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 is: $2Br^- → Br_2 + 2e^-$

For copper nitrate:

• Product at positive electrode: oxygen (O_2)
• Product at negative electrode: copper (Cu)

For potassium iodide:

• Product at positive electrode: iodine (I_2)
• Product at negative electrode: hydrogen (H_2)

The students should filter the mixture to collect the copper that did not adhere to the electrode. Following filtration, they should wash and dry the copper residue before weighing it. Finally, they should add this mass to the increase in mass of the electrode to obtain the total mass.

The graph indicates a linear relationship; as time increases, the total mass of copper produced increases proportionally. For instance, when the empirical times of 10, 20, and 30 minutes are compared, the mass of copper approximately doubles as time doubles.

According to the graph, for a given time, an increase in current leads to a corresponding increase in the total mass of copper produced. For example, at 20 minutes, the mass of copper produced at 0.3 A, 0.6 A, and 0.9 A demonstrates this proportionality.

The blue colour of the copper nitrate solution fades because copper ions (Cu²⁺) are discharged at the negative electrode during electrolysis, resulting in a lower concentration of copper ions in the solution as they are removed.

First, calculate the number of moles of copper produced using Faraday's laws of electrolysis:

1. Total charge (Q) = current (I) × time (t), therefore: $Q = 0.6 A × 1200 s = 720 C$
2. Using the formula: 1 mole of Cu requires 2 × 96500 C (2 Faradays). Thus, $ext{Number of moles} = \frac{720}{2 × 96500} = 0.00372 \text{ moles}$
3. Finally, to find the number of atoms: $0.00372 imes 6.02 × 10^{23} = 2.24 × 10^{21} ext{ atoms}$