Electron transfer

Definition

Oxidation

The loss of electrons by a substance.

Definition

Reduction

The gain of electrons by a substance.

Many redox reactions involve no oxygen at all, so a more general definition is based on electrons.
Oxidation is the loss of electrons, and reduction is the gain of electrons.
The phrase OIL RIG helps you remember this: Oxidation Is Loss, Reduction Is Gain.
Electrons lost by one substance are gained by another, so oxidation and reduction still happen together.
A redox reaction can be identified as a reaction in which one substance loses electrons and another substance gains electrons.
When magnesium is added to copper(II) sulfate solution, magnesium atoms lose electrons and copper(II) ions gain them: $Mg(s)+Cu2+(aq)→Mg2+(aq)+Cu(s)\mathrm{Mg(s) + Cu^{2+}(aq) \rightarrow Mg^{2+}(aq) + Cu(s)}$ $Mg (s) + C u^{2 +} (aq) \to M g^{2 +} (aq) + Cu (s)$ .
1. Magnesium is oxidised because it loses two electrons: $Mg→Mg2++2e−\mathrm{Mg \rightarrow Mg^{2+} + 2e^{-}}$ .
2. Copper(II) ions are reduced because they gain two electrons: $Cu2++2e−→Cu\mathrm{Cu^{2+} + 2e^{-} \rightarrow Cu}$ .
The sulfate ions do not change, so they are spectator ions and are left out of the ionic equation.

Example

Sodium burning in chlorine

\mathrm{2Na(s) + Cl_{2}(g) \rightarrow 2NaCl(s)}

No oxygen is involved, but each sodium atom loses one electron to become a sodium ion.

\mathrm{Na \rightarrow Na^{+} + e^{-}}

Sodium loses electrons, so sodium is oxidised.
Each chlorine molecule gains two electrons to become two chloride ions.

\mathrm{Cl_{2} + 2e^{-} \rightarrow 2Cl^{-}}

Chlorine gains electrons, so chlorine is reduced, and the reaction is a redox reaction.

Oxidation numbers

Definition

Oxidation number

A number given to each atom in a substance that shows how many electrons it has lost, as a positive value, or gained, as a negative value, compared with the uncombined element.

Definition

Oxidation (in terms of oxidation number)

An increase in the oxidation number of an element.

Definition

Reduction (in terms of oxidation number)

A decrease in the oxidation number of an element.

Oxidation numbers let you track electrons in any reaction, including reactions between molecules.
Uncombined elements
1. The oxidation number of an element in its uncombined state is zero, as in $Mg\mathrm{Mg}$ , $O2\mathrm{O_{2}}$ and $Cl2\mathrm{Cl_{2}}$ .
Monatomic ions
1. The oxidation number of an ion made of a single atom is the same as its charge, so $Na+\mathrm{Na^{+}}$ is $+ 1$ , $Cu2+\mathrm{Cu^{2+}}$ is $+ 2$ and $O2−\mathrm{O^{2-}}$ is $- 2$ .
Compounds
1. The oxidation numbers of all the atoms in a compound add up to zero.
2. In $H2O\mathrm{H_{2}O}$ , $\times (+1) + (-2) = 0$ .
Ions containing more than one atom
1. The oxidation numbers of all the atoms in the ion add up to the charge on the ion.
2. In $OH−\mathrm{OH^{-}}$ , $(- 2) + (+ 1) = - 1$ .
Useful values in compounds are oxygen at $- 2$ , hydrogen at $+ 1$ , Group I metals at $+ 1$ and Group II metals at $+ 2$ .
Oxidation numbers are written with the sign in front, such as $+ 3$ , while ionic charges are written with the sign after, such as $3 +$ .

Example

Finding oxidation numbers

Sulfur in the sulfate ion, $SO42−\mathrm{SO_{4}^{2-}}$ , with oxygen at $- 2$ .

