A cell Ag|`Ag^+||Cu^(++)|Cu` initially contains `2M Ag^+` and `2M Cu^(++)` ions. The change in cell potential after the passage of 10 amp current for 4825 sec is:

Options

A. `-0.0074 V`
B. `-1.00738 V`
C. `-0.0038 V`
D. none

Explanation

To solve the problem, we need to calculate the change in cell potential after passing a current through the electrochemical cell. Here’s a step-by-step breakdown of the solution:

### Step 1: Write the half-reaction equations
The cell consists of the following half-reactions:
- Oxidation: \( \text{Ag}^+ + e^- \rightarrow \text{Ag} \)
- Reduction: \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \)

The overall cell reaction can be written as:
\[ \text{Ag}^+ + \frac{1}{2} \text{Cu}^{2+} + e^- \rightarrow \text{Ag} + \frac{1}{2} \text{Cu} \]

### Step 2: Calculate the initial cell potential (E1)
Using the Nernst equation:
\[ E = E^\circ - \frac{0.059}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]

Where:
- \( E^\circ \) is the standard cell potential (not given, but will cancel out later).
- \( n \) is the number of electrons transferred (1 for Ag and 2 for Cu, but we will use the effective n for the overall reaction, which is 1).
- Initial concentrations: \( [\text{Ag}^+] = 2 \, M \) and \( [\text{Cu}^{2+}] = 2 \, M \).

Substituting into the Nernst equation:
\[ E_1 = E^\circ - \frac{0.059}{1} \log \left( \frac{2}{(2)^{1/2}} \right) \]
\[ E_1 = E^\circ - 0.059 \log(2^{1/2}) \]
\[ E_1 = E^\circ - 0.059 \cdot \frac{1}{2} \log(2) \]

### Step 3: Calculate the amount of substance changed after current passage
Using Faraday's law of electrolysis:
\[ W = \frac{E}{96500} \cdot I \cdot T \]

Where:
- \( I = 10 \, A \)
- \( T = 4825 \, s \)

Calculating the total charge:
\[ Q = I \cdot T = 10 \, A \cdot 4825 \, s = 48250 \, C \]

Now, using the number of moles of electrons transferred:
\[ n = \frac{Q}{F} = \frac{48250}{96500} = 0.5 \, moles \]

### Step 4: Calculate the change in concentration
For \( \text{Cu}^{2+} \):
- \( n = 2 \) (2 electrons for Cu)
- Change in concentration: \( \Delta [\text{Cu}^{2+}] = \frac{0.5}{2} = 0.25 \, M \)

For \( \text{Ag}^+ \):
- \( n = 1 \) (1 electron for Ag)
- Change in concentration: \( \Delta [\text{Ag}^+] = 0.5 \, M \)

New concentrations:
- \( [\text{Cu}^{2+}] = 2 - 0.25 = 1.75 \, M \)
- \( [\text{Ag}^+] = 2 + 0.5 = 2.5 \, M \)

### Step 5: Calculate the final cell potential (E2)
Using the Nernst equation again:
\[ E_2 = E^\circ - \frac{0.059}{1} \log \left( \frac{2.5}{(1.75)^{1/2}} \right) \]

### Step 6: Calculate the change in cell potential (ΔE)
\[ \Delta E = E_2 - E_1 \]
Substituting the expressions for \( E_2 \) and \( E_1 \):
\[ \Delta E = \left(E^\circ - 0.059 \log \left( \frac{2.5}{(1.75)^{1/2}} \right)\right) - \left(E^\circ - 0.059 \log \left( \frac{2}{(2)^{1/2}} \right)\right) \]
The \( E^\circ \) terms cancel out:
\[ \Delta E = 0.059 \left( \log \left( \frac{2}{(2)^{1/2}} \right) - \log \left( \frac{2.5}{(1.75)^{1/2}} \right) \right) \]

### Step 7: Calculate the numerical values
Using logarithmic properties:
\[ \Delta E = 0.059 \left( \log(2) - \log(2.5) + \frac{1}{2} \log(1.75) \right) \]
Calculating the logarithms and substituting will yield:
\[ \Delta E \approx -0.0074 \, V \]

### Final Answer
The change in cell potential after the passage of 10 amp current for 4825 seconds is approximately:
\[ \Delta E \approx -0.0074 \, V \]