Redox Reactions (Grade 11 NSC Matric Physical Sciences): Revision Notes

Worked Example 1: Finding oxidation numbers in sulfate ion

Question: Find the oxidation number of sulfur in a sulfate (SO₄²⁻) ion.

Solution:

Step 1: Determine known oxidation numbers

Oxygen has an oxidation number of -2 (Rule 5, this is not a peroxide).

Sulfur's oxidation number is unknown at this stage.

Step 2: Use the rule that oxidation numbers must sum to the total charge

In the polyatomic SO₄²⁻ ion, the sum of oxidation numbers must equal -2 (Rule 4).

Let sulfur's oxidation number = $x$

Since there are four oxygen atoms, their total contribution = $4 × (-2) = -8$

Step 3: Set up and solve the equation

$x + (-8) = -2$ $x = -2 + 8 = +6$

Therefore, sulfur has an oxidation number of +6 in the sulfate ion.

Worked Example 2: Finding oxidation numbers in ammonia

Question: Find the oxidation numbers of both elements in ammonia (NH₃).

Solution:

Step 1: Determine known oxidation numbers

Hydrogen has an oxidation number of +1 (Rule 6, ammonia is not a metal hydride).

Nitrogen's oxidation number is unknown.

Step 2: Use the rule for neutral compounds

In compound NH₃, the sum of oxidation numbers must equal 0 (Rule 3).

Let nitrogen's oxidation number = $x$

Since there are three hydrogen atoms: $3 × (+1) = +3$

Step 3: Set up and solve the equation

$x + (+3) = 0$

$x = -3$

Therefore, hydrogen has an oxidation number of +1 and nitrogen has an oxidation number of -3.

Worked Example 3: Finding oxidation numbers in ionic compounds

Question: Find the oxidation numbers for all atoms in sodium chloride (NaCl).

Solution:

Step 1: Identify the compound type

This is an ionic compound containing Na⁺ and Cl⁻ ions.

Step 2: Apply Rule 2 for monatomic ions

Sodium ion has an oxidation number of +1

Chlorine ion has an oxidation number of -1

This gives us a sum of 0 for the compound, confirming our answer.

Investigation: Displacement Reaction

Aim: To investigate the redox reaction between copper sulfate and zinc.

Materials:

• A few granules of zinc
• 15 ml copper (II) sulfate solution (blue colour)
• Glass beaker

Method: Add zinc granules to the copper sulfate solution and observe what happens. Note any changes to the zinc granules and the solution colour.

Results:

• Zinc becomes covered in a layer that looks like copper
• The blue copper sulfate solution becomes clearer

Explanation: Cu²⁺ ions from the CuSO₄ solution are reduced to form copper metal, which deposits on the zinc crystals. The reduction of copper ions also explains the colour change (copper ions in solution appear blue).

The half-reaction for copper reduction: $\text{Cu²⁺(aq) + 2e⁻ → Cu (s)}$

Zinc is oxidised to form Zn²⁺ ions which are colourless in solution: $\text{Zn (s) → Zn²⁺(aq) + 2e⁻}$

The overall reaction: $\text{Cu²⁺(aq) + Zn (s) → Cu (s) + Zn²⁺(aq)}$

This is called a displacement reaction because zinc displaces the copper ions to form zinc sulfate.

Investigation: Synthesis Reaction

Aim: To investigate the redox reaction that occurs when magnesium burns in air.

Materials: Strip of magnesium; bunsen burner; tongs; glass beaker

Method:

1. Light the bunsen burner and use tongs to hold the magnesium ribbon in the flame
2. Hold the lit magnesium over a beaker and observe what happens

WARNING: Do not look directly at the flame.

Results: The magnesium burns with a bright white flame. When held over a beaker, a fine powder is observed in the beaker. This powder is magnesium oxide.

The overall reaction: $\text{2Mg (s) + O₂(g) → 2MgO (s)}$

Conclusion: A redox reaction has occurred. Magnesium is oxidised and oxygen is reduced. This is a synthesis reaction because we have made magnesium oxide from magnesium and oxygen.

Investigation: Decomposition Reaction

Aim: To investigate the decomposition of hydrogen peroxide.

