According to my textbook, element $\ce{X}$ can oxidize element $\ce{Y}$ if $\ce{X}$ was lower in the activity series. Lead is lower than Tin in the activity series table in the textbook, but the following question contradict that rule and I can't figure out why:
Will the reaction $\ce{Sn(s) + Pb(NO3)2(aq) -> Sn(NO3)2(aq) + Pb(s)}$ occur?
The book reveals at the end that the answer is no, but doesn't explain why.
Are you certain that your statement from the textbook is correct? The cation of X will oxidize element Y if element Y is more active than X, which is usually labeled (may be towards the top or the bottom of the list). For reference, see the activity series at Wikipedia, where the more active metals are above the less active metals in the list.
How to use the activity series:
For example, what happens when iron metal is placed into an aqueous solution of copper (II) nitrate?
Let's write the potential equation:
$$\ce{2Fe(s) + 3Cu^{2+} -> 2Fe^{3+} + 3Cu(s)}$$
Since iron is more active than copper, iron will reduce copper cations (and copper cations will oxidize iron metal). The iron will dissolve.
What happens if copper metal is placed into an aqueous solution of iron (III) nitrate?
Since copper is less active than ion, copper metal will not reduce iron cations. No reaction occurs and the copper will not dissolve.
Your question
Lead is less active than tin in all activity series that I have seen. Lead metal should reduce tin cations. This reaction should work. It is possible that the textbook/solutions manual has a typo. I find them in my books all the time.
This is a nice example: the standard reduction potentials are close, so the Nernst equation ultimately reveals what is favored to happen spontaneously, ignoring all real world complications.
Consider a galvanic cell having a tin $(\ce{Sn})$ metal electrode in $x~\pu{M}$ aqueous $\ce{Sn(NO3)2}$ and a lead $(\ce{Pb})$ metal electrode in $y~\pu{M}$ aqueous $\ce{Pb(NO3)2}.$ Assume the salt bridge is an aqueous solution (very roughly $\pu{1 M})$ of $\ce{KNO3}.$ Under standard conditions, $x = y = \pu{1 M},$ and the standard reduction potentials are
$$ \begin{align} \ce{Pb^2+(aq) + 2 e- &<=> Pb(s)} &\quad E^\circ &= \pu{-0.126 V} \\ \ce{Sn^2+(aq) + 2 e- &<=> Sn(s)} &\quad E^\circ &= \pu{-0.14 V} \end{align} $$
So $\ce{Pb^2+}$ is spontaneously reduced to $\ce{Pb(s)}$ and $\ce{Sn(s)}$ is spontaneously oxidised to $\ce{Sn^2+}$. Hence, $\ce{Sn(s)}$ is the anode (where oxidation occurs) and $\ce{Pb(s)}$ is the cathode (where reduction occurs).
The galvanic cell, in standard notation, is
$$\ce{Sn(s)|Sn^2+(aq;\pu{1 M})||Pb^2+(aq;\pu{1 M})|Pb(s)}$$
and the standard cell potential is
$$E_\mathrm{cell}^\circ = E_\mathrm{cathode}^\circ – E_\mathrm{anode}^\circ = \pu{-0.126 V} – (\pu{-0.14 V}) = \pu{+0.014 V}$$
Note that the anode is at the left and it is $\ce{Sn(s)},$ so tin metal gets oxidized.
If $x = \pu{0.6 M}$ and $y = \pu{0.1 M},$ then the Nernst equation is used in the obvious way and the non-standard reduction potentials are computed to be $E_\ce{Pb} = \pu{-0.156 V}$ and $E_\ce{Sn} = \pu{-0.147 V}$, so
$$E_\mathrm{cell} = E_\mathrm{cathode} – E_\mathrm{anode} = \pu{-0.147 V} – (\pu{-0.156 V}) = \pu{+0.009 V}$$
and the galvanic cell notation is
$$\ce{Pb(s)|Pb^2+(aq;\pu{0.1 M})||Sn^2+(aq;\pu{0.6 M})|Sn(s)}$$
Again, the anode is at the left, but now it is $\ce{Pb(s)}$ that gets oxidized.
