P₄ and OH⁻
P₄ + 3H₂O + 3OH⁻ → PH₃ + 3H₂PO₂⁻
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.
P₄+3H₂O+3OH⁻→PH₃+3H₂PO₂⁻
- Oxidised and reducedP₄disproportionation
- Electrons3
OxidationP 0 → +1
3P₄+24OH⁻→12H₂PO₂⁻+12e⁻
loses 12 e⁻ReductionP 0 → −3
P₄+12H₂O+12e⁻→4PH₃+12OH⁻
gains 12 e⁻- Write the skeleton in ionic form
P₄ → PH₃ + H₂PO₂⁻ - Split into two half reactions
Oxidation: P₄ → H₂PO₂⁻ (P 0 → +1)
Reduction: P₄ → PH₃ (P 0 → −3) - Balance atoms other than O and H
Oxidation: P₄ → 4H₂PO₂⁻
Reduction: P₄ → 4PH₃ - Balance O by adding H₂O
Oxidation: P₄ + 8H₂O → 4H₂PO₂⁻ - Balance H by adding H⁺
Oxidation: P₄ + 8H₂O → 4H₂PO₂⁻ + 8H⁺
Reduction: P₄ + 12H⁺ → 4PH₃ - Basic solution: add OH⁻ to both sides for each H⁺, make water and cancel
Oxidation: P₄ + 8OH⁻ → 4H₂PO₂⁻
Reduction: P₄ + 12H₂O → 4PH₃ + 12OH⁻ - Balance the charge with electrons
Oxidation: P₄ + 8OH⁻ → 4H₂PO₂⁻ + 4e⁻
Reduction: P₄ + 12H₂O + 12e⁻ → 4PH₃ + 12OH⁻ - Make the electrons equal
Oxidation × 3: 3P₄ + 24OH⁻ → 12H₂PO₂⁻ + 12e⁻
Reduction: P₄ + 12H₂O + 12e⁻ → 4PH₃ + 12OH⁻ - Add the halves and cancel what appears on both sides
P₄ + 3H₂O + 3OH⁻ → PH₃ + 3H₂PO₂⁻ - Check
Each side has 4 P, 6 O, 9 H and a total charge of −3.
- Write the oxidation numbers and find the change
P: 0 in P₄ → +1 in H₂PO₂⁻, a rise of 4 for P₄ → 4H₂PO₂⁻ (oxidation)
P: 0 in P₄ → −3 in PH₃, a fall of 12 for P₄ → 4PH₃ (reduction) - Make the total rise equal the total fall (12 each)
4P₄ → 4PH₃ + 12H₂PO₂⁻ - Balance the charge with 12 OH⁻ (basic solution)
4P₄ + 12OH⁻ → 4PH₃ + 12H₂PO₂⁻ - Balance H with 12 H₂O
4P₄ + 12H₂O + 12OH⁻ → 4PH₃ + 12H₂PO₂⁻ - Check
P₄ + 3H₂O + 3OH⁻ → PH₃ + 3H₂PO₂⁻
Each side has 4 P, 6 O, 9 H and a total charge of −3.
Step by step
Ion-electron method
- Write the skeleton in ionic form
P₄ → PH₃ + H₂PO₂⁻ - Split into two half reactions
Oxidation: P₄ → H₂PO₂⁻ (P 0 → +1)
Reduction: P₄ → PH₃ (P 0 → −3) - Balance atoms other than O and H
Oxidation: P₄ → 4H₂PO₂⁻
Reduction: P₄ → 4PH₃ - Balance O by adding H₂O
Oxidation: P₄ + 8H₂O → 4H₂PO₂⁻ - Balance H by adding H⁺
Oxidation: P₄ + 8H₂O → 4H₂PO₂⁻ + 8H⁺
Reduction: P₄ + 12H⁺ → 4PH₃ - Basic solution: add OH⁻ to both sides for each H⁺, make water and cancel
Oxidation: P₄ + 8OH⁻ → 4H₂PO₂⁻
Reduction: P₄ + 12H₂O → 4PH₃ + 12OH⁻ - Balance the charge with electrons
Oxidation: P₄ + 8OH⁻ → 4H₂PO₂⁻ + 4e⁻
Reduction: P₄ + 12H₂O + 12e⁻ → 4PH₃ + 12OH⁻ - Make the electrons equal
Oxidation × 3: 3P₄ + 24OH⁻ → 12H₂PO₂⁻ + 12e⁻
Reduction: P₄ + 12H₂O + 12e⁻ → 4PH₃ + 12OH⁻ - Add the halves and cancel what appears on both sides
P₄ + 3H₂O + 3OH⁻ → PH₃ + 3H₂PO₂⁻ - Check
Each side has 4 P, 6 O, 9 H and a total charge of −3.
Oxidation number method
- Write the oxidation numbers and find the change
P: 0 in P₄ → +1 in H₂PO₂⁻, a rise of 4 for P₄ → 4H₂PO₂⁻ (oxidation)
P: 0 in P₄ → −3 in PH₃, a fall of 12 for P₄ → 4PH₃ (reduction) - Make the total rise equal the total fall (12 each)
4P₄ → 4PH₃ + 12H₂PO₂⁻ - Balance the charge with 12 OH⁻ (basic solution)
4P₄ + 12OH⁻ → 4PH₃ + 12H₂PO₂⁻ - Balance H with 12 H₂O
4P₄ + 12H₂O + 12OH⁻ → 4PH₃ + 12H₂PO₂⁻ - Check
P₄ + 3H₂O + 3OH⁻ → PH₃ + 3H₂PO₂⁻
Each side has 4 P, 6 O, 9 H and a total charge of −3.
One species, both ways
- P 0 → +1P₄ to H₂PO₂⁻: oxidised
- P 0 → −3P₄ to PH₃: reduced
- P₄is both oxidised and reduced: a disproportionation reaction
P₄ and OH⁻: quick answers
What is the balanced equation for white phosphorus in alkali (disproportionation)?
P₄ + 3H₂O + 3OH⁻ → PH₃ + 3H₂PO₂⁻.
Which is the oxidising agent and which is the reducing agent?
P goes from 0 to +1 and P goes from 0 to −3. P₄ is both oxidised and reduced, so this is a disproportionation reaction.
Why is this a disproportionation reaction?
The same species, P₄, is oxidised in one half reaction and reduced in the other. P goes from 0 to +1 and P goes from 0 to −3.