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5 L volumes of 1.0 M KNO₃(aq), 1.0 M AgNO₃(aq) and 1.0 M Cu(NO₃)₂(aq) are placed in three separate electrolytic cells using platinum electrodes - VCE - SSCE Chemistry - Question 20 - 2002 - Paper 1

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5-L-volumes-of-1.0-M-KNO₃(aq),-1.0-M-AgNO₃(aq)-and-1.0-M-Cu(NO₃)₂(aq)-are-placed-in-three-separate-electrolytic-cells-using-platinum-electrodes-VCE-SSCE Chemistry-Question 20-2002-Paper 1.png

5 L volumes of 1.0 M KNO₃(aq), 1.0 M AgNO₃(aq) and 1.0 M Cu(NO₃)₂(aq) are placed in three separate electrolytic cells using platinum electrodes. Over a period of 30 ... show full transcript

Worked Solution & Example Answer:5 L volumes of 1.0 M KNO₃(aq), 1.0 M AgNO₃(aq) and 1.0 M Cu(NO₃)₂(aq) are placed in three separate electrolytic cells using platinum electrodes - VCE - SSCE Chemistry - Question 20 - 2002 - Paper 1

Step 1

n(K)

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Answer

In the case of KNO₃, potassium does not get deposited at the cathode during electrolysis. Hence, the amount of potassium deposited, n(K), is 0 moles.

Step 2

n(Ag)

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Answer

For AgNO₃, silver ions gain one electron to form solid silver. The molar mass of silver is approximately 107.87 g/mol. The number of moles of silver deposited can be calculated using Faraday's laws of electrolysis. The charge needed to deposit 1 mole of silver is 96500 C (1 Faraday). Therefore, the total charge of 96500 C corresponds to 1 mole of silver. Hence, n(Ag) = 1 mole.

Step 3

n(Cu)

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Answer

For Cu(NO₃)₂, copper ions also gain two electrons to form solid copper. The total charge can similarly be utilized to find the number of moles of copper deposited. Each mole of copper requires 2 Faradays (2 × 96500 C = 193000 C) for deposition. Given that 96500 C is passed, half a mole of copper is deposited. Therefore, n(Cu) = 0.5 moles.

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