The complete balanced equation is:

$H_{2} S + K_{2} C r O_{4} + H_{2} S O_{4} \to$

3 H_{2} S + 2 K_{2} C r O_{4} + 5 H_{2} S O_{4} \to C r_{2} (S O_{4})_{3} + 2 K_{2} S O_{4} + 8 H_{2} O + 3 S

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3 H_{2} S + K_{2} C r O_{4} + 5 H_{2} S O_{4} \to C r_{2} (S O_{4})_{3} + K_{2} S O_{4} + 8 H_{2} O + 3 S

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H_{2} S + 2 K_{2} C r O_{4} + 3 H_{2} S O_{4} \to C r_{2} (S O_{4})_{3} + 2 K_{2} S O_{4} + 8 H_{2} O + 3 S

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none of these

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Solution

The correct option is A $3 H_{2} S + 2 K_{2} C r O_{4} + 5 H_{2} S O_{4} \to C r_{2} (S O_{4})_{3} + 2 K_{2} S O_{4} + 8 H_{2} O + 3 S$
The unbalanced redox equation is as follows:

$H_{2} S + K_{2} C r O_{4} + H_{2} S O_{4} \to C r_{2} (S O_{4})_{3} + K_{2} S O_{4} + S$

Balance all atoms other than $H$ and $O .$

$H_{2} S + 2 K_{2} C r O_{4} + 5 H_{2} S O_{4} \to C r_{2} (S O_{4})_{3} + 2 K_{2} S O_{4} + S$

The oxidation number of $C r$ changes from 6 to 3. The change in the oxidation number of $C r$ is 3. The total change in the oxidation number for 2 $C r$ atoms is 6.

The oxidation number of $S$ changes from -2 to 0. The change in the oxidation number of $S$ is 2.

The increase in the oxidation number is balanced with a decrease in the oxidation number by multiplying $H_{2} S$ and $S$ with 3.

$3 H_{2} S + 2 K_{2} C r O_{4} + 5 H_{2} S O_{4} \to C r_{2} (S O_{4})_{3} + 2 K_{2} S O_{4} + 3 S$

$O$ atoms are balanced by adding 8 water molecules on RHS.

$3 H_{2} S + 2 K_{2} C r O_{4} + 5 H_{2} S O_{4} \to C r_{2} (S O_{4})_{3} + 2 K_{2} S O_{4} + 3 S + 8 H_{2} O$

$H$ atoms are balanced. This is the balanced chemical equation.

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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

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