Unit AS 1: Redox Chemistry
Welcome to Redox! This is one of the most fundamental chapters in CCEA AS Chemistry. Whether it is rust forming on iron, bleach sanitising water, or batteries powering your phone, redox reactions are happening all around us. In this chapter, we will master how electrons move during chemical reactions, how to track them using oxidation numbers, and how to write and balance complex redox equations step by step.
---1. Fundamental Definitions: Oxidation and Reduction
The word redox is simply a combination of two words: Reduction and Oxidation. These two processes always occur together in a chemical reaction because electrons cannot just appear from nowhere or vanish into thin air—if one species loses electrons, another must gain them.
Core Definitions
There are two primary ways to define oxidation and reduction in AS Chemistry:
In terms of electron transfer:
• Oxidation is the loss of electrons.
• Reduction is the gain of electrons.
In terms of oxidation numbers:
• Oxidation is an increase in oxidation number.
• Reduction is a decrease in oxidation number.
Memory Aid: The Classic Mnemonic
Remember the trusty mnemonic: OIL RIG
• Oxidation Is Loss (of electrons)
• Reduction Is Gain (of electrons)
Oxidising Agents and Reducing Agents
Don't worry if this seems a bit upside-down at first! Think of a travel agent: a travel agent doesn't travel themselves; they arrange travel for someone else. Similarly:
• An Oxidising Agent (Oxidant) oxidises another chemical species. To do that, it takes electrons away from that species. Therefore, the oxidising agent accepts/gains electrons and is itself reduced (its oxidation number decreases).
• A Reducing Agent (Reductant) reduces another chemical species. To do that, it gives electrons to that species. Therefore, the reducing agent donates/loses electrons and is itself oxidised (its oxidation number increases).
What is Disproportionation?
A disproportionation reaction is a special type of redox reaction in which the same element in a single chemical species is simultaneously oxidised and reduced.
Example: The reaction of chlorine with cold, dilute aqueous sodium hydroxide:
\(\text{Cl}_2 + 2\text{OH}^- \rightarrow \text{Cl}^- + \text{ClO}^- + \text{H}_2\text{O}\)
• Chlorine in \(\text{Cl}_2\) starts with an oxidation number of \(0\).
• It is reduced to \(-1\) in \(\text{Cl}^-\).
• It is oxidised to \(+1\) in \(\text{ClO}^-\).
Because the same element (chlorine) from the same reactant (\(\text{Cl}_2\)) is both oxidised and reduced, this is a textbook disproportionation reaction.
Key Takeaway: Oxidation is the loss of electrons (increase in oxidation number); reduction is the gain of electrons (decrease in oxidation number). An oxidising agent gets reduced, while a reducing agent gets oxidised.
---2. Rules for Assigning Oxidation Numbers
An oxidation number (or oxidation state) is a bookkeeping tool used by chemists to keep track of electron distribution in compounds and ions. It represents the charge an atom would have if all bonds were completely ionic.
CCEA Examiner Tip: Sign Before Number!
In CCEA examinations, convention is critical:
• Oxidation states are written with the sign before the number: \(+1\), \(+2\), \(+3\), \(-1\), \(-2\).
• Ionic charges are written with the number before the sign: \(1+\), \(2+\), \(3+\), \(1-\), \(2-\).
Always include the plus sign for positive oxidation states! Writing \(2\) instead of \(+2\) will lose marks.
The Hierarchy of Rules
Follow these established rules to assign oxidation numbers to any chemical species:
1. Uncombined Elements:
The oxidation number of any uncombined, free element is always \(0\).
Examples: \(\text{Na} = 0\), \(\text{Fe} = 0\), \(\text{Cl}_2 = 0\), \(\text{O}_2 = 0\), \(\text{S}_8 = 0\).
2. Simple Monatomic Ions:
The oxidation number is equal to the charge on the ion.
Examples: \(\text{Na}^+ = +1\), \(\text{Mg}^{2+} = +2\), \(\text{Al}^{3+} = +3\), \(\text{Cl}^- = -1\), \(\text{S}^{2-} = -2\).
3. Neutral Compounds:
The sum of the oxidation numbers of all atoms in a neutral molecule or formula unit is always \(0\).
