This seems like a stupid question, but how would I find the product of a single displacement reaction like the following?

$$\ce{Pb + Pb(NO3)2 -> \ ?}$$

Normally, a single displacement reaction follows the path of $\ce{A + BX -> AX + B}$. However, what will the equation look like when A and B are the same element?

  • $\begingroup$ Try this link. Maybe that helps some. (Don't know if it's valuable advice, though...) $\endgroup$ – tschoppi Mar 10 '14 at 21:06
  • $\begingroup$ If Pb(I) were stable, 2PbNO3. Nitrate itself might be a sufficient oxidant to prevent that. Cu(0) + CuCl2 gives 2CuCl when refluxed in acetonitrile (soft Lewis base ligand). A blue solution with red metal dust turns opaque brown-green, then goes clear and colorless. $\endgroup$ – Uncle Al Mar 12 '14 at 1:06

Nothing. All you would have is just a mixture of white powder and lead metal.

In general, lead (II) nitrate is synthesized by:

$$\ce{PbO (s) + 2 HNO3 (aq) -> Pb(NO3)2 (aq) + H2O (l)}$$

Lead (II) nitrate is soluble in water.

Alternatively, you could do:

$$\ce{PbCO3(s) + 2HNO3(aq) -> Pb(NO3)2(aq) + H2O(l) + CO2(g)} $$

Single displacement reactions are a kind of redox reaction. For single displacement reactions to occur:

1.) A and B must either be different metals OR

$$ \ce{2AgNO3(aq) + Zn(s) -> 2Ag(s) + Zn(NO3)2(aq)}$$

2.) A and B must be a halogen

$$\ce{Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)}$$

This is necessary as for the redox reaction to occur, you need two differing metals on the reactivity series in order for the single replacement to work, or alternatively, use compounds that contain halogens.

Even so, in combining two compounds with metals or halogens, if the element is less reactive than the element in the compound, the single replacement reaction will not proceed.

Some examples are:

$$\ce{Ag (s) + Cu(NO3)2 (aq) -> } \text{No reaction}$$

$$ \ce{I2 + 2KBr ->} \text{No reaction} $$

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