The standard e.m.f. of the cell`Zn| Zn^(2+) (0.01 M) || Fe^(2+) (0.001 M)`| Fe at 298 K is 0.02905 then the value of equilibrium constant for the cell reaction`10^(x)` . Find x.

Explanation

To solve the problem, we need to find the equilibrium constant \( K_{eq} \) for the cell reaction given the standard EMF of the cell. The cell reaction is represented as:

\[ \text{Zn} | \text{Zn}^{2+} (0.01 \, M) || \text{Fe}^{2+} (0.001 \, M) | \text{Fe} \]

### Step 1: Identify the half-reactions
In this cell, zinc is oxidized and iron is reduced. The half-reactions can be written as:

1. Oxidation:
\[ \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \]

2. Reduction:
\[ \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \]

### Step 2: Determine the number of electrons transferred (n)
From the half-reactions, we can see that 2 electrons are transferred in the overall reaction. Therefore, \( n = 2 \).

### Step 3: Use the Nernst equation
The Nernst equation relates the cell potential to the standard cell potential and the concentrations of the reactants and products:

\[ E = E^0 - \frac{0.0591}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]

For our cell, the standard EMF \( E^0 \) is given as \( 0.02905 \, V \).

### Step 4: Rearranging the Nernst equation
We can rearrange the Nernst equation to find the equilibrium constant \( K_{eq} \):

\[ E^0 = \frac{0.0591}{n} \log K_{eq} \]

### Step 5: Solve for \( K_{eq} \)
Substituting the known values into the equation:

\[ 0.02905 = \frac{0.0591}{2} \log K_{eq} \]

### Step 6: Isolate \( \log K_{eq} \)
First, multiply both sides by \( 2 \):

\[ 0.0581 = 0.0591 \log K_{eq} \]

Now, divide both sides by \( 0.0591 \):

\[ \log K_{eq} = \frac{0.0581}{0.0591} \]

### Step 7: Calculate \( \log K_{eq} \)
Calculating the right side:

\[ \log K_{eq} \approx 0.983 \]

### Step 8: Convert to \( K_{eq} \)
Since \( K_{eq} = 10^{\log K_{eq}} \):

\[ K_{eq} \approx 10^{0.983} \]

### Step 9: Find \( x \)
Since \( K_{eq} \) is expressed as \( 10^x \), we can see that:

\[ x \approx 0.983 \]

Thus, the final answer is:

\[ \boxed{0.983} \]