Show that the following reactions are oxidation-reduction reactions indicating with reasons the oxidizing and reducting
1>2H²S+SO² --->2H²O + 3S
In the given reaction:
2H²S + SO² ---> 2H²O + 3S
The oxidation states of the elements involved are:
H: +1 in H₂S and 0 in H₂O
S: -2 in H₂S and +6 in SO₂
O: -2 in SO₂ and -2 in H₂O
Upon balancing the reaction, we can break it down into two half-reactions:
1. Oxidation half-reaction: 2H₂S ---> 3S + 6H+ + 6e-
2. Reduction half-reaction: SO₂ + 6H+ + 6e- ---> H₂O
In the oxidation half-reaction, hydrogen sulfide (H₂S) is oxidized to sulfur (S), while in the reduction half-reaction, sulfur dioxide (SO₂) is reduced to water (H₂O).
- Oxidizing agent: SO₂ acts as the oxidizing agent since it accepts electrons (undergoes reduction) in the reaction.
- Reducing agent: H₂S acts as the reducing agent since it donates electrons (undergoes oxidation) in the reaction.
Therefore, the given reaction is an oxidation-reduction reaction, with SO₂ as the oxidizing agent and H₂S as the reducing agent.