Consider the following equations for a cell reaction,
`A + B rarr C + D , E^(@) = x "volt", Delta G= Delta G_(1)`
`2A + 2B rarr 2C + 2D, E^(@) = y "volt", DeltaG = DeltaG_(2)`
Then,

Options

A. `x=y, DeltaG_(1) = DeltaG_(2)`
B. `x gt y, DeltaG_(1) gt DeltaG_(2)`
C. `x=y, DeltaG_(2) = 2DeltaG_(1)`
D. `x lt y, 2DeltaG_(2) = DeltaG_(1)`

Answer

To solve the problem, we need to analyze the relationships between the standard cell potentials (E°) and the Gibbs free energy changes (ΔG) for the given reactions.

### Step-by-Step Solution:

1. **Understanding the Reactions**:
- We have two reactions:
1. \( A + B \rightarrow C + D \) with \( E^\circ = x \) volts and \( \Delta G = \Delta G_1 \)
2. \( 2A + 2B \rightarrow 2C + 2D \) with \( E^\circ = y \) volts and \( \Delta G = \Delta G_2 \)

2. **Relationship Between E° and ΔG**:
- The relationship between the standard cell potential (E°) and Gibbs free energy (ΔG) is given by the equation:
\[ \Delta G = -nFE^\circ \]
where \( n \) is the number of moles of electrons transferred, and \( F \) is Faraday's constant.

3. **Analyzing the First Reaction**:
- For the first reaction \( A + B \rightarrow C + D \):
- Let \( n_1 \) be the number of moles of electrons transferred.
- Thus, we have:
\[ \Delta G_1 = -n_1 F x \]

4. **Analyzing the Second Reaction**:
- For the second reaction \( 2A + 2B \rightarrow 2C + 2D \):
- The number of moles of electrons transferred is now \( n_2 = 2n_1 \).
- Therefore:
\[ \Delta G_2 = -n_2 F y = -2n_1 F y \]

5. **Comparing E° Values**:
- Since the reactions are essentially the same but scaled by a factor of 2, the standard cell potentials will remain the same:
\[ x = y \]

6. **Comparing ΔG Values**:
- From the equations derived, we can relate ΔG values:
\[ \Delta G_2 = -2n_1 F y = -2 \left( -n_1 F x \right) = 2 \Delta G_1 \]
- Thus, we conclude:
\[ \Delta G_2 = 2 \Delta G_1 \]

7. **Final Relationships**:
- We have established that:
- \( x = y \)
- \( \Delta G_2 = 2 \Delta G_1 \)

### Conclusion:
From the analysis, we can conclude that:
- \( x = y \)
- \( \Delta G_2 = 2 \Delta G_1 \)

Thus, the correct option is that \( x = y \) and \( \Delta G_2 = 2 \Delta G_1 \).