EMF of a cell is related to standard EMF,`E^(@)` by the equation
`E=E^(@)-(0.0591)/(n)logk`
Where k is reaction quotient and n the volume of electrons which is correct graphical representation

Options

A.
B.
C.
D.

Explanation

To solve the problem, we need to analyze the equation given and understand how to represent it graphically. The equation provided is:

\[ E = E^\circ - \frac{0.0591}{n} \log K \]

Where:
- \( E \) is the EMF of the cell.
- \( E^\circ \) is the standard EMF.
- \( n \) is the number of electrons transferred in the reaction.
- \( K \) is the reaction quotient.

### Step-by-Step Solution:

1. **Identify the Variables:**
- From the equation, we can identify that \( E \) (EMF of the cell) is dependent on \( \log K \) (the logarithm of the reaction quotient).

2. **Rearranging the Equation:**
- The equation can be rearranged to fit the linear equation format \( Y = C - MX \):
\[ E = E^\circ - \left(\frac{0.0591}{n}\right) \log K \]
- Here, we can identify:
- \( Y \) (dependent variable) = \( E \) (EMF of the cell)
- \( C \) (intercept) = \( E^\circ \) (standard EMF)
- \( M \) (slope) = \(-\frac{0.0591}{n}\)
- \( X \) (independent variable) = \( \log K \)

3. **Determine the Slope:**
- The slope \( M \) is negative since it is represented as \(-\frac{0.0591}{n}\). This indicates that as \( \log K \) increases, \( E \) decreases.

4. **Graphical Representation:**
- On a graph, the Y-axis will represent the EMF of the cell (\( E \)), and the X-axis will represent \( \log K \).
- The graph will have a negative slope, indicating that as the reaction quotient \( K \) increases, the EMF of the cell decreases.

5. **Finding the Correct Option:**
- We need to identify the option that shows a negative slope. Based on the analysis:
- Option A: Positive slope
- Option B: Negative slope (correct option)
- Option C: Zero slope
- Option D: Positive slope
- Therefore, the correct option is **Option B**.