\quad \Rightarrow \quad x = +6

Manganese in the manganate(VII) ion, $MnO4−\mathrm{MnO_{4}^{-}}$ , which explains the numeral in its name.

\quad \Rightarrow \quad x = +7

Iron in iron(III) oxide, $Fe2O3\mathrm{Fe_{2}O_{3}}$ .

\quad \Rightarrow \quad x = +3

Nitrogen in the ammonium ion, $NH4+\mathrm{NH_{4}^{+}}$ , with hydrogen at $+ 1$ .

\quad \Rightarrow \quad x = -3

Identifying redox from oxidation numbers

Write the oxidation number of every element on both sides of the equation.
An element whose oxidation number increases has been oxidised.
An element whose oxidation number decreases has been reduced.
If any oxidation numbers change, the reaction is a redox reaction.
If no oxidation numbers change, the reaction is not a redox reaction.
Neutralisation, such as $NaOH(aq)+HCl(aq)→NaCl(aq)+H2O(l)\mathrm{NaOH(aq) + HCl(aq) \rightarrow NaCl(aq) + H_{2}O(l)}$ , is not a redox reaction because every element keeps the same oxidation number.

Example

Iron(III) oxide and carbon monoxide

\mathrm{Fe_{2}O_{3} + 3CO \rightarrow 2Fe + 3CO_{2}}

Iron goes from $+ 3$ in $Fe2O3\mathrm{Fe_{2}O_{3}}$ to $0$ in $Fe\mathrm{Fe}$ , a decrease, so iron is reduced.
Carbon goes from $+ 2$ in $CO\mathrm{CO}$ to $+ 4$ in $CO2\mathrm{CO_{2}}$ , an increase, so carbon is oxidised.
Oxygen stays at $- 2$ throughout, and the changes for iron and carbon show the reaction is redox.

Example

Chlorine and potassium bromide

\mathrm{Cl_{2}(aq) + 2KBr(aq) \rightarrow 2KCl(aq) + Br_{2}(aq)}

Chlorine goes from $0$ in $Cl2\mathrm{Cl_{2}}$ to $- 1$ in $KCl\mathrm{KCl}$ , so chlorine is reduced.
Bromine goes from $- 1$ in $KBr\mathrm{KBr}$ to $0$ in $Br2\mathrm{Br_{2}}$ , so bromine is oxidised.
Potassium stays at $+ 1$ , so the potassium ions are spectator ions.

Colour changes in redox reactions

Two reagents change colour when they take part in a redox reaction, so they can show that a redox reaction has happened.
Acidified aqueous potassium manganate(VII) is purple.
When it reacts with a substance that can be oxidised, the manganate(VII) ions are reduced and the solution changes from purple to colourless.
Manganese is reduced from $+ 7$ in $MnO4−\mathrm{MnO_{4}^{-}}$ to $+ 2$ in $Mn2+\mathrm{Mn^{2+}}$ , which is almost colourless in dilute solution.
Iron(II) ions and sulfur dioxide both decolourise acidified potassium manganate(VII), because they are oxidised by it.
Aqueous potassium iodide is colourless.
When it reacts with a substance that can oxidise it, iodide ions are oxidised to iodine and the solution changes from colourless to brown.
Iodine goes from $- 1$ in $I−\mathrm{I^{-}}$ to $0$ in $I2\mathrm{I_{2}}$ : $2I−(aq)→I2(aq)+2e−\mathrm{2I^{-}(aq) \rightarrow I_{2}(aq) + 2e^{-}}$ .
Chlorine water, hydrogen peroxide and iron(III) ions all turn aqueous potassium iodide brown.

