Oxidation and Reduction Revision Notes for HSC SSCE Chemistry

Oxidation and Reduction

Introduction

Understanding how atoms gain or lose electrons is fundamental to chemistry. This process, known as oxidation and reduction, occurs in countless chemical reactions, from rusting iron to the reactions in batteries. Before exploring these concepts, it's helpful to understand that electronegativity and metal reactivity are inversely related - as electronegativity increases, metal reactivity decreases. This is because highly electronegative elements attract electrons, whilst reactive metals readily lose electrons.

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The relationship between electronegativity and reactivity is crucial: elements with high electronegativity (like oxygen and chlorine) tend to gain electrons, while elements with low electronegativity (like metals) tend to lose electrons. This difference drives many chemical reactions.

What are oxidation and reduction?

When metals react with oxygen, water, or acids, a common feature emerges: metal atoms lose electrons to form positive ions. If metal atoms lose electrons, other substances must gain those electrons. These are called electron transfer reactions - reactions where one or more electrons move from one atom to another.

In electron transfer reactions, electrons move from the metal to:

Oxygen atoms forming $\text{O}^{2-}$ ions (e.g., $\text{Mg} + \text{O}_2$ )
Hydrogen ions forming $\text{H}_2$ molecules (e.g., $\text{Zn} + \text{H}_2\text{SO}_4$ )
Ions of other metals (e.g., $\text{Zn} + \text{Cu}^{2+}$ )

Key definitions

chatImportant

Oxidation occurs when an atom loses one or more electrons.

Reduction occurs when an atom gains one or more electrons.

Memory aids

Two helpful mnemonics can help you remember these definitions:

OXLOSS - oxidation is loss of electrons
OILRIG - oxidation is loss, reduction is gain

Redox reactions

In normal chemical reactions, electrons cannot simply disappear or appear from nowhere. Therefore, oxidation and reduction must always occur together. We call such reactions redox reactions (short for reduction-oxidation reactions). Another name for these is electron transfer reactions.

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Example 1: Magnesium and oxygen

When magnesium metal reacts with oxygen gas, white solid magnesium oxide forms:

$2\text{Mg}(s) + \text{O}_2(g) \rightarrow 2\text{MgO}(s)$

In this reaction:

Magnesium loses two electrons to become $\text{Mg}^{2+}$ (in ionic $\text{MgO}$ ), so magnesium is oxidised to $\text{Mg}^{2+}$
Oxygen gains two electrons to become oxide, $\text{O}^{2-}$ , so oxygen is reduced to oxide
This is a redox reaction because both oxidation and reduction occur

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Example 2: Zinc and hydrochloric acid

When zinc reacts with hydrochloric acid, hydrogen gas forms. The net ionic equation shows:

$\text{Zn}(s) + 2\text{H}^+(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{H}_2(g)$

In this reaction:

Zinc has been oxidised to $\text{Zn}^{2+}$
Hydrogen ions have been reduced to $\text{H}_2$

Half equations

To identify oxidation and reduction clearly, chemists write half equations (also called half reactions). These describe oxidation and reduction processes separately, showing electrons lost or gained.

Writing half equations

For the zinc and acid reaction above, we can write two half equations:

Oxidation half equation: $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$

Reduction half equation: $2\text{H}^+ + 2e^- \rightarrow \text{H}_2$

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In half equations, electrons are shown explicitly as $e^-$ . The oxidation half equation shows electrons as products (being lost), while the reduction half equation shows electrons as reactants (being gained).

Balancing electrons in half equations

When combining half equations into complete reactions, the number of electrons must balance. This is because electrons cannot be left over in a complete reaction.

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Worked Example: Aluminium reacting with dilute acid

For aluminium reacting with dilute acid, the half equations are:

$\text{Al} \rightarrow \text{Al}^{3+} + 3e^-$

$2\text{H}^+ + 2e^- \rightarrow \text{H}_2$

To combine these:

Multiply the first equation by $2$ (giving $6$ electrons)
Multiply the second equation by $3$ (giving $6$ electrons)
Add them together:

$2\text{Al}(s) + 6\text{H}^+(aq) \rightarrow 2\text{Al}^{3+}(aq) + 3\text{H}_2(g)$

Lithium and water example

For lithium reacting with water:

Oxidation half equation: $\text{Li} \rightarrow \text{Li}^+ + e^-$

Reduction half equation: $2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^-$

Water molecules gain electrons to form hydrogen gas and hydroxide ions. We say water has been reduced to hydrogen gas and hydroxide ions.

