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6 Electrochemistry

6.1 Redox processes

Redox Processes: Electron Transfer and Changes in Oxidation Number

Specification Reference Physical Chemistry: Electrochemistry 6.1

Quick Notes

  • Oxidation number (state): An assigned number that represents the "charge" an atom would have if the bonding in the compound was fully ionic.
  • 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.
  • Oxidation = increase in oxidation number / loss of electrons
  • Reduction = decrease in oxidation number / gain of electrons
  • Redox reaction = both oxidation and reduction occur.
  • Disproportionation = one species is both oxidised and reduced.
  • Oxidising agent = causes oxidation (is reduced)
  • Reducing agent = causes reduction (is oxidised)
  • Roman numerals are used to show oxidation number in names (e.g. iron(III), manganese(VII)).

Full Notes

Oxidation Numbers (Oxidation States)

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!

CIE A-Level Chemistry diagram showing carbon combusting to carbon dioxide with oxidation change.

To help, we consider each atom to have an ‘imaginary’ charge, described as its oxidation number (or state).

Rules for assigning oxidation states:

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).

CIE A-Level Chemistry diagram showing carbon changing from 0 in C(s) to +4 in CO2(g).
Worked Example

Assign oxidation states in H2SO4 (sulfuric acid).

  1. H = +1 (there are 2 H, total +2).
  2. O = −2 (there are 4 O, total −8).
  3. The total charge must be 0, so S must be +6 to balance the equation:
    2(+1) + S + 4(−2) = 0 → S = +6.

Changes in Oxidation Number and Redox

Oxidation: increase in oxidation number
(e.g. Fe2+ → Fe3+, Fe goes from +2 to +3)

Reduction: decrease in oxidation number
(e.g. Cl2 → 2Cl−, Cl goes from 0 to −1)

Redox reaction = both oxidation and reduction occur.

For Example Magnesium and 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 two half equations combine together to give the overall reaction.
(Mg + Cl2 + 2e− → Mg2+ + 2e− + 2Cl−)
Mg + Cl2 → MgCl2

Half-Equations and Balancing Redox Reactions

Half-equations show the oxidation and reduction happening in a redox reaction 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.

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.

For any redox reaction the total number of electrons lost = electrons gained.

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.

Example Reaction of chlorine with water

Cl2 + H2O → HCl + HClO

Chlorine is simultaneously oxidised and reduced.:

Cl2 → Cl− (oxidation state 0 to −1, reduction)
Cl2 → ClO− (oxidation state 0 to +1, oxidation)

Oxidising and Reducing Agents

Oxidising agent: accepts electrons → it is reduced

Reducing agent: donates electrons → it is oxidised

Example Cu + 2Ag+ → Cu2+ + 2Ag

Ag+ is the oxidising agent

Cu is the reducing agent

Using Roman Numerals

When naming compounds, Roman numerals are used to show the oxidation state of transition metals and other elements that can have variable oxidation states.

Summary