`50mL` of `0.1M CuSO_(4)` solution is electrolysed with a current of `0.965A` for a period of 200sec. The reactions at electrodes are:
Cathode: `Cu^(2+) +2e^(-) rarr Cu(s)`
Anode: `2H_(2)O rarr O_(2) +4H^(+) +4e`.
Assuming no change in volume during electrolysis, calculate the molar concentration of `Cu^(2+),H^(+)` and `SO_(4)^(2-)` at the end of electrolysis.

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the total charge (Q) passed through the solution
Using the formula:
\[ Q = I \times t \]
Where:
- \( I = 0.965 \, A \) (current)
- \( t = 200 \, s \) (time)

Calculating:
\[ Q = 0.965 \, A \times 200 \, s = 193 \, C \]

### Step 2: Calculate the number of moles of electrons (n)
Using Faraday's law:
\[ Q = n \times F \]
Where:
- \( F = 96500 \, C/mol \) (Faraday's constant)

Rearranging the formula to find \( n \):
\[ n = \frac{Q}{F} = \frac{193 \, C}{96500 \, C/mol} \approx 0.002 \, mol \]

### Step 3: Determine the moles of \( Cu^{2+} \) consumed
From the cathode reaction:
\[ Cu^{2+} + 2e^- \rightarrow Cu(s) \]
This indicates that 1 mole of \( Cu^{2+} \) consumes 2 moles of electrons. Therefore, the moles of \( Cu^{2+} \) consumed is:
\[ \text{Moles of } Cu^{2+} = \frac{n}{2} = \frac{0.002 \, mol}{2} = 0.001 \, mol \]

### Step 4: Calculate the initial moles of \( Cu^{2+} \)
The initial concentration of \( Cu^{2+} \) is given as \( 0.1 \, M \) and the volume is \( 50 \, mL \) (or \( 0.050 \, L \)):
\[ \text{Initial moles of } Cu^{2+} = \text{Concentration} \times \text{Volume} = 0.1 \, mol/L \times 0.050 \, L = 0.005 \, mol \]

### Step 5: Calculate the remaining moles of \( Cu^{2+} \)
\[ \text{Remaining moles of } Cu^{2+} = \text{Initial moles} - \text{Consumed moles} = 0.005 \, mol - 0.001 \, mol = 0.004 \, mol \]

### Step 6: Calculate the final concentration of \( Cu^{2+} \)
Using the remaining moles and the original volume:
\[ \text{Final concentration of } Cu^{2+} = \frac{\text{Remaining moles}}{\text{Volume}} = \frac{0.004 \, mol}{0.050 \, L} = 0.08 \, M \]

### Step 7: Calculate the concentration of \( H^+ \)
From the anode reaction:
\[ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \]
This indicates that 4 moles of \( H^+ \) are produced per 4 moles of electrons. Therefore, the moles of \( H^+ \) produced is equal to the moles of electrons:
\[ \text{Moles of } H^+ = n = 0.002 \, mol \]

### Step 8: Calculate the final concentration of \( H^+ \)
Using the moles of \( H^+ \) produced and the volume:
\[ \text{Final concentration of } H^+ = \frac{0.002 \, mol}{0.050 \, L} = 0.04 \, M \]

### Step 9: Determine the concentration of \( SO_4^{2-} \)
Since there is no change in the amount of \( SO_4^{2-} \) during the electrolysis, its concentration remains the same as the initial concentration:
\[ \text{Final concentration of } SO_4^{2-} = 0.1 \, M \]

### Summary of Final Concentrations:
- Final concentration of \( Cu^{2+} = 0.08 \, M \)
- Final concentration of \( H^+ = 0.04 \, M \)
- Final concentration of \( SO_4^{2-} = 0.1 \, M \)