What are the steps of the Half-reaction method for Balancing redox Reactions?

Balancing oxidation-reduction reactions?

• It needs to be balanced using redox methods, in an acidic solution: Fe{3+} (aq) + H2S (g) = Fe{2+} (aq) + S (s) WO3 (s) + Sn{2+} (aq) + Cl{1-} (aq) = W3O8 (s) + SnCl6{2-} (aq) We use the half reaction method, which is different from our textbooks. I've tried several times and gone over my notes but at the end my charges aren't the same on both sides.

• Answer:

  Fe 3+ + H2S --> Fe 2+ + S S in H2S = 2- and becomes 0 as solid S, lost 2 e-, oxidized Fe 3+ becomes Fe 2+, gained 1 e-, reduced H2S --> 2H+ + S + 2e- Fe 3+ + 1e- --> Fe 2+ balance the electrons H2S --> 2H+ + S + 2e- 2Fe 3+ + 2e- --> 2Fe 2+ combine H2s + 2Fe 3+ --> 2H+ + S + 2Fe 2+ 2. W in WO3 = 6+ and becomes 5 1/3 + in W3O8, each W gains 1/3 of an e- for a total of 1 e- Sn 2+ becomes Sn 4+ in SnCl6 2- 3WO3 + 5H2O + 1e- --> W3O8 + 10H+ Sn 2+ --> Sn4+ + 2e- balance the e- 6WO3 + 2e- + 4H+ --> 2W3O8 + 2H2O Sn2+ --> Sn4+ + 2e- combine and eliminate 6WO3 + Sn2+ + 6Cl- + 4H+ --> 2W3O8 + SnCl6 2- + 2H2O