50 mL of 0.1 M `CuSO_4` solution is electrolysed using Pt electrodes with a current of 0.965 ampere for a period of 1 minute. Assuming that volume of solution does not change during electrolysis, calculate `[Cu^(2+)]. [H^+]` and `[SO_4^(2–)]` after electrolysis. What will be the concentration of each species if current is passed using Cu electrodes ?

Answer

To solve the problem step-by-step, we will analyze the electrolysis of the `CuSO4` solution and calculate the concentrations of the species involved after electrolysis.

### Step 1: Calculate the Total Charge (Q)
Given:
- Current (I) = 0.965 A
- Time (t) = 1 minute = 60 seconds

Using the formula for charge:
\[ Q = I \times t \]
\[ Q = 0.965 \, \text{A} \times 60 \, \text{s} = 57.9 \, \text{C} \]

### Step 2: Calculate the Number of Moles of Electrons (n)
Using Faraday's constant (F = 96500 C/mol):
\[ n = \frac{Q}{F} \]
\[ n = \frac{57.9 \, \text{C}}{96500 \, \text{C/mol}} \approx 0.0006 \, \text{mol} \]

### Step 3: Determine the Reaction at the Cathode
The reduction reaction at the cathode for `Cu^2+` is:
\[ Cu^{2+} + 2e^- \rightarrow Cu \]

From the reaction, 2 moles of electrons are required to reduce 1 mole of `Cu^2+`. Therefore, the moles of `Cu^2+` reduced can be calculated as:
\[ \text{Moles of } Cu^{2+} = \frac{n}{2} = \frac{0.0006}{2} = 0.0003 \, \text{mol} \]

### Step 4: Calculate Initial Moles of `Cu^2+`
The initial concentration of `CuSO4` is 0.1 M in a volume of 50 mL (0.050 L):
\[ \text{Initial moles of } Cu^{2+} = \text{Concentration} \times \text{Volume} = 0.1 \, \text{mol/L} \times 0.050 \, \text{L} = 0.005 \, \text{mol} \]

### Step 5: Calculate Final Moles of `Cu^2+`
The final moles of `Cu^{2+}` after electrolysis:
\[ \text{Final moles of } Cu^{2+} = \text{Initial moles} - \text{Moles reduced} \]
\[ \text{Final moles of } Cu^{2+} = 0.005 - 0.0003 = 0.0047 \, \text{mol} \]

### Step 6: Calculate Final Concentration of `Cu^{2+}`
The final concentration of `Cu^{2+}` is:
\[ [Cu^{2+}] = \frac{\text{Final moles of } Cu^{2+}}{\text{Volume}} = \frac{0.0047 \, \text{mol}}{0.050 \, \text{L}} = 0.094 \, \text{M} \]

### Step 7: Calculate the Concentration of `H^+`
At the anode, water is oxidized to produce `H^+` ions:
\[ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \]

From the reaction, 4 moles of electrons produce 4 moles of `H^+`. Therefore, the moles of `H^+` produced are equal to the moles of electrons:
\[ \text{Moles of } H^+ = n = 0.0006 \, \text{mol} \]

Calculating the concentration of `H^+`:
\[ [H^+] = \frac{0.0006 \, \text{mol}}{0.050 \, \text{L}} = 0.012 \, \text{M} \]

### Step 8: Calculate the Concentration of `SO_4^{2-}`
Since `SO_4^{2-}` ions are not consumed during the electrolysis, their concentration remains the same:
\[ [SO_4^{2-}] = 0.1 \, \text{M} \]

### Summary of Concentrations After Electrolysis
- \([Cu^{2+}] = 0.094 \, \text{M}\)
- \([H^+] = 0.012 \, \text{M}\)
- \([SO_4^{2-}] = 0.1 \, \text{M}\)