4. Polyatomic Ions:
The sum of the oxidation numbers of all constituent atoms is equal to the overall charge on the ion.
Example: In \(\text{SO}_4^{2-}\), the sum of all oxidation numbers equals \(-2\).
5. Specific Element Rules (and their essential exceptions):
• Group 1 metals: Always \(+1\) in all compounds.
• Group 2 metals: Always \(+2\) in all compounds.
• Fluorine: Always \(-1\) in all compounds (it is the most electronegative element).
• Hydrogen: Almost always \(+1\).
Exception: In metal hydrides (such as \(\text{NaH}\) or \(\text{CaH}_2\)), hydrogen is \(-1\).
• Oxygen: Almost always \(-2\).
Exception 1: In peroxides (such as \(\text{H}_2\text{O}_2\) or \(\text{BaO}_2\)), oxygen is \(-1\).
Exception 2: In bonded compounds with fluorine (such as \(\text{OF}_2\)), oxygen is \(+2\) (because fluorine is more electronegative).
• Halogens (Chlorine, Bromine, Iodine): Usually \(-1\).
Exception: When combined with a more electronegative element (oxygen or fluorine), halogens have positive oxidation states (e.g., \(+1\) in \(\text{ClO}^-\), \(+5\) in \(\text{ClO}_3^-\)).
Step-by-Step Worked Examples
Example A: Finding the oxidation state of Sulfur in the Sulfate ion, \(\text{SO}_4^{2-}\)
1. Let the oxidation number of sulfur be \(x\).
2. Oxygen is in a standard compound, so its oxidation state is \(-2\).
3. The overall charge of the ion is \(-2\).
4. Set up the equation: \(x + 4(-2) = -2\)
5. Solve: \(x - 8 = -2 \implies x = +6\)
Therefore, the oxidation state of sulfur in \(\text{SO}_4^{2-}\) is \(+6\).
Example B: Finding the oxidation state of Nitrogen in the Nitrate ion, \(\text{NO}_3^-\)
1. Let the oxidation number of nitrogen be \(x\).
2. Oxygen is \(-2\).
3. The overall charge is \(-1\).
4. Equation: \(x + 3(-2) = -1\)
5. Solve: \(x - 6 = -1 \implies x = +5\)
Therefore, the oxidation state of nitrogen in \(\text{NO}_3^-\) is \(+5\).
Key Takeaway: Oxidation numbers follow strict rules. Always watch out for the exceptions: \(\text{H}\) is \(-1\) in metal hydrides, and \(\text{O}\) is \(-1\) in peroxides or \(+2\) in \(\text{OF}_2\).
---3. IUPAC Nomenclature and Systematic Naming
Many transition metals and non-metals can exist in several different oxidation states. To avoid ambiguity, the IUPAC (International Union of Pure and Applied Chemistry) uses Roman numerals placed in parentheses directly after the element to state its exact oxidation number.
Common Systematic Names
• \(\text{FeCl}_2\): Iron(II) chloride (iron has an oxidation state of \(+2\))
• \(\text{FeCl}_3\): Iron(III) chloride (iron has an oxidation state of \(+3\))
• \(\text{CuSO}_4\): Copper(II) sulfate (copper has an oxidation state of \(+2\))
• \(\text{ClO}^-\): Chlorate(I) ion (chlorine has an oxidation state of \(+1\))
• \(\text{ClO}_3^-\): Chlorate(V) ion (chlorine has an oxidation state of \(+5\))
Key Takeaway: Roman numerals represent the oxidation state of the specific element, not the number of atoms in the formula.
---4. Writing and Balancing Half-Equations
A redox reaction can be split into two halves: an oxidation half-equation and a reduction half-equation. Each shows the movement of electrons explicitly.
