Given `:`
`A^(2+)+2e^(-) rarr A(s)" "E^(c-)=0.08V`
`B^(o+)+e^(-)rarr B(s)" "E^(c-)=-0.64V`
`X_(2)(g)+2e^(-) rarr 2X^(c-)" "E^(c-)=1.03V`
Which of the following statements is `//` are correct ?

Options

A. `X_(2)(g)` will oxidize both `(A)` and `(B)`.
B. `A^(2+)` will oxidize `(B)`
C. The reaction
`2X^(c-)(1.0M)+A^(2+)(1.0M) rarr X_(2)(1 atm)+A(s)`
will be spontaneous.
D. The oxidizing power of `A^(2+),B^(o+),` and `X_(2)(g)` is in the order `X_(2)gtA^(2+)gtB^(o+)`

Explanation

To solve the problem, we need to analyze the given half-reactions and their standard reduction potentials. The goal is to determine which statements about the reactions are correct based on the reduction potentials provided.

### Step-by-Step Solution:

1. **Identify the Half-Reactions and Their Potentials:**
- For A: \( A^{2+} + 2e^- \rightarrow A(s) \) with \( E^\circ = 0.08 \, V \)
- For B: \( B^{+} + e^- \rightarrow B(s) \) with \( E^\circ = -0.64 \, V \)
- For X: \( X_2(g) + 2e^- \rightarrow 2X^{-} \) with \( E^\circ = 1.03 \, V \)

2. **Determine the Oxidizing and Reducing Agents:**
- The species with the highest reduction potential acts as the strongest oxidizing agent. Here, \( X_2 \) has the highest potential (1.03 V), followed by \( A^{2+} \) (0.08 V), and \( B^{+} \) (-0.64 V).
- Thus, \( X_2 \) will oxidize both \( A \) and \( B \).

3. **Evaluate Statement A:**
- Statement A claims that \( X_2 \) will oxidize both \( A \) and \( B \). Since \( X_2 \) has a higher reduction potential than both \( A^{2+} \) and \( B^{+} \), this statement is **correct**.

4. **Evaluate Statement B:**
- Statement B claims that \( A^{2+} \) can oxidize \( B \). Since \( A^{2+} \) (0.08 V) has a higher reduction potential than \( B^{+} \) (-0.64 V), \( A^{2+} \) can indeed oxidize \( B \). Therefore, this statement is also **correct**.

5. **Evaluate Statement C:**
- Statement C involves the reaction where \( X^{-} \) is oxidized and \( A^{2+} \) is reduced. The cell potential \( E_{cell} \) can be calculated as follows:
\[ E_{cell} = E^\circ_{reduction} - E^\circ_{oxidation} = 0.08 \, V - 1.03 \, V = -0.95 \, V \]
- Since \( E_{cell} \) is negative, the reaction is non-spontaneous. Thus, this statement is **incorrect**.

6. **Evaluate Statement D:**
- Statement D discusses the order of oxidizing power. The order based on reduction potentials is:
- \( X_2 \) (1.03 V) > \( A^{2+} \) (0.08 V) > \( B^{+} \) (-0.64 V)
- This order is correct, making this statement **correct**.

### Conclusion:
The correct statements are A, B, and D. Statement C is incorrect.