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This question is about electrolysis - AQA - GCSE Chemistry: Combined Science - Question 5 - 2019 - Paper 1

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This question is about electrolysis. Some metals are extracted from molten compounds using electrolysis. Why is electrolysis used to extract some metals? Aluminium... show full transcript

Worked Solution & Example Answer:This question is about electrolysis - AQA - GCSE Chemistry: Combined Science - Question 5 - 2019 - Paper 1

Step 1

Why is electrolysis used to extract some metals?

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Answer

Electrolysis is used to extract some metals because these metals are usually too reactive to be extracted by carbon reduction methods. For example, metals like sodium and aluminum cannot be easily obtained by heating with carbon as they react with it.

Step 2

What two substances does the molten mixture contain?

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Answer

  1. Aluminium oxide
  2. Cryolite

Step 3

Half equation at negative electrode

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Answer

Cu<sup>2+</sup> + 2e<sup>-</sup> → Cu

Step 4

Half equation at positive electrode

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Answer

2 Cl<sup>-</sup> → Cl<sub>2</sub> + 2e<sup>-</sup>

Step 5

Suggest two reasons why the mass deposited was different from the expected value.

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Answer

  1. The concentration or volume of the copper chloride solution may have been different from expected.
  2. There might have been errors in timing during the electrolysis process.

Step 6

Determine the expected mass of copper after 24 hours.

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Answer

Using Figure 5, we can see that the mass of copper produced is a linear function of time. If the expected mass produced each minute is around 5.4 mg, then over 24 hours (1440 minutes), the expected mass would be:

extExpectedmass=5.4extmgimes1440extminutes=2592extmg ext{Expected mass} = 5.4 ext{ mg} imes 1440 ext{ minutes} = 2592 ext{ mg}

Step 7

Determine the mass of the negative electrode at the start of the experiment.

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Answer

From Figure 6, the mass of the negative electrode at the start of the experiment is determined to be 5 grams.

Step 8

Calculate the gradient of the line in Figure 6.

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Answer

To calculate the gradient of the line in Figure 6, we can use the formula:

ext{Gradient} = rac{Y}{X} Where Y is the change in mass (in grams) over the change in time (in hours). Given that the mass increased from 5 grams to 10 grams over 10 hours:

ext{Gradient} = rac{10 - 5}{10 - 0} = 0.5 ext{ grams per hour}

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