The relationship between standard reduction potential of a cell and equilibrium constant is shown by

Options

A. `E_("Cell")^(0)=0.059nlogK_c`
B. `E_("Cell")^(0)=(0.059)/(n)logK_c`
C. `E_("Cell")^(0)=(n)/(0.059)logK_c`
D. `E_("Cell")^(0)=(logK_c)/n`

Answer

To solve the question about the relationship between standard reduction potential of a cell and the equilibrium constant, we can follow these steps:

### Step 1: Understand the relationship
The relationship between the standard reduction potential (E°) of a cell and the equilibrium constant (K) can be derived from the Nernst equation. At equilibrium, the cell potential (E) is zero.

### Step 2: Write the Nernst equation
The Nernst equation is given by:
\[ E = E° - \frac{0.059}{n} \log Q \]
where:
- \( E \) is the cell potential,
- \( E° \) is the standard cell potential,
- \( n \) is the number of moles of electrons transferred in the reaction,
- \( Q \) is the reaction quotient.

### Step 3: Set the cell potential to zero at equilibrium
At equilibrium, \( E = 0 \). Therefore, we can rewrite the Nernst equation as:
\[ 0 = E° - \frac{0.059}{n} \log K \]
where \( K \) is the equilibrium constant.

### Step 4: Rearrange the equation
Rearranging the equation gives us:
\[ E° = \frac{0.059}{n} \log K \]

### Step 5: Identify the correct option
From the rearranged equation, we can see that the correct relationship is:
\[ E° = \frac{0.059}{n} \log K \]
This corresponds to option number 2 from the provided choices.

### Final Answer
The correct answer is:
**E° cell = 0.059 / n * log K**

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