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3 years ago

Balance the following equation by redox method H2S + H2 O2 gives S + H2O

1 answer:

5 0

$\text{H}_2\stackrel{-2}{\text{S}} + \text{H}_2\stackrel{-1}{\text{O}_2} \to \; \stackrel{0}{\text{S}} + 2\; \text{H}_2\stackrel{-2}{\text{O}}$

<h3>Explanation</h3>

Find the oxidation state for each element.

The oxidation state of H is +1 in most compounds.

The S atom in H₂S is bonded to two H atoms. The oxidation state of S in H₂S will be -(2 × (+1)) = -2.
Two O atom in H₂O₂ are bonded to two H atoms. On average, the oxidation state for each O atom in H₂O₂ will be - 1/2 × (2 × (+1)) = -1.
S has an oxidation state of 0 when it is not bonded to any other element;
The O atom in H₂O is bonded to two H atoms. The oxidation state of O in H₂O will be -(2 × (+1)) = -2.

Changes in oxidation states:

S: from -2 to 0.
O: from -1 to -2.

Each S atom will reduce two O atoms. There are two O atoms in each H₂O₂ molecule. There's only one S atom in each H₂S molecule. As a result, pairing one H₂O₂ with one H₂S will balance the change in oxidation state.

Each H₂O₂ contains two O atoms. The two O atoms will produce two H₂O molecules. Each H₂S molecule will lead to one S molecule. Thus the coefficient for H₂O will be two and the coefficient for the rest three species will be one.

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Answer:

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