Practical

Testing for redox with potassium manganate(VII) and potassium iodide

Aim: to use colour changes to show whether a substance takes part in a redox reaction with acidified aqueous potassium manganate(VII) or aqueous potassium iodide.
Apparatus: test tubes, test-tube rack, dropping pipettes, $cm310\ \text{cm}^3$ measuring cylinder, white tile, dilute sulfuric acid, very dilute aqueous potassium manganate(VII), aqueous potassium iodide, freshly made aqueous iron(II) sulfate, dilute hydrogen peroxide solution, distilled water and eye protection.
Method, acidified potassium manganate(VII):
- Pour about $cm32\ \text{cm}^3$ of very dilute potassium manganate(VII) into a test tube and add an equal volume of dilute sulfuric acid to acidify it.
- Hold the tube in front of the white tile so that the purple colour is easy to see.
- Add aqueous iron(II) sulfate one drop at a time, shaking gently after each drop, and record any colour change.
- Repeat with distilled water in place of the iron(II) sulfate as a control.
Method, aqueous potassium iodide:
- Pour about $cm32\ \text{cm}^3$ of aqueous potassium iodide into a clean test tube and add three drops of dilute sulfuric acid.
- Add five drops of dilute hydrogen peroxide solution and shake gently.
- Record the colour after one minute, against the white tile.
- Repeat with distilled water in place of the hydrogen peroxide as a control.
Results: iron(II) sulfate turns the purple manganate(VII) solution colourless, and hydrogen peroxide turns the colourless potassium iodide solution brown, while neither control changes colour.
Explanation: the loss of purple shows that manganate(VII) ions were reduced as iron(II) ions were oxidised to iron(III) ions, and the brown colour shows that iodide ions were oxidised to iodine.
Safety: wear eye protection, avoid skin contact with potassium manganate(VII) because it is an oxidiser that stains, and handle dilute sulfuric acid and hydrogen peroxide with care because they irritate the skin and eyes.

Exam technique

Using oxidation numbers and colour changes

Write every oxidation number with a sign, such as $+ 2$ or $- 1$ , and write $0$ for an uncombined element.
When you identify oxidation or reduction, quote the change for the element, such as “iron decreases from $+ 3$ to $0$ , so it is reduced”.
When you use electron transfer, state which species loses electrons and which species gains them.
For the colour tests, give the starting and finishing colours: purple to colourless for acidified potassium manganate(VII), and colourless to brown for aqueous potassium iodide.
Write “colourless” rather than “clear”, because a clear solution can still be coloured.

Common Mistake

Do not write “clear” when you mean colourless.
Do not give oxygen in a compound an oxidation number of $+ 2$ , because it is almost always $- 2$ .
Do not mix up the two tests, because manganate(VII) is decolourised by a substance that is oxidised, while iodide turns brown when something oxidises it.

Self review

Define oxidation in terms of electrons.
What is the oxidation number of an element in its uncombined state?
What is the oxidation number of sulfur in $SO42−\mathrm{SO_4^{2-}}$ ?
How do oxidation numbers show that a reaction is a redox reaction?
What colour change is seen when acidified aqueous potassium manganate(VII) reacts with a substance that is oxidised?

Electron transfer

Definition

Oxidation

The loss of electrons by a substance.

Definition

Reduction

The gain of electrons by a substance.

Many redox reactions involve no oxygen at all, so a more general definition is based on electrons.
Oxidation is the loss of electrons, and reduction is the gain of electrons.
The phrase OIL RIG helps you remember this: Oxidation Is Loss, Reduction Is Gain.
Electrons lost by one substance are gained by another, so oxidation and reduction still happen together.
A redox reaction can be identified as a reaction in which one substance loses electrons and another substance gains electrons.
When magnesium is added to copper(II) sulfate solution, magnesium atoms lose electrons and copper(II) ions gain them: $Mg(s)+Cu2+(aq)→Mg2+(aq)+Cu(s)\mathrm{Mg(s) + Cu^{2+}(aq) \rightarrow Mg^{2+}(aq) + Cu(s)}$ $Mg (s) + C u^{2 +} (aq) \to M g^{2 +} (aq) + Cu (s)$ .
1. Magnesium is oxidised because it loses two electrons: $Mg→Mg2++2e−\mathrm{Mg \rightarrow Mg^{2+} + 2e^{-}}$ .
2. Copper(II) ions are reduced because they gain two electrons: $Cu2++2e−→Cu\mathrm{Cu^{2+} + 2e^{-} \rightarrow Cu}$ .
The sulfate ions do not change, so they are spectator ions and are left out of the ionic equation.

