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

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