[awsalazar]

The Problem: The percentage silver impurity in a lead sample was determined by the following procedure: A sample of lead weighing 1.050 grams was dissolved in nitric acid to produce an aqueous solution of Pb²⁺ and Ag⁺. The solution was made up to a volume of 350.0 mL with water; a pure silver electrode was immersed in the solution and the potential difference between this electrode and the standard hydrogen electrode was found to be 0.503 volts. What was the percent silver impurity in the lead metal?
Answer: 0.0360% Ag

Attempted method: The first thing I did was convert the grams of lead to moles, and then divided by 0.350 L to get a molarity of 0.0145 M, which I assumed to be the concentration of Pb²⁺ in the solution. I have a feeling this process might be wrong, since it does not take into account the silver impurity. I know that the voltage for the reaction is E = 0.503 V. Then I assumed the reaction to be Pb(s) + 2Ag⁺(aq)  Pb²⁺(aq) + Ag(s). I have a feeling that this might not be the reaction. Should I use the reaction of lead dissolving into nitric acid? That is, 3Pb(s) + 8H⁺(aq) + 2NO₃^-(aq)  3Pb²⁺(aq) + 2NO(g) + 4H₂O. If so, how do I account for the silver ion? From there I calculated E° to be 0.925 V because E°_red is 0.799 and E°_ox is 0.126. Note: Every table I refer to, the standard reduction values are always slightly different, and this can drastically affect the answer. I then solved for Q in the equation E = E° - RTln(Q)/(nF) and got 1.83E14. I then solved for the concentration of silver ion using Q = [Pb²⁺]/[Ag⁺]², which I found to be 8.895E-9 M. I then converted this to moles using the volume and then to grams using molar mass, which I found to be 3.36E-7 grams. To find the percentage of impurity, I divided the mass of silver by the mass of lead, and I obtained 3.2E-5%, which is not the right answer. I have been thinking about this problem for days, and cannot think of the right method.

Note: I know that the use of different reduction potentials will produce different answers, so I'm not so much concerned about obtaining 0.0360% as I am about having the right method.

[mjc123]

Ignore the lead; it does nothing. (If it did, this would be a bad way to measure the silver!) The reaction you need is
Ag⁺(aq) + 1/2H₂(g)  Ag + H⁺(aq)
for which E° = 0.799V. From E you can calculate [Ag⁺].

For your suggested reaction Pb(s) + 2Ag⁺(aq)  Pb²⁺(aq) + 2Ag(s), your expression for Q is wrong. Strictly, it is
Q = [Pb²⁺][Ag]²/[Ag⁺]²[Pb]
We can take [Ag] as 1 in the conventional way, but not [Pb] - because there is no solid lead present! If you started with only Pb²⁺ and Ag, you might imagine forming a tiny amount of lead, along with (by your calculation) 8.895e-9 M Ag⁺. But you actually have much more Ag⁺ than this to start with, so this will not happen - any Pb formed would be immediately re-oxidised. So Q has no value. Essentially, Pb²⁺ is a spectator ion that plays no part - which is just what you want for a method to determine impurity!

[awsalazar]

For your suggested reaction Pb(s) + 2Ag⁺(aq)  Pb²⁺(aq) + 2Ag(s), your expression for Q is wrong. Strictly, it is
Q = [Pb²⁺][Ag]²/[Ag⁺]²[Pb]

I thought that only substances in aqueous solution or gaseous state are included in the expression for Q, at least that is what it said in my general chemistry textbook. Anyways, thanks for the help.

[awsalazar]

The reaction you need is Ag⁺(aq) + 1/2H₂(g)  Ag + H⁺(aq)
for which E° = 0.799V. From E you can calculate [Ag⁺].

If I were to set up the Nernst equation as 0.503 = 0.799 - 0.0257*lnQ I get Q ≈ 1.0E5 = [H⁺]^1/2/[Ag⁺]. How can I find [Ag⁺] if I don't know [H⁺] or [Ag⁺]_o?

[mjc123]

I thought that only substances in aqueous solution or gaseous state are included in the expression for Q, at least that is what it said in my general chemistry textbook.

Normally, yes. But see what I said in the next sentence. The activity of solid reagents does matter, but usually we can take it as a constant, by convention 1.

How can I find [Ag+] if I don't know [H+] or [Ag+]o?

You do know [H⁺]. The clue is in the question - STANDARD hydrogen electrode, so [H⁺] = 1 M. (And P_H2, which ought also to be included in Q, is 1 atm.) And [Ag⁺] = [Ag⁺]₀. We assume no significant reaction takes place - that, again, is necessary for this to be a good method for measuring the silver content.