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

H $_2$ S

HCl

Cl $_2$

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Correct answer is D

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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H2_2 S(g)_{(g)} + Cl2(g)_{2(g)} → 2HCl(g)_{(g)} + S(g)_{(g)} In the reaction above, the substance that is reduced is

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