Chemistry 9701/12 — May/June 2013

Background Concept

Electrolysis is the decomposition of a compound by passing an electric current through it. In molten aluminium oxide ($\text{Al}_2\text{O}_3$), the ionic lattice breaks down and the ions become free to move and carry charge. At the anode (positive electrode), oxidation occurs — electrons are removed from the negative ions. The oxide ions ($\text{O}^{2-}$) are discharged, forming oxygen gas. However, because the anode is made of carbon, the hot oxygen gas reacts with the carbon electrode, producing carbon dioxide. This is why the carbon anode is gradually consumed during the industrial extraction of aluminium and must be replaced.

Oxidation number is a bookkeeping tool that tracks electron transfer in redox reactions. For oxygen, the key values in this question are:

• In oxide ions ($\text{O}^{2-}$): oxidation number is $-2$
• In elemental oxygen ($\text{O}_2$): oxidation number is $0$
• In carbon dioxide ($\text{CO}_2$): oxidation number is $-2$ (oxygen is more electronegative than carbon)

Understanding the Question

The question describes the electrolysis of molten aluminium oxide with carbon electrodes and states that two consecutive reactions occur at the anode, each producing a different gas. The task is to determine how the oxidation number of oxygen changes across these two reactions. The key is to identify both anode reactions and track oxygen's oxidation number through each step in sequence.

Approach

1. Identify the two anode reactions: first the discharge of oxide ions, then the reaction of the oxygen gas with the carbon anode.
2. Determine the oxidation number of oxygen in each species involved.
3. Calculate the change in oxidation number for each step and compare with the options.

Step-by-Step Reasoning

Step 1: First anode reaction — discharge of oxide ions

At the anode, oxide ions are oxidised:

$$2\text{O}^{2-} \rightarrow \text{O}_2 + 4\text{e}^-$$

• In $\text{O}^{2-}$, oxygen has oxidation number $-2$ (matching the ionic charge).
• In $\text{O}_2$, each oxygen atom has oxidation number $0$ (elemental form).
• Change: $0 - (-2) = +2$, so the oxidation number increases by 2.

This is an oxidation — electrons are lost.

Step 2: Second anode reaction — reaction with carbon

The oxygen gas produced at the anode reacts with the hot carbon electrode:

$$\text{C} + \text{O}_2 \rightarrow \text{CO}_2$$

• In $\text{O}_2$, oxygen has oxidation number $0$.
• In $\text{CO}_2$, each oxygen atom has oxidation number $-2$ (carbon is less electronegative than oxygen, so the shared electrons are counted as belonging to oxygen).
• Change: $-2 - 0 = -2$, so the oxidation number decreases by 2.

Therefore, the oxidation number of oxygen increases by 2, then decreases by 2 — option B.

Why the distractors are wrong:

• A (decreases by 2, then increases by 2): This reverses the order. The first step is an oxidation (increase in oxidation number), not a reduction.
• C (increases by 2, then decreases by 4): The second change is $-2$ per oxygen atom, not $-4$. The total change for both oxygen atoms combined in $\text{CO}_2$ would be $-4$, but the question asks about the change in oxidation number of oxygen as an element — i.e., per atom.
• D (no change, then decreases by 2): The first step does involve a change — oxygen goes from $-2$ in $\text{O}^{2-}$ to $0$ in $\text{O}_2$.

Key Takeaways

• In electrolysis, oxidation always occurs at the anode; reduction at the cathode.
• The oxidation number of an element can change across consecutive reactions — track it step by step.
• When tracking oxidation number changes, consider each atom individually, not the sum for the whole molecule.
• Elemental oxygen ($\text{O}_2$) has oxidation number $0$, while oxygen in most compounds has $-2$.

Common Mistakes

• Confusing the order of the reactions: the discharge of $\text{O}^{2-}$ comes first, producing $\text{O}_2$; only then does the oxygen react with the carbon anode.
• Thinking the second change is $-4$: In $\text{CO}_2$ there are two oxygen atoms, each with oxidation number $-2$. The question asks about the change in oxidation number of oxygen (per atom), which is $-2$, not $-4$.
• Forgetting that $\text{O}_2$ has oxidation number $0$: elemental oxygen is always $0$, regardless of the molecule.
• Reversing the direction: the first step is oxidation (increase), the second is reduction (decrease).

Things to Be Careful About

• The oxidation number of oxygen is $-2$ in most compounds, $0$ in elemental form ($\text{O}_2$), and $-1$ in peroxides (not relevant here but worth remembering).
• In $\text{O}^{2-}$, the oxidation number equals the ionic charge: $-2$.
• The carbon anode is consumed because it reacts with the oxygen produced — this is a key industrial detail.
• The question asks specifically about the change in oxidation number of oxygen, so track oxygen through both reactions, not carbon or aluminium.