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.