`E^@` values for the half cell reactions are given below :
`Cu^(2+) + e^(-) to Cu^(+) , E^@`=0.15 V
`Cu^(2+) + 2e^(-) to Cu, E^@`=0.34 V
What will be the `E^@` of the half-cell : `Cu^(+) + e^(-) to Cu` ?

Options

A. `+0.49 V`
B. `+0.19 V`
C. `+0.53V`
D. `+0.30V`

Explanation

To find the standard electrode potential \( E^\circ \) for the half-cell reaction \( \text{Cu}^+ + e^- \rightarrow \text{Cu} \), we can use the given half-cell reactions and their standard electrode potentials:

1. \( \text{Cu}^{2+} + e^- \rightarrow \text{Cu}^+ \), \( E^\circ = 0.15 \, \text{V} \)
2. \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \), \( E^\circ = 0.34 \, \text{V} \)

### Step 1: Write down the reactions and their potentials

We have the following reactions:
- Reaction 1: \( \text{Cu}^{2+} + e^- \rightarrow \text{Cu}^+ \) with \( E^\circ_1 = 0.15 \, \text{V} \)
- Reaction 2: \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \) with \( E^\circ_2 = 0.34 \, \text{V} \)

### Step 2: Reverse Reaction 1

To find the potential for the reaction \( \text{Cu}^+ + e^- \rightarrow \text{Cu} \), we need to reverse Reaction 1. When we reverse a reaction, the sign of \( E^\circ \) changes.

Reversed Reaction 1:
- \( \text{Cu}^+ \rightarrow \text{Cu}^{2+} + e^- \) with \( E^\circ = -0.15 \, \text{V} \)

### Step 3: Combine the reactions

Now we can combine the reversed Reaction 1 with Reaction 2:

1. Reversed Reaction 1: \( \text{Cu}^+ \rightarrow \text{Cu}^{2+} + e^- \) (with \( E^\circ = -0.15 \, \text{V} \))
2. Reaction 2: \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \) (with \( E^\circ = 0.34 \, \text{V} \))

When we add these two reactions, the \( \text{Cu}^{2+} \) cancels out:

\[ \text{Cu}^+ + e^- + \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}^{2+} + \text{Cu} \]

This simplifies to:

\[ \text{Cu}^+ + e^- \rightarrow \text{Cu} \]

### Step 4: Calculate the overall \( E^\circ \)

Now we can calculate the overall standard electrode potential \( E^\circ \) for the reaction \( \text{Cu}^+ + e^- \rightarrow \text{Cu} \):

\[ E^\circ = E^\circ_2 + E^\circ_{\text{reversed 1}} = 0.34 \, \text{V} + (-0.15 \, \text{V}) = 0.34 \, \text{V} - 0.15 \, \text{V} = 0.19 \, \text{V} \]

### Final Result

Thus, the standard electrode potential \( E^\circ \) for the half-cell reaction \( \text{Cu}^+ + e^- \rightarrow \text{Cu} \) is:

\[ E^\circ = 0.19 \, \text{V} \]