Writing Redox and Half-Reactions Revision Notes for Grade 12 NSC Matric Physical Sciences

Writing Redox and Half-Reactions

Understanding redox reactions

Redox reactions involve the transfer of electrons between different atoms or compounds. The word "redox" combines reduction and oxidation, which always occur together in these reactions.

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Redox reactions are fundamental to many chemical processes, including combustion, corrosion, and biological processes like cellular respiration. Understanding electron transfer helps explain how these reactions occur at the molecular level.

The OIL RIG mnemonic

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A simple way to remember oxidation and reduction is the OIL RIG mnemonic:

OIL: Oxidation Is Loss (of electrons)
RIG: Reduction Is Gain (of electrons)

This mnemonic is essential for understanding all redox chemistry and should be memorised as your first step in mastering this topic.

Key definitions

Oxidation occurs when an element or compound loses electrons. For example:

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

Reduction occurs when an element or compound gains electrons. For example:

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

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Important distinctions:

Oxidising agent is the substance that gains electrons and causes another substance to be oxidised. The oxidising agent itself is reduced in the process.

Reducing agent is the substance that loses electrons and causes another substance to be reduced. The reducing agent itself is oxidised in the process.

Identifying redox reactions

To determine if a reaction is a redox reaction, ask these two questions:

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Quick Check for Redox Reactions:

Is there a compound or atom being oxidised?
Is there a compound or atom being reduced?

If both answers are "yes", then the reaction is a redox reaction.

Writing half-reactions

A half-reaction shows either the oxidation process or the reduction process separately. Every redox reaction can be written as two half-reactions.

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Worked Example: Iron and tin reaction

Overall reaction: $2\text{Fe}^{3+}\text{(aq)} + \text{Sn}^{2+}\text{(aq)} \rightarrow 2\text{Fe}^{2+}\text{(aq)} + \text{Sn}^{4+}\text{(aq)}$

This can be written as two half-reactions:

Reduction half-reaction: $\text{Fe}^{3+}\text{(aq)} + e^- \rightarrow \text{Fe}^{2+}\text{(aq)}$
Oxidation half-reaction: $\text{Sn}^{2+}\text{(aq)} \rightarrow \text{Sn}^{4+}\text{(aq)} + 2e^-$

Notice that in the overall reaction, the reduction half-reaction is multiplied by two so that the number of electrons gained equals the number of electrons lost.

Balancing redox reactions

Half-reactions can be used to balance complex redox reactions. The key principle is that electrons lost must equal electrons gained.

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Seven-Step Balancing Method:

Step 1: Write down the unbalanced oxidation half-reaction

Step 2: Balance the number of atoms on both sides of the equation

Step 3: Once the atoms are balanced, check that the charges balance by adding electrons

Step 4: Repeat steps 1-3 with the reduction half-reaction

Step 5: Compare the number of electrons in each equation and multiply each half-reaction by a suitable number so that electrons released equals electrons accepted

Step 6: Combine the two half-reactions to get the final equation for the overall reaction

Step 7: Do a final check to ensure the equation is balanced

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Worked Example 1: Basic redox balancing

Question: Chlorine gas oxidises $\text{Fe}^{2+}$ ions to $\text{Fe}^{3+}$ ions. In the process, chlorine is reduced to chloride ions. Write a balanced equation for this reaction.

Solution:

Step 1: Write the unbalanced oxidation half-reaction $\text{Fe}^{2+}\text{(aq)} \rightarrow \text{Fe}^{3+}\text{(aq)}$

Step 2: Balance atoms (already balanced)

Step 3: Balance charges by adding electrons $\text{Fe}^{2+}\text{(aq)} \rightarrow \text{Fe}^{3+}\text{(aq)} + e^-$

Step 4: Write the unbalanced reduction half-reaction $\text{Cl}_2\text{(g)} \rightarrow \text{Cl}^-\text{(aq)}$

Balance atoms: $\text{Cl}_2\text{(g)} \rightarrow 2\text{Cl}^-\text{(aq)}$ Balance charges: $\text{Cl}_2\text{(g)} + 2e^- \rightarrow 2\text{Cl}^-\text{(aq)}$

Step 5: Compare electrons

Oxidation: $\text{Fe}^{2+}\text{(aq)} \rightarrow \text{Fe}^{3+}\text{(aq)} + e^-$
Reduction: $\text{Cl}_2\text{(g)} + 2e^- \rightarrow 2\text{Cl}^-\text{(aq)}$

Multiply oxidation by 2: $2\text{Fe}^{2+}\text{(aq)} \rightarrow 2\text{Fe}^{3+}\text{(aq)} + 2e^-$

Step 6: Combine half-reactions $2\text{Fe}^{2+}\text{(aq)} + \text{Cl}_2\text{(g)} \rightarrow 2\text{Fe}^{3+}\text{(aq)} + 2\text{Cl}^-\text{(aq)}$

Step 7: Check balance (atoms and charges are balanced)

Balancing in acidic medium

When reactions occur in acidic medium, water molecules ( $\text{H}_2\text{O}$ ) and hydrogen ions ( $\text{H}^+$ ) can be added to balance oxygen and hydrogen atoms.

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Key steps for acidic medium:

Add $\text{H}_2\text{O}$ molecules to balance oxygen atoms
Add $\text{H}^+$ ions to balance hydrogen atoms
Add electrons to balance charges

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Worked Example 2: Acidic medium

Question: Balance the reaction in acidic medium: $\text{Cr}_2\text{O}_7^{2-}\text{(aq)} + \text{H}_2\text{S(g)} \rightarrow \text{Cr}^{3+}\text{(aq)} + \text{S(s)}$

The balanced equation becomes: $\text{Cr}_2\text{O}_7^{2-}\text{(aq)} + 8\text{H}^+\text{(aq)} + 3\text{H}_2\text{S(g)} \rightarrow 3\text{S(s)} + 2\text{Cr}^{3+}\text{(aq)} + 7\text{H}_2\text{O(ℓ)}$

Balancing in alkaline medium

When reactions occur in alkaline medium, hydroxide ions ( $\text{OH}^-$ ) are present instead of $\text{H}^+$ ions.

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Key steps for alkaline medium:

Add $\text{OH}^-$ ions to balance charges
Use $\text{H}_2\text{O}$ and $\text{OH}^-$ to balance hydrogen and oxygen
The final balanced equation includes spectator ions

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Worked Example 3: Alkaline medium

Question: Balance the oxidation of $\text{Co(NH}_3\text{)}_6^{2+}$ by $\text{H}_2\text{O}_2$ in alkaline medium.

This requires careful consideration of the hydroxide ions present in the alkaline solution and their effect on the balancing process.

Exam tips

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Essential Exam Strategies:

Always check that both atoms and charges are balanced in your final equation
Remember that electrons lost in oxidation must equal electrons gained in reduction
In acidic medium, use $\text{H}^+$ and $\text{H}_2\text{O}$ for balancing
In alkaline medium, use $\text{OH}^-$ and $\text{H}_2\text{O}$ for balancing
Practice identifying which substance is oxidised and which is reduced

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Key Points to Remember:

OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons)
Oxidising agents gain electrons and cause other substances to lose electrons
Reducing agents lose electrons and cause other substances to gain electrons
Half-reactions show the oxidation and reduction processes separately
Always balance both atoms and charges in redox equations
The number of electrons lost must equal the number of electrons gained in any redox reaction

Writing Redox and Half-Reactions (Grade 12 NSC Matric Physical Sciences): Revision Notes