Topic 16 · page 13 of 19

Redox titration reactions and ratios

Know the equation and the mole ratio of each titration reaction. When iodine is made first from an oxidising agent and excess iodide, chain the ratios together to find how much thiosulfate it needs.

14 min read16.15

Read first: Predicting feasibility, Building redox equations

The two titration reactions 16.15

A redox titration works like any other titration, except that the reaction is a redox reaction. Two reactions matter in this topic.

Manganate(VII) with iron(II) is the reaction you built on page 2:

MnOX4X−(aq)+8 HX+(aq)+5 FeX2+(aq)→MnX2+(aq)+4 HX2O(l)+5 FeX3+(aq) \begin{aligned} &\ce{MnO4-(aq) + 8H+(aq) + 5Fe^2+(aq)} \\ &\ce{-> Mn^2+(aq) + 4H2O(l) + 5Fe^3+(aq)} \end{aligned}

The ratio is MnO₄⁻ : Fe²⁺ = 1 : 5. Ecell⊖E^\ominus_\text{cell} = +1.51 − (+0.77) = +0.74 V, so the reaction is feasible. Manganate(VII) is purple and manganese(II) is almost colourless, so the titration is self-indicating.

Iodine with thiosulfate:

IX2(aq)+2 SX2OX3X2−(aq)→2 IX−(aq)+SX4OX6X2−(aq) \small \ce{I2(aq) + 2S2O3^2-(aq) -> 2I-(aq) + S4O6^2-(aq)}

The ratio is I₂ : S₂O₃²⁻ = 1 : 2. The couples are I₂/I⁻ (+0.54 V) and S₄O₆²⁻/S₂O₃²⁻ (+0.09 V), so Ecell⊖E^\ominus_\text{cell} = +0.54 − (+0.09) = +0.45 V. It is feasible. Starch is added near the end point as the indicator.

Finding a ratio from the electrons 16.15

MethodDeriving a mole ratio

  1. Write the half-equation for the reduction and the half-equation for the oxidation.
  2. Note the number of electrons in each.
  3. Choose multiples so that the electrons are equal. The multiples are the mole ratio.

Worked exampleManganate(VII) with hydrogen peroxide

MnOX4X−+8 HX++5 eX−→MnX2++4 HX2O\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O} and HX2OX2→OX2+2 HX++2 eX−\ce{H2O2 -> O2 + 2H+ + 2e-}.

  1. The lowest common multiple of 5 and 2 is 10 electrons.
  2. 2 MnO₄⁻ gain 10 electrons and 5 H₂O₂ lose 10 electrons.
  3. The ratio is MnO₄⁻ : H₂O₂ = 2 : 5.
2 MnOX4X−+6 HX++5 HX2OX2→2 MnX2++8 HX2O+5 OX2 \begin{aligned} &\ce{2MnO4- + 6H+ + 5H2O2} \\ &\ce{-> 2Mn^2+ + 8H2O + 5O2} \end{aligned}

Iodine made first 16.15

Many oxidising agents cannot be titrated directly with thiosulfate. Instead, add them to excess iodide, which releases iodine, then titrate the iodine with thiosulfate. The amount of iodine formed equals what the oxidising agent can make, so thiosulfate measures the oxidising agent indirectly.

Oxidising agent Equation Ratio to S₂O₃²⁻
Cu²⁺ (for example in brass) 2Cu²⁺ + 4I⁻ → 2CuI + I₂ Cu²⁺ : S₂O₃²⁻ = 1 : 1
IO₃⁻ (iodate(V)) IO₃⁻ + 5I⁻ + 6H⁺ → 3I₂ + 3H₂O IO₃⁻ : S₂O₃²⁻ = 1 : 6
IO₄⁻ (iodate(VII)) IO₄⁻ + 7I⁻ + 8H⁺ → 4I₂ + 4H₂O IO₄⁻ : S₂O₃²⁻ = 1 : 8

Each ratio comes from chaining two steps. For copper: 2Cu²⁺ make 1 I₂, and 1 I₂ needs 2S₂O₃²⁻, so 2Cu²⁺ need 2S₂O₃²⁻, which simplifies to 1 : 1. For iodate(V): 1 IO₃⁻ makes 3 I₂, and 3 I₂ need 6S₂O₃²⁻, so 1 : 6.

The route for every calculation 16.15

Page 16 applies this route to full calculations.

Check yourself

  1. State the mole ratio MnO₄⁻ : Fe²⁺ and show where it comes from, using oxidation numbers.
  2. Calculate Ecell⊖E^\ominus_\text{cell} for the iodine and thiosulfate reaction, given +0.54 V for I₂/I⁻ and +0.09 V for S₄O₆²⁻/S₂O₃²⁻.
  3. Explain why no indicator is needed in a manganate(VII) titration.
  4. Use the equation 2Cu²⁺ + 4I⁻ → 2CuI + I₂ and I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻ to find the mole ratio Cu²⁺ : S₂O₃²⁻.
  5. Use IO₃⁻ + 5I⁻ + 6H⁺ → 3I₂ + 3H₂O to find the mole ratio IO₃⁻ : S₂O₃²⁻.
  6. What is a primary standard? Why is potassium iodate(V) suitable?

Answers cover these until you have tried the questions

  1. 1 : 5. Mn falls from +7 to +2, a change of 5. Each Fe rises from +2 to +3, a change of 1. So 5 Fe²⁺ are needed per MnO₄⁻.
  2. Ecell⊖E^\ominus_\text{cell} = +0.54 − (+0.09) = +0.45 V. Positive, so feasible.
  3. MnO₄⁻ is purple and Mn²⁺ is almost colourless. While Fe²⁺ remains, each drop is decolourised; the first excess drop turns the solution pink, which marks the end point.
  4. 2Cu²⁺ produce 1 I₂. 1 I₂ reacts with 2S₂O₃²⁻. So 2Cu²⁺ ≡ 2S₂O₃²⁻, giving Cu²⁺ : S₂O₃²⁻ = 1 : 1.
  5. 1 IO₃⁻ produces 3 I₂. 3 I₂ react with 6S₂O₃²⁻. So IO₃⁻ : S₂O₃²⁻ = 1 : 6.
  6. A pure, stable solid that can be weighed accurately to make a solution of known concentration. Potassium iodate(V) can be obtained pure and is stable, so the amount of iodine it releases is accurately known.