ involves neither oxidation nor reduction.
• Oxidation or reduction involves a change in the oxidation state of an element. If there is no change in the oxidation state of an element, it means the reaction undergoes neither oxidation nor reduction.
• Let us calculate the oxidation states of both reactant and product in each case.
(a) CrO₄²⁻ → Cr₂O₇²⁻
In CrO₄²⁻, the oxidation state of Cr can be calculated as :
x + { (-2) × 4 } = -2
=> x - 8 = -2
=> x = -2 + 8
=> x = +6
In Cr₂O₇²⁻, the oxidation state of Cr can be calculated as :
2x + { (-2) × 7 } = -2
=> 2x - 14 = -2
=> 2x = -2 + 14
=> 2x = 12
=> x = 12 / 2
=> x = +6
∴ The oxidation state of Cr in both reactant and product is +6. So, no oxidation or reduction is involved in the reaction.
(b) Cr → CrCl₃
In Cr, the oxidation state of Cr is 0.
In CrCl₃, the oxidation state of Cr can be calculated as :
x + { (-1) × 3 } = 0
=> x - 3 = 0
=> x = +3
The oxidation state changes from 0 to +3 (loss of 3 electrons). Therefore, the reaction undergoes oxidation.
(c) Na → Na⁺
In Na, the oxidation state of Na is 0.
In Na⁺, the oxidation state of Na is +1.
Again, a loss of one electron is involved as the oxidation state changes from 0 to +1. Therefore, the reaction involves oxidation.
(d) 2S₂O₃²⁻ → S₄O₆²⁻
In 2S₂O₃²⁻, the oxidation state of S can be calculated as :
2x + { (-2) × 3 } = -2
=> 2x - 6 = -2
=> 2x = -2 + 6
=> 2x = +4
=> x = +4 / 2
=> x = +2
In S₄O₆²⁻, the oxidation state of S can be calculated as :
4x + { (-2) × 6 } = -2
=> 4x - 12 = -2
=> 4x = -2 + 12
=> 4x = +10
=> x = +10 / 4
=> x = +5 / 2
This reaction also undergoes oxidation by a change in oxidation state of S from +2 to +5 / 2.
• Hence, our answer is (a).