Oxidization and reduction chemistry questions

combine equations eliminating itesm that appear on both sides
4MnO4- + 12H2O + 3ClO2- + 6H2O + 12OH- --> 4MnO2 + 8H2O + 16OH- + 3ClO4- + 12H2O
4MnO4 + 3ClO2- --> 4MnO2 + 2H2O + 4OH- + 3ClO4

b) Mn 3+ + I- --> Mn 2+ + IO3-
Mn 3+ to Mn 2+ gained 1-, reduced
I- --> IO3- (I = 5+) lost 6 e-, oxidized
Mn 3+ + 1e- --> Mn 2+
I- + 3H2O + 6OH- --> IO3- + 6H2O + 6e-
multilpy by common factor to get rid of the electrons
6Mn 3+ + 6e- --> 6Mn 2+
I- + 3H2O + 6OH- --> IO3- + 6H2O + 6e-
remove electrons and combine equations
6Mn 3+ + I- + 6OH- --> 6Mn 2+ + IO3- + 3H2O

3. Cu + HNO3 --> Cu(NO3)2 + NO2 + H2O
Cu = 0 becomes Cu 2+, oxidized, lost 2 e-, reducing agnet
N in HNO3 = 5+ becomes 4+ in NO2, gained 1e-, oxidized, reducing agent

Cu --> Cu2+ + 2e-
NO3 + 1e- + 2H+ --> NO2 + H2O
Cu + 2NO3- + 2H+ --> Cu 2+ + 2NO2 + 2H2O


Bi(OH)3 + Na2SnO2 --> Bi + Na2SnO3 + H2O
Bi in Bi(OH)3 = 3+ becomes 0 gained electrons, reduced, oxidizing agnet
Sn in Na2SnO2 = 2+ becomes 4+ in Na2SnO3, lost electrons, oxidized, reducing agent

Bi 3+ + 3e- --> Bi
SnO2 2- + H2O --> SnO3 2- + 2e- + 2H+
2Bi 3+ + 6e- --> 2Bi
3SnO2 2- + 3H2O + 6OH- --> 3SnO3 2- + 6e- + 6H2O
2Bi 3+ + 3SnO2 2- + 6OH- --> 2Bi + 3SnO3 2- + 3H2O

P + HNO3 + H2O --> NO + H3PO4
P = 0 becomes 5+ in PO43-, oxidized, reducing agnet, lost electrons
N in HNO3 = 5+ becomes 2+ in NO, reduced, oxidizing agnet, gained electrons
3(P + 4H2O --> PO4 3- + 8H+ + 5e-)
5(NO3- + 3e- + H+ --> NO2 + H2O)

3P + 12H2O --> 3PO4 3- + 24H+ + 15e-
5NO3- + 15e- + 5H+ --> 5NO2 + 5H2O
3P + 7H2O + 5NO3- --> 3PO4 3- + 14H+ + 5NO2