`Cu^(2+)(aq.)` is unstable in solution and under goes simultaneous oxidation and reduction according to the reaction
`2Cu^(+)(aq.)hArr Cu^(2+)(aq.)+Cu(s)`
Choose the correct `E^(@)` for the above reaction if
`E_(Cu^(2+))^(@)//Cu = 0.34 V and E_(Cu^(2+))^(@)//Cu^(+) = 0.15 V`

Options

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

Explanation

To solve the problem, we need to analyze the given electrochemical reactions and their standard reduction potentials. The overall reaction given is:

\[ 2Cu^{+}(aq.) \rightleftharpoons Cu^{2+}(aq.) + Cu(s) \]

### Step 1: Identify the half-reactions
We can break the overall reaction into two half-reactions:

1. **Oxidation half-reaction**:
\[ Cu^{+} \rightarrow Cu^{2+} + e^{-} \]
The standard reduction potential for this half-reaction can be derived from the given data. The reduction potential for \( Cu^{2+} + e^{-} \rightarrow Cu^{+} \) is \( E^\circ = 0.15 \, V \). Therefore, the oxidation potential for \( Cu^{+} \rightarrow Cu^{2+} + e^{-} \) is:
\[ E_{ox} = -0.15 \, V \]

2. **Reduction half-reaction**:
\[ Cu^{2+} + 2e^{-} \rightarrow Cu(s) \]
The standard reduction potential for this half-reaction is given as \( E^\circ = 0.34 \, V \).

### Step 2: Calculate the overall cell potential
To find the standard cell potential \( E^\circ_{cell} \) for the overall reaction, we use the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]

In our case:
- The cathode reaction (reduction) is \( Cu^{2+} + 2e^{-} \rightarrow Cu(s) \) with \( E^\circ_{cathode} = 0.34 \, V \).
- The anode reaction (oxidation) is \( Cu^{+} \rightarrow Cu^{2+} + e^{-} \) with \( E^\circ_{anode} = -0.15 \, V \).

So, we can calculate:
\[ E^\circ_{cell} = 0.34 \, V - (-0.15 \, V) \]
\[ E^\circ_{cell} = 0.34 \, V + 0.15 \, V \]
\[ E^\circ_{cell} = 0.49 \, V \]

### Step 3: Final answer
Thus, the correct standard cell potential \( E^\circ \) for the reaction is:
\[ E^\circ = 0.49 \, V \]