Given : `E^(@) (Cu^(2+//Cu)` = 0.337V` and `E^(@) (Sn^(2+//Sn) = -0.136V`. Which of the following statements is correct?

Options

A. `Cu^(2+)` ions can be reduced by `H_(2)(g)`
B. `Cu` can be oxidized by `H^(+)`
C. `Sn^(2+)` ions can be reduced by `H_(2)(g)`
D. `Cu` can reduce `Sn^(2+)`

Explanation

To solve the question, we need to analyze the given standard reduction potentials for the two half-reactions:

1. \( E^\circ (Cu^{2+} / Cu) = 0.337 \, V \)
2. \( E^\circ (Sn^{2+} / Sn) = -0.136 \, V \)

### Step-by-Step Solution:

**Step 1: Understand the Standard Reduction Potentials**
- The standard reduction potential indicates the tendency of a species to gain electrons (be reduced). A higher (more positive) value means a greater tendency to be reduced.

**Step 2: Compare the Given Potentials**
- We compare the two given potentials:
- \( E^\circ (Cu^{2+} / Cu) = 0.337 \, V \)
- \( E^\circ (Sn^{2+} / Sn) = -0.136 \, V \)

**Step 3: Determine Which Species is Reduced**
- Since \( 0.337 \, V > -0.136 \, V \), it indicates that copper (Cu) has a higher tendency to gain electrons compared to tin (Sn). Therefore, Cu will be reduced, and Sn will be oxidized.

**Step 4: Identify the Anode and Cathode**
- In an electrochemical cell:
- The electrode where reduction occurs is called the cathode.
- The electrode where oxidation occurs is called the anode.
- Since Cu is reduced, it acts as the cathode.
- Consequently, Sn acts as the anode.

**Step 5: Conclusion**
- From the analysis, we can conclude that Cu can reduce \( H^+ \) ions (from \( H_2 \)) because it is a stronger oxidizing agent than Sn. Therefore, the correct statement is that Cu can reduce \( H^+ \) ions.

### Final Answer:
The correct statement is that Cu can reduce \( H^+ \) ions.