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
Where is copper formed during the electrolysis?
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Answer
During electrolysis, copper is formed at the cathode. The cathode is where reduction occurs, meaning that copper ions gain electrons to become neutral copper atoms.
Step 2
State and explain the trend shown in these results.
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The trend shows that as the current increases, the mass of copper produced also increases. This is because the mass of copper formed is proportional to the electric current according to Faraday's law of electrolysis. The relationship can be expressed as:
ext{Mass} acksim ext{Current}
This means that the greater the current, the more mass of copper is deposited.
Step 3
Describe how, after the power supply has been switched off, the mass of copper formed can be measured.
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To measure the mass of copper formed:
Rinse and dry the electrode at the cathode to remove any impurities.
Weigh the electrode (cathode) on a balance to find the initial mass.
After switching off the power supply, allow the electrode to cool down if necessary.
Weigh the electrode again to get the final mass.
Subtract the initial mass from the final mass to determine the mass of copper deposited.
Step 4
Calculate the number of copper atoms in 74 mg of copper.
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To calculate the number of copper atoms in 74 mg, we first need to convert the mass from milligrams to grams:
74extmg=0.074extg
Then, using the formula:
ext{Number of atoms} = rac{ ext{mass in g}}{ ext{relative atomic mass}} imes ext{Avogadro's constant}
Substituting the values:
ext{Number of atoms} = rac{0.074 ext{ g}}{63.5} imes 6.02 imes 10^{23} \ ext{Number of atoms} \ = 0.001165 imes 6.02 imes 10^{23} \ ext{Number of atoms} \ ext{≈ } 7.006 imes 10^{20}