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 find the equilibrium constant (K) for the redox reaction of a galvanic cell with a standard EMF (E°) of 0.59 V and involving 2 moles of electrons, we can use the relationship between the standard EMF and the equilibrium constant given by the Nernst equation.

### Step-by-Step Solution:

1. **Understand the Relationship**: The relationship between the standard EMF (E°) of a cell and the equilibrium constant (K) is given by the equation:
\[ E° = \frac{RT}{nF} \ln K \]
where:
- \(E°\) = standard EMF of the cell (in volts)
- \(R\) = universal gas constant (8.314 J/(mol·K))
- \(T\) = temperature in Kelvin (assume standard temperature, 298 K)
- \(n\) = number of moles of electrons transferred in the reaction
- \(F\) = Faraday's constant (96485 C/mol)

2. **Substitute Known Values**: Given that \(E° = 0.59\) V and \(n = 2\), we can rearrange the equation to solve for \(K\):
\[ K = e^{\frac{nFE°}{RT}} \]

3. **Convert Constants**: Substitute the values of \(R\), \(F\), and \(T\):
- \(R = 8.314 \, \text{J/(mol·K)}\)
- \(F = 96485 \, \text{C/mol}\)
- \(T = 298 \, \text{K}\)

4. **Calculate the Exponent**:
\[ \frac{nFE°}{RT} = \frac{2 \times 96485 \times 0.59}{8.314 \times 298} \]
Calculate the numerator:
\[ = 2 \times 96485 \times 0.59 = 114,178.3 \, \text{J} \]
Calculate the denominator:
\[ = 8.314 \times 298 = 2477.572 \, \text{J} \]
Now, divide the two:
\[ \frac{114178.3}{2477.572} \approx 46.0 \]

5. **Calculate K**:
\[ K = e^{46.0} \]
Using a calculator:
\[ K \approx 10^{20.0} \]

### Final Answer:
The equilibrium constant \(K\) for the redox reaction of the cell is approximately:
\[ K \approx 10^{20} \]