Example

Sodium burning in chlorine

\mathrm{2Na(s) + Cl_{2}(g) \rightarrow 2NaCl(s)}

No oxygen is involved, but each sodium atom loses one electron to become a sodium ion.

\mathrm{Na \rightarrow Na^{+} + e^{-}}

Sodium loses electrons, so sodium is oxidised.
Each chlorine molecule gains two electrons to become two chloride ions.

\mathrm{Cl_{2} + 2e^{-} \rightarrow 2Cl^{-}}

Chlorine gains electrons, so chlorine is reduced, and the reaction is a redox reaction.

Oxidation numbers

Definition

Oxidation number

A number given to each atom in a substance that shows how many electrons it has lost, as a positive value, or gained, as a negative value, compared with the uncombined element.

Definition

Oxidation (in terms of oxidation number)

An increase in the oxidation number of an element.

Definition

Reduction (in terms of oxidation number)

A decrease in the oxidation number of an element.

Oxidation numbers let you track electrons in any reaction, including reactions between molecules.
Uncombined elements
1. The oxidation number of an element in its uncombined state is zero, as in $Mg\mathrm{Mg}$ , $O2\mathrm{O_{2}}$ and $Cl2\mathrm{Cl_{2}}$ .
Monatomic ions
1. The oxidation number of an ion made of a single atom is the same as its charge, so $Na+\mathrm{Na^{+}}$ is $+ 1$ , $Cu2+\mathrm{Cu^{2+}}$ is $+ 2$ and $O2−\mathrm{O^{2-}}$ is $- 2$ .
Compounds
1. The oxidation numbers of all the atoms in a compound add up to zero.
2. In $H2O\mathrm{H_{2}O}$ , $\times (+1) + (-2) = 0$ .
Ions containing more than one atom
1. The oxidation numbers of all the atoms in the ion add up to the charge on the ion.
2. In $OH−\mathrm{OH^{-}}$ , $(- 2) + (+ 1) = - 1$ .
Useful values in compounds are oxygen at $- 2$ , hydrogen at $+ 1$ , Group I metals at $+ 1$ and Group II metals at $+ 2$ .
Oxidation numbers are written with the sign in front, such as $+ 3$ , while ionic charges are written with the sign after, such as $3 +$ .

Example

Finding oxidation numbers

Sulfur in the sulfate ion, $SO42−\mathrm{SO_{4}^{2-}}$ , with oxygen at $- 2$ .

\quad \Rightarrow \quad x = +6

Manganese in the manganate(VII) ion, $MnO4−\mathrm{MnO_{4}^{-}}$ , which explains the numeral in its name.

\quad \Rightarrow \quad x = +7

Iron in iron(III) oxide, $Fe2O3\mathrm{Fe_{2}O_{3}}$ .

\quad \Rightarrow \quad x = +3

Nitrogen in the ammonium ion, $NH4+\mathrm{NH_{4}^{+}}$ , with hydrogen at $+ 1$ .

\quad \Rightarrow \quad x = -3

Identifying redox from oxidation numbers

Write the oxidation number of every element on both sides of the equation.
An element whose oxidation number increases has been oxidised.
An element whose oxidation number decreases has been reduced.
If any oxidation numbers change, the reaction is a redox reaction.
If no oxidation numbers change, the reaction is not a redox reaction.
Neutralisation, such as $NaOH(aq)+HCl(aq)→NaCl(aq)+H2O(l)\mathrm{NaOH(aq) + HCl(aq) \rightarrow NaCl(aq) + H_{2}O(l)}$ , is not a redox reaction because every element keeps the same oxidation number.

