Question #9720b

1 Answer
Mar 27, 2016

Answer:

Here's what's going on here.

Explanation:

Lead(II, IV) oxide, #"Pb"_3"O"_4#, commonly called red lead, is actually a mixture of two lead oxides, lead(II) oxide, #"PbO"#, and lead(IV) oxide, #"PbO"_2#.

This means that in chemical reactions, you can treat red lead as a mixture of these two oxides. More specifically, you have

#"Pb"_3"O"_4 <=> 2"PbO" * "PbO"_2#

Now, here is where things get interesting. Both of these oxides will react with concentrated hydrochloric acid to produce aqueous lead(II) chloride, #"PbCl"_2#, and water.

However, the difference between the two lead oxides is that lead(IV) oxide will also oxidize hydrochloric acid to chlorine gas, #"Cl"_2#.

You can thus say that

#2stackrel(color(blue)(+2))("Pb")"O"_text((s]) + 4"H"stackrel(color(blue)(-1))("Cl")_text((aq]) -> 2stackrel(color(blue)(+2))("Pb")stackrel(color(blue)(-1))("Cl")_text(2(aq]) + 2"H"_2"O"_text((l])#

This is not a redox reaction. Here you start with lead in the #color(blue)(+2)# oxidation state and chlorine in the #color(blue)(-1)# oxidation state, and end up with the same oxidation states for both elements.

The second reaction will be a redox reaction.

#stackrel(color(blue)(+4))("Pb")"O"_text(2(s]) + 4"H"stackrel(color(blue)(-1))("Cl")_text((aq]) -> stackrel(color(blue)(+2))("Pb")"Cl"_text(2(aq]) + 2"H"_2"O"_text((l]) + stackrel(color(blue)(0))("Cl")_text(2(g])# #uarr#

The lead(IV) cations gets reduced to lead(II) cations, and the hydrochloric acid gets oxidized to chlorine gas.

Add these two reactions to get the overall view of what's going on here

#overbrace(2"PbO"_text((s]) * "PbO"_text(2(s]))^(color(blue)("Pb"_3"O"_4)) + 8"HCl"_text((aq]) -> 3"PbCl"_text(2(aq]) + 4"H"_2"O"_text((l]) + "Cl"_text(2(g])# #uarr#

Now, when red lead reacts with nitric acid, #"HNO"_3#, you don't get two separate reactions because lead(IV) oxide will not react with the acid.

In this case, only lead(II) oxide will react with the nitric acid to form aqueous lead(II) nitrate, #"Pb"("NO"_3)_2#. The leead(IV0 oxide will remain unreacted, which is why you find it on the products' side in this reaction

What actually goes on here is

#2"PbO"_text((s]) + 4"HNO"_text(3(aq]) -> 2"Pb"("NO"_3)_text(2(aq]) + 2"H"_2"O"_text((l])#

This is a classic double replacement reaction in which the hydrogen cations exchange places with the lead(II)cations.

You can say that

#"Pb"_3"O"_text(4(s]) + 4"HNO"_text(3(aq]) -> 2"Pb"("NO"_3)_text(2(aq]) + overbrace("PbO"_text(2(s]))^(color(red)("from Pb"_3"O"_4)) + 2"H"_2"O"_text((l])#