Earlier definitions of oxidation and reduction

Originally, chemists defined oxidation and reduction differently:

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Historical definitions:

Oxidation was defined as gain of oxygen or loss of hydrogen.

Reduction was defined as loss of oxygen or gain of hydrogen.

These definitions are equivalent to the modern electron-transfer definitions. Sometimes thinking about oxygen and hydrogen changes is simpler than considering electrons.

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Example: Copper oxide and carbon

$\text{CuO}(s) + \text{C}(s) \rightarrow \text{Cu}(s) + \text{CO}(g)$

Using the historical definitions:

Copper in copper oxide has been reduced to metallic copper (lost oxygen)
Carbon has been oxidised to carbon monoxide (gained oxygen)

Oxidants and reductants

Two important terms used with redox reactions are oxidant (or oxidising agent) and reductant (or reducing agent).

Definitions

chatImportant

An oxidant or oxidising agent is a substance that causes oxidation of another substance and is itself reduced.

A reductant or reducing agent is a substance that causes reduction of another substance and is itself oxidised.

Understanding oxidants and reductants

Consider this displacement reaction:

$\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Cu}(s) + \text{Zn}^{2+}(aq)$

Copper ions oxidise zinc metal to zinc ions (i.e., $\text{Cu}^{2+}$ takes electrons from $\text{Zn}$ )
Therefore, $\text{Cu}^{2+}$ is an oxidising agent
Note: the oxidising agent gets reduced ( $\text{Cu}^{2+}$ to $\text{Cu}$ )
Zinc reduces $\text{Cu}^{2+}$ to $\text{Cu}$ (gives it electrons)
Therefore, zinc is a reducing agent
Note: the reducing agent gets oxidised ( $\text{Zn}$ to $\text{Zn}^{2+}$ )

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Another example: Magnesium and chlorine

$\text{Mg}(s) + \text{Cl}_2(g) \rightarrow \text{MgCl}_2(s)$

( $\text{MgCl}_2$ exists as a lattice of $\text{Mg}^{2+}$ and $\text{Cl}^-$ ions)

Magnesium metal reduces chlorine to chloride ions (gives electrons to $\text{Cl}_2$ )
Magnesium is therefore a reducing agent
The reducing agent gets oxidised ( $\text{Mg}$ to $\text{Mg}^{2+}$ )

Key distinction

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Oxidants and reductants are substances or chemical species (like chlorine gas, copper ions, or copper sulfate solution). Oxidation and reduction are processes that happen to elements within substances.

Example: In the reaction below, the element iron is reduced whilst the element carbon is oxidised. Carbon monoxide is the reducing agent:

$\text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) \rightarrow 2\text{Fe}(s) + 3\text{CO}_2(g)$

Oxidation numbers

Sometimes it's not easy to tell whether oxidation and reduction occur in a reaction. When magnesium reacts with chlorine, oxidation and reduction are obvious because ions form. But what about these reactions?

$2\text{P}(s) + 3\text{Cl}_2(g) \rightarrow 2\text{PCl}_3(l)$

$\text{C}(s) + 2\text{S}(s) \rightarrow \text{CS}_2(l)$

The compounds formed are covalent, so there's no clear loss or gain of electrons. To solve this problem, chemists developed oxidation numbers (or oxidation states).

What is an oxidation number?

chatImportant

The oxidation number (or oxidation state) of an element in a molecule or ion is the charge the atom would carry if the molecule or ion were completely ionic.

Although oxidation numbers are somewhat arbitrary, the sum of all oxidation numbers in a species (molecule or polyatomic ion) must equal the net charge on that species.

Example 1: For nitric acid ( $\text{HNO}_3$ ), if it were completely ionic, it would be $\text{H}^+\text{N}^{w+}(\text{O}^{2-})_3$ . To give a net charge of zero, $w$ must be $+5$ . Hence nitrogen's oxidation number in nitric acid is $+5$ .

