Standard Gibb's free energy change for a cell
`Ni_((s))|Ni_((aq))^(++)||Ag_((aq))^(+)|Ag`
`E_((cell))^(@) = 1.049 V`

Options

A. `-202.5 J`
B. 20245 J
C. `-202.5 kJ`
D. 2025 kJ

Explanation

To find the standard Gibbs free energy change (ΔG°) for the given electrochemical cell reaction, we can use the relationship between Gibbs free energy and the cell potential. The formula is:

\[ \Delta G^\circ = -nFE^\circ_{cell} \]

Where:
- \( \Delta G^\circ \) is the standard Gibbs free energy change,
- \( n \) is the number of moles of electrons transferred,
- \( F \) is the Faraday constant (approximately \( 96500 \, C/mol \)),
- \( E^\circ_{cell} \) is the standard cell potential.

### Step 1: Identify the half-reactions and determine \( n \)

In the given cell notation:
\[ \text{Ni}_{(s)} | \text{Ni}^{2+}_{(aq)} || \text{Ag}^{+}_{(aq)} | \text{Ag}_{(s)} \]

The half-reactions are:
1. \( \text{Ni} \rightarrow \text{Ni}^{2+} + 2e^- \) (oxidation)
2. \( \text{Ag}^{+} + e^- \rightarrow \text{Ag} \) (reduction)

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

### Step 2: Substitute values into the formula

Now we can substitute the known values into the Gibbs free energy formula:
- \( n = 2 \)
- \( F = 96500 \, C/mol \)
- \( E^\circ_{cell} = 1.049 \, V \)

Substituting these values gives:

\[ \Delta G^\circ = -nFE^\circ_{cell} = -2 \times 96500 \, C/mol \times 1.049 \, V \]

### Step 3: Calculate \( \Delta G^\circ \)

Now we perform the calculation:

\[ \Delta G^\circ = -2 \times 96500 \times 1.049 \]

Calculating this step-by-step:
1. Calculate \( 2 \times 96500 = 193000 \)
2. Then, calculate \( 193000 \times 1.049 = 202457 \, J \)

Thus,

\[ \Delta G^\circ = -202457 \, J \]

### Step 4: Convert to kilojoules

To express this in kilojoules, we divide by 1000:

\[ \Delta G^\circ = -202.457 \, kJ \]

Rounding this gives:

\[ \Delta G^\circ \approx -202.5 \, kJ \]

### Final Answer

Thus, the standard Gibbs free energy change for the cell reaction is:

\[ \Delta G^\circ \approx -202.5 \, kJ \]