A cell `Cu|Cu^(2+)||Ag^(+)|Ag` inintially contains `2 M Ag^(+)` 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
a. − 0.003 V
b. − 0.02 V
c. − 0.04 V
d. None of these

Options

A. `-0.003V`
B. `-0.02V`
C. `-0.04V`
D. None of these

Explanation

To solve the problem, we need to determine the change in cell potential after a current of 1 A has been supplied for 96500 seconds in the electrochemical cell `Cu|Cu^(2+)||Ag^(+)|Ag`. Here are the steps to arrive at the solution:

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

### Step 2: Calculate the number of Faradays
Using the relation:
\[ \text{Number of Faradays} = \frac{Q}{F} \]
where \( F = 96500 \, \text{C/mol} \):
\[ \text{Number of Faradays} = \frac{96500 \, \text{C}}{96500 \, \text{C/mol}} = 1 \, \text{mol} \]

### Step 3: Determine the change in concentration of Cu²⁺ ions
At the anode, copper is oxidized:
\[ \text{Cu} \rightarrow \text{Cu}^{2+} + 2e^- \]
1 mole of Cu gives 2 moles of electrons. Therefore, 1 Faraday (1 mole of electrons) will oxidize:
\[ \frac{1 \, \text{mol}}{2} = 0.5 \, \text{mol} \, \text{of Cu} \]
Since the initial concentration of Cu²⁺ is 2 M in 1 L, the new concentration of Cu²⁺ after oxidation becomes:
\[ \text{New } [\text{Cu}^{2+}] = 2 \, \text{M} + 0.5 \, \text{mol} = 2.5 \, \text{M} \]

### Step 4: Determine the change in concentration of Ag⁺ ions
At the cathode, silver ions are reduced:
\[ \text{Ag}^{+} + e^- \rightarrow \text{Ag} \]
1 mole of Ag⁺ consumes 1 mole of electrons. Therefore, 1 Faraday will reduce:
\[ 1 \, \text{mol} \, \text{of Ag}^{+} \]
The initial concentration of Ag⁺ is 2 M in 1 L, so the new concentration of Ag⁺ after reduction becomes:
\[ \text{New } [\text{Ag}^{+}] = 2 \, \text{M} - 1 \, \text{mol} = 1 \, \text{M} \]

### Step 5: Apply the Nernst equation
The Nernst equation is given by:
\[ E = E^\circ - \frac{0.059}{n} \log \left( \frac{[\text{Ox}]}{[\text{Red}]} \right) \]
For the cell reaction:
\[ \text{Cu}^{2+} + 2\text{Ag} \rightarrow \text{Cu} + 2\text{Ag}^{+} \]
Here, \( n = 2 \) (2 electrons involved).

The initial cell potential \( E^\circ \) is not given but we are interested in the change in potential:
\[ \Delta E = E_2 - E_1 \]

Calculating for \( E_1 \) (initial state):
\[ E_1 = E^\circ - \frac{0.059}{2} \log \left( \frac{2}{2} \right) = E^\circ \]

Calculating for \( E_2 \) (after current):
\[ E_2 = E^\circ - \frac{0.059}{2} \log \left( \frac{2.5}{1} \right) \]

### Step 6: Calculate the change in potential
\[ \Delta E = E_2 - E_1 = -\frac{0.059}{2} \log(2.5) \]
Using \( \log(2.5) \approx 0.39794 \):
\[ \Delta E = -\frac{0.059}{2} \times 0.39794 \approx -0.0117 \, \text{V} \]

### Step 7: Final calculation
This value is approximately:
\[ \Delta E \approx -0.02 \, \text{V} \]

### Conclusion
Thus, the change in cell potential after supplying the current is approximately:
\[ \Delta E \approx -0.02 \, \text{V} \]
The correct answer is **b. −0.02 V**.