Write the Nernst equation and calculate the value of `Delta G^(Theta)` for the galvanic cell
`Cu (s) |Cu^2+ (0.130 M) || Ag^(+) (1.00 xx 10^(-4) M) |Ag (s)`
Given `E_(Cu^(2+) |Cu)^(Theta) = 0.34 V , E_(Ag^(+)|Ag)^(Theta) = 0.80 V`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the half-reactions for the galvanic cell.
The galvanic cell consists of two half-reactions: oxidation at the anode and reduction at the cathode.

**Anode (Oxidation)**:
\[ \text{Cu (s)} \rightarrow \text{Cu}^{2+} (aq) + 2e^- \]

**Cathode (Reduction)**:
\[ \text{Ag}^{+} (aq) + e^- \rightarrow \text{Ag (s)} \]

Since we need to balance the number of electrons, we multiply the cathodic reaction by 2:

\[ 2\text{Ag}^{+} (aq) + 2e^- \rightarrow 2\text{Ag (s)} \]

### Step 2: Write the overall reaction.
Combining the two half-reactions gives us the overall reaction:

\[ \text{Cu (s)} + 2\text{Ag}^{+} (aq) \rightarrow \text{Cu}^{2+} (aq) + 2\text{Ag (s)} \]

### Step 3: Calculate the standard cell potential (E°cell).
The standard cell potential is calculated using the standard reduction potentials given:

- For Cu²⁺/Cu: \( E^\circ_{\text{Cu}^{2+}/\text{Cu}} = 0.34 \, \text{V} \) (oxidation potential will be -0.34 V)
- For Ag⁺/Ag: \( E^\circ_{\text{Ag}^{+}/\text{Ag}} = 0.80 \, \text{V} \)

Using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

Substituting the values:
\[ E^\circ_{\text{cell}} = 0.80 \, \text{V} - 0.34 \, \text{V} = 0.46 \, \text{V} \]

### Step 4: Write the Nernst equation.
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[\text{Products}]}{[\text{Reactants}]} \right) \]

Where:
- \( n = 2 \) (number of electrons transferred)
- Concentrations:
- \([\text{Cu}^{2+}] = 0.130 \, \text{M}\)
- \([\text{Ag}^{+}] = 1.00 \times 10^{-4} \, \text{M}\)

### Step 5: Substitute values into the Nernst equation.
Substituting the values into the Nernst equation:
\[ E_{\text{cell}} = 0.46 \, \text{V} - \frac{0.0591}{2} \log \left( \frac{(1)^2}{(0.130)(1.00 \times 10^{-4})^2} \right) \]

### Step 6: Calculate the logarithm term.
Calculating the logarithm term:
\[ \log \left( \frac{1}{0.130 \times (1.00 \times 10^{-4})^2} \right) = \log \left( \frac{1}{0.130 \times 1.00 \times 10^{-8}} \right) = \log \left( \frac{1}{1.30 \times 10^{-9}} \right) \approx 9.886 \]

### Step 7: Substitute back into the Nernst equation.
Now substituting back:
\[ E_{\text{cell}} = 0.46 \, \text{V} - \frac{0.0591}{2} \times 9.886 \]
\[ E_{\text{cell}} = 0.46 \, \text{V} - 0.292 \]
\[ E_{\text{cell}} \approx 0.168 \, \text{V} \]

### Step 8: Calculate ΔG° using the equation.
The Gibbs free energy change can be calculated using:
\[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \]

Where:
- \( F = 96500 \, \text{C/mol} \) (Faraday's constant)
- \( n = 2 \)

Substituting the values:
\[ \Delta G^\circ = -2 \times 96500 \times 0.46 \]
\[ \Delta G^\circ = -88880 \, \text{J} \]
\[ \Delta G^\circ \approx -88.88 \, \text{kJ} \]

### Final Answer:
\[ \Delta G^\circ \approx -88.88 \, \text{kJ} \]

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