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Oxidation Number Method

Balance the f...

Question

Balance the following reaction by oxidation number method:
$K_{2} C r_{2} O_{7} + F e S O_{4} + H_{2} S O_{4} \to K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + F e_{2} (S O_{4})_{3} + H_{2} O$ .

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Solution

The unbalanced equation is
$K_{2} C r_{2} O_{7} + F e S O_{4} + H_{2} S O_{4} \to K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + F e_{2} (S O_{4})_{3} + H_{2} O$ .
Balance all atoms except H and O.
$K_{2} C r_{2} O_{7} + 2 F e S O_{4} + 5 H_{2} S O_{4} \to K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + F e_{2} (S O_{4})_{3} + H_{2} O$ .
The oxidation number of Cr decreases from +6 to +3. The decrease in oxidation number of 1 Cr atom is 3. The decrease in oxidation number of 2 Cr atoms is 6.
The oxidation number of Fe increases from +2 to +3. The increase in oxidation number of 1 Fe atom is 1. The increase in oxidation number of 2 Fe atom is 2.
Multiply species containing Fe with 3. It will balance increase in oxidation number with decrease in oxidation number.
$K_{2} C r_{2} O_{7} + 6 F e S O_{4} + 7 H_{2} S O_{4} \to K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + 3 F e_{2} (S O_{4})_{3} + H_{2} O$ .
Balance O atoms by adding water molecules.
$K_{2} C r_{2} O_{7} + 6 F e S O_{4} + 7 H_{2} S O_{4} \to K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + 3 F e_{2} (S O_{4})_{3} + 7 H_{2} O$ .
H atoms are balanced. This is balanced chemical equation.

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Similar questions

Q. Balance the following equation by oxidation number method.

K_{2} C r_{2} O_{7} + F e S O_{4} + H_{2} S O_{4} \to C r_{2} (S O_{4})_{3} + F e_{2} (S O_{4})_{3} + K_{2} S O_{4} + H_{2} O

Q. Balance the following equation by oxidation number method.

P b C r O_{4} + H_{2} S O_{4} + F e S O_{4} \to F e_{2} (S O_{4})_{3} + P b S O_{4} + C r_{2} (S O_{4})_{3} + H_{2} O

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