Topic 17 · page 20 of 20

Homogeneous catalysis in solution

Fe²⁺ speeds up the reaction of two negative ions by reacting with each in turn, with Fe³⁺ as the intermediate. Mn²⁺ is a product that catalyses its own formation, so the reaction starts slowly and then speeds up.

13 min read17.3017.3117.32

Read first: Two kinds of catalyst

Fe²⁺ with iodide and peroxodisulfate 17.3117.30

Iodide ions reduce peroxodisulfate ions to sulfate ions, producing iodine:

SX2OX8X2−(aq)+2 IX−(aq)→2 SOX4X2−(aq)+IX2(aq) \ce{S2O8^2-(aq) + 2I-(aq) -> 2SO4^2-(aq) + I2(aq)}

The data are E°(S₂O₈²⁻/SO₄²⁻) = +2.01 V and E°(I₂/I⁻) = +0.54 V, so E°cell = 2.01 − 0.54 = +1.47 V: the reaction is feasible.

But it is slow. Both reactants are negative ions, and negative ions repel each other, so the activation energy is high.

Add a little Fe²⁺(aq) and the reaction speeds up. The iron(II) ions react with each reactant in turn.

Step 1. Peroxodisulfate oxidises Fe²⁺ to Fe³⁺:

SX2OX8X2−+2 FeX2+→2 SOX4X2−+2 FeX3+ \ce{S2O8^2- + 2Fe^2+ -> 2SO4^2- + 2Fe^3+}

E°cell = +2.01 − (+0.77) = +1.24 V.

Step 2. Fe³⁺ oxidises iodide to iodine, and Fe²⁺ is regenerated:

2 FeX3++2 IX−→2 FeX2++IX2 \ce{2Fe^3+ + 2I- -> 2Fe^2+ + I2}

E°cell = +0.77 − (+0.54) = +0.23 V.

Add the two steps. The Fe²⁺ and Fe³⁺ cancel, and the overall equation is the one at the top.

A vertical scale of electrode potentials from 0.4 volts at the top to 2.2 volts at the bottom. I2/I minus at plus 0.54, Fe3+/Fe2+ at plus 0.77 and S2O8 2-/SO4 2- at plus 2.01. An arrow labelled step 1 runs from Fe2+ down to S2O8 2- with cell potential plus 1.24 volts. An arrow labelled step 2 runs from I minus down to Fe3+ with plus 0.23 volts.
The iron couple lies between the iodine and peroxodisulfate couples, so both steps are feasible.

Each step is between ions of opposite charge (a 2− ion with a 2+ ion, then a 3+ ion with a 1− ion). They attract, the activation energy is lower, and the reaction is faster. Fe³⁺ is the intermediate species, formed and used up again.

Fe³⁺ will also catalyse the reaction. The steps simply happen in the other order: Fe³⁺ oxidises I⁻ first, and the Fe²⁺ formed then reduces S₂O₈²⁻.

Autocatalysis by Mn²⁺ 17.32

In the oxalate titration, manganate(VII) oxidises ethanedioate (oxalate) ions in acid:

2 MnOX4X−+16 HX++5 CX2OX4X2−→2 MnX2++8 HX2O+10 COX2 \begin{aligned} &\ce{2MnO4- + 16H+ + 5C2O4^2-} \\ &\ce{-> 2Mn^2+ + 8H2O + 10CO2} \end{aligned}

The reaction is between two negative ions, which repel each other, so it is slow at first. But one of the products, Mn²⁺, is itself a catalyst for the reaction.

A graph of the concentration of MnO4 minus against time. The curve starts at a high value and falls slowly, then falls steeply in the middle where the catalyst has built up, then flattens as the reactants run out. The three parts are labelled slow, fast and slowing.
Concentration of MnO4- against time for an autocatalysed reaction (illustrative). The steepest part is in the middle, not at the start.

Most reactions are fastest at the start, because the reactant concentrations are highest. The curve for an autocatalysed reaction is different: the steepest part is in the middle.

Check yourself

  1. Define homogeneous catalyst and autocatalysis.
  2. Write the two equations that show how Fe²⁺ catalyses the reaction between I⁻ and S₂O₈²⁻. Show that they add up to the overall equation.
  3. Explain why the uncatalysed reaction between I⁻ and S₂O₈²⁻ is slow.
  4. Use E° values to show that both steps in the Fe²⁺ mechanism are feasible. (S₂O₈²⁻/SO₄²⁻ +2.01 V; Fe³⁺/Fe²⁺ +0.77 V; I₂/I⁻ +0.54 V)
  5. Name the intermediate in the Fe²⁺ mechanism.
  6. In the reaction of MnO₄⁻ with C₂O₄²⁻, the purple colour fades slowly at first and then quickly. Explain.

Answers cover these until you have tried the questions

  1. A homogeneous catalyst is in the same phase as the reactants and works via an intermediate species. Autocatalysis is catalysis of a reaction by one of its own products.
  2. S₂O₈²⁻ + 2Fe²⁺ → 2SO₄²⁻ + 2Fe³⁺ and 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂. Adding, 2Fe²⁺ and 2Fe³⁺ appear on both sides and cancel: S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂.
  3. Both reactants are negative ions, which repel each other, so the activation energy is high.
  4. Step 1: +2.01 − 0.77 = +1.24 V. Step 2: +0.77 − 0.54 = +0.23 V. Both are positive, so both steps are feasible.
  5. Fe³⁺.
  6. At first there is no Mn²⁺ catalyst, so the reaction between two negative ions is slow. As Mn²⁺ forms it catalyses (autocatalysis) the reaction, so the rate increases and the colour fades quickly. It slows again when the reactants are used up.