The emf of the cell:
`Cu(s)|Cu^(2+) (1M)||Ag^(+)(1M)|Ag`
is 0.46V. The standard reduction potential for `Ag^(+)//Ag` is 0.80V. The standard reduction potential of `Cu^(2+)//Cu` is:

Options

A. `-0.34V`
B. 1.26V
C. 0.34V
D. `-1.26V`

Explanation

To find the standard reduction potential of the copper half-cell (Cu²⁺/Cu), we can use the information provided about the cell and the standard reduction potential of the silver half-cell (Ag⁺/Ag).

### Step-by-Step Solution:

1. **Identify the Components of the Cell:**
- The cell is represented as: `Cu(s)|Cu²⁺(1M)||Ag⁺(1M)|Ag(s)`.
- The given emf (electromotive force) of the cell is 0.46 V.
- The standard reduction potential for the Ag⁺/Ag half-cell is given as 0.80 V.

2. **Use the Formula for Cell EMF:**
- The formula for the emf of a galvanic cell is:
\[ E_{cell} = E_{reduction} - E_{oxidation} \]
- In this cell, Ag⁺ is being reduced (gaining electrons) and Cu is being oxidized (losing electrons).

3. **Assign the Potentials:**
- Let \( E^\circ_{Cu} \) be the standard reduction potential for the Cu²⁺/Cu half-cell.
- The reduction potential for Ag⁺/Ag is given as:
\[ E^\circ_{Ag} = 0.80 \, V \]

4. **Set Up the Equation:**
- According to the formula:
\[ E_{cell} = E^\circ_{Ag} - E^\circ_{Cu} \]
- Substituting the known values:
\[ 0.46 \, V = 0.80 \, V - E^\circ_{Cu} \]

5. **Solve for \( E^\circ_{Cu} \):**
- Rearranging the equation gives:
\[ E^\circ_{Cu} = 0.80 \, V - 0.46 \, V \]
- Calculating the right side:
\[ E^\circ_{Cu} = 0.34 \, V \]

6. **Conclusion:**
- The standard reduction potential for the Cu²⁺/Cu half-cell is:
\[ E^\circ_{Cu} = 0.34 \, V \]

### Final Answer:
The standard reduction potential of \( Cu^{2+}/Cu \) is **0.34 V**.