Example 2: For permanganate ion ( $\text{MnO}_4^-$ ), if completely ionic, it would be $\text{Mn}^{z+}(\text{O}^{2-})_4^-$ . To give a net charge of $-1$ , $z$ must be $+7$ . Hence manganese's oxidation number in permanganate is $+7$ .

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Important note: The oxidation number is not necessarily the actual charge on the atom. In nitric acid and permanganate, covalent bonding occurs between the central atom and oxygen atoms. The actual charges are certainly not $7+$ or $5+$ .

Rules for assigning oxidation numbers

Because oxidation numbers are arbitrarily assigned, they must be assigned in order, following these rules:

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Rule 1: A substance in its elemental state has an oxidation number of zero, regardless of its formula. This applies to copper, oxygen ( $\text{O}_2$ ), hydrogen ( $\text{H}_2$ ), phosphorus ( $\text{P}_4$ ), sulfur ( $\text{S}_8$ ), etc.

The remaining rules apply to elements in compounds:

Rule 2: The oxidation number of a monatomic ion equals the charge on the ion.

Rule 3: Fluorine always has an oxidation number of $-1$ .

Rule 4: Hydrogen has an oxidation number of $+1$ except where rule 2 requires it to be $-1$ (as in $\text{NaH}$ ).

Rule 5: Oxygen has an oxidation number of $-2$ unless rules 2, 3, and 4 dictate a different value (such as $+2$ in $\text{F}_2\text{O}$ and $-1$ in hydrogen peroxide, $\text{H}_2\text{O}_2$ ).

Rule 6: Chlorine, bromine, and iodine have oxidation numbers of $-1$ unless rule 5 requires a different value.

Rule 7: The algebraic sum of oxidation numbers in a molecule or ion must equal the net charge on the species.

Worked examples

Let's calculate oxidation numbers for several compounds:

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Worked Example 1: Sulfur in sulfuric acid ( $\text{H}_2\text{SO}_4$ )

By rules 4 and 5:

$\text{H}$ is $+1$
$\text{O}$ is $-2$

Let $z$ be the oxidation number of $\text{S}$ .

Using rule 7: $2(+1) + z + 4(-2) = 0$ $z = +6$

The oxidation number of sulfur in $\text{H}_2\text{SO}_4$ is $+6$ .

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Worked Example 2: Chlorine in perchlorate ion ( $\text{ClO}_4^-$ )

By rule 5: $\text{O}$ is $-2$

Let $y$ be the oxidation number of $\text{Cl}$ .

Using rule 7: $y + 4(-2) = -1$ $y = +7$

The oxidation number of chlorine in $\text{ClO}_4^-$ is $+7$ .

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Worked Example 3: Chromium in dichromate ion ( $\text{Cr}_2\text{O}_7^{2-}$ )

By rule 5: $\text{O}$ is $-2$

Let $w$ be the oxidation number of $\text{Cr}$ .

Using rule 7 (note the $2w$ because there are two chromium atoms): $2w + 7(-2) = -2$ $w = +6$

The oxidation state of chromium in $\text{Cr}_2\text{O}_7^{2-}$ is $+6$ .

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Important notes:

When halogens are present as oxides, oxyacids, or oxyanions, assign oxygen a value of $-2$ first, then calculate the halogen's value.
Exam tip: The coefficients in a balanced equation play no part in determining oxidation numbers. Only the formulae themselves matter.

Oxidation state and valency

When naming simple ionic compounds, we sometimes need to specify the valency of metal ions using Roman numerals in brackets - for example, tin(II) oxide, iron(III) chloride, and copper(I) sulfide. What we call valency can equally be called oxidation state.

Instead of saying copper has a valency of $1$ in $\text{Cu}_2\text{S}$ , chemists often say copper is in the $+1$ oxidation state in $\text{Cu}_2\text{S}$ . Similarly, titanium is in the $+3$ oxidation state in $\text{TiCl}_3$ , and manganese exists in the $+2$ , $+4$ , and $+7$ oxidation states in $\text{MnCl}_2$ , $\text{MnO}_2$ , and $\text{KMnO}_4$ respectively.

