$S O_{2}$ reduces ${C r_{2} O_{7}}^{2 -}$ to $C r^{2 +}$ . The change in oxidation number of $C r$ is:

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Solution

The balanced chemical equation for the reduction of dichromate with sulphur dioxide is shown.
$K_{2} C r_{2} O_{7} + 3 S O_{2} + H_{2} S O_{4} \to K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + H_{2} O$
In $K_{2} C r_{2} O_{7}$ , the oxidation number of Cr is +6.
In $C r_{2} (S O_{4})_{3}$ , the oxidation number of Cr is +3.
The change in the oxidation number of each Cr atom is 3.

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C r_{2} O 7^{2}

P O_{4}^{3 -}

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How many electrons should be added to the reactant to balance the atom whose oxidation number is changing?

$C r_{2} O_{7}^{2 -}$ $⟶$ $C r^{+ 3}$

How many electrons should be added to the reactant to balance the atom whose oxidation number is changing?

$C r_{2} O_{7}^{2 -}$ $⟶$ $C r^{+ 3}$

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