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

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

1. Copper nitrate:

   • Positive electrode: oxygen (O2)
   • Negative electrode: copper (Cu)

2. Potassium iodide:

   • Positive electrode: iodine (I2)
   • Negative electrode: hydrogen (H2)

The students can filter the mixture to collect the copper residue that settled at the bottom of the beaker, wash and dry the copper, and then weigh it to add to the increase in mass of the negative electrode.

The graph in Figure 5 shows a straight line through the origin for the current values, indicating that when the time doubles, the total mass of copper produced also doubles, demonstrating direct proportionality.

When the current increases, the slope of the graph also increases correspondingly. For example, when the current is 0.6 A, the mass is greater than when the current is 0.3 A, supporting the proportional relationship.

The blue color fades because copper ions are discharged from the solution during electrolysis; as copper ions are removed, the concentration of the blue-colored copper nitrate decreases.

Using the formula:

Number of moles =

rac{Q}{F} where:

• Q is the charge (in coulombs),
• F is the Faraday constant (96485 C/mol).

First calculate Q:

$Q = I imes t = 0.6 ext{ A} imes (20 imes 60 ext{ s}) = 720 C$

Now, calculating moles:

$ext{moles} = \frac{720}{96485} = 0.00748 ext{ mol}$

Finally, determine the number of atoms:

Number of atoms = moles × Avogadro's number =

$0.00748 imes 6.02 imes 10^{23} = 4.50 imes 10^{21} ext{ atoms}$.

The answer is 4.50 x 10²¹ atoms.