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

Home » Past Questions » Chemistry » Jamb » 2018 » H\(_2\) S\(_{(g)}\) + Cl\(_{2(g)}\) → 2HCl\(_{(g)}\) + S\(_{(g)}\) In the reaction above, the substance that…
Chemistry JAMB 2018

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

  • H\(_2\)S
  • S
  • HCl
  • Cl\(_2\) checkmark

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

There is an explanation video available .

Previous Question Next Question

Leave A Comment