Answer:
In the above reaction, the oxidation state of tin changes from 2+ to 4+.
10 moles of electrons are transferred in the reaction
Explanation:
Redox reaction is:
2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻
SnO₂²⁻ → SnO₃²⁻
Tin changes the oxidation state from +2 to +4. It has increased it so this is the oxidation from the redox (it released 2 e⁻). We are in basic medium, so we add water in the side of the reaction where we have the highest amount of oxygen. We have 2 O on left side and 3 O on right side so we add 1 water on the right and we complete with OH⁻ in the opposite side to balance the H.
SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O Oxidation
BrO₃⁻ → Br₂
First of all, we have unbalance the bromine, so we add 2 on the BrO₃⁻. We have 6 O in left side and there are no O on the right, so we add 6 H₂O on the left. To balance the H, we must complete with 12OH⁻. Bromate reduces to bromine at ground state, so it gained 5e⁻. We have 2 atoms of Br, so finally it gaines 10 e⁻.
6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ Reduction
In order to balance the main reaction and balance the electrons we multiply (x5) the oxidation and (x1) the reduciton
(SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O) . 5
(6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻) . 1
5SnO₂²⁻ + 10OH⁻ + 6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ + 5SnO₃²⁻ + 10e⁻ + 5H₂O
We can cancel the e⁻ and we substract:
12OH⁻ - 10OH⁻ = 2OH⁻ (on the right side)
6H₂O - 5H₂O = H₂O (on the left side)
2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻
