How to Balance Redox Equations?

How to use the half-reaction method to balance redox equations in acidic solution?

• The following is a question on my chemistry homework and I don't get it at all. Ag (s) + NO3- (aq) + H+ (aq) ---> Ag+ (aq) + H2O (l) + NO (g)

• Answer:

  This is a long, multi-step process that's not particularly well suited to this format, but I'll give it a try. Step 1: Break the reaction into half reactions. Ag --> Ag+ and NO3- --> NO The half reactions contain all of the non-water, non-acid/base bits in the reaction. The interesting parts. Group up the non-H/O elements into half reactions. 2. Balance oxygen atoms using water. (Don't need to mess with the silver half reaction yet.) NO3- --> NO + 2 H2O 3. Balance oxygen atoms using hydrogen ions. (Still don't need to mess with silver.) NO3- + 4 H+ --> NO + 2 H2O 4. Balance charges with electrons. Ag --> Ag+ + e- NO3- + 4 H+ + 3 e- --> NO + 2 H2O 5. Multiply each half reaction by an integer so that the same number of electrons is present in each. Add the half reactions and cancel any species that appear on the left and right hand sides. (Electrons should always cancel completely.) 3(Ag --> Ag+ + e-) Add: 3Ag --> 3Ag+ + 3e- NO3- + 4 H+ + 3 e- --> NO + 2 H2O 3Ag + NO3- + 4 H+ + 3 e- --> 3 Ag+ + 3 e- + NO + 2 H2O Cancel. 3Ag + NO3- + 4H+ --> 3Ag+ + NO + 2 H2O Done. Believe it or not this one was fairly easy.