The standard Gibbs energy for the given cell reaction is `-3.84 xx 10^(x)` J/mole at 298 K. The numerical value of x is______.
`Zn(s)+Cu^(2+)(aq) rarr Zn^(2+)(aq)+Cu(s)`
`E^(@)=2V ` at 298 K
(Faraday’s constant, `F = 96000 C "mol"^(-1)`)

Explanation

To find the numerical value of \( x \) in the standard Gibbs energy for the given cell reaction, we will follow these steps:

### Step 1: Write down the given cell reaction and identify the components.
The cell reaction is:
\[ \text{Zn(s)} + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu(s)} \]

### Step 2: Identify the standard cell potential \( E^0 \) and Faraday's constant \( F \).
From the question, we have:
- \( E^0 = 2 \, \text{V} \)
- \( F = 96000 \, \text{C/mol} \)

### Step 3: Determine the number of electrons transferred \( n \) in the reaction.
In the reaction:
- Zinc is oxidized:
\[ \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \]
- Copper is reduced:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]

Thus, \( n = 2 \) (the number of electrons transferred).

### Step 4: Use the formula for Gibbs free energy change.
The formula to calculate the standard Gibbs free energy change \( \Delta G^0 \) is:
\[ \Delta G^0 = -nFE^0 \]

### Step 5: Substitute the values into the equation.
Substituting the values we have:
\[ \Delta G^0 = -2 \times 96000 \, \text{C/mol} \times 2 \, \text{V} \]

### Step 6: Calculate \( \Delta G^0 \).
Calculating the above expression:
\[ \Delta G^0 = -2 \times 96000 \times 2 = -384000 \, \text{J/mol} \]
\[ \Delta G^0 = -3.84 \times 10^5 \, \text{J/mol} \]

### Step 7: Compare with the given expression.
The problem states that:
\[ \Delta G^0 = -3.84 \times 10^x \, \text{J/mol} \]

Comparing both expressions:
\[ -3.84 \times 10^5 = -3.84 \times 10^x \]

### Step 8: Solve for \( x \).
From the comparison, we can see that:
\[ x = 5 \]

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

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