On passing C ampere of current for time t sec through 1 litre of 2(M) `CuSO_(4)` solution (atomic weight of Cu=63.5), the amount of m of Cu (in g) deposited on cathode will be:-

Options

A. `m=(Ct)/((63.5xx96500))`
B. `m=(Ct)/((31.25xx96500))`
C. `m(Cxx96500)/((31.25xxt))`
D. `m=(31.75xxCxxt)/(96500)`

Explanation

To find the amount of copper (Cu) deposited on the cathode when passing C ampere of current for time t seconds through a 1-liter solution of 2 M CuSO₄, we can use Faraday's laws of electrolysis. Here’s a step-by-step solution:

### Step 1: Determine the total charge (Q)
The total charge (Q) passed through the solution can be calculated using the formula:
\[ Q = I \times t \]
where:
- \( I \) is the current in amperes (C ampere),
- \( t \) is the time in seconds.

### Step 2: Calculate the number of moles of electrons (n)
Using Faraday's law, the number of moles of electrons (n) can be calculated by:
\[ n = \frac{Q}{F} \]
where \( F \) is Faraday's constant, approximately \( 96500 \, C/mol \).

### Step 3: Determine the moles of copper deposited
The reduction half-reaction for copper ions (Cu²⁺) at the cathode is:
\[ Cu^{2+} + 2e^- \rightarrow Cu \]
From this reaction, we see that 2 moles of electrons are required to deposit 1 mole of copper. Therefore, the moles of copper (Cu) deposited (moles of Cu) can be calculated as:
\[ \text{Moles of Cu} = \frac{n}{2} \]

### Step 4: Calculate the mass of copper deposited
To find the mass (m) of copper deposited, we can use the formula:
\[ m = \text{Moles of Cu} \times \text{Molar mass of Cu} \]
Given that the atomic weight of Cu is 63.5 g/mol, we can substitute:
\[ m = \left(\frac{n}{2}\right) \times 63.5 \]

### Step 5: Substitute the value of n
Substituting the expression for n from Step 2 into the mass equation:
\[ m = \left(\frac{Q}{F \times 2}\right) \times 63.5 \]

### Step 6: Substitute the value of Q
Now substituting \( Q = I \times t \):
\[ m = \left(\frac{I \times t}{F \times 2}\right) \times 63.5 \]

### Final Expression
Thus, the final expression for the mass of copper deposited is:
\[ m = \frac{C \times t \times 63.5}{2 \times 96500} \]