Does this equation - involving calcium carbonate and a lead (II) cation - balance?
Unfortunately your equation does not add up, let me briefly explain why.

Calcium carbonate is a not very soluble salt. There is ionic binding between the calcium cation and the molecular carbonate anion. If you dissolve it in water you, the following equilibrium will establish: $$\ce{CaCO3~(s) &lt;=&gt;[\ce{H2O}] Ca^{2+}~(aq) + CO3^{2-}~(aq)}$$

Lead carbonate is also a salt, it is even less soluble. Only a little bit will dissolve according to the following equilibrium:$$\ce{PbCO3~(s) &lt;=&gt;[\ce{H2O}] Pb^{2+}~(aq) + CO3^{2-}~(aq)}$$

Calcium hydroxide is also a salt. It is a little bit more soluble than calcium carbonate. The following equilibrium is established when put in water:$$\ce{Ca(OH)2~(s) &lt;=&gt;[\ce{H2O}] Ca^{2+}~(aq) + 2{}^{-}OH~(aq)}$$

You write the following:


  $$\ce{CaCO3 + 2H2O + Pb^{2+} -&gt; Pb(CO3) + Ca(OH)2 + H2}$$


So in principle you have $\ce{Ca^{2+}}$, $\ce{Pb^{2+}}$ and $\ce{CO3^{2-}}$ on both sides of the equation. What is left is not a balanced equation, since there are two electrons missing:
$$\ce{2H2O ~\nrightarrow 2{}^{-}OH + H2}$$

You need a source for the electrons, one way of balancing the equation is
$$\ce{CaCO3 + 2H2O + Pb -&gt; Pb(CO3) + Ca(OH)2 + H2}$$

Given the insolubilities of all the species involved, it might be hard to see any reaction at all. And I have not checked if the redox reaction is favourable, with potentials.