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

The balanced redox is:

14 H⁺ + Cr₂O₇²⁻ + 3Sn²⁺ → 3Sn4⁺ + 2Cr³⁺ + 7H₂O

So the coefficient for the H⁺ is greater than 4 (option E)

Explanation:

This is the redox reaction:

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

First of all, we must determine the half reactions:

In dycromate, Cr acts with + 6 in the oxidation state → Cr cation has + 3 in product side - Oxidation state, has decreased so this is the reduction.

In reactant side Sn cation acts with + 2 → In product side Sn acts with + 4

The oxidation state has increased, so this is the oxidation.

Cr₂O₇²⁻ → Cr³⁺

We have to add 2, to Cr in reactant side, and as we are in adicid medium we add water in the opposite side of oxygen. The same amount of oxgen, that we have.

Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O

Finally, as we have 14 H in product side, we must add 14 H⁺ to the reactant side. Cr+⁶ in dycromate to change to Cr³⁺, gained 3 e⁻, but we have 2 Cr, so in total the Cr gained 6e⁻. The balanced half reaction is:

14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O

Sn²⁺ to change the oxidation state, to + 4 had to release electrons:

Sn²⁺ → Sn4⁺ + 2e⁻

The electrons are unbalanced, so we must to multiply the half reactions:

(14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O) x1

(Sn²⁺ → Sn4⁺ + 2e⁻) x3

And we sum both:

14 H⁺ + Cr₂O₇²⁻ + 6e⁻ + 3Sn²⁺ → 3Sn4⁺ + 6e⁻ + 2Cr³⁺ + 7H₂O