Numford
Balanced in acidic solution

S₂O₃²⁻ and I₂

2S₂O₃²⁻ + I₂ → S₄O₆²⁻ + 2I⁻

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.

2S₂O₃²⁻+I₂→S₄O₆²⁻+2I⁻

  • Oxidising agentI₂
  • Reducing agentS₂O₃²⁻
  • Electrons2

OxidationS +2 → +5/2

2S₂O₃²⁻→S₄O₆²⁻+2e⁻

loses 2 e⁻

ReductionI 0 → −1

I₂+2e⁻→2I⁻

gains 2 e⁻
  1. Write the skeleton in ionic form
    S₂O₃²⁻ + I₂ → S₄O₆²⁻ + I⁻
  2. Split into two half reactions
    Oxidation: S₂O₃²⁻ → S₄O₆²⁻ (S +2 → +5/2)
    Reduction: I₂ → I⁻ (I 0 → −1)
  3. Balance atoms other than O and H
    Oxidation: 2S₂O₃²⁻ → S₄O₆²⁻
    Reduction: I₂ → 2I⁻
  4. Balance the charge with electrons
    Oxidation: 2S₂O₃²⁻ → S₄O₆²⁻ + 2e⁻
    Reduction: I₂ + 2e⁻ → 2I⁻
  5. The electrons are already equal
    Oxidation: 2S₂O₃²⁻ → S₄O₆²⁻ + 2e⁻
    Reduction: I₂ + 2e⁻ → 2I⁻
  6. Add the halves and cancel what appears on both sides
    2S₂O₃²⁻ + I₂ → S₄O₆²⁻ + 2I⁻
  7. Check
    Each side has 4 S, 6 O, 2 I and a total charge of −4.
Both methods

Step by step

Ion-electron method

  1. Write the skeleton in ionic form
    S₂O₃²⁻ + I₂ → S₄O₆²⁻ + I⁻
  2. Split into two half reactions
    Oxidation: S₂O₃²⁻ → S₄O₆²⁻ (S +2 → +5/2)
    Reduction: I₂ → I⁻ (I 0 → −1)
  3. Balance atoms other than O and H
    Oxidation: 2S₂O₃²⁻ → S₄O₆²⁻
    Reduction: I₂ → 2I⁻
  4. Balance the charge with electrons
    Oxidation: 2S₂O₃²⁻ → S₄O₆²⁻ + 2e⁻
    Reduction: I₂ + 2e⁻ → 2I⁻
  5. The electrons are already equal
    Oxidation: 2S₂O₃²⁻ → S₄O₆²⁻ + 2e⁻
    Reduction: I₂ + 2e⁻ → 2I⁻
  6. Add the halves and cancel what appears on both sides
    2S₂O₃²⁻ + I₂ → S₄O₆²⁻ + 2I⁻
  7. Check
    Each side has 4 S, 6 O, 2 I and a total charge of −4.

Oxidation number method

  1. Write the oxidation numbers and find the change
    S: +2 in S₂O₃²⁻ → +5/2 in S₄O₆²⁻, a rise of 2 for 2S₂O₃²⁻ → S₄O₆²⁻ (oxidation)
    I: 0 in I₂ → −1 in I⁻, a fall of 2 for I₂ → 2I⁻ (reduction)
  2. Make the total rise equal the total fall (2 each)
    2S₂O₃²⁻ + I₂ → S₄O₆²⁻ + 2I⁻
  3. Check
    2S₂O₃²⁻ + I₂ → S₄O₆²⁻ + 2I⁻
    Each side has 4 S, 6 O, 2 I and a total charge of −4.
Who gives, who takes

Oxidising and reducing agents

  • S +2 → +5/2S₂O₃²⁻ to S₄O₆²⁻: oxidised
  • I 0 → −1I₂ to I⁻: reduced
  • I₂oxidising agent, gains electrons
  • S₂O₃²⁻reducing agent, loses electrons
Questions

S₂O₃²⁻ and I₂: quick answers

What is the balanced equation for thiosulphate and iodine?

2S₂O₃²⁻ + I₂ → S₄O₆²⁻ + 2I⁻.

Which is the oxidising agent and which is the reducing agent?

S goes from +2 to +5/2 and I goes from 0 to −1. I₂ is the oxidising agent (it is reduced) and S₂O₃²⁻ is the reducing agent (it is oxidised).

How many electrons are transferred?

2 electrons pass from S₂O₃²⁻ to I₂ in the balanced equation as written.

Balance your own equation