Respuesta :
Answer:
SO₃²⁻ is the reducing agent and Cr₂O₇²⁻ is the oxidizing agent.
Explanation:
Oxidation reaction:
3SO₃²⁻ (aq) + 3H₂O (l) → 3SO₄²⁻ (aq) + 6H⁺ (aq) + 6e⁻
Reduction reaction:
Cr₂O₇²⁻ (aq) + 14H⁺ (aq) + 6e⁻ → 2Cr ³⁺ (aq) + 7H₂O (l)
Now, adding the oxidation and the reduction reactions we get the full net reaction:
Cr₂O₇²⁻ (aq) + 3SO₃²⁻ (aq) + 8H⁺ (aq) → 2Cr ³⁺ (aq) + 3SO₄²⁻ (aq) + 4H₂O(l)
Since, the S in SO₃²⁻, present in the +4 oxidation state is oxidized to +6 oxidation state in SO₄²⁻, by the loss of 2e⁻.
Therefore, SO₃²⁻ is the reducing agent.
And, the Cr in Cr₂O₇²⁻, present in the +6 oxidation state is getting reduced to +3 oxidation state, Cr ³⁺, by the gain of 6e⁻.
Therefore, Cr₂O₇²⁻ is the oxidizing agent.
