Acidified potassium dichromate reacts with potassium iodide and oxidizes it to $I_{2}$ . What is the oxidation state of chromium in the products of the reaction?

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Solution

The correct option is C 3
$K_{2} C r_{2} O_{7} + 7 H_{2} S O_{4} + 6 K I \to 4 K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + 7 H_{2} O + 3 I_{2}$
We have $C r_{2} (S O_{4})_{3}$ in the product in which the oxidation state of Cr is calculated as:
Let, x is the oxidation state of Cr, so,
2x +3(-2) =0
x=+3

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K I

S O_{2}

The products formed when $S O_{2}$ reacts with acidified potassium dichromate solution is :

K_{2} C r_{2} O_{7}

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