2. Steps in Balancing Redox 1.Determine the oxidation number of all elements in the compounds 2. Identify which species have undergone oxidation and reduction 3. Balance the number of atoms by placing the appropriate coefficients on each species 4. Balance the number of electrons. Write the oxidation half-reaction by placing the appropriate number of electrons on the right side of the equation. The reduction half reaction is written by placing the appropriate number of electrons on the left side of the equation 5. Check the balanced overall reaction by adding the two half reactions Example: Mg (s) + HCl (aq) ---> MgCl 2(aq) + H 2 1. 0 +1-1 +2 -1 0 2. Oxidation: Mg  Mg +2 reduction: H +  H 2 3. Mg  Mg +2, 2 H +  H 2 4. Mg  Mg +2 + 2e 2e - + 2H +  H 2 5. Mg + 2H+  Mg +2 + H 2 note: the overall redox reaction does not include the electrons on both sides because they cancel each other

3. More Examples Na (s) + O 2(g)  Na 2 O (s) 1. 0 0 +1 -2 2. oxidation: Na  Na+ / reduction: O 2  O 2- 3. Na  Na + / O 2  2 O -2 4. Na  Na+ + e- 4e- + O 2  2 O -2 5. Note: first equation should be multiplied by 4 so that 4e- in the second equation will be cancelled. 4 Na  4Na+ + 4e - 4e- + O -2  2 O -2 4Na + O -2  4Na + + 2O -2

4. Steps in Balancing Redox Half Reaction in acidic solution p.586-588 1. Write the equation for the oxidation and reduction half reaction 2. For each half reaction a) balance all elements except H and O b) Balance oxygen using H2O c) Balance hydrogen using H+ d) Balance the charge using electrons 3. If necessary, multiply one or both balanced half-reactions by an integer to equalize the number of electrons transferred in the two half-reactions 4. Add the two half reactions, and cancel identical species that appear on both sides. Ex: acid MnO 4 - + Fe 2+ ----> Fe 3+ + Mn 2+ 1. (Red) MnO 4 -  Mn 2+ / (Ox) Fe 2+  Fe 3+ +7 -2 +2 +2 +3 2.MnO4-  Mn2+ Fe 2+  Fe 3+ a) Mn is balanced one e- is needed to the right to give a 2+ net b) O can be balanced by adding 4H 2 O to the right charge: Fe 2+  Fe 3+ + e- c) Balance H by 8H+ to the left d) Balance the charge by adding 5e- to the left 5e- + 8H+ MnO 4 -  Mn 2+ + 4 H 2 O 5 - + 8 + + 1 - 2 + + 0 3. Equalize the number of electrons transferred in two reactions 5e- + 8H + MnO 4 -  Mn 2+ + 4 H 2 O 5Fe 2+  5Fe 3+ + 5e - 8H + + MnO 4 - + 5Fe 2+  Mn 2+ + 4H 2 O + 5Fe 3+

5. For more examples check examples in the book pp. 588-589 and answer self check 18.4 on p. 590