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
Elements From the Sea
Oxidation state calculations, and defining oxidation and reduction
- 3 things to know
- Components 01, 02 and 03
Overview
Recall questions on solving for an unknown oxidation state, why one can be fractional, and defining oxidation and reduction by the change.
What you need to know
Answer each one before you reveal it.
In 2Na + Cl₂ → 2NaCl, which element is oxidised and by how much does its oxidation state change?
How is oxidation defined in terms of oxidation states?
Questions from this subtopic
Each one with the model answer underneath.
- 1
Calculate the oxidation state of chlorine in ClO⁻.
Model answer
Oxygen is −2 and the ion carries −1, so chlorine is +1.
- 2
Calculate the oxidation state of chlorine in ClO₃⁻.
Model answer
Three oxygens give −6 and the ion carries −1, so chlorine is +5.
- 3
Calculate the oxidation state of chlorine in HClO₄.
Model answer
Hydrogen is +1 and the four oxygens give −8, so chlorine must be +7 for the total to be zero.
- 4
Calculate the oxidation state of iodine in IO₃⁻.
Model answer
The three oxygens give −6 and the ion carries −1, so iodine is +5.
- 5
Calculate the oxidation state of sulfur in the thiosulfate ion, S₂O₃²⁻.
Model answer
Three oxygens give −6, so the two sulfur atoms must total +4, an average of +2 each.
- 6
Calculate the average oxidation state of sulfur in the tetrathionate ion, S₄O₆²⁻.
Model answer
Six oxygens give −12, so the four sulfur atoms must total +10, an average of +2.5 each.
Other ways to work this
The same content, four other ways.
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Six ways to work the same material.
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