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A student investigates the mass of copper produced when copper chloride solution in a beaker is electrolyzed using inert electrodes - Edexcel - GCSE Chemistry Combined Science - Question 4 - 2023 - Paper 1
Question 4
A student investigates the mass of copper produced when copper chloride solution in a beaker is electrolyzed using inert electrodes. (a) Where is copper formed duri... show full transcript
Worked Solution & Example Answer:A student investigates the mass of copper produced when copper chloride solution in a beaker is electrolyzed using inert electrodes - Edexcel - GCSE Chemistry Combined Science - Question 4 - 2023 - Paper 1
Step 1
Step 2
State and explain the trend shown in these results.
Answer
As the current increases, the mass of copper produced also increases. This is because the mass of copper formed is directly proportional to the current. The higher the current, the more copper ions are discharged at the cathode, leading to more copper being deposited. A positive correlation exists between the current and the mass of copper formed.
Step 3
Describe how, after the power supply has been switched off, the mass of copper formed can be measured.
Answer
To measure the mass of copper after switching off the power supply, first remove the electrode where copper has been deposited and rinse it with distilled water to remove any impurities. Then, dry the electrode thoroughly. Finally, weigh the electrode with the deposited copper on a balance and subtract the original mass of the electrode to find the mass of copper formed.
Step 4
Calculate the number of copper atoms in 74 mg of copper.
Answer
First, convert the mass of copper to grams: 74 mg = 0.074 g. Then, use the formula:
ext{Number of moles of copper} = rac{ ext{mass}}{ ext{relative atomic mass}} = rac{0.074 ext{ g}}{63.5 ext{ g/mol}} \ ext{Number of moles of copper} \ ext{Number of atoms} = ext{Number of moles} imes ext{Avogadro's constant} \ = rac{0.074}{63.5} imes 6.02 imes 10^{23} \ = 0.000116 imes 6.02 imes 10^{23} \ ext{ atoms} \ ext{Number of atoms} \ ext{Number of atoms} \approx 4.23 imes 10^{19} ext{ atoms.}