Cl₂ and I⁻
Cl₂ + 2I⁻ → 2Cl⁻ + I₂
Redox balancer
Type the reactants and products with an arrow, without numbers: MnO4- + Fe2+ -> Mn2+ + Fe3+. Leave out H+, OH- and water; they are added for you. Salts and acids such as KMnO4 or HNO3 are fine.
Cl₂+2I⁻→2Cl⁻+I₂
- Oxidising agentCl₂
- Reducing agentI⁻
- Electrons2
OxidationI −1 → 0
2I⁻→I₂+2e⁻
loses 2 e⁻ReductionCl 0 → −1
Cl₂+2e⁻→2Cl⁻
gains 2 e⁻- Write the skeleton in ionic form
Cl₂ + I⁻ → Cl⁻ + I₂ - Split into two half reactions
Oxidation: I⁻ → I₂ (I −1 → 0)
Reduction: Cl₂ → Cl⁻ (Cl 0 → −1) - Balance atoms other than O and H
Oxidation: 2I⁻ → I₂
Reduction: Cl₂ → 2Cl⁻ - Balance the charge with electrons
Oxidation: 2I⁻ → I₂ + 2e⁻
Reduction: Cl₂ + 2e⁻ → 2Cl⁻ - The electrons are already equal
Oxidation: 2I⁻ → I₂ + 2e⁻
Reduction: Cl₂ + 2e⁻ → 2Cl⁻ - Add the halves and cancel what appears on both sides
Cl₂ + 2I⁻ → 2Cl⁻ + I₂ - Check
Each side has 2 I, 2 Cl and a total charge of −2.
- Write the oxidation numbers and find the change
I: −1 in I⁻ → 0 in I₂, a rise of 2 for 2I⁻ → I₂ (oxidation)
Cl: 0 in Cl₂ → −1 in Cl⁻, a fall of 2 for Cl₂ → 2Cl⁻ (reduction) - Make the total rise equal the total fall (2 each)
Cl₂ + 2I⁻ → 2Cl⁻ + I₂ - Check
Cl₂ + 2I⁻ → 2Cl⁻ + I₂
Each side has 2 I, 2 Cl and a total charge of −2.
Step by step
Ion-electron method
- Write the skeleton in ionic form
Cl₂ + I⁻ → Cl⁻ + I₂ - Split into two half reactions
Oxidation: I⁻ → I₂ (I −1 → 0)
Reduction: Cl₂ → Cl⁻ (Cl 0 → −1) - Balance atoms other than O and H
Oxidation: 2I⁻ → I₂
Reduction: Cl₂ → 2Cl⁻ - Balance the charge with electrons
Oxidation: 2I⁻ → I₂ + 2e⁻
Reduction: Cl₂ + 2e⁻ → 2Cl⁻ - The electrons are already equal
Oxidation: 2I⁻ → I₂ + 2e⁻
Reduction: Cl₂ + 2e⁻ → 2Cl⁻ - Add the halves and cancel what appears on both sides
Cl₂ + 2I⁻ → 2Cl⁻ + I₂ - Check
Each side has 2 I, 2 Cl and a total charge of −2.
Oxidation number method
- Write the oxidation numbers and find the change
I: −1 in I⁻ → 0 in I₂, a rise of 2 for 2I⁻ → I₂ (oxidation)
Cl: 0 in Cl₂ → −1 in Cl⁻, a fall of 2 for Cl₂ → 2Cl⁻ (reduction) - Make the total rise equal the total fall (2 each)
Cl₂ + 2I⁻ → 2Cl⁻ + I₂ - Check
Cl₂ + 2I⁻ → 2Cl⁻ + I₂
Each side has 2 I, 2 Cl and a total charge of −2.
Oxidising and reducing agents
- I −1 → 0I⁻ to I₂: oxidised
- Cl 0 → −1Cl₂ to Cl⁻: reduced
- Cl₂oxidising agent, gains electrons
- I⁻reducing agent, loses electrons
Cl₂ and I⁻: quick answers
What is the balanced equation for chlorine and iodide?
Cl₂ + 2I⁻ → 2Cl⁻ + I₂.
Which is the oxidising agent and which is the reducing agent?
I goes from −1 to 0 and Cl goes from 0 to −1. Cl₂ is the oxidising agent (it is reduced) and I⁻ is the reducing agent (it is oxidised).
How many electrons are transferred?
2 electrons pass from I⁻ to Cl₂ in the balanced equation as written.