A cell, `Ag|Ag^(o+)||Cu^(2+)|Cu` , initially contains `1 M Ag^(o+)` and `1M Cu^(2+)` ions. Calculate the change in the cell the potential after the passage of `9.65A` of current for `1h.`

Explanation

To solve the problem step by step, we will follow the outlined process to calculate the change in cell potential after the passage of a current through the electrochemical cell.

### Step 1: Identify the reactions and calculate the number of moles of electrons transferred.
The given cell is `Ag|Ag^(o+)||Cu^(2+)|Cu`. The half-reactions involved are:
- Reduction: \( Cu^{2+} + 2e^- \rightarrow Cu \)
- Oxidation: \( Ag \rightarrow Ag^{+} + e^- \)

From the reaction, we can see that 1 mole of copper ions (\( Cu^{2+} \)) gains 2 moles of electrons, while 1 mole of silver (\( Ag \)) loses 1 mole of electrons.

### Step 2: Calculate the total charge passed through the cell.
The current \( I \) is given as \( 9.65 A \) and the time \( t \) is \( 1 \, hour = 3600 \, seconds \).

Using the formula for charge:
\[ Q = I \times t = 9.65 \, A \times 3600 \, s = 34740 \, C \]

### Step 3: Calculate the number of moles of electrons transferred.
Using Faraday's constant \( F = 96500 \, C/mol \):
\[ n = \frac{Q}{F} = \frac{34740 \, C}{96500 \, C/mol} \approx 0.36 \, mol \]

### Step 4: Calculate the change in concentration of \( Cu^{2+} \) ions.
From the reaction, 1 mole of \( Cu^{2+} \) ions consumes 2 moles of electrons. Therefore, the moles of \( Cu^{2+} \) ions that reacted:
\[ \text{Moles of } Cu^{2+} = \frac{0.36 \, mol}{2} = 0.18 \, mol \]

The initial concentration of \( Cu^{2+} \) is \( 1 \, M \). After the reaction, the new concentration of \( Cu^{2+} \):
\[ \text{New concentration of } Cu^{2+} = 1 \, M - 0.18 \, M = 0.82 \, M \]

### Step 5: Calculate the change in concentration of \( Ag^{+} \) ions.
The moles of \( Ag^{+} \) ions produced will be equal to the moles of electrons transferred:
\[ \text{Moles of } Ag^{+} = 0.36 \, mol \]

The initial concentration of \( Ag^{+} \) is \( 1 \, M \). After the reaction, the new concentration of \( Ag^{+} \):
\[ \text{New concentration of } Ag^{+} = 1 \, M + 0.36 \, M = 1.36 \, M \]

### Step 6: Apply the Nernst equation to find the change in cell potential.
The Nernst equation is given by:
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \left( \frac{[Ag^{+}]}{[Cu^{2+}]} \right) \]

Assuming \( E^{\circ}_{cell} \) is the standard cell potential (which we need to know for a complete calculation), we can denote it as \( E^{\circ} \).

Substituting the concentrations:
\[ E_{cell} = E^{\circ} - \frac{0.0591}{2} \log \left( \frac{1.36}{\sqrt{0.82}} \right) \]

### Step 7: Calculate the change in potential \( \Delta E \).
\[ \Delta E = E_{cell} - E^{\circ}_{cell} = - \frac{0.0591}{2} \log \left( \frac{1.36}{\sqrt{0.82}} \right) \]

Calculating the logarithm:
\[ \Delta E \approx 0.0591 \times \frac{1}{2} \times \log \left( \frac{1.36}{0.905} \right) \]
\[ \Delta E \approx 0.0591 \times 0.5 \times 0.051 = 0.0010 \, V \]

### Final Answer:
The change in cell potential after the passage of \( 9.65 A \) for \( 1 h \) is approximately \( +0.0010 \, V \).