For the reaction taking place in the cell :
`Pt(s) H_2 (g) H^+ (aq) || Ag^+ (aq) |Ag(s)`
`E_("cell")^@ = + 0.5332 V`.
The value of `DeltafG^(ө ) ` is ______________ KJ `"mol"^(-1)` . (in nearest integer)

Explanation

To calculate the standard Gibbs free energy change (ΔfG°) for the given electrochemical cell reaction, we can use the relationship between the cell potential (E°cell) and Gibbs free energy:

\[ \Delta G° = -nFE°_{cell} \]

Where:
- \( n \) = number of moles of electrons transferred in the reaction
- \( F \) = Faraday's constant (approximately \( 96500 \, C/mol \))
- \( E°_{cell} \) = standard cell potential (given as \( +0.5332 \, V \))

### Step-by-Step Solution:

**Step 1: Identify the number of electrons transferred (n)**
In the given cell reaction:
\[ \frac{1}{2} H_2(g) + Ag^+(aq) \rightarrow Ag(s) + H^+(aq) \]
One mole of \( H_2 \) produces 2 moles of electrons (since it is oxidized), and one mole of \( Ag^+ \) accepts one mole of electrons. Thus, the total number of electrons transferred in this half-reaction is:
\[ n = 1 \]

**Step 2: Substitute values into the Gibbs free energy equation**
Using \( n = 1 \), \( F = 96500 \, C/mol \), and \( E°_{cell} = 0.5332 \, V \):
\[ \Delta G° = -nFE°_{cell} = -1 \times 96500 \, C/mol \times 0.5332 \, V \]

**Step 3: Calculate ΔG° in Joules**
Calculating the above expression:
\[ \Delta G° = -96500 \times 0.5332 = -51453.8 \, J/mol \]

**Step 4: Convert Joules to Kilojoules**
To convert the result from Joules to Kilojoules, divide by 1000:
\[ \Delta G° = -51453.8 \, J/mol \div 1000 = -51.4538 \, kJ/mol \]

**Step 5: Round to the nearest integer**
The problem asks for the answer in the nearest integer:
\[ \Delta G° \approx -51 \, kJ/mol \]

### Final Answer:
The value of \( \Delta G° \) is approximately **-51 kJ/mol**.