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22 September, 00:07

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)

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  1. 22 September, 03:00
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    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.
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