Redox
The Core Concept of Redox
Redox is a portmanteau of reduction and oxidation. These two processes always happen simultaneously; you cannot have one without the other. This is because electrons (or oxygen) lost by one substance must be gained by another.
We can understand redox in two ways:
- Oxygen Transfer (Core): Focuses on the movement of oxygen atoms.
- Electron/Oxidation Number Transfer (Supplement): Focuses on the movement of electrons and changes in oxidation states.
Oxidation and Reduction (Oxygen Definition)
Oxidation is the gain of oxygen.
Reduction is the loss of oxygen.
Reduction is the loss of oxygen.
Example: Copper(II) Oxide and Carbon
Consider the reaction:
2CuO + C \rightarrow 2Cu + CO_2
- Copper(II) oxide (CuO) loses oxygen to become copper (Cu). Therefore, CuO is reduced.
- Carbon (C) gains oxygen to become carbon dioxide (CO_2). Therefore, C is oxidised.
⚠︎ Confusing the Substance with the Process
Mistake: Saying 'Copper is reduced' in the equation above.
Correction: Copper (Cu) is the product of reduction. The substance being reduced is copper(II) oxide (CuO). Reduction is what happens to a reactant, not what the product does.
Correction: Copper (Cu) is the product of reduction. The substance being reduced is copper(II) oxide (CuO). Reduction is what happens to a reactant, not what the product does.
Oxidation Numbers and Naming
An oxidation number (often denoted as ON or simply the oxidation state) represents the charge an atom would have if the compound was composed of ions. We use Roman numerals in names to indicate this.
Rules for Assigning Oxidation Numbers:
- The oxidation number of an element in its uncombined state (e.g., Fe, O_2, Cl_2) is zero.
- The oxidation number of a monatomic ion is the same as its charge (e.g., Na^+ has ON = +1; Cl^- has ON = -1).
- The sum of oxidation numbers in a neutral compound is zero.
- The sum of oxidation numbers in an ion is equal to the charge on the ion.
Roman Numerals in Names
The Roman numeral indicates the oxidation number of the element.
- Example: In iron(III) oxide, the (III) means the oxidation number of iron is +3.
Oxidation and Reduction (Electron Definition)
Oxidation is the loss of electrons or an increase in oxidation number.
Reduction is the gain of electrons or a decrease in oxidation number.
Reduction is the gain of electrons or a decrease in oxidation number.
Example: Sodium and Chlorine
Reaction: 2Na + Cl_2 \rightarrow 2NaCl
- Sodium (Na) starts as an element (ON=0) and becomes Na^+ (ON=+1). The oxidation number increases. Sodium is oxidised.
- Chlorine (Cl_2) starts as an element (ON=0) and becomes Cl^- (ON=-1). The oxidation number decreases. Chlorine is reduced.
⚠︎ Ignoring Signs in Oxidation Numbers
Mistake: Stating the oxidation number of chloride is just '1'.
Correction: You must include the sign. The oxidation number is -1. An increase from 0 to +1 is oxidation; a decrease from 0 to -1 is reduction.
Correction: You must include the sign. The oxidation number is -1. An increase from 0 to +1 is oxidation; a decrease from 0 to -1 is reduction.
Identifying Redox via Oxidation Numbers
To identify if a reaction is redox without obvious oxygen transfer, calculate the oxidation number of each element in the reactants and products. If any element changes its oxidation number, it is a redox reaction.
Why this matters: In the reaction 3SO_2 + Cr_2O_7^{2-} + 2H^+ \rightarrow 3SO_4^{2-} + 2Cr^{3+} + H_2O:
- Chromium goes from +6 (in Cr_2O_7^{2-}) to +3 (in Cr^{3+}). Decrease = Reduction.
- Sulfur goes from +4 (in SO_2) to +6 (in SO_4^{2-}). Increase = Oxidation.
This confirms it is a redox reaction.
Worked Example: Calculating Oxidation Number in Dichromate Ion (Cr_2O_7^{2-})
We need to find the oxidation number of Chromium (Cr) in Cr_2O_7^{2-}.
- Oxygen usually has an oxidation number of -2.
- The sum of all oxidation numbers must equal the charge of the ion, which is -2.
- Let x be the oxidation number of one Chromium atom.
- Equation: 2(x) + 7(-2) = -2
- Solve: 2x - 14 = -2 \Rightarrow 2x = +12 \Rightarrow x = +6
The oxidation number of Cr is +6.
Oxidising and Reducing Agents
Oxidising Agent: A substance that oxidises another substance. It does this by taking electrons away (or giving oxygen). Because it takes electrons, the oxidising agent itself is reduced.
Reducing Agent: A substance that reduces another substance. It does this by giving electrons away (or taking oxygen away). Because it gives electrons, the reducing agent itself is oxidised.
⚠︎ Confusing Agents with Products
Mistake: Identifying the product (e.g., Cu) as the reducing agent.
Correction: The reactant is the agent. In 2CuO + C \rightarrow 2Cu + CO_2, Carbon (C) is the reducing agent because it causes CuO to lose oxygen.
Correction: The reactant is the agent. In 2CuO + C \rightarrow 2Cu + CO_2, Carbon (C) is the reducing agent because it causes CuO to lose oxygen.
Defining Agents Correctly
When asked to define an oxidising or reducing agent, you must mention both parts of the definition: what it does to the other substance AND what happens to itself.
Correct phrasing: 'An oxidising agent is a substance that gains electrons (or gives oxygen) and is itself reduced.'
Why this works: Examiners look for the reciprocal nature of redox. Stating only 'it oxidises another substance' is incomplete because it doesn't demonstrate understanding of electron conservation.
Identifying Redox via Colour Changes
Certain ions are used as indicators for redox reactions due to distinct colour changes.
Acidified aqueous potassium manganate(VII) (KMnO_4):
- Contains MnO_4^- ions which are purple.
- When reduced (gains electrons), it forms Mn^{2+} ions which are colourless (or very pale pink).
- Use: Used to test for reducing agents. The purple solution decolourises.
Aqueous potassium iodide (KI):
- Contains I^- ions which are colourless.
- When oxidised (loses electrons), it forms iodine (I_2) which is brown (in solution) or orange-brown.
- Use: Used to test for oxidising agents. The solution turns brown.
Example: Chlorine and Iodide
Reaction: Cl_2 + 2I^- \rightarrow 2Cl^- + I_2
- Chlorine (Cl_2) is the oxidising agent because it oxidises iodide ions to iodine.
- The solution turns from colourless to brown, confirming oxidation has occurred.
Describing Colour Changes Accurately
When describing the test with acidified KMnO_4, always specify 'acidified'. Without acid, the reduction product is manganese(IV) oxide (MnO_2), which is a brown precipitate, not colourless.
Correct phrasing: 'The purple solution turns colourless.'
Why this works: It shows you know the specific conditions required for the standard redox indicator reaction. Mentioning 'acidified' demonstrates precise experimental knowledge.
Practice Questions
Q:
Define the term oxidation in terms of electron transfer.
A:
Oxidation is the loss of electrons.
Q:
In the reaction Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2, identify the oxidising agent and explain your answer.
A:
The oxidising agent is iron(III) oxide (Fe_2O_3). It is the oxidising agent because it loses oxygen (or causes CO to gain oxygen).
Q:
Calculate the oxidation number of nitrogen in NH_3 and in NO.
A:
In NH_3: H is +1, so N is -3. In NO: O is -2, so N is +2.
Q:
State the colour change observed when acidified potassium manganate(VII) acts as an oxidising agent.
A:
Purple to colourless.