# Adding powdered Pb and Fe to a solution

My book states that if we add powdered $\ce{Pb}$ (lead) and $\ce{Fe}$ (iron) to a solution containing $1\ \mathrm{M}$ each of $\ce{Pb^2+}$ and $\ce{Fe^2+}$ it would result in formation of more $\ce{Pb}$ and $\ce{Fe^2+}$
What I think:

1) Since both $\ce{Pb}$ and $\ce{Fe}$ have negitive standard electrode poential both will have tendency to form $\ce{Pb^2+}$ and $\ce{Fe^2+}$, then as a result $\ce{Pb^2+}$ and $\ce{Fe^2+}$ will increase in solution which is against the result given by book.

2) Another thing which comes to my mind is that since $\ce{Pb}$ has greater standard reduction potential than $\ce{Fe}$ so only added Fe will convert into $\ce{Fe^2+}$ And the electrons released by it will reduce pre-existing $\ce{Pb^2+}$ to $\ce{Pb}$ and hence the result as given by book

So my question is what exactly will happen if we conduct the following experiment.

Your second thought is the right one. As for the first, it implies the reaction of both metals with $\ce{H+}$ which releases $\ce{H2}$; for that to happen, the solution must contain some significant amount of $\ce{H+}$, i.e., be acidic, which is not the case.

• You should elaborate. – Mithoron Oct 20 '15 at 9:57

Let us consider everything we have before and after. Before adding the powdered substances, we have a solution containing an iron(II) salt and a lead(II) salt. Both are stable by themselves and they are even stable when mixed — this is because lead(II) is the most stable form of lead in aquaeous solution. Iron(II) might want to oxidise itself to iron(III) but cannot since there is no electron acceptor (hydrogen atoms don’t count, they are too bad an oxidising agent when only present as water).

Consider having pure water and adding powdered lead(± 0) and/or iron(± 0). By a simple check of standard potentials we would assume both to be oxidised to lead(II) and iron(II) respectively. But standard potentials are measured at $\mathrm{pH}\ 0$; at neutral pH the potential of hydrogen equals $-0.41~\mathrm{V}$ — not enough to oxidise lead and only just enough for oxidising iron, but that is easily passivised. Of course, solid lead won’t react with solid iron at room temperature in any way. So no reaction here, either.

(This also means that your first attempt is incorrect: it needs a certain concentration of acid to oxidatively dissolve lead or iron to their respective metal(II) ions. Iron would also further be affected by passivation; not sure about lead.)

Meaning that all we truely need to consider is the reaction between the two lead species and the two iron species. There are six equations we might want to consider:

• $\ce{Fe^2+ + 2 e- <=> Fe} ~~~~ E^0 = -0.41~\mathrm{V}$

• $\ce{Fe^3+ + 3 e- <=> Fe} ~~~~ E^0 = -0.04~\mathrm{V}$

• $\ce{Fe^3+ + e- <=> Fe^2+} ~~~~ E^0 = +0.77~\mathrm{V}$

• $\ce{Pb^2+ + 2 e- <=> Pb} ~~~~ E^0 = -0.13~\mathrm{V}$

• $\ce{Pb^4+ + 2 e- <=> Pb^2+} ~~~~ E^0 = +1.69~\mathrm{V}$

• $\ce{Pb^4+ + 4 e- <=> Pb}$ — value not given.

Of course, reactions will happen if $\Delta G < 0$ and for electrochemical cells we have $\Delta G = - z F \Delta E$ where $z$ is the number of electrons transferred and $F$ is Faraday’s constant. Since $\Delta E = E_\mathrm{red} - E_\mathrm{ox}$ and we want that value to be as large as possible, we need to look for high values for the reducing half-cell and low values for the oxidising one. And finally, since $E = E^0 + \frac{0.059}{z} \lg \frac{[\mathrm{ox}]}{[\mathrm{red}]}$ and the activities of $\ce{Fe^{II}, Pb^{II}, Fe^{\pm 0}}$ and $\ce{Pb^{\pm 0}}$ are all $1$, for half-cells using only those four components, we can simply use the standard potentials.

We immediately see that an oxidation of lead(II) to lead(IV) is not possible: We cannot overcome the $+ 1.69~\mathrm{V}$ required. For the same reason, it is not possible to oxidise iron(II) to iron(III).

When considering something to be reduced, we only have the choice between lead(II) and iron(II) — there is no iron(III) in solution (see above). Out of those two, lead(II) is reduced more easily (higest standard potential), so let’s take that as a starting point. And for the necessary oxidation, we want the lowest possible standard potential which only leaves us with oxidation of iron to iron(II). Thankfully, the overall $\Delta G$ is negative:

$$\ce{Fe + Pb^2+ <=>> Pb + Fe^2+} \\ ~ \\ \Delta E = \Delta E^0 = E^0(\ce{Pb}) - E^0(\ce{Fe}) \\ = -0.13~\mathrm{V} - (-0.41~\mathrm{V}) = +0.28~\mathrm{V} \\ \Delta G = - z F \Delta E = - 2 \cdot 96485~\mathrm{\frac{C}{mol}} \cdot (+0.28~\mathrm{V}) = - 54.0~\mathrm{\frac{kJ}{mol}}$$

It’s nice that this is also the answer your book gives; hence your second attempt is correct. In a nutshell, your reasoning was shorter.