Redox

1. Overview

Redox reactions (Reduction-Oxidation) are fundamental chemical processes where the transfer of oxygen or electrons occurs between substances. These reactions are essential in industrial processes, such as the extraction of metals in a blast furnace and the generation of energy in batteries and fuel cells.

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Key Definitions

• Redox Reaction: A reaction where oxidation and reduction occur simultaneously.
• Oxidation: The gain of oxygen, the loss of electrons, or an increase in oxidation number.
• Reduction: The loss of oxygen, the gain of electrons, or a decrease in oxidation number.
• Oxidising Agent: A substance that oxidises another substance and is itself reduced.
• Reducing Agent: A substance that reduces another substance and is itself oxidised.
• Oxidation Number: A number assigned to an atom in a compound that represents the number of electrons lost or gained by that atom.

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Core Content

Simultaneous Reactions

In any redox reaction, one substance is always oxidised while another is reduced. They cannot happen in isolation.

Oxidation and Reduction in terms of Oxygen

• Oxidation is the gain of oxygen.
• Reduction is the loss of oxygen.

Example: Reaction between Copper(II) Oxide and Hydrogen

• Word Equation: copper(II) oxide + hydrogen → copper + water
• Symbol Equation: CuO (s) + H₂ (g) → Cu (s) + H₂O (l)
• Analysis:
  • Copper(II) oxide loses oxygen to become copper; therefore, CuO is reduced.
  • Hydrogen gains oxygen to become water; therefore, H₂ is oxidised.

Using Roman Numerals

Roman numerals are used in the names of compounds to indicate the oxidation number of an element that can have multiple oxidation states (usually transition metals).

• Iron(II) chloride: FeCl₂ (Iron has an oxidation state of +2)
• Iron(III) chloride: FeCl₃ (Iron has an oxidation state of +3)
• Manganese(IV) oxide: MnO₂ (Manganese has an oxidation state of +4)

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Extended Content (Extended Curriculum Only)

Oxidation and Reduction in terms of Electrons

To remember the transfer of electrons, use the mnemonic OIL RIG:

• Oxidation Is Loss (of electrons)
• Reduction Is Gain (of electrons)

Example: Magnesium reacting with Oxygen 2Mg (s) + O₂ (g) → 2MgO (s)

• Magnesium atoms lose electrons: Mg → Mg²⁺ + 2e⁻ (Oxidation)
• Oxygen atoms gain electrons: O₂ + 4e⁻ → 2O²⁻ (Reduction)

Oxidation Numbers

You can identify redox reactions by tracking changes in oxidation numbers using these rules:

1. Elements: The oxidation number of an element in its uncombined state is 0 (e.g., Zn, O₂, Cl₂).
2. Monatomic Ions: The oxidation number is the same as the charge (e.g., Na⁺ is +1, S²⁻ is -2).
3. Compounds: The sum of oxidation numbers in a neutral compound is 0.
4. Polyatomic Ions: The sum of oxidation numbers equals the charge on the ion (e.g., in SO₄²⁻, the sum is -2).

• Oxidation = An increase in oxidation number.
• Reduction = A decrease in oxidation number.

Worked Example: Finding an unknown oxidation number What is the oxidation number of sulfur in $SO_3$?

1. Oxygen is always $-2$ in compounds.
2. There are 3 oxygen atoms, so total oxygen contribution = $3 \times (-2) = -6$.
3. The compound is neutral (total = 0), so: $S + (-6) = 0$, therefore $S = +6$.

Worked Example: Oxidation number in a polyatomic ion What is the oxidation number of manganese in $MnO_4^-$?

1. Oxygen = $-2$ each, so 4 oxygens = $-8$.
2. The ion has a charge of $-1$, so: $Mn + (-8) = -1$, therefore $Mn = +7$.

When the oxidation number of an element increases during a reaction, that element has been oxidised. When it decreases, it has been reduced. This is how you identify which species is the oxidising agent (it gets reduced itself) and which is the reducing agent (it gets oxidised itself).

