An alloy of `Pb-Ag` weighing 54 mg was dissolved in desired amount of `HNO_(3)` & volume was made upto `500ml`. An Ag electrode was dipped in solution and then connected to standard hydrogen electrode anode. Then calculate `%` of Ag in alloy.
Given: `E_(cell) = 0.5V, E_(Ag^(+)//Ag)^(@) = 0.8V (2.303RT)/(F) = 0.06`

Explanation

To solve the problem of determining the percentage of silver (Ag) in a Pb-Ag alloy weighing 54 mg, we will use the Nernst equation and the provided data. Here’s a step-by-step breakdown of the solution:

### Step 1: Write the Nernst Equation
The Nernst equation is given by:
\[ E_{cell} = E^{\circ}_{cell} - \frac{0.06}{n} \log \left( \frac{[Ag^+]}{1} \right) \]
Where:
- \(E_{cell}\) is the cell potential (0.5 V)
- \(E^{\circ}_{cell}\) is the standard reduction potential for Ag (0.8 V)
- \(n\) is the number of electrons transferred in the reaction (1 for Ag)
- \([Ag^+]\) is the concentration of silver ions in the solution.

### Step 2: Substitute Known Values into the Nernst Equation
Substituting the known values into the Nernst equation:
\[ 0.5 = 0.8 - 0.06 \log \left( [Ag^+] \right) \]

### Step 3: Rearrange the Equation
Rearranging the equation to solve for \([Ag^+]\):
\[ 0.5 - 0.8 = -0.06 \log \left( [Ag^+] \right) \]
\[ -0.3 = -0.06 \log \left( [Ag^+] \right) \]
\[ \log \left( [Ag^+] \right) = \frac{-0.3}{-0.06} = 5 \]

### Step 4: Calculate the Concentration of Ag+
Taking the antilogarithm:
\[ [Ag^+] = 10^{-5} \text{ mol/L} \]

### Step 5: Calculate the Moles of Ag+ in 500 mL
Since the total volume of the solution is 500 mL (0.5 L):
\[ \text{Moles of } Ag^+ = [Ag^+] \times \text{Volume} = 10^{-5} \times 0.5 = 5 \times 10^{-6} \text{ mol} \]

### Step 6: Calculate the Mass of Ag
Using the molar mass of Ag (approximately 108 g/mol):
\[ \text{Mass of } Ag = \text{Moles} \times \text{Molar Mass} = 5 \times 10^{-6} \text{ mol} \times 108 \text{ g/mol} = 5.4 \times 10^{-4} \text{ g} = 0.54 \text{ mg} \]

### Step 7: Calculate the Percentage of Ag in the Alloy
To find the percentage of Ag in the alloy:
\[ \text{Percentage of } Ag = \left( \frac{\text{Mass of } Ag}{\text{Mass of Alloy}} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage of } Ag = \left( \frac{0.54 \text{ mg}}{54 \text{ mg}} \right) \times 100 = 1\% \]

### Final Answer
The percentage of silver (Ag) in the alloy is **1%**.

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