Consider the following equations for a cell reaction
`A+BiffC+D, E^(@)=x"volt",K_(eq)=K_1`
`2A+2Biff2C+2D, E^(@)=y"volt",K_(eq)=K_2` then:

Options

A. (a) `x=y,k_1+k_2`
B. (b) `x=2y,K_1=2K_2`
C. (c) `x=y,K_1^2=K_2`
D. (d) `x^2=y,K_1^2=K_2`

Answer

To solve the problem, we need to establish the relationship between the equilibrium constants \( K_1 \) and \( K_2 \) and the standard cell potentials \( E^0 \) for the two given reactions.

### Step 1: Write the reactions and their equilibrium constants
1. The first reaction is:
\[ A + B \iff C + D \]
- Standard cell potential: \( E^0 = x \)
- Equilibrium constant: \( K_1 \)

2. The second reaction is:
\[ 2A + 2B \iff 2C + 2D \]
- Standard cell potential: \( E^0 = y \)
- Equilibrium constant: \( K_2 \)

### Step 2: Write the expressions for the equilibrium constants
- For the first reaction:
\[ K_1 = \frac{[C][D]}{[A][B]} \]

- For the second reaction:
\[ K_2 = \frac{[C]^2[D]^2}{[A]^2[B]^2} \]

### Step 3: Relate \( K_2 \) to \( K_1 \)
Since the second reaction is just the first reaction multiplied by 2, we can express \( K_2 \) in terms of \( K_1 \):
\[ K_2 = (K_1)^2 \]

### Step 4: Use the Nernst equation to relate \( E^0 \) and \( K \)
The Nernst equation at equilibrium states:
\[ E^0 = \frac{0.0591}{n} \log K \]
where \( n \) is the number of moles of electrons transferred.

### Step 5: Apply the Nernst equation to both reactions
1. For the first reaction:
\[ x = \frac{0.0591}{n} \log K_1 \]

2. For the second reaction (where \( n' = 2n \)):
\[ y = \frac{0.0591}{2n} \log K_2 \]

### Step 6: Substitute \( K_2 \) into the equation for \( y \)
Substituting \( K_2 = (K_1)^2 \) into the equation for \( y \):
\[ y = \frac{0.0591}{2n} \log (K_1^2) \]
Using the logarithmic identity \( \log(a^b) = b \log a \):
\[ y = \frac{0.0591}{2n} \cdot 2 \log K_1 \]
This simplifies to:
\[ y = \frac{0.0591}{n} \log K_1 \]

### Step 7: Compare \( x \) and \( y \)
From the equations for \( x \) and \( y \):
\[ x = \frac{0.0591}{n} \log K_1 \]
\[ y = \frac{0.0591}{n} \log K_1 \]
Thus, we can conclude:
\[ x = y \]

### Final Result
The relationship we have established is:
- \( K_2 = K_1^2 \)
- \( x = y \)