How to balance redox equations using the half-reaction method

1.Sn^0 + Ag+ ---> Sn+2 + Ag^0


2. PbS + H2O2 ---> PbSO4 + H2O

Sn + Ag+ --> Sn 2+ + Ag
Sn is oxidized, Ag is reduced

Sn --> Sn 2+ + 2e-
2Ag+ + 2e- --> 2Ag
Sn + 2Ag+ --> Sn 2+ + 2Ag


balance in acidic conditions
PbS + H2O2 --> PbSO4 + H2O
S in PbS = 2- and becomes 6+ in PbSO4, oxidized, loses 8e-
O in H2O2 = 1- becomes 2- in H2O, reduced gains 1e-

S2- + 4H2O --> SO4 2- + 8e- + 8H+
8O 1- + 8e- --> 8O 2-
S2- + 4H2O + 8O- --> SO4 2- + 8O 2- + 8H+