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Re: Net Ionic Equations for Redox Reactions?

When dealing with redox reactions, are they special in the ways of breaking apart? For instance:

2MnO(s) + 5PbO2(s) + 10 HNO3(aq)----> 2HMnO4(aq) + 5Pb(NO3)2(aq) + 4H2O(l)

If they are solid, the do not come apart correct? and the 2HmnO4(aq) I was told doesn't come apart either and same goes for the water. I can't narrow it down to a net ionic equations from my total ionic equation because none of them are spectator ions... Confused... Thanks in advance!

Ions exist in solid as well as solutions.
In the equation MnO has Mn in the +2 oxidation state (as O is in the -2)
In the equation PbO2 has Pb in the +4 oxidation state (as O = +2 x2 = +4) - equivalent to Pb4+
Ten NO3- ions in the HNO3 on the LHS remain as  ten NO3- ions on the RHS - not involved in the redox reaction (=spectator ions)

On the RHS Mn in HMnO4 has Mn in the +7 oxidation state ( +1 for H, +7 for Mn and -2 x 4 = -8 for O)
On the RHS Pb on Pb(NO3)2 has Pb in the +2 oxidation state

So Pb is reduced (a decrease in oxidation state) while Mn is oxidised (increase in oxidation state)
MnO  ---> MnO4-
MnO + 3H2O ----> MnO4- + 6H+
MnO + 3H2O ----> MnO4- + 6H+ + 5e-

PbO2 ----> Pb2+
PbO2 ----> Pb2+ + 2H2O
PbO2 + 4H+ + 2e- ----> Pb2+ + 2H2O

Adding the two equations gives
2MnO + 6H2O ----> 2MnO4- + 12H+ + 10e-
5PbO2 + 20H+ + 10e- ----> 5Pb2+ + 10H2O

2MnO + 6H2O + 5PbO2 + 20H+ + 10e- ----> 2MnO4- + 12H+ + 10e- + 5Pb2+ + 10H2O
2MnO + 5PbO2 + 8H+ ----> 2MnO4- + 5Pb2+ + 4H2O

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