A piece of silver jewellery is coated with gold (Au) in an electrolytic cell that contains gold ions in an aqueous solution. Use the information below to determine the oxidation number of the gold ions in the aqueous solution. - volume of gold deposited 0.150 cm³ density of gold 19.3 g/cm³ cell current 4.00 A time taken to plate jewellery 17.75 minutes a. Calculate the amount of electrons, in mole, passed through the cell: b. Calculate the amount of gold, in mole, deposited on the jewellery: c. i. Determine the ratio \( \frac{n(e^{-})}{n(Au)} \) ii. Hence give the oxidation state of gold in the gold ions in this solution.

SSCE VCE Chemistry Production of Chemicals Using Electrolysis: A piece of silver jewellery is c

A piece of silver jewellery is coated with gold (Au) in an electrolytic cell that contains gold ions in an aqueous solution - VCE - SSCE Chemistry - Question 4 - 2003 - Paper 1

Question 4

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A piece of silver jewellery is coated with gold (Au) in an electrolytic cell that contains gold ions in an aqueous solution. Use the information below to determine t... show full transcript

Worked Solution & Example Answer:A piece of silver jewellery is coated with gold (Au) in an electrolytic cell that contains gold ions in an aqueous solution - VCE - SSCE Chemistry - Question 4 - 2003 - Paper 1

Step 1

Calculate the amount of electrons, in mole, passed through the cell:

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Answer

To calculate the amount of electrons, we use the charge formula:

[ Q = I \times t ]\nWhere:

( Q ) is the total charge in coulombs,
( I ) is the current in amperes (A), and
( t ) is the time in seconds (s).

First, convert time from minutes to seconds:

[ t = 17.75 \text{ minutes} \times 60 \text{ s/min} = 1065 \text{ s} ]

Now, substituting the values into the charge formula:

[ Q = 4.00 \text{ A} \times 1065 \text{ s} = 4260 \text{ C} ]

Next, calculate the number of moles of electrons (using Faraday's constant, 1 mol of charge = 96485 C):

[ n(e^{-}) = \frac{Q}{F} = \frac{4260}{96485} \approx 0.0442 \text{ mol} ]

Step 2

Calculate the amount of gold, in mole, deposited on the jewellery:

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Answer

To find the number of moles of gold deposited, we first need to find its mass using its density and volume:

[ \text{mass} = \text{density} \times \text{volume} ]

Substituting the values:

[ \text{mass} = 19.3 \text{ g/cm}^3 \times 0.150 \text{ cm}^3 = 2.895 \text{ g} ]

Next, convert the mass of gold into moles:

Molar mass of gold (Au) = 197 g/mol

[ n(Au) = \frac{\text{mass}}{\text{molar mass}} = \frac{2.895}{197} \approx 0.0147 \text{ mol} ]

Step 3

Determine the ratio \( \frac{n(e^{-})}{n(Au)} \):

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Using the previously calculated moles:

[ \frac{n(e^{-})}{n(Au)} = \frac{0.0442}{0.0147} \approx 3.00 ]

Step 4

Hence give the oxidation state of gold in the gold ions in this solution:

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Answer

The ratio of ( \frac{n(e^{-})}{n(Au)} ) tells us the number of electrons transferred per mole of gold deposited. Since 3 electrons are involved in the deposition of one mole of Au, we can infer that the oxidation state of gold in solution is +3.

A piece of silver jewellery is coated with gold (Au) in an electrolytic cell that contains gold ions in an aqueous solution - VCE - SSCE Chemistry - Question 4 - 2003 - Paper 1

Worked Solution & Example Answer:A piece of silver jewellery is coated with gold (Au) in an electrolytic cell that contains gold ions in an aqueous solution - VCE - SSCE Chemistry - Question 4 - 2003 - Paper 1

Calculate the amount of electrons, in mole, passed through the cell:

Calculate the amount of gold, in mole, deposited on the jewellery:

Determine the ratio \( \frac{n(e^{-})}{n(Au)} \):

Hence give the oxidation state of gold in the gold ions in this solution:

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