A cell `Cu|Cu^(2+)||Ag^(o+)|Ag` inintially contains `2 M Ag^(o+)` and `2 M Cu^(2+)` ion in `1L` solution each . The change in cell potential after it has supplied `1A` current for `96500 s` is

Options

A. `(-0.003V)`
B. `(-0.0266V)`
C. `(-0.04V)`
D. None of these

Explanation

To solve the problem step by step, we need to determine the change in cell potential after a current of 1 A has been supplied for 96,500 seconds in the given electrochemical cell.

### Step 1: Calculate the total charge passed
Using the formula for charge (Q):
\[ Q = I \times t \]
Where:
- \( I = 1 \, \text{A} \) (current)
- \( t = 96500 \, \text{s} \)

Calculating:
\[ Q = 1 \, \text{A} \times 96500 \, \text{s} = 96500 \, \text{C} \]

### Step 2: Calculate the number of Faradays (n)
Using Faraday's constant (\( F = 96500 \, \text{C/mol} \)):
\[ n = \frac{Q}{F} \]
Calculating:
\[ n = \frac{96500 \, \text{C}}{96500 \, \text{C/mol}} = 1 \, \text{mol} \]

### Step 3: Determine the change in concentration of \( \text{Cu}^{2+} \)
At the anode, the reaction is:
\[ \text{Cu} \rightarrow \text{Cu}^{2+} + 2e^- \]
From the reaction, 1 mole of Cu produces 1 mole of \( \text{Cu}^{2+} \) for every 2 Faradays. Since we have 1 Faraday:
- Moles of \( \text{Cu}^{2+} \) produced = \( \frac{1}{2} \, \text{mol} \)

Initial concentration of \( \text{Cu}^{2+} \):
- Initial concentration = 2 M in 1 L = 2 moles

After the reaction:
\[ \text{Final concentration of } \text{Cu}^{2+} = 2 + 0.5 = 2.5 \, \text{M} \]

### Step 4: Determine the change in concentration of \( \text{Ag}^{+} \)
At the cathode, the reaction is:
\[ \text{Ag}^{+} + e^- \rightarrow \text{Ag} \]
From the reaction, 1 mole of \( \text{Ag}^{+} \) is consumed for every 1 Faraday. Since we have 1 Faraday:
- Moles of \( \text{Ag}^{+} \) consumed = 1 mol

Initial concentration of \( \text{Ag}^{+} \):
- Initial concentration = 2 M in 1 L = 2 moles

After the reaction:
\[ \text{Final concentration of } \text{Ag}^{+} = 2 - 1 = 1 \, \text{M} \]

### Step 5: Calculate the cell potential (E_cell)
Using the Nernst equation:
\[ E_{\text{cell}} = E^{\circ} - \frac{0.059}{n} \log \left( \frac{[\text{Cu}^{2+}]}{[\text{Ag}^{+}]^2} \right) \]

For the first calculation (initial conditions):
\[ E_{\text{cell1}} = E^{\circ} - \frac{0.059}{2} \log \left( \frac{2}{2^2} \right) \]
\[ E_{\text{cell1}} = E^{\circ} - \frac{0.059}{2} \log \left( \frac{2}{4} \right) \]
\[ E_{\text{cell1}} = E^{\circ} - \frac{0.059}{2} \log \left( 0.5 \right) \]

For the second calculation (after the reaction):
\[ E_{\text{cell2}} = E^{\circ} - \frac{0.059}{2} \log \left( \frac{2.5}{1^2} \right) \]
\[ E_{\text{cell2}} = E^{\circ} - \frac{0.059}{2} \log \left( 2.5 \right) \]

### Step 6: Calculate the change in cell potential
The change in cell potential is given by:
\[ \Delta E_{\text{cell}} = E_{\text{cell2}} - E_{\text{cell1}} \]
\[ \Delta E_{\text{cell}} = \left( E^{\circ} - \frac{0.059}{2} \log(2.5) \right) - \left( E^{\circ} - \frac{0.059}{2} \log(0.5) \right) \]
\[ \Delta E_{\text{cell}} = - \frac{0.059}{2} \left( \log(2.5) - \log(0.5) \right) \]
Using properties of logarithms:
\[ \Delta E_{\text{cell}} = - \frac{0.059}{2} \log \left( \frac{2.5}{0.5} \right) \]
\[ \Delta E_{\text{cell}} = - \frac{0.059}{2} \log(5) \]

### Step 7: Final calculation
Calculating the numerical value:
Assuming \( \log(5) \approx 0.699 \):
\[ \Delta E_{\text{cell}} = - \frac{0.059}{2} \times 0.699 \]
\[ \Delta E_{\text{cell}} \approx -0.0206 \, \text{V} \]

### Final Answer
The change in cell potential after supplying the current is approximately:
\[ \Delta E_{\text{cell}} \approx -0.02 \, \text{V} \]