A current of `2.68 A` is passed for `1.0` hour through an aqueous solution of `CuSO_(4)` using copper electrodes.
Which of the following statements is `//` are correct ?

Options

A. Increase in the mass of cathode `=3.174g`
B. Decrease in the mass of anode `=3.174g`
C. No change in the mass of electrodes
D. The ration between the change in the mass of cathode to anode is `1:2`

Explanation

To solve the problem step by step, we will analyze the electrochemical process occurring when a current is passed through an aqueous solution of copper sulfate (CuSO₄) using copper electrodes.

### Step 1: Calculate the total charge passed
We know that:
- Current (I) = 2.68 A
- Time (t) = 1 hour = 3600 seconds

The total charge (Q) passed can be calculated using the formula:
\[ Q = I \times t \]
\[ Q = 2.68 \, \text{A} \times 3600 \, \text{s} = 9648 \, \text{C} \]

### Step 2: Calculate the number of Faradays
1 Faraday (F) = 96500 C. Therefore, the number of Faradays (n) can be calculated as:
\[ n = \frac{Q}{F} = \frac{9648 \, \text{C}}{96500 \, \text{C/F}} \approx 0.1 \, \text{F} \]

### Step 3: Determine the amount of copper deposited at the cathode
At the cathode, copper ions (Cu²⁺) are reduced to copper metal (Cu). The half-reaction is:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]

This means that 2 Faradays are required to deposit 1 mole of copper. Therefore, the moles of copper deposited (moles of Cu) can be calculated as:
\[ \text{moles of Cu} = \frac{n}{2} = \frac{0.1 \, \text{F}}{2} = 0.05 \, \text{moles} \]

### Step 4: Calculate the mass of copper deposited
The molar mass of copper (Cu) is approximately 63.5 g/mol. Thus, the mass of copper deposited can be calculated as:
\[ \text{mass of Cu} = \text{moles of Cu} \times \text{molar mass of Cu} \]
\[ \text{mass of Cu} = 0.05 \, \text{moles} \times 63.5 \, \text{g/mol} = 3.175 \, \text{g} \]

### Step 5: Determine the change in mass at the anode
At the anode, copper metal (Cu) is oxidized to copper ions (Cu²⁺). The half-reaction is:
\[ \text{Cu} \rightarrow \text{Cu}^{2+} + 2e^- \]

The same amount of copper that is deposited at the cathode will be oxidized at the anode. Therefore, the mass of copper lost at the anode is also:
\[ \text{mass of Cu lost} = 3.175 \, \text{g} \]

### Step 6: Analyze the statements
1. **Increase in the mass of cathode = 3.175 g** - Correct.
2. **Decrease in the mass of anode = 3.175 g** - Correct.
3. **No change in the mass of electrodes** - Incorrect (there is a change).
4. **Ratio between the change in the mass of cathode to anode = 1:2** - Incorrect (the ratio is 1:1).

### Conclusion
The correct statements are:
- Increase in the mass of cathode = 3.175 g
- Decrease in the mass of anode = 3.175 g