Two Faraday of electricity is passed through a solution of $$CuSO_4$$. The mass of copper deposited at the cathode is: (Atomic mass of Cu = 63.5 amu)

First, we recall Faraday’s First Law of Electrolysis, which states:

$$\text{Mass deposited} = Z \times Q,$$

where $$Z$$ is the electro‐chemical equivalent and $$Q$$ is the total charge passed. The electro‐chemical equivalent $$Z$$ is further given by

$$Z = \dfrac{E}{F},$$

where $$E$$ is the equivalent weight of the ion being discharged and $$F$$ is one Faraday $$\left( F = 96\,500 \, \text{C} \right).$$ Hence we can also write

$$\text{Mass deposited} = \dfrac{E}{F} \times Q.$$

Now we identify the ion involved. In the electrolysis of $$CuSO_4$$, copper is present as $$Cu^{2+}$$. At the cathode the reduction reaction is

$$Cu^{2+} + 2e^- \rightarrow Cu.$$

From this half-reaction we see that two electrons are needed to deposit one atom of copper. Therefore the valency $$n$$ of the ion is $$2$$.

The equivalent weight $$E$$ is defined as

$$E = \dfrac{\text{Atomic (molar) mass}}{\text{valency}}.$$

Substituting the given atomic mass of copper, $$63.5 \, \text{g mol}^{-1},$$ and the valency $$2,$$ we get

$$E = \dfrac{63.5}{2} = 31.75 \, \text{g per equivalent}.$$

The question tells us that two Faradays of electricity are passed, so the total charge is

$$Q = 2F = 2 \times 96\,500 \, \text{C}.$$

Using the mass formula, we have

$$\text{Mass deposited} = \dfrac{E}{F} \times Q.$$

Substituting $$E = 31.75 \, \text{g}$$ and $$Q = 2F$$, we obtain

$$\text{Mass deposited} = \dfrac{31.75}{F} \times 2F.$$

Because $$F$$ appears in both numerator and denominator, it cancels out, leaving

$$\text{Mass deposited} = 31.75 \times 2 = 63.5 \, \text{g}.$$

So the cathode will receive a deposit of exactly one mole of copper, which is $$63.5 \, \text{g}$$.

Hence, the correct answer is Option C.