The nature of curve of `E_(cell)^(@)` vs. log `K_c` is :

Options

A. straight line
B. parabola
C. hyperbola
D. elliptical curve

Answer

To determine the nature of the curve of \( E_{cell}^\circ \) vs. \( \log K_c \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship**:
We start with the Nernst equation, which relates the cell potential to the reaction quotient:
\[ E_{cell} = E_{cell}^\circ - \frac{0.0591}{n} \log Q \]
Here, \( E_{cell} \) is the cell potential, \( E_{cell}^\circ \) is the standard cell potential, \( n \) is the number of moles of electrons exchanged, and \( Q \) is the reaction quotient.

2. **Identify \( Q \) at Equilibrium**:
At equilibrium, the reaction quotient \( Q \) becomes equal to the equilibrium constant \( K_c \). Therefore, we can rewrite the equation as:
\[ E_{cell} = E_{cell}^\circ - \frac{0.0591}{n} \log K_c \]

3. **Set \( E_{cell} \) to Zero**:
At equilibrium, \( E_{cell} \) becomes zero. Thus, we can set the equation to:
\[ 0 = E_{cell}^\circ - \frac{0.0591}{n} \log K_c \]

4. **Rearranging the Equation**:
Rearranging the equation gives us:
\[ E_{cell}^\circ = \frac{0.0591}{n} \log K_c \]

5. **Identify the Form of the Equation**:
The equation \( E_{cell}^\circ = \frac{0.0591}{n} \log K_c \) can be recognized as the equation of a straight line in the form \( y = mx \), where:
- \( y \) is \( E_{cell}^\circ \)
- \( x \) is \( \log K_c \)
- \( m \) (the slope) is \( \frac{0.0591}{n} \)

6. **Conclusion**:
Since the relationship between \( E_{cell}^\circ \) and \( \log K_c \) is linear, the nature of the curve of \( E_{cell}^\circ \) vs. \( \log K_c \) is a straight line.

### Final Answer:
The nature of the curve of \( E_{cell}^\circ \) vs. \( \log K_c \) is a **straight line**.