Identifying Redox using Color Changes

Two specific reagents are used in the lab to test for oxidising or reducing agents:

1. Acidified Aqueous Potassium Manganate(VII):

   • This is a strong oxidising agent.
   • Colour change: Purple → Colourless.
   • If it turns colourless, it means it has reacted with a reducing agent.

2. Aqueous Potassium Iodide:

   • This is a reducing agent.
   • Colour change: Colourless → Brown (due to the formation of I₂).
   • If it turns brown, it means it has reacted with an oxidising agent.

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Key Equations

Reaction Type	Equation	Notes
Metal Displacement	Zn (s) + CuSO₄ (aq) → ZnSO₄ (aq) + Cu (s)	Zn is oxidised (0 to +2), Cu is reduced (+2 to 0)
Extraction of Iron	Fe₂O₃ (s) + 3CO (g) → 2Fe (l) + 3CO₂ (g)	Fe₂O₃ is reduced, CO is the reducing agent
Half-equation (Ox)	Mg (s) → Mg²⁺ (aq) + 2e⁻	Shows loss of electrons
Half-equation (Red)	Cl₂ (g) + 2e⁻ → 2Cl⁻ (aq)	Shows gain of electrons

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Common Mistakes to Avoid

• ❌ Wrong: Saying "The reducing agent is reduced."
• ✅ Right: The reducing agent is the substance that gets oxidised (it gives electrons away to reduce something else).
• ❌ Wrong: Forgetting state symbols in redox equations.
• ✅ Right: Always include (s), (l), (g), or (aq) as they often indicate the type of reaction taking place.
• ❌ Wrong: Thinking oxidation is only about oxygen.
• ✅ Right: In the extended curriculum, always check electron transfer or oxidation numbers first.

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Exam Tips

• Command Words:
  • "State": Give a brief answer (e.g., "State the colour change" -> "Purple to colourless").
  • "Explain": You must give a reason (e.g., "Explain why this is redox" -> "Because magnesium loses electrons and copper gains electrons").
  • "Identify": Pick the specific substance from the equation.
• Typical Contexts: Questions often use the context of vehicle catalytic converters (reducing NOₓ), chemical manufacturing, or thermal decomposition in a blast furnace.
• Observation Questions: If you see Potassium Manganate(VII) mentioned, the answer almost always involves the word "Purple." If you see Potassium Iodide, the answer almost always involves "Brown."
• Oxidation Numbers: Remember that Oxygen is almost always -2 (except in peroxides) and Hydrogen is +1 (except in metal hydrides). Use these as anchors to calculate unknown oxidation numbers.

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0620 Theory papers.

Exam-Style Question 1 — Short Answer [5 marks]

Question:

(a) Define the term oxidation in terms of electron transfer. [1]

(b) Copper(II) oxide reacts with hydrogen gas according to the following equation: $CuO(s) + H_2(g) \rightarrow Cu(s) + H_2O(g)$

Identify which substance is oxidised and which substance is reduced in this reaction. [2]

(c) State the colour change that would be observed if acidified aqueous potassium manganate(VII) is added to a solution of iron(II) sulfate. [2]

Worked Solution:

(a)

1. Oxidation is the loss of electrons. [Defining oxidation in terms of electron transfer]

How to earn full marks:

• State that oxidation is the loss of electrons.

(b)

1. Hydrogen is oxidised. [Hydrogen gains oxygen, therefore it is oxidised]
2. Copper(II) oxide is reduced. [Copper(II) oxide loses oxygen, therefore it is reduced]

How to earn full marks:

• Correctly identify hydrogen as being oxidised.
• Correctly identify copper(II) oxide as being reduced.

(c)

1. The colour changes from purple to colourless. [Potassium manganate(VII) is reduced by iron(II) ions, causing it to decolourise]

How to earn full marks:

• State that the initial colour is purple (or pink).
• State that the final colour is colourless.

