Question

Identify the oxidizing agent and the reducing agent in the reaction. 8H +( aq) + Cr 2O 7 2–( aq) + 3SO 3 2–( aq) → 2Cr 3+( aq) + 3SO 4 2–( aq) + 4H 2O( l)

Answer 1

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.