Which is/are correct among the following? Given the half cell EMFs
`E_(Cu^(2+)//Cu)^(@)=0.337V, E_(Cu^(+)|Cu)^(@)=0.521V`

Options

A. `Cu^(+1)` disproportionates.
B. Cu and `Cu^(2+)` comproportionates (reverse of disproportionation into `Cu^(+)`).
C. `E_(Cu|Cu^(+2))^(0)+E_(Cu^(+1)|Cu)^(0)` is positive
D. All of these

Explanation

To determine which statements are correct regarding the given half-cell EMFs for copper, we will analyze the provided information step by step.

### Step 1: Understanding the Half-Cell Reactions
We have two half-cell reactions with their standard electrode potentials (E°):
1. \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \) with \( E° = 0.337 \, \text{V} \)
2. \( \text{Cu}^+ + e^- \rightarrow \text{Cu} \) with \( E° = 0.521 \, \text{V} \)

### Step 2: Identifying the Oxidation and Reduction Processes
- In the first half-cell, \( \text{Cu}^{2+} \) is reduced to copper (Cu), meaning it gains electrons.
- In the second half-cell, \( \text{Cu}^+ \) can either be oxidized to \( \text{Cu}^{2+} \) or reduced to copper (Cu).

### Step 3: Disproportionation of \( \text{Cu}^+ \)
Disproportionation occurs when a species is both oxidized and reduced in the same reaction. For \( \text{Cu}^+ \):
- It can be oxidized to \( \text{Cu}^{2+} \) (losing an electron).
- It can be reduced to Cu (gaining an electron).

Thus, the overall reaction can be written as:
\[ 2 \text{Cu}^+ \rightarrow \text{Cu}^{2+} + \text{Cu} \]
This confirms that \( \text{Cu}^+ \) does indeed undergo disproportionation.

### Step 4: Comproportionation of Cu and \( \text{Cu}^{2+} \)
Comproportionation is the reverse of disproportionation, where two different oxidation states combine to form a species of an intermediate oxidation state. In this case:
- \( \text{Cu} + \text{Cu}^{2+} \) can combine to form \( 2 \text{Cu}^+ \):
\[ \text{Cu} + \text{Cu}^{2+} \rightarrow 2 \text{Cu}^+ \]
This confirms that Cu and \( \text{Cu}^{2+} \) do indeed undergo comproportionation.

### Step 5: Calculating the Standard Electrode Potential for \( \text{Cu} \) to \( \text{Cu}^+ \)
To find the standard electrode potential for the reaction \( \text{Cu} \rightarrow \text{Cu}^+ + e^- \):
- We take the negative of the reduction potential for \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \):
\[ E°(\text{Cu} \rightarrow \text{Cu}^{2+}) = -0.337 \, \text{V} \]
- Then we add the potential for \( \text{Cu}^+ + e^- \rightarrow \text{Cu} \):
\[ E°(\text{Cu}^+ \rightarrow \text{Cu}) = 0.521 \, \text{V} \]
- Therefore, the overall potential for \( \text{Cu} \rightarrow \text{Cu}^+ \) is:
\[ E° = -0.337 + 0.521 = 0.184 \, \text{V} \]
This is a positive value, confirming that the reaction is feasible.

### Conclusion
All statements regarding the reactions of \( \text{Cu}^+ \) and the calculated potentials are correct. Therefore, the final answer is that all statements are correct.