The standard e.m.f. of a cell, invoving one electron change is found to be 0.591 V at `25^0C`. The equilibrium constant of the reaction is `( F= 96500 mol^-1, R=8.314 JK^-1mol^-1)`

Options

A. `1.0xx10^1`
B. `1.0xx10^5`
C. `1.0xx10^10`
D. `1.0xx10^30`

Answer

To solve the problem, we will use the relationship between the standard EMF of the cell (E°), the number of electrons transferred (n), the Faraday constant (F), the gas constant (R), and the equilibrium constant (Kc). The relevant equations are:

1. ΔG° = -nFE°
2. ΔG° = -2.303RT log Kc

### Step-by-Step Solution:

**Step 1: Write down the equations.**
We have two equations for the standard Gibbs free energy change (ΔG°):
- ΔG° = -nFE°
- ΔG° = -2.303RT log Kc

**Step 2: Set the equations equal to each other.**
Since both expressions equal ΔG°, we can set them equal to each other:
- -nFE° = -2.303RT log Kc

**Step 3: Cancel the negative signs.**
Removing the negative signs from both sides gives us:
- nFE° = 2.303RT log Kc

**Step 4: Substitute known values.**
Given:
- E° = 0.591 V
- n = 1 (since it involves one electron change)
- F = 96500 C/mol
- R = 8.314 J/(K·mol)
- T = 25°C = 298 K

Substituting these values into the equation:
- 1 × 96500 C/mol × 0.591 V = 2.303 × 8.314 J/(K·mol) × 298 K × log Kc

**Step 5: Calculate the left side.**
Calculating the left side:
- 96500 × 0.591 = 57073.5 J/mol

**Step 6: Calculate the right side.**
Now, calculate the right side:
- 2.303 × 8.314 × 298 = 5730.2 J/mol

So we have:
- 57073.5 = 5730.2 × log Kc

**Step 7: Solve for log Kc.**
Now, isolate log Kc:
- log Kc = 57073.5 / 5730.2

Calculating this gives:
- log Kc ≈ 9.95

**Step 8: Find Kc.**
To find Kc, we take the antilog:
- Kc = 10^(log Kc) = 10^(9.95) ≈ 1.12 × 10^10

### Final Answer:
Thus, the equilibrium constant Kc is approximately:
- Kc ≈ 1.12 × 10^10