The electrode potentials for
` Cu^(2+) (aq) +e^(-) rarr Cu^+ (aq) `
and ` Cu^+ (aq) + e^(-) rarr Cu (s)`
are `+ 0.15 V` and ` +0. 50 V` repectively. The value of `E_(Cu^(2+)//Cu)^@` will be.

Options

A. (a) 0.325V
B. (b) 0.650V
C. (c) 0.150 V
D. (d) 0.500 V

Explanation

To find the standard electrode potential \( E^\circ_{Cu^{2+}/Cu} \), we will use the given half-reactions and their respective electrode potentials.

### Step 1: Write down the half-reactions and their potentials.
1. The first half-reaction is:
\[ Cu^{2+} (aq) + e^- \rightarrow Cu^+ (aq) \quad E^\circ_1 = +0.15 \, V \]
2. The second half-reaction is:
\[ Cu^+ (aq) + e^- \rightarrow Cu (s) \quad E^\circ_2 = +0.50 \, V \]

### Step 2: Combine the half-reactions.
To find the overall reaction from \( Cu^{2+} \) to \( Cu \), we need to add the two half-reactions. We can do this by ensuring that the electrons cancel out.

- The first half-reaction provides \( Cu^+ \) and consumes 1 electron.
- The second half-reaction consumes \( Cu^+ \) and also consumes 1 electron.

When we add these two half-reactions together, we get:
\[ Cu^{2+} (aq) + 2e^- \rightarrow Cu (s) \]

### Step 3: Relate the Gibbs free energy changes.
The Gibbs free energy change for the overall reaction can be expressed as:
\[ \Delta G^\circ = -n F E^\circ_{Cu^{2+}/Cu} \]
where \( n \) is the number of electrons transferred (which is 2 in this case) and \( F \) is Faraday's constant.

For the two half-reactions, we can write:
\[ \Delta G^\circ_1 = -n_1 F E^\circ_1 \]
\[ \Delta G^\circ_2 = -n_2 F E^\circ_2 \]
where \( n_1 = 1 \) and \( n_2 = 1 \).

### Step 4: Set up the equation.
Since the overall Gibbs free energy change is the sum of the individual changes:
\[ \Delta G^\circ = \Delta G^\circ_1 + \Delta G^\circ_2 \]
Substituting the expressions for Gibbs free energy:
\[ -n F E^\circ_{Cu^{2+}/Cu} = -n_1 F E^\circ_1 - n_2 F E^\circ_2 \]

### Step 5: Substitute values.
Substituting \( n = 2 \), \( n_1 = 1 \), and \( n_2 = 1 \):
\[ -2 F E^\circ_{Cu^{2+}/Cu} = -1 F (0.15) - 1 F (0.50) \]
This simplifies to:
\[ -2 E^\circ_{Cu^{2+}/Cu} = -0.15 - 0.50 \]
\[ -2 E^\circ_{Cu^{2+}/Cu} = -0.65 \]

### Step 6: Solve for \( E^\circ_{Cu^{2+}/Cu} \).
Dividing both sides by -2:
\[ E^\circ_{Cu^{2+}/Cu} = \frac{0.65}{2} = 0.325 \, V \]

### Final Answer:
Thus, the value of \( E^\circ_{Cu^{2+}/Cu} \) is:
\[ \boxed{0.325 \, V} \]