Balancing Redox Reactions

Steps: 1.Assign oxidation numbers 2.Write half reactions 3.Balance electrons to determine coefficients needed

2HgO  O 2 + 2Hg Hg e -  Hg 0 Reduction Problem 1: O -2  O 0 + 2e - Oxidation

2KClO 3  2KCl + 3O e- + Cl +5  Cl -1 Reduction Problem 2: 3(O -2  O 0 + 2e - ) Oxidation

2HCl + Zn  ZnCl 2 + H e- + 2H +1  H 2 0 Reduction Problem 3: Zn 0  Zn e-Oxidation

2Ag NO 3 + Cu  Cu(NO 3 ) 2 + 2Ag e- + 2Ag +1  2Ag 0 Reduction Problem 4: Cu 0  Cu e-Oxidation

4NH 3 + 3O 2  2N 2 + 6H 2 O e- + 3O 2 0  6O -2 Reduction Problem 5: 4N -3  2N e-Oxidation

3Mg + 2Al(NO 3 ) 3  2Al + 3Mg(NO 3 ) e- + 2Al +3  2Al 0 Reduction Problem 6: 3Mg 0  3Mg e-Oxidation

3Ca + 2Al 3+  3Ca Al e- + 2Al +3  2Al 0 Reduction Problem 6: 3Ca 0  3Ca e-Oxidation

2Fe 3+ + Sn 2+  2Fe 2+ + Sn 4+ 2e- + 2Fe +3  2Fe +2 Reduction Problem 7: Sn +2  Sn e-Oxidation

2HNO 3 + 3MnCl 2 + 6HCl  2NO + 3MnCl 4 + 4H e- + 2N +5  2N +2 Reduction Problem 8: 3Mn +2  3Mn e-Oxidation

K 2 (Cr 2 O 7 ) + 14HCl  3Cl 2 + 7H 2 O + 2CrCl 3 + 2KCl e- + Cr 2 +6  2Cr +3 Reduction Problem 9: 6Cl -1  3Cl e-Oxidation