Example

Iron(III) oxide and carbon monoxide

\mathrm{Fe_{2}O_{3} + 3CO \rightarrow 2Fe + 3CO_{2}}

Iron goes from $+ 3$ in $Fe2O3\mathrm{Fe_{2}O_{3}}$ to $0$ in $Fe\mathrm{Fe}$ , a decrease, so iron is reduced.
Carbon goes from $+ 2$ in $CO\mathrm{CO}$ to $+ 4$ in $CO2\mathrm{CO_{2}}$ , an increase, so carbon is oxidised.
Oxygen stays at $- 2$ throughout, and the changes for iron and carbon show the reaction is redox.

Example

Chlorine and potassium bromide

\mathrm{Cl_{2}(aq) + 2KBr(aq) \rightarrow 2KCl(aq) + Br_{2}(aq)}

Chlorine goes from $0$ in $Cl2\mathrm{Cl_{2}}$ to $- 1$ in $KCl\mathrm{KCl}$ , so chlorine is reduced.
Bromine goes from $- 1$ in $KBr\mathrm{KBr}$ to $0$ in $Br2\mathrm{Br_{2}}$ , so bromine is oxidised.
Potassium stays at $+ 1$ , so the potassium ions are spectator ions.

Colour changes in redox reactions

Two reagents change colour when they take part in a redox reaction, so they can show that a redox reaction has happened.
Acidified aqueous potassium manganate(VII) is purple.
When it reacts with a substance that can be oxidised, the manganate(VII) ions are reduced and the solution changes from purple to colourless.
Manganese is reduced from $+ 7$ in $MnO4−\mathrm{MnO_{4}^{-}}$ to $+ 2$ in $Mn2+\mathrm{Mn^{2+}}$ , which is almost colourless in dilute solution.
Iron(II) ions and sulfur dioxide both decolourise acidified potassium manganate(VII), because they are oxidised by it.
Aqueous potassium iodide is colourless.
When it reacts with a substance that can oxidise it, iodide ions are oxidised to iodine and the solution changes from colourless to brown.
Iodine goes from $- 1$ in $I−\mathrm{I^{-}}$ to $0$ in $I2\mathrm{I_{2}}$ : $2I−(aq)→I2(aq)+2e−\mathrm{2I^{-}(aq) \rightarrow I_{2}(aq) + 2e^{-}}$ .
Chlorine water, hydrogen peroxide and iron(III) ions all turn aqueous potassium iodide brown.

Practical

Testing for redox with potassium manganate(VII) and potassium iodide

Aim: to use colour changes to show whether a substance takes part in a redox reaction with acidified aqueous potassium manganate(VII) or aqueous potassium iodide.
Apparatus: test tubes, test-tube rack, dropping pipettes, $cm310\ \text{cm}^3$ measuring cylinder, white tile, dilute sulfuric acid, very dilute aqueous potassium manganate(VII), aqueous potassium iodide, freshly made aqueous iron(II) sulfate, dilute hydrogen peroxide solution, distilled water and eye protection.
Method, acidified potassium manganate(VII):
- Pour about $cm32\ \text{cm}^3$ of very dilute potassium manganate(VII) into a test tube and add an equal volume of dilute sulfuric acid to acidify it.
- Hold the tube in front of the white tile so that the purple colour is easy to see.
- Add aqueous iron(II) sulfate one drop at a time, shaking gently after each drop, and record any colour change.
- Repeat with distilled water in place of the iron(II) sulfate as a control.
Method, aqueous potassium iodide:
- Pour about $cm32\ \text{cm}^3$ of aqueous potassium iodide into a clean test tube and add three drops of dilute sulfuric acid.
- Add five drops of dilute hydrogen peroxide solution and shake gently.
- Record the colour after one minute, against the white tile.
- Repeat with distilled water in place of the hydrogen peroxide as a control.
Results: iron(II) sulfate turns the purple manganate(VII) solution colourless, and hydrogen peroxide turns the colourless potassium iodide solution brown, while neither control changes colour.
Explanation: the loss of purple shows that manganate(VII) ions were reduced as iron(II) ions were oxidised to iron(III) ions, and the brown colour shows that iodide ions were oxidised to iodine.
Safety: wear eye protection, avoid skin contact with potassium manganate(VII) because it is an oxidiser that stains, and handle dilute sulfuric acid and hydrogen peroxide with care because they irritate the skin and eyes.

