mf of the following cell at 298 K in V is `x""xx10^(-2).Zn|Zn^(2+)(0.1M)||Ag^(+)(0.01M)|Ag` The value of x is ______ Rounded off the the nearest integer)
Givne `E_(Zn^(+)//Zn)^(0)=-0.76V,E_(Ag^(+)//Ag)^(0)=+0.80V,(2.303RT)/(F)=0.059`]

Answer

To solve the given electrochemical cell problem, we will follow these steps:

### Step 1: Identify the half-reactions
The cell notation given is:
\[ \text{Zn} | \text{Zn}^{2+} (0.1 \, M) || \text{Ag}^{+} (0.01 \, M) | \text{Ag} \]

The half-reactions are:
- **Anode (oxidation)**:
\[ \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^{-} \]
- **Cathode (reduction)**:
\[ \text{Ag}^{+} + e^{-} \rightarrow \text{Ag} \]

### Step 2: Balance the overall reaction
To balance the electrons transferred, we multiply the cathode reaction by 2:
\[ 2 \text{Ag}^{+} + 2e^{-} \rightarrow 2 \text{Ag} \]

The overall reaction becomes:
\[ \text{Zn} + 2 \text{Ag}^{+} \rightarrow \text{Zn}^{2+} + 2 \text{Ag} \]

### Step 3: Calculate the standard cell potential (E°cell)
Using the standard reduction potentials:
- \( E^{\circ}_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \, V \)
- \( E^{\circ}_{\text{Ag}^{+}/\text{Ag}} = +0.80 \, V \)

The standard cell potential is calculated as:
\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \]
\[ E^{\circ}_{\text{cell}} = 0.80 \, V - (-0.76 \, V) = 0.80 \, V + 0.76 \, V = 1.56 \, V \]

### Step 4: Apply the Nernst equation
The Nernst equation is given by:
\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{2.303RT}{nF} \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Ag}^{+}]^2} \right) \]
Where:
- \( R = 8.314 \, J/(mol \cdot K) \)
- \( T = 298 \, K \)
- \( n = 2 \) (number of electrons transferred)
- \( F = 96500 \, C/mol \)

Given \( \frac{2.303RT}{F} = 0.059 \, V \) at 298 K, we can substitute this into the equation.

### Step 5: Substitute the concentrations
Substituting the values:
- \( [\text{Zn}^{2+}] = 0.1 \, M \)
- \( [\text{Ag}^{+}] = 0.01 \, M \)

The Nernst equation becomes:
\[ E_{\text{cell}} = 1.56 \, V - 0.059 \log \left( \frac{0.1}{(0.01)^2} \right) \]

Calculating the logarithm:
\[ \frac{0.1}{(0.01)^2} = \frac{0.1}{0.0001} = 1000 \]
Thus,
\[ \log(1000) = 3 \]

### Step 6: Calculate Ecell
Substituting back into the equation:
\[ E_{\text{cell}} = 1.56 \, V - 0.059 \cdot 3 \]
\[ E_{\text{cell}} = 1.56 \, V - 0.177 \, V \]
\[ E_{\text{cell}} \approx 1.383 \, V \]

### Step 7: Express in the required format
We need to express \( E_{\text{cell}} \) in the form \( x \times 10^{-2} \, V \):
\[ 1.383 \, V = 138.3 \times 10^{-2} \, V \]

### Final Answer
Rounding off \( 138.3 \) to the nearest integer gives:
\[ x = 138 \]