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

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