Exam technique

Using oxidation numbers and colour changes

Write every oxidation number with a sign, such as $+ 2$ or $- 1$ , and write $0$ for an uncombined element.
When you identify oxidation or reduction, quote the change for the element, such as “iron decreases from $+ 3$ to $0$ , so it is reduced”.
When you use electron transfer, state which species loses electrons and which species gains them.
For the colour tests, give the starting and finishing colours: purple to colourless for acidified potassium manganate(VII), and colourless to brown for aqueous potassium iodide.
Write “colourless” rather than “clear”, because a clear solution can still be coloured.

Common Mistake

Do not write “clear” when you mean colourless.
Do not give oxygen in a compound an oxidation number of $+ 2$ , because it is almost always $- 2$ .
Do not mix up the two tests, because manganate(VII) is decolourised by a substance that is oxidised, while iodide turns brown when something oxidises it.

Self review

Define oxidation in terms of electrons.
What is the oxidation number of an element in its uncombined state?
What is the oxidation number of sulfur in $SO42−\mathrm{SO_4^{2-}}$ ?
How do oxidation numbers show that a reaction is a redox reaction?
What colour change is seen when acidified aqueous potassium manganate(VII) reacts with a substance that is oxidised?

6.4.2 Oxidation and reduction as electron transfer and change in oxidation number

Electron transfer

Sodium burning in chlorine

Oxidation numbers

Finding oxidation numbers

Identifying redox from oxidation numbers

Iron(III) oxide and carbon monoxide

Chlorine and potassium bromide

Colour changes in redox reactions

Testing for redox with potassium manganate(VII) and potassium iodide

Using oxidation numbers and colour changes

Review 6.4.2 Oxidation and reduction as electron transfer and change in oxidation number by teaching Genie

6.4.2 Oxidation and reduction as electron transfer and change in oxidation number

Electron transfer

Sodium burning in chlorine

Oxidation numbers

Finding oxidation numbers

Identifying redox from oxidation numbers

Iron(III) oxide and carbon monoxide

Chlorine and potassium bromide

Colour changes in redox reactions

Testing for redox with potassium manganate(VII) and potassium iodide

Using oxidation numbers and colour changes

Review 6.4.2 Oxidation and reduction as electron transfer and change in oxidation number by teaching Genie

6.4.2 Oxidation and reduction as electron transfer and change in oxidation number

Electron transfer

Sodium burning in chlorine

Oxidation numbers

Finding oxidation numbers

Identifying redox from oxidation numbers

Iron(III) oxide and carbon monoxide

Chlorine and potassium bromide

Colour changes in redox reactions

Testing for redox with potassium manganate(VII) and potassium iodide

Using oxidation numbers and colour changes

Review 6.4.2 Oxidation and reduction as electron transfer and change in oxidation number by teaching Genie

6.4.2 Oxidation and reduction as electron transfer and change in oxidation number Revision Guide

6.4.2 Oxidation and reduction as electron transfer and change in oxidation number

Electron transfer

Sodium burning in chlorine

Oxidation numbers

Finding oxidation numbers

Identifying redox from oxidation numbers

Iron(III) oxide and carbon monoxide

Chlorine and potassium bromide

Colour changes in redox reactions

Testing for redox with potassium manganate(VII) and potassium iodide

Using oxidation numbers and colour changes

Review 6.4.2 Oxidation and reduction as electron transfer and change in oxidation number by teaching Genie

6.4.2 Oxidation and reduction as electron transfer and change in oxidation number Revision Guide