For the reduction of silver ions with copper metal, the standard cell potential was foound to be `+0.46 V` at `25^(@) C`. The value of standard Gibbs energy, `DeltaG^(@)` will be `(F = 96,500C mol^(-1))`:

Options

A. –89.0 kJ
B. –89.0 J
C. –44.5 kJ
D. –98.0 kJ

Explanation

To calculate the standard Gibbs energy change (ΔG°) for the reduction of silver ions with copper metal, we will use the formula:

\[ \Delta G° = -nFE°_{cell} \]

Where:
- \( n \) = number of moles of electrons transferred in the reaction
- \( F \) = Faraday's constant (approximately \( 96,500 \, C \, mol^{-1} \))
- \( E°_{cell} \) = standard cell potential (given as \( +0.46 \, V \))

### Step 1: Identify the number of electrons transferred (n)
In the reaction where copper is oxidized to \( Cu^{2+} \) and silver ions \( Ag^{+} \) are reduced to silver metal \( Ag \), the balanced half-reactions are:

1. \( Cu \rightarrow Cu^{2+} + 2e^{-} \) (oxidation)
2. \( 2Ag^{+} + 2e^{-} \rightarrow 2Ag \) (reduction)

From the oxidation half-reaction, we see that 2 moles of electrons are transferred. Therefore, \( n = 2 \).

### Step 2: Substitute the values into the Gibbs energy formula
Now we can substitute the values into the Gibbs energy formula:

\[ \Delta G° = -nFE°_{cell} \]

Substituting \( n = 2 \), \( F = 96,500 \, C \, mol^{-1} \), and \( E°_{cell} = 0.46 \, V \):

\[ \Delta G° = -2 \times 96,500 \, C \, mol^{-1} \times 0.46 \, V \]

### Step 3: Calculate the value
Now we perform the calculation:

\[ \Delta G° = -2 \times 96,500 \times 0.46 \]
\[ \Delta G° = -2 \times 44,390 \]
\[ \Delta G° = -88,780 \, J/mol \]

### Step 4: Convert to kJ
To convert this value to kilojoules, we divide by 1000:

\[ \Delta G° = -88.78 \, kJ/mol \approx -89 \, kJ/mol \]

### Conclusion
Thus, the value of standard Gibbs energy \( \Delta G° \) is approximately \( -89 \, kJ/mol \).

### Final Answer
The correct option is \( -89 \, kJ/mol \).