Common Pitfall: Remember that oxidation and reduction always occur together in a redox reaction. If one substance is being oxidised, another substance must be reduced at the same time. Also, be precise with your colour descriptions – "clear" is not the same as "colourless".

Exam-Style Question 2 — Short Answer [6 marks]

Question:

(a) Define the term reducing agent. [1]

(b) Consider the following reaction: $Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)$

Determine the oxidation number of magnesium before and after the reaction. [2]

(c) Explain, in terms of oxidation numbers, why this reaction is a redox reaction. [3]

Worked Solution:

(a)

1. A reducing agent is a substance that reduces another substance and is itself oxidised. [Definition of a reducing agent]

How to earn full marks:

• State that a reducing agent reduces another substance.
• State that a reducing agent is itself oxidised.

(b)

1. Oxidation number of magnesium before the reaction: 0. [The oxidation number of an element in its uncombined state is zero]
2. Oxidation number of magnesium after the reaction: +2. [Magnesium forms a 2+ ion]

How to earn full marks:

• State that the initial oxidation number is 0.
• State that the final oxidation number is +2.

(c)

1. The oxidation number of magnesium increases from 0 to +2. [Magnesium is oxidised because its oxidation number increases]
2. This indicates that magnesium has been oxidised. [Explaining that magnesium is oxidised]
3. Since oxidation has occurred, reduction must also have occurred, making it a redox reaction. [Redox reactions involve simultaneous oxidation and reduction]

How to earn full marks:

• State that the oxidation number of magnesium increases.
• Explain that this increase in oxidation number indicates oxidation.
• State that both oxidation and reduction occur in redox reactions.

Common Pitfall: Students often forget that elements in their uncombined state have an oxidation number of zero. Also, make sure you clearly state that both oxidation and reduction must be present for a reaction to be classified as redox.

Exam-Style Question 3 — Extended Response [8 marks]

Question:

Iron can be extracted from its ore, hematite ($Fe_2O_3$), by heating it with carbon monoxide in a blast furnace. The overall reaction is:

$Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(l) + 3CO_2(g)$

(a) State the oxidation number of iron in $Fe_2O_3$. [1]

(b) Explain, in terms of electron transfer, why the conversion of $Fe_2O_3$ to $Fe$ is a reduction reaction. [3]

(c) Carbon monoxide acts as a reducing agent in this process. Explain what is meant by the term reducing agent. [2]

(d) Identify one environmental problem associated with the large-scale extraction of iron from its ore. [1]

(e) Suggest one method to mitigate the environmental problem you identified in part (d). [1]

Worked Solution:

(a)

1. The oxidation number of iron in $Fe_2O_3$ is +3. [Oxygen has an oxidation number of -2. 3 oxygen atoms contribute -6. To balance, 2 iron atoms must contribute +6, so each iron atom has an oxidation number of +3]

How to earn full marks:

• State that the oxidation number of iron is +3.

(b)

1. In $Fe_2O_3$, the iron has an oxidation number of +3. [Restating the oxidation number of iron in iron(III) oxide]
2. In elemental iron (Fe), the oxidation number is 0. [Oxidation number of an element in its uncombined state]
3. The oxidation number of iron decreases from +3 to 0, which means it gains electrons. [Linking the decrease in oxidation number to the gain of electrons]

How to earn full marks:

• State the initial oxidation number of iron (+3).
• State the final oxidation number of iron (0).
• Explain that the decrease in oxidation number means that iron gains electrons, therefore it is reduced.

(c)

1. A reducing agent is a substance that reduces another substance. [Part 1 of the definition]
2. A reducing agent is itself oxidised in the process. [Part 2 of the definition]

How to earn full marks:

• State that a reducing agent reduces another substance.
• State that a reducing agent is itself oxidised.