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Key difference: Oxidation states always carry a plus or minus sign. For valency, giving a sign is optional.

Using oxidation numbers to identify redox reactions

An increase in oxidation number represents an increase in positive charge (or decrease in negative charge) on an atom. This must correspond to loss of electrons.

chatImportant

An increase in oxidation number corresponds to oxidation.

A decrease in oxidation number represents a decrease in positive charge (or increase in negative charge). This corresponds to gain of electrons.

A decrease in oxidation number corresponds to reduction.

By calculating oxidation numbers, we can determine whether oxidation or reduction has occurred.

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Example 1: Phosphorus and chlorine

$2\text{P}(s) + 3\text{Cl}_2(g) \rightarrow 2\text{PCl}_3(l)$

Oxidation number of $\text{P}$ changes from $0$ to $+3$ (oxidation)
Oxidation number of $\text{Cl}$ changes from $0$ to $-1$ (reduction)
Therefore, oxidation and reduction are involved

Example 2: Identifying non-redox reactions

Consider these two reactions:

$2\text{CrO}_4^{2-} + 2\text{H}^+ \rightarrow \text{Cr}_2\text{O}_7^{2-} + \text{H}_2\text{O}$

$2\text{CrO}_2^- + 3\text{H}_2\text{O}_2 + 2\text{OH}^- \rightarrow 2\text{CrO}_4^{2-} + 4\text{H}_2\text{O}$

For the first reaction:

Oxidation number of $\text{Cr}$ in $\text{CrO}_4^{2-}$ : $y + 4(-2) = -2$ , so $y = +6$
Oxidation number of $\text{Cr}$ in $\text{Cr}_2\text{O}_7^{2-}$ : $2z + 7(-2) = -2$ , so $z = +6$
Chromium has not been oxidised or reduced
Oxygen and hydrogen oxidation numbers haven't changed either
This is not a redox reaction

For the second reaction:

Oxidation number of $\text{Cr}$ in $\text{CrO}_2^-$ : $w + 2(-2) = -1$ , so $w = +3$
Oxidation number of $\text{Cr}$ in $\text{CrO}_4^{2-}$ : $+6$ (from above)
Chromium has been oxidised from $+3$ to $+6$
Oxidation number of $\text{O}$ in $\text{H}_2\text{O}_2$ is $-1$ (rule 5)
Oxidation number of $\text{O}$ in $\text{H}_2\text{O}$ is $-2$
Oxygen in hydrogen peroxide has been reduced
This is a redox reaction: hydrogen peroxide has oxidised chromium from $+3$ to $+6$

Oxidation numbers and electron transfer

The change in oxidation number represents the number of electrons gained or lost per atom in the redox half equation.

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Example 1: $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$

Manganese's oxidation number changed from $+7$ to $+2$ . This change of $-5$ corresponds to gaining five electrons.

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Example 2: $\text{HNO}_2 + \text{H}_2\text{O} \rightarrow \text{NO}_3^- + 3\text{H}^+ + 2e^-$

Nitrogen's oxidation number changed from $+3$ to $+5$ . This change of $+2$ corresponds to losing two electrons.

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Important: The change is per atom of the element involved.

Example 3: $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$

Chromium's oxidation number changed from $+6$ to $+3$ , a change of $-3$ per atom. Since two chromium atoms are reduced, the total number of electrons in the half equation is six.

Remember!

bookmarkSummary

Key Points to Remember:

Oxidation means loss of electrons; reduction means gain of electrons (remember OILRIG!)
Oxidation and reduction always occur together in redox reactions - electrons lost by one species are gained by another
Half equations show oxidation and reduction separately, with electrons explicitly shown as $e^-$
When combining half equations, balance the number of electrons
An oxidant (oxidising agent) causes oxidation and is itself reduced
A reductant (reducing agent) causes reduction and is itself oxidised
Oxidation numbers help identify redox reactions in covalent compounds where electron transfer isn't obvious
Use the seven rules to assign oxidation numbers in order, ensuring the sum equals the net charge
An increase in oxidation number indicates oxidation; a decrease indicates reduction
The change in oxidation number equals the number of electrons transferred per atom in half equations

Oxidation and Reduction (HSC SSCE Chemistry): Revision Notes