`Zn|Zn^(2+)(a= 0.1M) || Fe^(2+) (a=0.01M)|Fe`
The emf of the above cell is 0.2905 V. Equilibrium constant for the cell reaction is:

Options

A. `10^(0.32//0.059)`
B. `10^(0.32//0.0295)`
C. `10^(0.26//0.0295)`
D. `10^(0.32//0.295)`

Explanation

To find the equilibrium constant for the given electrochemical cell reaction, we will follow these steps:

### Step 1: Identify the half-reactions and the overall cell reaction
The cell notation provided is:
\[ \text{Zn} | \text{Zn}^{2+} (a = 0.1M) || \text{Fe}^{2+} (a = 0.01M) | \text{Fe} \]

From this notation, we can identify the half-reactions:
- Oxidation at the anode:
...