Find the the reducation potential of the half-cell
`Pt(s)|Cu^(2+)(aq., 0.22 M), Cu^(+)(aq., 0.043 M)`

Options

A. `0.20 V`
B. `-0.20 V`
C. `0.30 V`
D. `-0.30 V`

Explanation

To find the reduction potential of the half-cell `Pt(s)|Cu^(2+)(aq., 0.22 M), Cu^(+)(aq., 0.043 M)`, we will use the Nernst equation. Here are the steps to solve the problem:

### Step 1: Identify the half-reaction
The half-reaction for the reduction process is:
\[ \text{Cu}^{2+} + e^- \rightarrow \text{Cu}^{+} \]

### Step 2: Determine the standard reduction potential (E°)
The standard reduction potential (E°) for the half-reaction is given as:
\[ E° = 0.158 \, \text{V} \]

### Step 3: Write the Nernst equation
The Nernst equation relates the cell potential (E) to the standard potential (E°) and the concentrations of the reactants and products:
\[ E = E° - \frac{RT}{nF} \ln Q \]

Where:
- \( R \) is the universal gas constant (8.314 J/(mol·K))
- \( T \) is the temperature in Kelvin (assume 298 K if not provided)
- \( n \) is the number of electrons transferred (1 for this half-reaction)
- \( F \) is Faraday's constant (96485 C/mol)
- \( Q \) is the reaction quotient

### Step 4: Calculate the reaction quotient (Q)
The reaction quotient \( Q \) for the half-reaction is given by:
\[ Q = \frac{[\text{Cu}^{+}]}{[\text{Cu}^{2+}]} \]
Substituting the concentrations:
\[ Q = \frac{0.043}{0.22} \]

### Step 5: Substitute values into the Nernst equation
Using \( n = 1 \), \( R = 8.314 \, \text{J/(mol·K)} \), \( T = 298 \, \text{K} \), and \( F = 96485 \, \text{C/mol} \):
\[ E = 0.158 - \frac{(8.314)(298)}{(1)(96485)} \ln\left(\frac{0.043}{0.22}\right) \]

### Step 6: Calculate the logarithmic term
Calculate \( \ln\left(\frac{0.043}{0.22}\right) \):
\[ \frac{0.043}{0.22} \approx 0.1955 \]
\[ \ln(0.1955) \approx -1.626 \]

### Step 7: Substitute the logarithmic value into the Nernst equation
Now substitute the logarithmic value back into the equation:
\[ E = 0.158 - \frac{(8.314)(298)}{(96485)} (-1.626) \]

### Step 8: Calculate the potential
Calculate the term:
\[ \frac{(8.314)(298)}{(96485)} \approx 0.008314 \times 298 / 96485 \approx 0.0257 \]
Now substitute:
\[ E = 0.158 + (0.0257)(1.626) \]
\[ E \approx 0.158 + 0.0419 \]
\[ E \approx 0.1999 \, \text{V} \]

### Step 9: Final answer
The reduction potential of the half-cell is approximately:
\[ E \approx 0.20 \, \text{V} \]