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H\(_2\) S\(_{(g)}\) + Cl\(_{2(g)}\) → 2HCl\(_{(g)}\) + S\(_{(g)}\) In the reactio...


Question

H\(_2\) S\(_{(g)}\) + Cl\(_{2(g)}\) → 2HCl\(_{(g)}\) + S\(_{(g)}\) In the reaction above, the substance that is reduced is

Options

A) H\(_2\)S

B) S

C) HCl

D) Cl\(_2\)


The correct answer is D.

Explanation:

H \(_2\) S + Cl \(_2\) → 2HCl + S
Please note: substance oxidized or reduced are always the reactants. So, one can easily cross out the two products.
Reduction is
1. Addition of hydrogen
2. Removal of Oxygen
3. Addition of Electron
Oxidation is
1. Removal of hydrogen
2. Addition of hydrogen
3. Removal of Electron
From the question we can see that H was added to Cl in the product, thereby making hydrogen to be removed from S in the reactant hence S is oxidized as a result of the removal of hydrogen and Cl \(_2\) has been reduced due to addition of hydrogen.
Cl \(_2\) is reduced.


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Dicussion (1)

  • H \(_2\) S + Cl \(_2\) → 2HCl + S
    Please note: substance oxidized or reduced are always the reactants. So, one can easily cross out the two products.
    Reduction is
    1. Addition of hydrogen
    2. Removal of Oxygen
    3. Addition of Electron
    Oxidation is
    1. Removal of hydrogen
    2. Addition of hydrogen
    3. Removal of Electron
    From the question we can see that H was added to Cl in the product, thereby making hydrogen to be removed from S in the reactant hence S is oxidized as a result of the removal of hydrogen and Cl \(_2\) has been reduced due to addition of hydrogen.
    Cl \(_2\) is reduced.

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