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The actual reaction is:

Pb(s) + 4HNO3(aq) -----> Pb(NO3)2(aq) + 2H2O(l) + 2NO2(g)

This is interesting for a number of reasons. Since the activity of lead is less than H2, (The Eo for the reduction of Pb2+ to the metal is -0.125 V, compared to the minimum -0.6 V or less that is required) the reaction cannot occur as a single replacement, e.g.

Pb(s) + 2HNO3(aq) -----> Pb(NO3)2(aq) + H2(g) does not occur. There are really two reactions at work.

1.) HNO3 oxidizes Pb to form PbO:

Pb(s) + 2HNO3(aq) -----> PbO(s) + H2O(l) + 2NO2(g)

Lead is oxidized from 0 to +2 and two moles of N are reduced from +5 to +4. This is a redox reaction.

2.) HNO3 then reacts with PbO and to form Pb(NO3)2:

PbO(s) + 2HNO3(aq) -----> Pb(NO3)2(aq) + H2O(l)

This is a double replacement reaction. This sequence of events is seen (albeit with a change in stoichiometry) for other some other metals as well, most notably copper and silver.

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