The standard electrode potential `E^(Theta)` and its temperature coefficent `((dE^Theta)/(d T))` for a cell are 2V and `-5xx10^(-4)VK^(-1)` at 300 K respectively. The cell reaction is `Zn(s)+Cu^(2+)(aq)to Zn^(2+)(aq) +Cu(s)`.
The standard reaction enthalpy `(Delta_(t)H^(Theta))` at 300 K is `4.12 xx10^(x) J//"mol"`. Numerical value of x is _________________ [Use `R=8JK^(-1)mol^(-1) and F=96,000 C "mol"^(-1)`].

Explanation

To solve the problem, we need to find the standard reaction enthalpy \((\Delta H^\Theta)\) at 300 K for the given cell reaction and express it in the form \(4.12 \times 10^x \, \text{J/mol}\).

### Step-by-Step Solution:

1. **Identify the given values**:
- Standard electrode potential \(E^\Theta = 2 \, \text{V}\)
- Temperature coefficient \(\frac{dE^\Theta}{dT} = -5 \times 10^{-4} \, \text{V/K}\)
- Temperature \(T = 300 \, \text{K}\)
- \(R = 8 \, \text{J/(K mol)}\)
- \(F = 96000 \, \text{C/mol}\)

2. **Determine the number of electrons transferred (n)**:
- From the cell reaction \( \text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s) \), we see that:
- Zinc is oxidized from 0 to +2, losing 2 electrons.
- Copper is reduced from +2 to 0, gaining 2 electrons.
- Therefore, \(n = 2\).

3. **Use the Gibbs free energy equation**:
- The relationship between Gibbs free energy change \((\Delta G^\Theta)\), enthalpy change \((\Delta H^\Theta)\), and entropy change \((\Delta S^\Theta)\) is given by:
\[ \Delta G^\Theta = \Delta H^\Theta - T \Delta S^\Theta \]
- We can express \(\Delta G^\Theta\) in terms of the cell potential:
\[ \Delta G^\Theta = -nFE^\Theta \]

4. **Substitute \(\Delta G^\Theta\) into the equation**:
- Rearranging the equation gives:
\[ \Delta H^\Theta = \Delta G^\Theta + T \Delta S^\Theta \]
- Since \(\Delta S^\Theta\) can be expressed using the temperature coefficient:
\[ \Delta S^\Theta = -nF \frac{dE^\Theta}{dT} \]

5. **Calculate \(\Delta H^\Theta\)**:
- Substitute the values into the equations:
\[ \Delta G^\Theta = -nFE^\Theta = -2 \times 96000 \times 2 \]
\[ \Delta G^\Theta = -384000 \, \text{J/mol} \]
- Now calculate \(\Delta S^\Theta\):
\[ \Delta S^\Theta = -nF \frac{dE^\Theta}{dT} = -2 \times 96000 \times (-5 \times 10^{-4}) \]
\[ \Delta S^\Theta = 96 \, \text{J/K mol} \]
- Now substitute back to find \(\Delta H^\Theta\):
\[ \Delta H^\Theta = -384000 + 300 \times 96 \]
\[ \Delta H^\Theta = -384000 + 28800 = -355200 \, \text{J/mol} \]

6. **Express \(\Delta H^\Theta\) in the required form**:
- Since the problem states that the standard reaction enthalpy is given as \(4.12 \times 10^x \, \text{J/mol}\), we take the absolute value:
\[ |\Delta H^\Theta| = 355200 \, \text{J/mol} = 3.552 \times 10^5 \, \text{J/mol} \]
- Therefore, \(x = 5\).

### Final Answer:
The numerical value of \(x\) is \(5\).