This question is about chemical reactions and electricity - AQA - GCSE Chemistry - Question 7 - 2021 - Paper 1

The balanced half equation for the production of bromine at the positive electrode is:

$2Br^- \rightarrow Br_2 + 2e^-$

To accurately determine the total mass of copper produced, students should filter the mixture to separate the copper from any other residues, wash and dry the copper collected, and then weigh it. Finally, they should add this mass to the increase in the mass of the electrode.

The graph in Figure 5 shows a straight line that passes through the origin when plotting the total mass of copper produced against time. This indicates that as the time of electrolysis increases, the mass of copper produced increases proportionally, confirming that these two variables are directly related.

The data from Figure 5 illustrates that for the same duration of electrolysis, a higher current results in a greater mass of copper produced. For example, at a current of 0.6 A, the mass produced is double that at 0.3 A, demonstrating that the total mass of copper is directly proportional to the current.

The blue color of the copper nitrate solution fades as the copper ions are discharged during electrolysis. As copper ions are removed from the solution to deposit on the negative electrode, the concentration of copper ions decreases, leading to a lighter color.

First, calculate the number of moles of copper produced using the formula:

$ext{moles} = \frac{Q}{nF}$

where:

• $Q = I \times t = 0.6 \text{ A} \times 1200 ext{ s} = 720 ext{ C}$
• $n = 2$ (from the half equation)
• $F = 96485 ext{ C/mol}$ (Faraday's constant)

Calculating:

$ext{moles} = \frac{720}{2 \times 96485} \approx 0.00373 ext{ moles}$

Now, to find the number of copper atoms:

$\text{Number of atoms} = 0.00373 \text{ moles} \times 6.02 \times 10^{23} \approx 2.24 \times 10^{21}$

Therefore, the answer to 3 significant figures is: $2.24 \times 10^{21} \text{ atoms of copper}$