Question

The amount of copper deposited by the passage of 1 mole of electrons during the electrolysis of aqueous \(\mathrm{CuSO}_{4}\) solution is (a) \(63.5 \mathrm{~g}\) (b) \(31.75 \mathrm{~g}\) (c) \(159.5 \mathrm{~g}\) (d) \(79.75 \mathrm{~g}\)

Short answer

Answer: (b) \(31.75 \mathrm{~g}\).

Step-by-step solution

Identify the half-reaction occurring at the cathode

In electrolysis, copper is deposited at the cathode. The half-reaction occurring at the cathode is the reduction of copper ions. The ionic equation for the reduction half-reaction at the cathode is: $$\mathrm{Cu^{2+}(aq) + 2e^{-} \rightarrow Cu(s)}$$

Determine the number of moles of copper ions reduced and deposited

Since the passage of 1 mole of electrons is given, we have to determine how many moles of copper ions will be reduced and deposited. Using the stoichiometry of the half-reaction, we can see that 2 moles of electrons are needed to reduce 1 mole of copper ions. Therefore, with 1 mole of electrons, half a mole (0.5 moles) of copper ions will be reduced and deposited on the cathode.

Determine the mass of the deposited copper

To find the mass of the deposited copper, we'll use the molar mass of copper and the number of moles of copper ions reduced. The molar mass of copper (\(\mathrm{Cu}\)) is \(63.5 \mathrm{~g/mol}\). Now we will multiply the number of moles of copper ions reduced (0.5 moles) by the molar mass of copper: $$0.5 \mathrm{~mol} \times 63.5 \mathrm{~g/mol} = 31.75 \mathrm{~g}$$ Thus, the amount of copper deposited by the passage of 1 mole of electrons during the electrolysis of aqueous \(\mathrm{CuSO}_{4}\) solution is \(31.75 \mathrm{~g}\). The correct answer is (b) \(31.75 \mathrm{~g}\).

Key concepts

Copper Deposition

During electrolysis, copper deposition is a process where copper ions in solution are transformed into solid copper. This process occurs at the cathode, which is the negatively charged electrode. When an electric current is passed through an aqueous solution of copper sulfate (\(\mathrm{CuSO_{4}}\)), copper ions (\(\mathrm{Cu^{2+}}\)) move towards the cathode.
Here, they gain electrons (a process known as reduction) and are converted into solid copper metal. This transformation is crucial in metal refining and electroplating processes. By controlling the current, we can determine the exact amount of copper that will be deposited.

• The purity of the copper deposited depends on the composition of the electrolyte solution.
• Impurities in the solution can affect the efficiency of the deposition process.

Half-Reaction

A half-reaction in electrolysis is simply the part of the process that involves either the gain or loss of electrons. At the heart of copper deposition, the half-reaction represents the change where copper ions are reduced.
The specific half-reaction for copper deposition at the cathode can be represented as:\[\mathrm{Cu^{2+}(aq) + 2e^{-} \rightarrow Cu(s)}\]
This equation shows that two electrons are needed to convert one copper ion into solid copper. During electrolysis:

• The source of electrons is typically an external battery or power supply.
• The flow of electrons directly influences the speed at which reactions occur.

Understanding half-reactions helps us predict and calculate the amount of material deposited or transformed during electrolysis by using stoichiometric relationships.

Molar Mass Calculation

Molar mass is a fundamental concept used to relate the mass of a substance to the number of moles, which is crucial in calculating the result of chemical processes. When calculating how much copper is deposited in electrolysis, knowing the molar mass of copper is essential.
For copper, this value is:\[63.5 \, \text{g/mol}\]
To find how much copper will be deposited by 1 mole of electrons, first note that 1 mole of electrons will reduce half a mole of copper ions due to the requirement of 2 moles of electrons for each mole of copper. Therefore:

• The mass of deposited copper equals to 0.5 moles multiplied by the molar mass of copper.
• This calculation helps us determine that 31.75 g of copper is deposited.

Molar mass calculation is a straightforward way of converting moles to grams and is extremely useful in both chemistry and industry applications.