Cl₂ and cold NaOH
Cl₂ + 2NaOH → NaCl + NaClO + H₂O
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+2NaOH→NaCl+NaClO+H₂O
Ionic Cl₂+2OH⁻→Cl⁻+ClO⁻+H₂O
- Oxidised and reducedCl₂disproportionation
- Electrons1
OxidationCl 0 → +1
Cl₂+4OH⁻→2ClO⁻+2H₂O+2e⁻
loses 2 e⁻ReductionCl 0 → −1
Cl₂+2e⁻→2Cl⁻
gains 2 e⁻- Write the skeleton in ionic form
Cl₂ → Cl⁻ + ClO⁻
Written as ions: NaOH as Na⁺ + OH⁻; NaCl as Na⁺ + Cl⁻; NaClO as Na⁺ + ClO⁻.
Na⁺ does not change and is left out. - Split into two half reactions
Oxidation: Cl₂ → ClO⁻ (Cl 0 → +1)
Reduction: Cl₂ → Cl⁻ (Cl 0 → −1) - Balance atoms other than O and H
Oxidation: Cl₂ → 2ClO⁻
Reduction: Cl₂ → 2Cl⁻ - Balance O by adding H₂O
Oxidation: Cl₂ + 2H₂O → 2ClO⁻ - Balance H by adding H⁺
Oxidation: Cl₂ + 2H₂O → 2ClO⁻ + 4H⁺ - Basic solution: add OH⁻ to both sides for each H⁺, make water and cancel
Oxidation: Cl₂ + 4OH⁻ → 2ClO⁻ + 2H₂O - Balance the charge with electrons
Oxidation: Cl₂ + 4OH⁻ → 2ClO⁻ + 2H₂O + 2e⁻
Reduction: Cl₂ + 2e⁻ → 2Cl⁻ - The electrons are already equal
Oxidation: Cl₂ + 4OH⁻ → 2ClO⁻ + 2H₂O + 2e⁻
Reduction: Cl₂ + 2e⁻ → 2Cl⁻ - Add the halves and cancel what appears on both sides
Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O - Check
Each side has 2 Cl, 2 O, 2 H and a total charge of −2. - Put back the ions that do not change
Cl₂ + 2NaOH → NaCl + NaClO + H₂O
- Write the oxidation numbers and find the change
Cl: 0 in Cl₂ → +1 in ClO⁻, a rise of 2 for Cl₂ → 2ClO⁻ (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)
2Cl₂ → 2Cl⁻ + 2ClO⁻ - Balance the charge with 4 OH⁻ (basic solution)
2Cl₂ + 4OH⁻ → 2Cl⁻ + 2ClO⁻ - Balance H with 2 H₂O
2Cl₂ + 4OH⁻ → 2Cl⁻ + 2ClO⁻ + 2H₂O - Check
Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O
Each side has 2 Cl, 2 O, 2 H and a total charge of −2. - Put back the ions that do not change
Cl₂ + 2NaOH → NaCl + NaClO + H₂O
Step by step
Ion-electron method
- Write the skeleton in ionic form
Cl₂ → Cl⁻ + ClO⁻
Written as ions: NaOH as Na⁺ + OH⁻; NaCl as Na⁺ + Cl⁻; NaClO as Na⁺ + ClO⁻.
Na⁺ does not change and is left out. - Split into two half reactions
Oxidation: Cl₂ → ClO⁻ (Cl 0 → +1)
Reduction: Cl₂ → Cl⁻ (Cl 0 → −1) - Balance atoms other than O and H
Oxidation: Cl₂ → 2ClO⁻
Reduction: Cl₂ → 2Cl⁻ - Balance O by adding H₂O
Oxidation: Cl₂ + 2H₂O → 2ClO⁻ - Balance H by adding H⁺
Oxidation: Cl₂ + 2H₂O → 2ClO⁻ + 4H⁺ - Basic solution: add OH⁻ to both sides for each H⁺, make water and cancel
Oxidation: Cl₂ + 4OH⁻ → 2ClO⁻ + 2H₂O - Balance the charge with electrons
Oxidation: Cl₂ + 4OH⁻ → 2ClO⁻ + 2H₂O + 2e⁻
Reduction: Cl₂ + 2e⁻ → 2Cl⁻ - The electrons are already equal
Oxidation: Cl₂ + 4OH⁻ → 2ClO⁻ + 2H₂O + 2e⁻
Reduction: Cl₂ + 2e⁻ → 2Cl⁻ - Add the halves and cancel what appears on both sides
Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O - Check
Each side has 2 Cl, 2 O, 2 H and a total charge of −2. - Put back the ions that do not change
Cl₂ + 2NaOH → NaCl + NaClO + H₂O
Oxidation number method
- Write the oxidation numbers and find the change
Cl: 0 in Cl₂ → +1 in ClO⁻, a rise of 2 for Cl₂ → 2ClO⁻ (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)
2Cl₂ → 2Cl⁻ + 2ClO⁻ - Balance the charge with 4 OH⁻ (basic solution)
2Cl₂ + 4OH⁻ → 2Cl⁻ + 2ClO⁻ - Balance H with 2 H₂O
2Cl₂ + 4OH⁻ → 2Cl⁻ + 2ClO⁻ + 2H₂O - Check
Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O
Each side has 2 Cl, 2 O, 2 H and a total charge of −2. - Put back the ions that do not change
Cl₂ + 2NaOH → NaCl + NaClO + H₂O
One species, both ways
- Cl 0 → +1Cl₂ to ClO⁻: oxidised
- Cl 0 → −1Cl₂ to Cl⁻: reduced
- Cl₂is both oxidised and reduced: a disproportionation reaction
Cl₂ and cold NaOH: quick answers
What is the balanced equation for chlorine and cold dilute sodium hydroxide?
Cl₂ + 2NaOH → NaCl + NaClO + H₂O. In ionic form: Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O.
Which is the oxidising agent and which is the reducing agent?
Cl goes from 0 to +1 and Cl goes from 0 to −1. Cl₂ is both oxidised and reduced, so this is a disproportionation reaction.
Why is this a disproportionation reaction?
The same species, Cl₂, is oxidised in one half reaction and reduced in the other. Cl goes from 0 to +1 and Cl goes from 0 to −1.