Answer:
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
