An alloy of Pb-Ag weighing `1.08g` was dissolved in dilute `HNO_(3)` and the volume made to 100 mL.A ? Silver electrode was dipped in the solution and the emf of the cell dipped in the solution and the emf of the cell set-up as `Pt(s),H_(2)(g)|H^(+)(1M)||Ag^(+)(aq.)|Ag(s)` was `0.62 V` . If `E_("cell")^(@)` is `0.80 V`, what is the percentage of Ag in the alloy ? (At `25^(@)C, RT//F=0.06`)

Options

A. 25
B. `2.50`
C. 10
D. `1.0`

Explanation

To solve the problem step-by-step, we will follow these calculations:

### Step 1: Write the half-reactions
The half-reactions occurring at the electrodes are:
1. At the anode (oxidation):
\[ H_2(g) \rightarrow 2H^+(aq) + 2e^- \]
2. At the cathode (reduction):
\[ Ag^+(aq) + e^- \rightarrow Ag(s) \]

### Step 2: Write the overall cell reaction
Combining the half-reactions gives the overall cell reaction:
\[ H_2(g) + 2Ag^+(aq) \rightarrow 2H^+(aq) + 2Ag(s) \]

### Step 3: Use the Nernst equation
The Nernst equation is given by:
\[ E_{cell} = E_{cell}^0 - \frac{RT}{nF} \ln Q \]
Where:
- \(E_{cell}\) = 0.62 V (given)
- \(E_{cell}^0\) = 0.80 V (given)
- \(R = 8.314 \, J/(mol \cdot K)\)
- \(T = 298 \, K\) (25°C)
- \(n = 1\) (number of moles of electrons transferred)
- \(F = 96485 \, C/mol\)
- \(Q = \frac{[H^+]^2}{[Ag^+]^2}\) (reaction quotient)

### Step 4: Calculate \(RT/F\)
Using the value \(RT/F = 0.06\):
\[ E_{cell} = E_{cell}^0 - \frac{0.06}{n} \log Q \]
Substituting the known values:
\[ 0.62 = 0.80 - 0.06 \log Q \]

### Step 5: Solve for \(\log Q\)
Rearranging gives:
\[ 0.06 \log Q = 0.80 - 0.62 \]
\[ 0.06 \log Q = 0.18 \]
\[ \log Q = \frac{0.18}{0.06} = 3 \]

### Step 6: Calculate \(Q\)
Taking the antilogarithm:
\[ Q = 10^3 = 1000 \]
Thus,
\[ Q = \frac{[H^+]^2}{[Ag^+]^2} = \frac{(1)^2}{[Ag^+]^2} \quad \text{(since [H+] is 1M)} \]
This implies:
\[ 1000 = \frac{1}{[Ag^+]^2} \]
\[ [Ag^+]^2 = \frac{1}{1000} \]
\[ [Ag^+] = 10^{-3} \, M \]

### Step 7: Calculate moles of Ag
Using the concentration to find moles in 100 mL:
\[ \text{Moles of } Ag^+ = [Ag^+] \times \text{Volume (L)} = 10^{-3} \times 0.1 = 10^{-4} \, moles \]

### Step 8: Calculate mass of Ag
Using the molar mass of Ag (108 g/mol):
\[ \text{Mass of } Ag = \text{Moles} \times \text{Molar Mass} = 10^{-4} \times 108 = 0.0108 \, g \]

### Step 9: Calculate percentage of Ag in the alloy
The percentage of Ag in the alloy is given by:
\[ \text{Percentage of } Ag = \left( \frac{\text{Mass of } Ag}{\text{Mass of alloy}} \right) \times 100 = \left( \frac{0.0108}{1.08} \right) \times 100 \approx 1\% \]

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