In the reaction below, identify the species undergoing oxidation and reduction. \( \text{H}_2\text{S}(\text{g})+\text{Cl}_2(\text{g})\rightarrow2\text{HCl}(\text{g})+\text{S}(\text{s}) \)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(H_2S\) is oxidized and \(Cl_2\) is reduced
In a redox reaction, the species whose oxidation number increases (it loses electrons) is oxidised, and the species whose oxidation number decreases (it gains electrons) is reduced. The two always happen together.
Following the oxidation numbers in the reaction:
Sulphur: −2 in \( \ce{H2S} \) (2(+1) + x = 0) and 0 in S. It rises, so \( \ce{H2S} \) is oxidised.
Chlorine: 0 in \( \ce{Cl2} \) and −1 in HCl. It falls, so \( \ce{Cl2} \) is reduced.
Hydrogen stays at +1 throughout.
The two half reactions show the electron transfer:
\( \displaystyle \text{Oxidation: } \ce{H2S -> S + 2H+ + 2e-} \)
\( \displaystyle \text{Reduction: } \ce{Cl2 + 2e- -> 2Cl^-} \)
So \( \ce{H2S} \) is oxidised and \( \ce{Cl2} \) is reduced; \( \ce{H2S} \) acts as the reducing agent and \( \ce{Cl2} \) as the oxidising agent.
Why the other options are incorrect
Option (2) \(H_2S\) and \(Cl_2\) both are reduced — Incorrect. The electrons gained in a reduction must come from a species that is oxidised; two reductions cannot take place alone.
Option (3) \(Cl_2\) and \(H_2S\) both are oxidized — Incorrect. For the same reason, two oxidations cannot take place without a reduction.
Option (4) \(H_2S\) is reduced and \(Cl_2\) is oxidized — Incorrect. This reverses the roles: the oxidation number of sulphur rises (−2 to 0) and that of chlorine falls (0 to −1).