Standard electrode potentials are given as,
`E_(Cu^(2+)//Cd)^(@)=0.34" V " " and " E_(Ag^(+)//Ag)^(@)=0.80" V "`
Calculate the cell potential, E for cell containing 0.100 M `Ag^(+)` and 4.00 M `Cu^(2+)` at `25^(@)C`.

Explanation

To calculate the cell potential (E) for the given electrochemical cell, we will follow these steps:

### Step 1: Identify the half-reactions and their standard electrode potentials.

The standard electrode potentials provided are:
- For the reduction of silver:
\[ E^\circ_{\text{Ag}^+/\text{Ag}} = 0.80 \, \text{V} \]
- For the reduction of copper:
\[ E^\circ_{\text{Cu}^{2+}/\text{Cu}} = 0.34 \, \text{V} \]

### Step 2: Determine the overall cell reaction.

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

### Step 3: Calculate the standard cell potential (E°cell).

The standard cell potential can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Here, the cathode is where reduction occurs (Ag+ to Ag) and the anode is where oxidation occurs (Cu to Cu2+).

Thus,
\[ E^\circ_{\text{cell}} = E^\circ_{\text{Ag}^+/\text{Ag}} - E^\circ_{\text{Cu}^{2+}/\text{Cu}} = 0.80 \, \text{V} - 0.34 \, \text{V} = 0.46 \, \text{V} \]

### Step 4: Use the Nernst equation to calculate the cell potential under non-standard conditions.

The Nernst equation is given by:
\[ E = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \frac{[\text{products}]}{[\text{reactants}]} \]
Where:
- \( n \) is the number of moles of electrons transferred (which is 2 for this reaction).
- The concentration of products is \([Ag^+]^2\) and the concentration of reactants is \([Cu^{2+}]\).

Substituting the concentrations:
- \([Ag^+] = 0.100 \, \text{M}\)
- \([Cu^{2+}] = 4.00 \, \text{M}\)

The reaction gives:
\[ E = 0.46 \, \text{V} - \frac{0.0591}{2} \log \frac{(0.100)^2}{(4.00)} \]

### Step 5: Calculate the logarithmic term.

Calculating the ratio:
\[ \frac{(0.100)^2}{(4.00)} = \frac{0.0100}{4.00} = 0.0025 \]
Now, calculate the logarithm:
\[ \log(0.0025) \approx -2.6 \]

### Step 6: Substitute the logarithm back into the Nernst equation.

Now, substitute back into the Nernst equation:
\[ E = 0.46 \, \text{V} - \frac{0.0591}{2} \times (-2.6) \]
Calculating the second term:
\[ E = 0.46 \, \text{V} + 0.02955 \times 2.6 \approx 0.46 \, \text{V} + 0.0768 \, \text{V} \approx 0.5368 \, \text{V} \]

### Step 7: Final calculation for cell potential.

Thus, the final cell potential is:
\[ E \approx 0.46 \, \text{V} + 0.0768 \, \text{V} \approx 0.5368 \, \text{V} \]

### Conclusion

The cell potential \( E \) for the cell containing 0.100 M \( \text{Ag}^+ \) and 4.00 M \( \text{Cu}^{2+} \) at \( 25^\circ C \) is approximately:
\[ E \approx 0.54 \, \text{V} \]