Redox

Introduction to Redox Chemistry

Welcome to the world of Redox! The word "Redox" is simply a combination of two words: Reduction and Oxidation. Whether you realize it or not, redox reactions are happening all around you every single day. When your smartphone battery powers your screen, when iron rusts in the rain, when bleach removes a stain from your clothes, or even when your body extracts energy from food, redox is at work.

In this chapter, we will master how electrons move from one substance to another, how to track them using oxidation numbers, how to name chemical compounds systematically, and how to balance complex redox reactions step-by-step. Don't worry if this seems like a lot to take in at first — we will break down every single rule and process into simple, bite-sized steps!

Key Takeaway: Redox reactions always involve two simultaneous processes: one chemical species loses electrons while another gains them.

1. Defining Oxidation and Reduction

At GCSE, you may have defined oxidation as the gain of oxygen (or loss of hydrogen) and reduction as the loss of oxygen (or gain of hydrogen). At AS Level, we focus primarily on the movement of electrons and changes in oxidation states.

Definitions in Terms of Electron Transfer

• Oxidation: The loss of electrons.
• Reduction: The gain of electrons.

Helpful Memory Aid: Use the classic mnemonic OIL RIG:
Oxidation Is Loss (of electrons)
Reduction Is Gain (of electrons)

For example, consider the reaction between magnesium and copper(II) sulfate:
\(\text{Mg(s)} + \text{Cu}^{2+}\text{(aq)} \rightarrow \text{Mg}^{2+}\text{(aq)} + \text{Cu(s)}\)

• Magnesium atom loses two electrons: \(\text{Mg} \rightarrow \text{Mg}^{2+} + 2\text{e}^-\) (Oxidation)
• Copper(II) ion gains two electrons: \(\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}\) (Reduction)

Definitions in Terms of Oxidation Number

• Oxidation: An increase in oxidation number (becoming more positive / less negative).
• Reduction: A decrease in oxidation number (becoming less positive / more negative).

Quick Review: Oxidation and reduction MUST occur together. You cannot have a substance losing electrons unless there is another substance ready to take them!

2. Oxidation Numbers (Oxidation States)

An oxidation number (or oxidation state) is a hypothetical charge assigned to an atom in a compound or ion, assuming all bonds were completely ionic. It acts as an essential bookkeeping tool for chemists to track where electrons have moved.

The Fundamental Rules for Assigning Oxidation Numbers

Always apply these rules in order of priority when working out oxidation numbers:

Rule 1: Uncombined Elements
The oxidation number of an atom in any uncombined pure element is always \(0\).
Examples: \(\text{Fe}\) has an oxidation state of \(0\); each atom in \(\text{O}_2\), \(\text{Cl}_2\), \(\text{H}_2\), \(\text{N}_2\), \(\text{P}_4\), and \(\text{S}_8\) is \(0\).

Rule 2: Simple Monatomic Ions
The oxidation number of a monatomic ion equals the charge on the ion.
Examples: \(\text{Na}^+\) is \(+1\); \(\text{Mg}^{2+}\) is \(+2\); \(\text{Al}^{3+}\) is \(+3\); \(\text{Cl}^-\) is \(-1\); \(\text{O}^{2-}\) is \(-2\).

Rule 3: Neutral Compounds
The sum of all oxidation numbers in a neutral compound must equal \(0\).

Rule 4: Polyatomic Ions
The sum of all oxidation numbers of all atoms in a polyatomic ion must equal the overall charge of the ion.

Rule 5: Elements with Fixed (or near-fixed) Oxidation States in Compounds:
• Group 1 metals: Always \(+1\) in compounds.
• Group 2 metals: Always \(+2\) in compounds.
• Fluorine: Always \(-1\) in all compounds (because it is the most electronegative element).
• Hydrogen: Almost always \(+1\), except in metal hydrides (such as \(\text{NaH}\) or \(\text{CaH}_2\)) where it is \(-1\).
• Oxygen: Almost always \(-2\), with two important exceptions:
   1. In peroxides (such as \(\text{H}_2\text{O}_2\)), oxygen is \(-1\).
   2. In compounds with fluorine (such as \(\text{OF}_2\)), oxygen is \(+2\).
• Chlorine, Bromine, Iodine: Usually \(-1\), except when bonded to a more electronegative element (such as oxygen or fluorine).

Step-by-Step Worked Examples

Example A: Finding the oxidation number of Sulfur in Sulfuric Acid, \(\text{H}_2\text{SO}_4\)
1. Identify the knowns: \(\text{H} = +1\), \(\text{O} = -2\).
2. Set up the equation for a neutral molecule (sum \(= 0\)):
\(2(\text{H}) + 1(\text{S}) + 4(\text{O}) = 0\)
\(2(+1) + \text{S} + 4(-2) = 0\)
\(+2 + \text{S} - 8 = 0\)
\(\text{S} - 6 = 0 \implies \text{S} = +6\)
Therefore, the oxidation number of sulfur in \(\text{H}_2\text{SO}_4\) is \(+6\).

