`(CU^(+) ((aq))` is unstable in solution and undergoes simultaneous oxidation and reduction, according to the reaction
`2Cu_((aq))^(+)` `Cu_((aq))^(2+) +Cu_((s))` choose `E^(@)` for the above reaction if
`E_(Cu^(2+)//Cu^(@))= 0.34V` and `E_(Cu^(2+)//Cu^(+) )= 0.15V`

Options

A. `(-0.38)`
B. `(+0.49)`
C. `(+0.38V)`
D. `(+0.19V)`

Explanation

To solve the problem, we need to find the standard electrode potential (E°) for the reaction:

\[ 2Cu^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + Cu_{(s)} \]

We are given the following standard electrode potentials:
- \( E_{Cu^{2+}/Cu} = 0.34 \, V \)
- \( E_{Cu^{2+}/Cu^{+}} = 0.15 \, V \)

### Step 1: Write the half-reactions
1. **Reduction of \( Cu^{2+} \) to \( Cu \)**:
\[ Cu^{2+} + 2e^{-} \rightarrow Cu \quad (E° = 0.34 \, V) \]

2. **Reduction of \( Cu^{2+} \) to \( Cu^{+} \)**:
\[ Cu^{2+} + e^{-} \rightarrow Cu^{+} \quad (E° = 0.15 \, V) \]

### Step 2: Determine the oxidation reaction
The oxidation reaction will be the reverse of the reduction of \( Cu^{+} \) to \( Cu^{2+} \):
\[ Cu^{+} \rightarrow Cu^{2+} + e^{-} \]
The standard potential for this oxidation reaction can be calculated as:
\[ E°_{Cu^{+}/Cu^{2+}} = -E°_{Cu^{2+}/Cu^{+}} = -0.15 \, V \]

### Step 3: Combine the half-reactions
Now we can combine the half-reactions to find the overall cell potential for the reaction:
\[ 2Cu^{+} \rightarrow Cu^{2+} + Cu \]

### Step 4: Calculate the overall cell potential
Using the formula for the cell potential:
\[ E°_{cell} = E°_{cathode} - E°_{anode} \]
In our case:
- \( E°_{cathode} = E°_{Cu^{2+}/Cu} = 0.34 \, V \)
- \( E°_{anode} = E°_{Cu^{+}/Cu^{2+}} = -0.15 \, V \)

Thus:
\[ E°_{cell} = 0.34 \, V - (-0.15 \, V) = 0.34 \, V + 0.15 \, V = 0.49 \, V \]

### Step 5: Final result
The standard electrode potential \( E° \) for the reaction \( 2Cu^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + Cu_{(s)} \) is:
\[ E° = 0.49 \, V \]