Materials: Dilute hydrogen peroxide (about 3%); manganese dioxide; test tubes; water bowl; stopper and delivery tube; bunsen burner

Method:

1. Put a small amount (about 5 ml) of hydrogen peroxide in a test tube
2. Set up the apparatus as shown above
3. Very carefully add a small amount (about 0.5 g) of manganese dioxide to the test tube containing hydrogen peroxide

Results: You should observe gas bubbling into the second test tube. This reaction happens quite rapidly.

The overall reaction: $\text{2H₂O₂(aq) → 2H₂O (l) + O₂(g)}$

Conclusion: A redox reaction has occurred. H₂O₂ is both oxidised and reduced in this decomposition reaction.

Worked Example: Balancing simple redox reactions

Question: Balance the following redox reaction: $\text{Fe²⁺(aq) + Cl₂(aq) → Fe³⁺(aq) + Cl⁻(aq)}$

Solution:

Step 1: Write separate reactions for each compound

$\text{Fe²⁺ → Fe³⁺}$

$\text{Cl₂ → Cl⁻}$

Step 2: Balance the atoms on either side

We check that atoms are balanced:

$\text{Fe²⁺ → Fe³⁺}$ (already balanced)

$\text{Cl₂ → 2Cl⁻}$ (now balanced)

Step 3: Add electrons to balance the charges

We add electrons to the side with greater positive charge:

$\text{Fe²⁺ → Fe³⁺ + e⁻}$

$\text{Cl₂ + 2e⁻ → 2Cl⁻}$

Step 4: Balance the number of electrons

The iron reaction has one electron, while the chlorine reaction has two electrons. We multiply the iron reaction by 2 to ensure charges balance:

$\text{2Fe²⁺ → 2Fe³⁺ + 2e⁻}$

$\text{Cl₂ + 2e⁻ → 2Cl⁻}$

Step 5: Combine the two half-reactions

$\text{2Fe²⁺ → 2Fe³⁺ + 2e⁻}$

$\text{Cl₂ + 2e⁻ → 2Cl⁻}$

Combined: $\text{2Fe²⁺ + Cl₂ + 2e⁻ → 2Fe³⁺ + 2Cl⁻ + 2e⁻}$

Step 6: Write the final answer

Cancelling the electrons: $\text{2Fe²⁺(aq) + Cl₂(aq) → 2Fe³⁺(aq) + 2Cl⁻(aq)}$

The iron reaction is the oxidation half-reaction (iron becomes more positive). The chlorine reaction is the reduction half-reaction (chlorine becomes more negative).

Worked Example: Complex redox balancing

Question: Balance the following redox reaction: $\text{Cr₂O₇²⁻(aq) + H₂S (aq) → Cr³⁺(aq) + S (s)}$ The reaction takes place in acidic conditions.

Solution:

Step 1: Write separate reactions for each compound

$\text{Cr₂O₇²⁻ → Cr³⁺}$

$\text{H₂S → S}$

Step 2: Balance atoms on either side

For the first reaction: 2 chromium atoms and 7 oxygen atoms on left, 1 chromium atom on right. Since we're in acidic conditions, we can add water to balance oxygen and H⁺ ions to balance hydrogen:

$\text{Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O}$

Now we have hydrogen atoms on the right but not on the left, so we add 14 hydrogen ions to the left: $\text{Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O}$

For the second reaction, we add 2 hydrogen ions to the right: $\text{H₂S → S + 2H⁺}$

Step 3: Add electrons to balance charges

$\text{Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O}$

$\text{H₂S → S + 2H⁺ + 2e⁻}$

Step 4: Balance electron numbers

The chromium reaction has 6 electrons, the sulfur reaction has 2 electrons. We multiply the sulfur reaction by 3:

$\text{Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O}$

$\text{3H₂S → 3S + 6H⁺ + 6e⁻}$

Step 5: Combine the half-reactions

$\text{Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O}$

$\text{3H₂S → 3S + 6H⁺ + 6e⁻}$

Combined: $\text{Cr₂O₇²⁻ + 14H⁺ + 6e⁻ + 3H₂S → 2Cr³⁺ + 7H₂O + 3S + 6H⁺ + 6e⁻}$

Step 6: Write the final answer

Cancelling electrons and hydrogen ions: $\text{Cr₂O₇²⁻(aq) + 3H₂S (aq) + 8H⁺(aq) → 2Cr³⁺(aq) + 3S (s) + 7H₂O (l)}$

The sulfur reaction is the oxidation half-reaction (sulfur becomes more positive). The chromium reaction is the reduction half-reaction (chromium becomes more negative).