If `DeltaG` for the reaction is `A^(+) + B^(-) rarr A^(2+) + B^(2-) ` is x, the `DeltaG` for the reaction is
`(1)/(2)A^(+) + (1)/(2)B^(-) rarr (1)/(2) A^(2+) + (1)/(2) B^(2-)` is

Options

A. `(x)/(2)`
B. 2x
C. `x^(2)`
D. `sqrtx`

Answer

To solve the problem, we need to analyze the given reactions and their corresponding Gibbs free energy changes (ΔG).

### Step-by-Step Solution:

1. **Identify the Given Reaction and ΔG**:
The first reaction is:
\[ A^+ + B^- \rightarrow A^{2+} + B^{2-} \]
The change in Gibbs free energy (ΔG) for this reaction is given as \( x \).

2. **Understand the Relationship Between ΔG and Electrons**:
For any redox reaction, the relationship between ΔG and the cell potential (E_cell) is given by:
\[ \Delta G = -nFE_{cell} \]
where \( n \) is the number of moles of electrons transferred, \( F \) is Faraday's constant, and \( E_{cell} \) is the cell potential.

3. **Determine the Number of Electrons Transferred in the First Reaction**:
In the first reaction:
- \( A^+ \) is oxidized to \( A^{2+} \) (loses 1 electron).
- \( B^- \) is reduced to \( B^{2-} \) (gains 1 electron).

Therefore, the total number of electrons transferred \( n \) in this reaction is 1.

4. **Write the Expression for ΔG of the First Reaction**:
From the relationship, we have:
\[ \Delta G_1 = -1 \cdot F \cdot E_{cell} = x \]

5. **Analyze the Second Reaction**:
The second reaction is:
\[ \frac{1}{2}A^+ + \frac{1}{2}B^- \rightarrow \frac{1}{2}A^{2+} + \frac{1}{2}B^{2-} \]
Here, we need to determine the number of electrons transferred.

6. **Determine the Number of Electrons Transferred in the Second Reaction**:
In this case:
- \( \frac{1}{2} A^+ \) loses \( \frac{1}{2} \) electron.
- \( \frac{1}{2} B^- \) gains \( \frac{1}{2} \) electron.

Thus, the total number of electrons transferred \( n \) in this reaction is \( \frac{1}{2} \).

7. **Write the Expression for ΔG of the Second Reaction**:
Using the same relationship:
\[ \Delta G_2 = -\frac{1}{2} \cdot F \cdot E_{cell} \]

8. **Relate ΔG of the Second Reaction to the First**:
Since we know that \( \Delta G_1 = x \), we can express \( \Delta G_2 \) in terms of \( x \):
\[ \Delta G_2 = -\frac{1}{2} \cdot F \cdot E_{cell} = \frac{1}{2} \Delta G_1 = \frac{1}{2} x \]

9. **Final Answer**:
Therefore, the change in Gibbs free energy for the second reaction is:
\[ \Delta G_2 = \frac{x}{2} \]

### Conclusion:
The answer is \( \frac{x}{2} \).