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

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

Step by step solution

01

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: $$\mathrm{Cu^{2+}(aq) + 2e^{-} \rightarrow Cu(s)}$$
02

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.
03

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) \(31.75 \mathrm{~g}\).

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