X and Y are the number of electrons involved, respectively during the oxidation of I⁻ to I₂ and S²⁻ to S by acidified K₂Cr₂O₇. The value of X + Y is ____

Electrons involved in oxidation of iodide and sulphide by acidified potassium dichromate

Q. X and Y are the number of electrons involved, respectively during the oxidation of I⁻ to I₂ and S²⁻ to S by acidified K₂Cr₂O₇. The value of X + Y is _____ .

Correct Answer: 12

Explanation

First, write the standard reduction half-reaction of acidified potassium dichromate:

$$ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O $$

This shows that one mole of acidified K₂Cr₂O₇ accepts 6 electrons.

Oxidation of iodide ion (I⁻ → I₂):

Basic oxidation half-reaction:

$$ 2I^- \rightarrow I_2 + 2e^- $$

Electrons released = 2

To supply 6 electrons, multiply the reaction by 3:

$$ 6I^- \rightarrow 3I_2 + 6e^- $$

Therefore,

$$ X = 6 $$

Oxidation of sulphide ion (S²⁻ → S):

Basic oxidation half-reaction:

$$ S^{2-} \rightarrow S + 2e^- $$

Electrons released = 2

To match 6 electrons accepted by dichromate:

$$ 3S^{2-} \rightarrow 3S + 6e^- $$

Hence,

$$ Y = 6 $$

Final Calculation:

$$ X + Y = 6 + 6 = \boxed{12} $$

Therefore, the correct answer is 12.