13. Calculate the mass of copper deposited by the electrolysis of copper (II) sulphate solution placed in series with a silver nitrate solution, when 0.108 g of silver is being deposited (1pt)

Answer 1

The first step is to write the balanced chemical equation for the electrolysis of copper (II) sulphate solution:

CuSO4 + 2e- -> Cu + SO4^2-

From the equation, it can be seen that 1 mol of copper is deposited for every 2 moles of electrons transferred.

Next, we need to determine the number of moles of electrons transferred in the deposition of 0.108g of silver.

The molar mass of silver (Ag) is 107.87 g/mol. Therefore:

0.108g Ag x (1 mol Ag / 107.87 g Ag) = 0.001 mol Ag

Since 1 mole of silver corresponds to 1 mole of electrons, there are 0.001 moles of electrons transferred.

As mentioned earlier, 2 moles of electrons are required for the deposition of 1 mole of copper. Therefore, for the deposition of 0.108g of silver, 0.002 moles of electrons are transferred.

Finally, the mass of copper deposited can be calculated by converting moles of electrons to moles of copper and then to grams of copper:

0.002 moles Cu x (64.546 g Cu / 1 mol Cu) = 0.129 g Cu

Therefore, the mass of copper deposited by the electrolysis of copper (II) sulphate solution in the given scenario is 0.129 grams.