(d)

1. Emission of greenhouse gases (e.g., $CO_2$) / air pollution (e.g., sulfur dioxide) / water pollution from mine runoff / deforestation. [Identifying a problem]

How to earn full marks:

• Identify a valid environmental problem.

(e)

1. Use carbon capture technology / use scrubbers to remove sulfur dioxide / treat mine runoff to remove pollutants / replant trees after deforestation. [Suggesting a method]

How to earn full marks:

• Suggest a realistic method to mitigate the problem identified in part (d).

Common Pitfall: When explaining reduction or oxidation in terms of electron transfer, remember to explicitly state that electrons are either gained (reduction) or lost (oxidation). Don't just say the oxidation number changes; link it to electron movement.

Exam-Style Question 4 — Extended Response [9 marks]

Question:

A student investigates the reaction between zinc metal and copper(II) sulfate solution. The equation for the reaction is:

$Zn(s) + CuSO_4(aq) \rightarrow ZnSO_4(aq) + Cu(s)$

(a) Define oxidation in terms of:

(i) Loss of electrons [1] (ii) Increase in oxidation number [1]

(b) Use the concepts in (a) to explain why zinc is oxidised in this reaction. [3]

(c) Copper(II) sulfate solution is blue. State what colour change would be observed during the reaction. [1]

(d) The student performs an experiment to investigate the rate of this reaction. They add excess zinc powder to $50.0 cm^3$ of $0.500 mol/dm^3$ copper(II) sulfate solution. The initial mass of zinc added is $2.00g$.

Calculate the maximum mass of copper that could be produced in this reaction. (Relative atomic masses: Cu = 63.5, Zn = 65.4) [3]

Worked Solution:

(a) (i) Oxidation is the loss of electrons. [Definition of oxidation]

How to earn full marks:

• State that oxidation is the loss of electrons.

(ii) Oxidation is an increase in oxidation number. [Definition of oxidation]

How to earn full marks:

• State that oxidation is an increase in oxidation number.

(b)

1. Zinc metal ($Zn$) has an oxidation number of 0. [Oxidation number of element in its uncombined state is zero]
2. In zinc sulfate ($ZnSO_4$), zinc has an oxidation number of +2. [Zinc forms a 2+ ion]
3. Therefore, the oxidation number of zinc increases, meaning it is oxidised, and zinc loses two electrons. [Linking the increase in oxidation number and electron loss to oxidation]

How to earn full marks:

• State the initial oxidation number of zinc (0).
• State the final oxidation number of zinc (+2).
• Explain that the increase in oxidation number and electron loss means zinc is oxidised.

(c)

1. The solution changes from blue to colourless (or less blue). [The blue copper(II) ions are being removed from the solution]

How to earn full marks:

• State that the initial colour is blue.
• State that the final colour is colourless or a lighter shade of blue.

(d)

1. Calculate the number of moles of $CuSO_4$: $moles = concentration \times volume = 0.500 mol/dm^3 \times (50.0 cm^3 / 1000 cm^3/dm^3) = 0.0250 mol$. [Using the formula to calculate moles]
2. The mole ratio of $CuSO_4$ to $Cu$ is 1:1, so the number of moles of $Cu$ produced is also 0.0250 mol. [Using the stoichiometry of the equation]
3. Calculate the mass of copper produced: $mass = moles \times Mr = 0.0250 mol \times 63.5 g/mol = 1.5875 g$. Therefore, the maximum mass of copper is $1.59g$. [Using the formula to calculate mass]

How to earn full marks:

• Correctly calculate the number of moles of copper(II) sulfate (0.0250 mol).
• Correctly identify the mole ratio and state the number of moles of copper produced is the same (0.0250 mol).
• Correctly calculate the mass of copper produced, including the correct units. $\boxed{1.59 g}$

Common Pitfall: In calculations involving mole ratios, always double-check that the equation is balanced first. Also, pay attention to units – make sure to convert $cm^3$ to $dm^3$ when using the formula moles = concentration x volume.