In a cell , the following reactions take place
`Fe^(2+)to Fe^(3+)+e^(-) " " E_(Fe^(3)//Fe^(2+))^(@)=0.77 V `
`2I^(-)toI_(2)+2e^(-) " " E_(I_(2)//I^(-))^(@)=0.54 V `
The standard electrode potentail for the spontaneous reaction in the cell is ` x xx 10^(-2)` V 298 K . The value of x is _________. (Nearest Integer)

Explanation

To solve the problem, we need to determine the standard electrode potential for the spontaneous reaction in the cell using the given half-reactions and their standard electrode potentials.

### Step-by-Step Solution:

1. **Identify the Half-Reactions and Their Potentials**:
- The first half-reaction is:
\[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^{-} \quad E^{\circ} = 0.77 \, \text{V} \]
- The second half-reaction is:
\[ 2\text{I}^{-} \rightarrow \text{I}_{2} + 2e^{-} \quad E^{\circ} = 0.54 \, \text{V} \]

2. **Determine the Cathode and Anode**:
- The half-reaction with the higher standard electrode potential will act as the cathode (reduction), and the one with the lower potential will act as the anode (oxidation).
- Here, \(E^{\circ}_{\text{Fe}^{3+}/\text{Fe}^{2+}} = 0.77 \, \text{V}\) (cathode) and \(E^{\circ}_{\text{I}_{2}/\text{I}^{-}} = 0.54 \, \text{V}\) (anode).

3. **Calculate the Standard Cell Potential**:
- The standard cell potential \(E^{\circ}_{\text{cell}}\) is calculated using the formula:
\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \]
- Substituting the values:
\[ E^{\circ}_{\text{cell}} = 0.77 \, \text{V} - 0.54 \, \text{V} = 0.23 \, \text{V} \]

4. **Convert to the Required Format**:
- The problem states that the standard electrode potential for the spontaneous reaction in the cell is in the form \(x \times 10^{-2} \, \text{V}\).
- We have:
\[ 0.23 \, \text{V} = 23 \times 10^{-2} \, \text{V} \]
- Thus, the value of \(x\) is \(23\).

### Final Answer:
The value of \(x\) is **23**.