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22 October, 04:41

Balance the following redox equation in acidic solution using the smallest integers possible and select the correct coefficient for the H + (aq) ion.

Cr2O72 - (aq) + Sn2 + (aq) → Cr3 + (aq) + Sn4 + (aq)

a. 1 (no coefficient written)

b. 2

c. 3

d. 4

e. More than 4

+3
Answers (1)
  1. 22 October, 05:42
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    3Sn²⁺ + 14 H⁺ + Cr₂O₇⁻² → 3Sn⁴⁺ + 2Cr³⁺ + 7H₂O

    Coefficient for the H⁺ is 14 (more than 4. - option e)

    Explanation:

    First of all, we should determine the half reactions

    Cr₂O₇⁻² in dichromate, chromium acts with + 6 → Cr³⁺

    Sn²⁺ → Sn⁴⁺

    So in the second element, Sn changed the oxidation state from + 2 to + 4. It has increased, so it has oxidized and it has released electrons.

    In dichromate, chromium decreased the oxidation state, from + 6 to + 3. It was reduced so it gained electrons.

    Sn²⁺ → Sn⁴⁺ + 2e⁻ oxidation

    Cr₂O₇⁻² + 6e⁻ → Cr³⁺ reduction

    We must balance the Cr with 2 and, in the opposite side of oxygens, we complete with the same amount of water. If we have 7 O, we must add 7 H₂O to the product side. Finally to balance the protons, we add 14 H⁺, to reactant side as we have 14 H in product side The half reaction of reduction will be:

    14 H⁺ + Cr₂O₇⁻² + 6e⁻ → 2Cr³⁺ + 7H₂O

    Now we have to balance the e⁻, so they can be cancelled. We multiply x3 half reaction of oxidation and x1 half reaction of reduction

    (Sn²⁺ → Sn⁴⁺ + 2e⁻) x3

    (14 H⁺ + Cr₂O₇⁻² + 6e⁻ → 2Cr³⁺ + 7H₂O). 1

    We sum each of them

    3Sn²⁺ + 14 H⁺ + Cr₂O₇⁻² + 6e⁻ → 3Sn⁴⁺ + 6e⁻ + 2Cr³⁺ + 7H₂O

    The balance reaction will be:

    3Sn²⁺ + 14 H⁺ + Cr₂O₇⁻² → 3Sn⁴⁺ + 2Cr³⁺ + 7H₂O
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