Standard Procedure for Balancing Half-Equations in Acidic Solution
Whenever you are asked to balance a complex half-equation, follow these four steps in exact order:
1. Balance all elements other than oxygen (\(\text{O}\)) and hydrogen (\(\text{H}\)).
2. Balance oxygen atoms by adding water molecules (\(\text{H}_2\text{O}\)) to the side deficient in oxygen.
3. Balance hydrogen atoms by adding hydrogen ions (\(\text{H}^+\)) to the side deficient in hydrogen.
4. Balance electrical charge by adding electrons (\(\text{e}^-\)) to the more positive side.
Step-by-Step Worked Example: Dichromate(VI) to Chromium(III)
Let's balance the reduction of \(\text{Cr}_2\text{O}_7^{2-}\) into \(\text{Cr}^{3+}\) in acidic solution:
Step 1: Balance atoms other than \(\text{O}\) and \(\text{H}\) (the chromium atoms):
\(\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+}\)
Step 2: Balance oxygen atoms by adding \(\text{H}_2\text{O}\):
There are 7 oxygen atoms on the left, so add \(7\text{H}_2\text{O}\) to the right:
\(\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\)
Step 3: Balance hydrogen atoms by adding \(\text{H}^+\):
There are 14 hydrogen atoms on the right (\(7 \times 2\)), so add \(14\text{H}^+\) to the left:
\(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\)
Step 4: Balance the electrical charge by adding electrons (\(\text{e}^-\)):
• Total charge on left side: \((-2) + 14(+1) = +12\)
• Total charge on right side: \(2(+3) + 7(0) = +6\)
To balance the charges, add \(6\text{e}^-\) to the more positive (left) side:
\(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\)
5. Combining Half-Equations into Full Overall Redox Equations
To produce an overall ionic redox equation, combine the oxidation and reduction half-equations such that the number of electrons lost equals the number of electrons gained.
Method:
1. Multiply one or both half-equations by whole numbers so that both have the same number of electrons.
2. Add the two equations together.
3. Cancel out common species that appear on both sides of the arrow (including electrons \(\text{e}^-\), \(\text{H}^+\) ions, and \(\text{H}_2\text{O}\) molecules).
Worked Example: Combining Dichromate(VI) and Iron(II)
Reduction half-equation:
\(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\)
Oxidation half-equation:
\(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-\)
Step 1: Equalise electrons
Multiply the iron oxidation equation by \(6\):
\(6\text{Fe}^{2+} \rightarrow 6\text{Fe}^{3+} + 6\text{e}^-\)
Step 2: Add equations and cancel species
\(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 6\text{Fe}^{3+} + 6\text{e}^-\)
Cancel the \(6\text{e}^-\) from both sides to give the final balanced overall equation:
\(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}\)
Check:
• Left charge: \((-2) + 14(+1) + 6(+2) = -2 + 14 + 12 = +24\)
• Right charge: \(2(+3) + 6(+3) = +6 + 18 = +24\)
Both mass and charge are balanced!
Key Takeaway: The total number of electrons lost in the oxidation half-reaction must strictly equal the total number of electrons gained in the reduction half-reaction. Overall ionic redox equations must never show free electrons.
---6. Summary & Common Pitfalls to Avoid
Common Exam Mistakes:
• Sign notation: Always write oxidation numbers as \(+2\) or \(-1\) (sign first), not \(2+\) or \(1-\) (charge format).
• Confusing the agent with the process: Always remember that the reducing agent is oxidised, and the oxidising agent is reduced.
• Forgetting exceptions: Always check for peroxides (\(\text{H}_2\text{O}_2\) where \(\text{O} = -1\)) and metal hydrides (\(\text{NaH}\) where \(\text{H} = -1\)).
• Unbalanced charges: Double-check that total electrical charge on the left equals total electrical charge on the right in both half-equations and overall equations.
• Disproportionation definitions: Ensure you clearly state that the same element in the same species is simultaneously oxidised and reduced.
Quick Review Summary:
• Oxidation: Loss of electrons / Increase in oxidation number.
• Reduction: Gain of electrons / Decrease in oxidation number.
• Oxidising Agent: Electron acceptor (gets reduced).
• Reducing Agent: Electron donor (gets oxidised).
• Balancing Half-Equations: Balance other atoms \(\rightarrow\) Balance \(\text{O}\) with \(\text{H}_2\text{O}\) \(\rightarrow\) Balance \(\text{H}\) with \(\text{H}^+\) \(\rightarrow\) Balance charge with \(\text{e}^-\).