Redox Equations (Oxford AQA International A Level Chemistry)

Revision Note

Oxidation states are used to:
- Tell if oxidation or reduction has taken place
- Work out what has been oxidised and/or reduced
- Construct half equations and balance redox equations
Oxidation states are also referred to as oxidation numbers
A positive oxidation states = loss of electrons
- The more positive the number, the more the element has been oxidised
A negative oxidation state = gain of electrons
- The more negative the number, the more the element has been reduced

Rule	Example
Uncombined elements have an oxidation state of 0	H₂ Cl₂ Na
Some elements have the same oxidation state in their compounds with some exceptions	Group 1 elements are always +1 Group 2 element are always +2 Aluminium is always +3 Hydrogen is always +1 except in metal hydrides where it is -1 Fluorine is always -1 Oxygen is always -2 except in peroxides where is it -1 and OF₂, where it is +2 Chlorine is always -1 except in compounds with F and O where it has a positive value
The sum of all of the oxidation states in a compound is equal to 0	In NaCl: Na= +1 Cl = -1 Sum of oxidation states= 0
The oxidation state of an element in a monoatomic ion is always equal to its charge	Zn²⁺= +2 Fe³⁺ = +3 Cl^-= -1
The sum of the oxidation states in an ion is equal to the charge on the ion	In SO₄^2- S= +6 O= (-2 x 4) = -8 Sum of oxidation states= -2
In a compound, the more electronegative element is given the negative oxidation state	In F₂O F is more electronegative so is (-1 x 2) = -2 O = +2 Sum of oxidation states= 0

Deducing oxidation states

Give the oxidation state of the bold atoms in these compounds or ions.

Answers:

P₂O₅
- 5 O atoms = 5 x (–2) = –10
- The overall charge of the compound = 0
- 2 P atoms = +10
- Oxidation state of 1 P atom = (+10) / 2 = +5
SO₄^2–
- 4 O atoms = 4 x (–2) = –8
- The overall charge of the compound = –2
- The oxidation state of 1 S atom = +6
H₂S
- 2 H atoms = 2 x (+1) = +2
- The overall charge of the compound = 0
- The oxidation state of 1 S atom = –2
Al₂Cl₆
- 6 Cl atoms = 6 x (–1) = –6
- The overall charge of the compound = 0
- 2 Al atoms = +6
- The oxidation state of 1 Al atom = (+6) / 2 = +3
NH₃
- 3 H atoms = 3 x (+1) = +3
- The overall charge of the compound = 0
- The oxidation state of 1 N atom = –3
ClO₂^–
- 2 O atoms = 2 x (–2) = –4
- The overall charge of the compound = –1
- The oxidation state of 1 Cl atom = +3
CaCO₃
- 3 O atoms = 3 x (–2) = –6
- 1 Ca atom = +2
- The overall charge of the compound = 0
- The oxidation state of 1 C atom = +4

Redox reactions are reactions in which oxidation and reduction take place together
- While one species is being oxidised, another is being reduced in the same reaction
For example:

Cu²⁺+ Zn → Zn²⁺ + Cu

Cu has been reduced from +2 to 0
Zn has been oxidised from 0 to +2
You should be able to split full or ionic equations into half equations and be able to identifying the species oxidised and reduced

In each of the following equations, write the half equations and state which reactant has been oxidised and which has been reduced.

Answers:

Answer 1:
- Na → Na⁺+ e^-
- Cl₂+ 2e^- → 2Cl^-
- Oxidised: Na as the oxidation state has increased from 0 to +1
- Reduced: Cl₂ as the oxidation state has decreased from 0 to –1
Answer 2:
- Mg + 2e^- → Mg²⁺ + 2e^-
- Fe²⁺ + 2e^- → Fe
- Oxidised: Mg as the oxidation state has increased by 2
- Reduced: Fe²⁺ as the oxidation state has decreased by 2
Answer 3:
- 2Ag⁺+ 2e^- → 2Ag
- C²⁺→ C⁴⁺ + 2e^-
- Oxidised: C as it has lost electrons
- Reduced: Ag as it has accepted electrons

Oxidation states can be used to balance chemical equations
Roman numerals between brackets are used to show the ox. state of an atom that can have multiple oxidation states, e.g.:
- Fe(II) = iron with an oxidation state of +2
- Fe(III) = iron with an oxidation state of +3
You should be able to combine half equations and produce an overall redox equation
Using simple half equations involves adjusting the number of electrons and other coefficients to produce the overall equation:

Example:

Al→ Al³⁺ + 3e-

O₂ + 4e^- → 2O^2-

4Al + 3O₂→ 2Al₂O₃

In this example it's easy to see that the lowest common factor is 12, so the first half equation is multiplied by 4 and the second one by 3, and then they are combined
More complicated examples involve being given an unbalanced redox equation and working through the redox changes, as well as having to add extra species such as H⁺ and H₂0 to balance the overall equation
Go through these steps to balance a redox equation:
- Write the unbalanced equation and identify the atoms which change in oxidation state
- Deduce the oxidation state changes
- Balance the oxidation state changes
- Balance the charges
- Balance the remaining atoms