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O-Level Redox Chemistry
oxidation and reduction, and electrochemistry
What the syllabus expects
- Define oxidation and reduction (redox) by the gain or loss of oxygen or hydrogen.
- Define redox using electron transfer and changes in oxidation state.
- Recognise redox reactions from oxygen or hydrogen gain/loss, electron gain/loss and shifts in oxidation state.
- Describe electrolysis as passing an electric current through an ionic compound, called the electrolyte, either molten or dissolved in water, which brings about chemical changes such as decomposition at the electrodes.
- Explain how electrolysis gives evidence for ions, which are locked in a lattice as a solid but able to move once the substance is molten or in solution.
- Describe the electrolysis of molten sodium chloride with inert electrodes, referring to how the ions move and what forms at each electrode.
- Predict the likely products when a molten binary ionic compound is electrolysed using inert electrodes.
- Apply the idea of selective discharge, considering cations in relation to the reactivity series, anions among halides, hydroxides and sulfates (as with dilute sodium chloride and aqueous copper(II) sulfate solutions, effectively electrolysing water), and the influence of concentration (as with concentrated versus dilute aqueous sodium chloride).
Scope: Inert electrodes are assumed throughout these cases. - Given the relevant details, predict the likely products from electrolysing an aqueous electrolyte.
- Write the ionic equations for the reactions at the electrodes during electrolysis when given suitable information.
- Describe how electrolysing aqueous copper(II) sulfate with copper electrodes serves to purify copper.
Scope: Technical details are not required. - Describe the electroplating of metals, such as copper plating, and give one use of electroplating.
- Describe how simple cells (two electrodes in an electrolyte) produce electrical energy, relating this to the reactivity series and to redox reactions in terms of electron transfer.
- Describe hydrogen, obtained from water or hydrocarbons, as a possible fuel that reacts with oxygen inside a hydrogen fuel cell to generate electricity directly.
Scope: How a fuel cell is constructed and operated is not required.
How it's examined
Questions on this topic most often ask you to explain, state, describe, suggest. About 6% of the past-paper style questions in Rae's bank for this subject sit in this topic.
Worked examples
Example 1 (3 marks)
B8 OR The head of a safety match holds potassium chlorate(V), KClO3, while the striking surface of the box contains red phosphorus, P4, together with an impurity. Striking the match on this surface triggers the reaction shown by the equation: 3P4 + 10KClO3 -> 3P4O10 + 10KCl Using changes in oxidation states, explain why this reaction is classed as a redox reaction.
Show the worked answer
Assign oxidation states. Phosphorus: 0 in P4 (element) rising to +5 in P4O10, so P is oxidised (oxidation state increases). Chlorine: +5 in KClO3 falling to -1 in KCl, so Cl is reduced (oxidation state decreases). Because oxidation and reduction occur together in the same reaction, it is a redox reaction.
Example 2 (2 marks)
Construct the half-equation for the conversion of ethanedioate ions, C2O4²⁻, into carbon dioxide, CO2.
Show the worked answer
Ethanedioate is oxidised to carbon dioxide, so electrons are released. Balance carbon: one C2O4²- gives 2 CO2. Oxygen is already balanced (4 on each side); no H or O species need adding. Balance charge: left side is 2-, right side (2CO2) is 0, so add 2 electrons to the right. C2O4²- -> 2CO2 + 2e^-
Example 3 (2 marks)
The zinc electrode was swapped for a silver one, after which the voltmeter read -0.52 V. Suggest why the voltmeter reading turned negative. [Total: 10]
Show the worked answer
In the original cell zinc was the more reactive electrode, so it was the negative terminal and set the direction of electron flow, giving a positive reading. Silver is less reactive than the other (fixed) electrode, so when it replaces zinc the roles of the two electrodes are reversed: the electrode that was positive is now the more reactive (negative) one. Electrons therefore flow through the meter in the opposite direction, so the voltmeter deflects the other way and reads a negative value.
More worked questions on this topic
- In another experiment, fluorine gas was passed through a separate solution of copper(II) chlori (2 marks)
More O-Level Pure Chemistry topics
Experimental Chemistry · The Particulate Nature of Matter · Chemical Bonding and Structure · Chemical Calculations · Acid-Base Chemistry · Qualitative Analysis · all of O-Level Pure Chemistry