Example B: Finding the oxidation number of Manganese in the Manganate(VII) ion, \(\text{MnO}_4^-\)
1. Identify the knowns: \(\text{O} = -2\). The overall charge is \(-1\).
2. Set up the equation (sum \(= -1\)):
\(1(\text{Mn}) + 4(\text{O}) = -1\)
\(\text{Mn} + 4(-2) = -1\)
\(\text{Mn} - 8 = -1\)
\(\text{Mn} = -1 + 8 \implies \text{Mn} = +7\)
Therefore, the oxidation number of manganese in \(\text{MnO}_4^-\) is \(+7\).

Example C: Finding the oxidation number of Chromium in the Dichromate(VI) ion, \(\text{Cr}_2\text{O}_7^{2-}\)
1. Identify the knowns: \(\text{O} = -2\). The overall charge is \(-2\).
2. Set up the equation:
\(2(\text{Cr}) + 7(\text{O}) = -2\)
\(2\text{Cr} + 7(-2) = -2\)
\(2\text{Cr} - 14 = -2\)
\(2\text{Cr} = +12 \implies \text{Cr} = +6\)
Therefore, each chromium atom has an oxidation number of \(+6\).

Common Mistake to Avoid: Always state the sign (+ or -) explicitly when writing oxidation numbers (write "\( +6 \)", not just "\( 6 \)"). Oxidation numbers are written with the sign first (\(+2\)), whereas ionic charges are traditionally written with the number first (\(2+\)).

3. Systematic Chemical Nomenclature (Stock Notation)

Many transition metals and non-metals can exist in multiple oxidation states. To avoid ambiguity, the IUPAC (International Union of Pure and Applied Chemistry) naming system uses Roman numerals in parentheses to specify the oxidation state of the key element.

Examples of Systematic Naming

• Iron compounds:
\(\text{FeCl}_2\) : Iron is \(+2\) \(\rightarrow\) Iron(II) chloride
\(\text{FeCl}_3\) : Iron is \(+3\) \(\rightarrow\) Iron(III) chloride

• Copper compounds:
\(\text{Cu}_2\text{O}\) : Copper is \(+1\) \(\rightarrow\) Copper(I) oxide
\(\text{CuO}\) : Copper is \(+2\) \(\rightarrow\) Copper(II) oxide

• Oxyanions (Polyatomic ions containing oxygen):
\(\text{SO}_3^{2-}\) : Sulfur is \(+4\) \(\rightarrow\) Sulfate(IV) ion (commonly called sulfite)
\(\text{SO}_4^{2-}\) : Sulfur is \(+6\) \(\rightarrow\) Sulfate(VI) ion (commonly called sulfate)
\(\text{NO}_2^-\) : Nitrogen is \(+3\) \(\rightarrow\) Nitrate(III) ion (commonly called nitrite)
\(\text{NO}_3^-\) : Nitrogen is \(+5\) \(\rightarrow\) Nitrate(V) ion (commonly called nitrate)
\(\text{ClO}^-\) : Chlorine is \(+1\) \(\rightarrow\) Chlorate(I) ion
\(\text{ClO}_3^-\) : Chlorine is \(+5\) \(\rightarrow\) Chlorate(V) ion

Key Takeaway: The Roman numeral tells you the oxidation number of that specific element, NOT how many atoms of that element are present!

4. Oxidising and Reducing Agents

In any redox reaction, the two participating species play complementary roles:

• Oxidising Agent (Oxidant):
A species that oxidises another substance and is itself reduced in the process. It is an electron acceptor. The oxidation number of the key atom in an oxidising agent decreases.

• Reducing Agent (Reductant):
A species that reduces another substance and is itself oxidised in the process. It is an electron donor. The oxidation number of the key atom in a reducing agent increases.

Everyday Analogy: Think of a travel agent. A travel agent helps you travel; they don't necessarily go on vacation themselves! Similarly, an oxidising agent oxidises something else by taking electrons from it.

Identifying Agents in a Reaction

Consider the reaction:
\(2\text{Fe}^{2+}\text{(aq)} + \text{Cl}_2\text{(g)} \rightarrow 2\text{Fe}^{3+}\text{(aq)} + 2\text{Cl}^-\text{(aq)}\)

• \(\text{Fe}^{2+}\) goes from \(+2\) to \(+3\) (Oxidation number increases \(\rightarrow\) \(\text{Fe}^{2+}\) is oxidised). Therefore, \(\text{Fe}^{2+}\) is the reducing agent.
• \(\text{Cl}_2\) goes from \(0\) to \(-1\) (Oxidation number decreases \(\rightarrow\) \(\text{Cl}_2\) is reduced). Therefore, \(\text{Cl}_2\) is the oxidising agent.

5. Disproportionation Reactions

Sometimes, a single chemical element can be both oxidised and reduced simultaneously within the very same chemical reaction. This special type of redox reaction has a specific name.

Definition: A disproportionation reaction is a reaction in which the same element in a single species is simultaneously oxidised and reduced.

