Question

An alloy of Pb-Ag weighing 54 mg was dissolved in desired amount of HNO3 & 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 : $E_{cell}$ = 0.5 V; $E^{\circ}_{Ag^{+}|Ag}$ = 0.8V, $\frac{2.303RT}{F}$=0.06

Answer 1

1%

Answer 2

The problem involves an electrochemical cell formed by an Ag electrode and a Standard Hydrogen Electrode (SHE). We need to use the Nernst equation to determine the concentration of Ag$^+$ ions in the solution, and then calculate the mass of Ag in the alloy to find its percentage.

1. Identify the Electrochemical Cell and Half-Reactions: The cell consists of a Standard Hydrogen Electrode (SHE) acting as the anode and an Ag electrode acting as the cathode.

• Anode (Oxidation): Standard Hydrogen Electrode $H_2(g) \rightarrow 2H^+(aq) + 2e^-$ (For SHE, $[H^+] = 1 M$ and $P_{H_2} = 1 \text{ atm}$, $E^\circ_{H^+|H_2} = 0 V$)
• Cathode (Reduction): Silver Electrode $Ag^+(aq) + e^- \rightarrow Ag(s)$ ($E^\circ_{Ag^+|Ag} = 0.8 V$)

2. Write the Overall Cell Reaction and Determine 'n': To balance the electrons, multiply the cathode reaction by 2: $2Ag^+(aq) + 2e^- \rightarrow 2Ag(s)$ Now, combine the anode and cathode reactions: $H_2(g) + 2Ag^+(aq) \rightarrow 2H^+(aq) + 2Ag(s)$ The number of electrons transferred in the balanced reaction, $n = 2$.

3. Calculate the Standard Cell Potential ($E^\circ_{cell}$): $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$ $E^\circ_{cell} = E^\circ_{Ag^+|Ag} - E^\circ_{H^+|H_2} = 0.8 V - 0 V = 0.8 V$

4. Apply the Nernst Equation: The Nernst equation for the cell is: $E_{cell} = E^\circ_{cell} - \frac{2.303RT}{nF} \log Q$ The reaction quotient $Q$ for the overall reaction is: $Q = \frac{[H^+]^2 [Ag(s)]^2}{[Ag^+]^2 P_{H_2}}$ Since $[H^+] = 1 M$ and $P_{H_2} = 1 \text{ atm}$ (for SHE), and $Ag(s)$ is a pure solid (its activity is 1), the expression for $Q$ simplifies to: $Q = \frac{1}{[Ag^+]^2}$

Now, substitute the given values into the Nernst equation: $E_{cell} = 0.5 V$ $E^\circ_{cell} = 0.8 V$ $\frac{2.303RT}{F} = 0.06$ $n = 2$

$0.5 = 0.8 - \frac{0.06}{2} \log \left(\frac{1}{[Ag^+]^2}\right)$ $0.5 - 0.8 = -0.03 \log ([Ag^+]^{-2})$ $-0.3 = -0.03 \times (-2 \log [Ag^+])$ $-0.3 = 0.06 \log [Ag^+]$ $\log [Ag^+] = \frac{-0.3}{0.06}$ $\log [Ag^+] = -5$ $[Ag^+] = 10^{-5} M$

5. Calculate the Moles and Mass of Ag in the Solution: The volume of the solution is 500 mL = 0.5 L. Moles of $Ag^+ = \text{Concentration} \times \text{Volume}$ Moles of $Ag^+ = 10^{-5} \text{ mol/L} \times 0.5 \text{ L} = 0.5 \times 10^{-5} \text{ mol} = 5 \times 10^{-6} \text{ mol}$

The molar mass of Ag is 108 g/mol. Mass of Ag = Moles of $Ag^+ \times \text{Molar mass of Ag}$ Mass of Ag = $5 \times 10^{-6} \text{ mol} \times 108 \text{ g/mol} = 540 \times 10^{-6} \text{ g} = 0.000540 \text{ g}$ Converting to milligrams: $0.000540 \text{ g} \times 1000 \text{ mg/g} = 0.54 \text{ mg}$

6. Calculate the Percentage of Ag in the Alloy: The total weight of the alloy is 54 mg. Percentage of Ag = $\frac{\text{Mass of Ag}}{\text{Total mass of alloy}} \times 100\%$ Percentage of Ag = $\frac{0.54 \text{ mg}}{54 \text{ mg}} \times 100\%$ Percentage of Ag = $0.01 \times 100\% = 1\%$

The final answer is $\boxed{1\%}$

Explanation of the solution:

1. Cell Setup: SHE (anode) and Ag electrode (cathode).
2. Reactions: Anode: $H_2 \rightarrow 2H^+ + 2e^-$. Cathode: $Ag^+ + e^- \rightarrow Ag$.
3. Overall Reaction: $H_2 + 2Ag^+ \rightarrow 2H^+ + 2Ag$. Number of electrons ($n$) = 2.
4. Standard Cell Potential: $E^\circ_{cell} = E^\circ_{Ag^+|Ag} - E^\circ_{H^+|H_2} = 0.8 V - 0 V = 0.8 V$.
5. Nernst Equation: $E_{cell} = E^\circ_{cell} - \frac{0.06}{n} \log Q$.
6. Reaction Quotient: $Q = \frac{[H^+]^2}{[Ag^+]^2 P_{H_2}} = \frac{1}{[Ag^+]^2}$ (for SHE).
7. Solve for $[Ag^+]$: $0.5 = 0.8 - \frac{0.06}{2} \log (\frac{1}{[Ag^+]^2})$ $-0.3 = -0.03 (-2 \log [Ag^+])$ $\log [Ag^+] = -5 \implies [Ag^+] = 10^{-5} M$.
8. Moles of Ag: Moles = Concentration $\times$ Volume = $10^{-5} M \times 0.5 L = 5 \times 10^{-6}$ mol.
9. Mass of Ag: Mass = Moles $\times$ Molar Mass = $5 \times 10^{-6}$ mol $\times 108$ g/mol = $0.00054$ g = $0.54$ mg.
10. Percentage of Ag: $\frac{0.54 \text{ mg}}{54 \text{ mg}} \times 100\% = 1\%$.