Question

(a) Copper oxide, CuO, is heated with carbon, C. Copper, Cu, and carbon dioxide, CO₂, are made as shown in the equation:

2CuO + C → 2Cu + CO₂

This reaction is an example of reduction. Use the equation to explain what reduction means. 

▶️Answer/Explanation

Reduction is the process in which a substance loses oxygen. In the given reaction, copper oxide (CuO) loses oxygen to form copper (Cu), which is why this reaction is an example of reduction.

(b) The copper made from copper oxide is not pure. A student purifies the impure copper using electrolysis. Fig. 11.1 shows the apparatus the student uses. 

(i) State the name of the electrolyte solution the student uses.

(ii) The student uses impure copper as the anode. State what the student uses as the cathode.

▶️Answer/Explanation

(i) The electrolyte solution used is copper(II) sulfate solution.

(ii) The student uses pure copper as the cathode.

(c) Copper atoms are formed from copper ions, Cu²⁺, at the cathode. Construct the balanced ionic half-equation for this reaction. Use the symbol e^– for an electron. 

▶️Answer/Explanation

The balanced ionic half-equation for the formation of copper atoms from copper ions at the cathode is:

Cu²⁺ + 2e^– → Cu

(d) Aluminium is a metal that is extracted by electrolysis. Fig. 11.2 shows the apparatus that is used. 

The word equation for the electrolysis of aluminium oxide is:

aluminium oxide → aluminium + oxygen

(i) State what is made at the cathode.

(ii) Oxide ions lose electrons to form oxygen molecules. The ionic half-equation for the reaction is:

2O^2- – 4e^– → O₂

Electrons are lost during this process. State the name of this type of reaction.

▶️Answer/Explanation

(i) Aluminium is made at the cathode.

(ii) This type of reaction is called oxidation.

(e) Aluminium reacts with oxygen to make aluminium oxide, Al₂O₃:

4Al + 3O₂ → 2Al₂O₃

Calculate the maximum mass of aluminium oxide that can be made from 1.35 g of aluminium. Show your working.

▶️Answer/Explanation

First, calculate the number of moles of aluminium:

Moles of Al = mass / molar mass = 1.35 g / 27 g/mol = 0.05 mol

From the balanced equation, 4 moles of Al produce 2 moles of Al₂O₃. Therefore, 0.05 moles of Al will produce:

0.05 mol Al × (2 mol Al₂O₃ / 4 mol Al) = 0.025 mol Al₂O₃

Now, calculate the mass of Al₂O₃:

Mass of Al₂O₃ = moles × molar mass = 0.025 mol × 102 g/mol = 2.55 g

Therefore, the maximum mass of aluminium oxide that can be made from 1.35 g of aluminium is 2.55 g.