Important Examples you must know:

1. Chlorine reacting with cold dilute sodium hydroxide:
\(\text{Cl}_2\text{(aq)} + 2\text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{NaClO(aq)} + \text{H}_2\text{O(l)}\)
• In \(\text{Cl}_2\), the oxidation number of chlorine is \(0\).
• In \(\text{NaCl}\), the oxidation number of chlorine is \(-1\) (Chlorine is reduced).
• In \(\text{NaClO}\) (sodium chlorate(I)), the oxidation number of chlorine is \(+1\) (Chlorine is oxidised).
Because chlorine goes from \(0\) to both \(-1\) and \(+1\), this is a classic disproportionation reaction.

2. Decomposition of Hydrogen Peroxide:
\(2\text{H}_2\text{O}_2\text{(aq)} \rightarrow 2\text{H}_2\text{O(l)} + \text{O}_2\text{(g)}\)
• In \(\text{H}_2\text{O}_2\), oxygen has an oxidation number of \(-1\).
• In \(\text{H}_2\text{O}\), oxygen has an oxidation number of \(-2\) (Oxygen is reduced).
• In \(\text{O}_2\), oxygen has an oxidation number of \(0\) (Oxygen is oxidised).
Oxygen is simultaneously oxidised and reduced.

6. Writing and Balancing Redox Half-Equations

To balance complex redox equations in acidic aqueous solution, we break the reaction into two half-equations (one for oxidation and one for reduction) and balance them systematically using water (\(\text{H}_2\text{O}\)), hydrogen ions (\(\text{H}^+\)), and electrons (\(\text{e}^-\)).

The 5-Step Method for Balancing a Half-Equation:

1. Balance the main element (all atoms other than \(\text{O}\) and \(\text{H}\)).
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 the overall electrical charge by adding electrons (\(\text{e}^-\)) to the more positive side.
5. Check that both the number of atoms and total charges balance on both sides.

Worked Example 1: Balancing the reduction of \(\text{MnO}_4^-\) to \(\text{Mn}^{2+}\)

• Step 1 (Main atom): \(\text{Mn}\) is already balanced: \(\text{MnO}_4^- \rightarrow \text{Mn}^{2+}\)
• Step 2 (Oxygen): Add \(4\text{H}_2\text{O}\) to the right: \(\text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\)
• Step 3 (Hydrogen): Add \(8\text{H}^+\) to the left: \(\text{MnO}_4^- + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\)
• Step 4 (Charge):
Left side charge \(= (-1) + 8(+1) = +7\)
Right side charge \(= +2 + 0 = +2\)
To balance, add \(5\text{e}^-\) to the left side:
\(\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\)

Worked Example 2: Balancing the reduction of \(\text{Cr}_2\text{O}_7^{2-}\) to \(\text{Cr}^{3+}\)

• Step 1 (Main atom): Balance \(\text{Cr}\) by placing a 2 in front of \(\text{Cr}^{3+}\):
\(\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+}\)
• Step 2 (Oxygen): 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 (Hydrogen): 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 (Charge):
Left side charge \(= (-2) + 14(+1) = +12\)
Right side charge \(= 2(+3) = +6\)
To balance, add \(6\text{e}^-\) to the left side:
\(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\)

7. Combining Half-Equations to Form an Overall Equation

When combining an oxidation half-equation and a reduction half-equation, the electrons must completely cancel out because free electrons cannot exist in an aqueous solution.

Step-by-Step Example: Reaction between \(\text{MnO}_4^-\) and \(\text{Fe}^{2+}\)

Step 1: Write down both balanced half-equations.
Reduction: \(\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\)
Oxidation: \(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-\)

Step 2: Equalise the number of electrons.
The reduction half-equation consumes \(5\text{e}^-\), but the oxidation half-equation only produces \(1\text{e}^-\).
Multiply the entire oxidation half-equation by \(5\):
\(5\text{Fe}^{2+} \rightarrow 5\text{Fe}^{3+} + 5\text{e}^-\)

Step 3: Add the two half-equations together and cancel out the electrons.
\(\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+} + 5\text{e}^-\)

Cancelling \(5\text{e}^-\) from both sides gives the overall balanced ionic equation:
\(\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}\)

Step 4: Final verification check.
• Atoms check: \(1\text{ Mn}\), \(4\text{ O}\), \(8\text{ H}\), \(5\text{ Fe}\) on both sides.
• Charge check:
Left side: \((-1) + 8(+1) + 5(+2) = -1 + 8 + 10 = +17\)
Right side: \((+2) + 5(+3) + 0 = +2 + 15 = +17\)
Both atoms and charges are balanced!

Summary & Revision Checklist

• Oxidation: Loss of electrons / Increase in oxidation state.
• Reduction: Gain of electrons / Decrease in oxidation state.
• Oxidising Agent: Gains electrons / Oxidation state decreases / Oxidises something else.
• Reducing Agent: Loses electrons / Oxidation state increases / Reduces something else.
• Disproportionation: The same element in a single chemical species is simultaneously oxidised and reduced.
• Half-Equations: Balance the key atom \(\rightarrow\) Balance \(\text{O}\) using \(\text{H}_2\text{O}\) \(\rightarrow\) Balance \(\text{H}\) using \(\text{H}^+\) \(\rightarrow\) Balance charge using \(\text{e}^-\) \(\rightarrow\) Multiply to cancel electrons when combining.