The standard emf of a galvanic cell involving 2 motes of electrons in its redox reaction is 0.59 V the equilibriun constant for the redox reaction of the cell is

Options

A. `10^(20)`
B. `10^(5)`
C. 10
D. `10^(10)`

Explanation

To solve the problem of finding the equilibrium constant (K_eq) for a redox reaction in a galvanic cell with a standard EMF (E°) of 0.59 V and involving 2 moles of electrons, we can use the Nernst equation in its equilibrium form.

### Step-by-Step Solution:

1. **Understand the relationship between E° and K_eq**:
The relationship between the standard EMF of the cell (E°) and the equilibrium constant (K_eq) is given by the equation:
\[ E° = \frac{0.0591}{n} \log K_{eq} \]
where:
- E° is the standard EMF of the cell,
- n is the number of moles of electrons transferred in the redox reaction,
- K_eq is the equilibrium constant.

2. **Substitute the known values**:
From the problem, we know:
- E° = 0.59 V
- n = 2 moles of electrons

Plugging these values into the equation gives:
\[ 0.59 = \frac{0.0591}{2} \log K_{eq} \]

3. **Rearrange the equation to solve for log K_eq**:
Multiply both sides of the equation by 2:
\[ 2 \times 0.59 = 0.0591 \log K_{eq} \]
\[ 1.18 = 0.0591 \log K_{eq} \]

4. **Divide both sides by 0.0591**:
\[ \log K_{eq} = \frac{1.18}{0.0591} \]
Calculate the right side:
\[ \log K_{eq} \approx 19.97 \]

5. **Convert log K_eq to K_eq**:
To find K_eq, we take the antilog (base 10) of both sides:
\[ K_{eq} = 10^{19.97} \]

6. **Approximate K_eq**:
Since \(10^{19.97} \approx 10^{20}\), we conclude:
\[ K_{eq} \approx 10^{20} \]

### Final Answer:
The equilibrium constant (K_eq) for the redox reaction of the cell is approximately \(10^{20}\).

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