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 energies (ΔG) for the two given reactions.

### Step-by-Step Solution:

1. **Identify the Given Reactions:**
- Reaction 1: \( A + B \rightarrow C + D \)
- \( E^{\circ} = x \) volts
- \( \Delta G = \Delta G_1 \)
- Reaction 2: \( 2A + 2B \rightarrow 2C + 2D \)
- \( E^{\circ} = y \) volts
- \( \Delta G = \Delta G_2 \)

2. **Understanding the Relationship of E°:**
- The standard cell potential \( E^{\circ} \) is an intensive property, meaning it does not change with the amount of substance involved in the reaction.
- Since Reaction 2 is simply a scaled version of Reaction 1 (multiplied by 2), the standard cell potentials for both reactions must be equal:
\[ x = y \]

3. **Understanding the Relationship of ΔG:**
- The Gibbs free energy change \( \Delta G \) is an extensive property, meaning it depends on the amount of substance involved.
- For the second reaction, since we have doubled the coefficients, the change in Gibbs free energy will also double:
\[ \Delta G_2 = 2 \Delta G_1 \]

4. **Summarize the Relationships:**
- From the above analysis, we conclude:
- \( x = y \) (E° values are equal)
- \( \Delta G_2 = 2 \Delta G_1 \) (Gibbs free energy for the second reaction is double that of the first)

5. **Final Conclusion:**
- The correct relationships derived from the reactions are:
- \( E^{\circ} \) values are equal: \( x = y \)
- \( \Delta G_2 \) is twice \( \Delta G_1 \): \( \Delta G_2 = 2 \Delta G_1 \)