Answer:
4Br₂+ 5H₂O+ S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8Br⁻
Step-by-step explanation:
Br₂ + S₂O₃²⁻ + H₂O → Br⁻ + SO₄²⁻ + H⁺
This is a redox reaction:
Br₂ changes the oxidation state from 0 to -1, so it was reduced
In the S₂O₃⁻² anion S changes the oxidation state from +2 to +6 in sulfate anion. (S₂O₃⁻², it is called thiosulfate)
We have protons in the main equation, so we assume we are in acidic medium:
Br₂ + 2e⁻ → 2Br⁻ Reduction
We balanced the bromide with 2, so the bromine has gained 2 electrons.
5H₂O + S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8e- Oxidation
First of all, we add 2 to the sulfate anion in the product side, in order to balance the S.
As we have 8 O in right side, and 3 O in left side, we must add 5 O. We add 5 water in the place where the O are lower (reactant side).
Now, we have 10 H, in the reactant side, so we balance the product side with protons (10 H⁺).
Sulfur changed the oxidation state from +2 to +6, so it released 4 electrons, but, if you see thiosulfate anion you have 2 sulfurs so finally it has released 8 electrons.
Electrons are unbalanced so we multiply reduction x4, and oxidation x1.
(Br₂ + 2e⁻ → 2Br⁻) . 4 = 4Br₂ + 8e⁻ → 8Br⁻
(5H₂O + S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8e-) . 1 = STAYS THE SAME.
We sum both half reactions, to cancel the elecetrons:
4Br₂ + 8e⁻ + 5H₂O + S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8e- + 8Br⁻
Finally the balanced reaction is: 4Br₂+ 5H₂O+ S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8Br⁻