OCR A-Level Chemistry B H433
Development of Practical Skills in Chemistry
Elements of Life
Developing Fuels
Elements From the Sea
The Ozone Story
What's in a Medicine?
The Chemical Industry
Polymers and Life
Oceans
Developing Metals
Colour by Design
Chemical Literacy
Mathematical Skills in Chemistry
Developing Metals
Catalytic activity of transition metals
- 3 things to know
- Components 01, 02 and 03
Overview
Recall questions on how a variable oxidation state allows catalysis, the two steps of the iodide and peroxodisulfate route, and adsorption strength.
What you need to know
Answer each one before you reveal it.
Why is the peroxodisulfate and iodide reaction slow uncatalysed, and what happens if adsorption is too strong or too weak?
Which transition metals are used as catalysts in a car exhaust catalytic converter?
Questions from this subtopic
Each one with the model answer underneath.
- 1
Which two features of transition metals explain their catalytic activity?
Model answer
They have variable oxidation states, and their 3d and 4s electrons can form weak bonds to reactant molecules at a surface.
- 2
Explain how a variable oxidation state allows a transition metal ion to catalyse a reaction in solution.
Model answer
The ion is oxidised by one reactant and then reduced back by the other, so it takes part in two faster steps and is regenerated unchanged.
- 3
Which transition metal ion catalyses the reaction between peroxodisulfate ions and iodide ions?
Model answer
Iron(II) ions catalyse it, cycling between Fe²⁺ and Fe³⁺.
- 4
Write the equation for the first step of that catalysed reaction.
Model answer
The equation is S₂O₈²⁻ + 2Fe²⁺ → 2SO₄²⁻ + 2Fe³⁺.
- 5
Write the equation for the second step of that catalysed reaction.
Model answer
The equation is 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂.
- 6
What is the overall equation, and what has happened to the catalyst by the end?
Model answer
The overall equation is S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂, and the iron(II) is regenerated in the second step so it is not used up.
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