Calculate the number of moles of Copper that will be deposited, if 2 Faraday of electricity is passed through…

Analysis

To solve this problem, we need to look at the electrolysis of copper(II) tetraoxosulphate(VI) ($CuSO_4$) and apply Faraday’s Laws of Electrolysis. The goal is to determine how many moles of copper are deposited when a specific quantity of electricity (in Faradays) is passed through the solution.

Concept: Faraday’s Laws of Electrolysis

During electrolysis, the amount of substance deposited at an electrode is directly proportional to the quantity of electricity passed.

The reaction for the deposition of copper from a copper(II) ion solution is:

From this balanced half-equation, we can see that 2 moles of electrons (which is equivalent to 2 Faradays of electricity) are required to discharge 1 mole of copper metal.

Step-by-Step Solution

1. Identify the half-equation: $Cu^{2+} + 2e^- \rightarrow Cu$

2. Interpret the stoichiometry: The equation shows that 2 moles of electrons ($2F$) are needed to deposit 1 mole of Copper ($Cu$).

3. Apply the given data: The question states that 2 Faradays of electricity are passed through the electrolyte. Since $2F$ is exactly the amount required to deposit 1 mole of copper according to our balanced equation, the amount of copper deposited is 1 mole.

Conclusion

Since the electricity provided ($2F$) matches the stoichiometric requirement of the copper ion reduction ($2e^-$), the number of moles deposited is 1.0 mole.

Exam Tip: Always write out the half-equation for the metal being deposited. The coefficient of the electron ($e^-$) tells you exactly how many Faradays are needed to produce 1 mole of that metal. For example, if it were $Al^{3+} + 3e^- \rightarrow Al$, you would know it takes 3 Faradays to deposit 1 mole of Aluminum.