Question

What is the balanced form of the following equation?
Br2 + S2O32- + H2O → Br1- + SO42- + H+

Answer 1

The balanced form of the equation is Br2 + 3S2O32- + 6H2O → 6Br1- + 3SO42- + 12H+.

Step-by-step explanation:

The balanced form of the equation is:

Br2 + 3S2O32- + 6H2O → 6Br1- + 3SO42- + 12H+

In the balanced equation, we have 2 bromine molecules on the reactant side and 6 bromide ions on the product side (Br1-). We also have 2 sulfate ions on the reactant side and 3 sulfate ions on the product side (SO42-). Lastly, we have 2 water molecules on the reactant side and 12 hydrogen ions on the product side (H+).

Answer 2

5 Br₂ + S₂O₃²⁻ + 5 H₂O ⇒ 10 Br⁻ + 2 SO₄²⁻ + 10 H⁺

Step-by-step explanation:

We will balance the redox reaction through the ion-electron method.

Step 1: Identify both half-reactions

Reduction: Br₂ ⇒ Br⁻

Oxidation: S₂O₃²⁻ ⇒ SO₄²⁻

Step 2: Perform the mass balance, adding H⁺ and H₂O where appropriate

Br₂ ⇒ 2 Br⁻

5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺

Step 3: Perform the charge balance, adding electrons where appropriate

2 e⁻ + Br₂ ⇒ 2 Br⁻

5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺ + 10 e⁻

Step 4: Make the number of electrons gained and lost equal

5 × (2 e⁻ + Br₂ ⇒ 2 Br⁻)

1 × (5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺ + 10 e⁻)

Step 5: Add both half-reactions

5 Br₂ + S₂O₃²⁻ + 5 H₂O ⇒ 10 Br⁻ + 2 SO₄²⁻ + 10 H⁺