Oxidation, Reduction, and Redox Equations
Quick Notes
- Redox reactions involve the transfer of electrons between species.
- Oxidation is loss of electrons
- Reduction is gain of electrons
- The oxidation state (or oxidation number) of an atom represents an imaginary (or real) charge based on a set of rules, assuming electrons in bonds are assigned to the more electronegative atoms — even if no actual electron transfer has occurred.
- Rules for oxidation states:
- Elements in their natural state have an oxidation number of 0.
- Oxygen is usually -2 (except in peroxides where it is -1).
- Hydrogen is usually +1 (except in metal hydrides where it is -1).
- Group 1 metals are +1, Group 2 metals are +2.
- The sum of oxidation states in a neutral compound is 0.
- The sum of oxidation states in an ion equals the charge of the ion.
- Half-equations show oxidation and reduction separately.
- Overall redox equations combine oxidation and reduction half-equations.
Full Notes
Definition of Redox
Redox reactions involve both oxidation and reduction.
- Oxidation means loss of electrons
- Reduction means gain of electrons
Oxidation and reduction always occur together in a reaction.
A species that loses electrons must give them to something else and so acts as a reducing agent.
A species that gains electrons must take them from something else and so acts as an oxidising agent.
Example: Magnesium + Chlorine
Mg + Cl₂ → MgCl₂
In this reaction, the magnesium (Mg) loses electrons and is oxidised and the chlorine (Cl2) gains electrons and is reduced. We can show this using half-equations (see below for more):
Mg → Mg²⁺ + 2e⁻
Cl₂ + 2e⁻ → 2Cl⁻
The Mg gives electrons to Cl, meaning it acts as a reducing agent and the Cl takes electrons from the Mg, meaning it acts asan oxidising agent.
The two half equations combine together to give the overall reaction.
(Mg + Cl₂ + 2e⁻ → Mg²⁺ + 2e⁻ + 2Cl⁻)
Mg + Cl₂ → MgCl₂
Oxidation States (Oxidation Numbers)
Oxidation states help track electron transfer in reactions. It is straightforward to see how atoms have lost or gained electrons when ions get formed, however it can be harder to see how atoms have lost or gained electron density when dealing with molecules.
For example:
Carbon is oxidised to form carbon dioxide when combusted. However, no ions get formed, meaning it isn’t immediately clear how electrons are involved!
To help, we consider each atom to have an ‘imaginary’ charge, described as its oxidation number (or state).
Rules for assigning oxidation states:
- Uncombined elements (e.g., O₂, N₂, Fe) have an oxidation state of 0.
- Group 1 metals = +1, Group 2 metals = +2.
- Oxygen is -2, except:
- In peroxides (O₂²⁻), oxygen is -1.
- With fluorine (OF₂), oxygen is +2.
- Hydrogen is +1, except in metal hydrides (e.g., NaH), where it is -1.
- In a neutral compound, the sum of oxidation states = 0.
- In polyatomic ions, the sum of oxidation states = charge of the ion.
An increase in oxidation number (gets more positive) means oxidation has occurred.
A decrease in oxidation number (gets more negative) means reduction has occurred.
Using these rules, we can see now how carbon gets oxidised from an oxidation state of 0 in C(s) to +4 in CO2(g).
Assign oxidation states in H₂SO₄ (sulfuric acid).
- H = +1 (there are 2 H, total +2).
- O = -2 (there are 4 O, total -8).
The total charge must be 0, so:
S must be +6 to balance the equation: 2(+1) + S + 4(-2) = 0 → S = +6.
Writing Half-Equations for Redox Reactions
When writing half-equations, we show the oxidation and reduction steps separately. This means one half equation will have electrons being gained by something on the left of the arrow (reduction happens) and the other will have electrons being lost by something (oxidation happens) on the right of the arrow.
For Example: Reaction of iron with chlorine:
2Fe + 3Cl₂ → 2FeCl₃
Step 1: Write oxidation equation (loss of electrons)
Fe → Fe³⁺ + 3e⁻
Step 2: Write reduction equation (gain of electrons)
Cl₂ + 2e⁻ → 2Cl⁻
The half-equations don’t have to match the ratios in the overall equation it is only when we combine the half equations that the electrons must balance.
Always remember that half equations don’t occur on their own; they’re a tool to represent just the oxidation or reduction part of a redox reaction, focusing only on the species gaining or losing electrons and ignoring everything else.
Combining Half-Equations to Form Full Redox Equations
For any redox reaction the total number of electrons lost = electrons gained.
For Example: Copper reacting with silver ions:
Cu + Ag⁺ → Cu²⁺ + Ag
Step 1: Write oxidation half-equation
Cu → Cu²⁺ + 2e⁻
Step 2: Write reduction half-equation
2Ag⁺ + 2e⁻ → 2Ag
Step 3: Combine to form full redox equation
Cu + 2Ag⁺ → Cu²⁺ + 2Ag
There must be 2Ag+ ions reacting because each copper atom will lose 2 electrons, meaning 2 Ag+ ions are needed (as each Ag+ can only gain 1 electron).
Disproportionation Reactions
A disproportionation reaction is when one species is both oxidised and reduced simultaneously.
For Example: Reaction of chlorine with water is a classic disproportionation reaction:
Cl₂ + H₂O → HCl + HClO
Chlorine is both oxidized and reduced:
Cl₂ → Cl⁻ (oxidation state 0 to -1, reduction)
Cl₂ → ClO⁻ (oxidation state 0 to +1, oxidation)
Summary
| Process | Definition | Example |
|---|---|---|
| Oxidation | Loss of electrons | Mg → Mg²⁺ + 2e⁻ |
| Reduction | Gain of electrons | Cl₂ + 2e⁻ → 2Cl⁻ |
| Oxidizing Agent | Accepts electrons (causes oxidation) | Cl₂ in Mg + Cl₂ |
| Reducing Agent | Donates electrons (causes reduction) | Mg in Mg + Cl₂ |
| Disproportionation | Same species oxidised and reduced simultaneously | Cl₂ + H₂O → HCl + HClO |