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 Redox Reactions: 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

a. Calculate the amount of electrons, in mole, passed through the cell.

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Answer

To calculate the amount of electrons that passed through the cell, we can use the formula:

$n(e^-) = \frac{Q}{F}$

where:

(Q) is the total charge in coulombs,
(F) is Faraday's constant, approximately (96485 , C , mol^{-1}).

First, we calculate the total charge, (Q):

[Q = I \times t]
where:

(I = 4.00 , A)
(t = 17.75 , minutes = 17.75 \times 60 , seconds = 1065 , seconds)

Now, substituting: [Q = 4.00 , A \times 1065 , s = 4260 , C]\

Now, substituting (Q) into the moles of electrons:

[n(e^-) = \frac{4260 , C}{96485 , C , mol^{-1}} \approx 0.0442 , mol]

Step 2

b. Calculate the amount of gold, in mole, deposited on the jewellery.

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Answer

To calculate the amount of gold deposited, we use the formula:

$n(Au) = \frac{m}{M}$

where:

(m) is the mass of gold,
(M) is the molar mass of gold (approximately (197 , g , mol^{-1})).

First, we need to calculate the mass of gold using the volume and density: [m = \text{volume} \times \text{density}]

First convert volume from cm³ to g:

[m = 0.150 , cm³ \times 19.3 , g/cm³ = 2.895 , g]\

Now, substituting this into the amount of gold formula:

[n(Au) = \frac{2.895 , g}{197 , g/mol} \approx 0.0147 , mol]

Step 3

c. i. Determine the ratio $\frac{n(e^-)}{n(Au)}$.

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Answer

To find the ratio of the amounts of electrons to gold, we use the previously calculated values:

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

Step 4

c. ii. Hence give the oxidation state of gold in the gold ions in this solution.

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Answer

From the ratio (\frac{n(e^-)}{n(Au)} = 3.00), this indicates that for every atom of gold that is deposited, three electrons are involved in the reduction process.

Thus, the oxidation state of gold in the gold ions must be +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

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 \(\frac{n(e^-)}{n(Au)}\).

c. ii. Hence give the oxidation state of gold in the gold ions in this solution.

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