Balancing Redox Equations

analyze: Mg + S  MgS Mg: 0 to +2 = oxidation S: 0 to -2 = reduction -2 +2 analyze: Mg + S  MgS 1. assign oxidation numbers to all elements 2. figure out change in oxidation numbers what is oxidized? what is reduced? Mg: 0 to +2 = oxidation S: 0 to -2 = reduction

Mg + S  MgS +2 -2 half-reactions: Mg is oxidized: 0 to +2 +2 -2 half-reactions: Mg is oxidized: 0 to +2 Mg  Mg+2 + 2e- S is reduced: 0 to -2 S + 2e-  S-2

electrons lost = electrons gained add half-reactions: Mg  Mg+2 + 2e- S + 2e-  S-2 ____+______________________________ Mg + S + 2e-  Mg+2 +2e- + S-2

Zn + 2HCl  H2 + ZnCl2 +1 -1 +2 -1 Zn goes from 0 to +2 = oxidation +1 -1 +2 -1 Zn goes from 0 to +2 = oxidation H goes from +1 to 0 = reduction Cl goes from -1 to -1; no change

Zn +2HCl  H2 + ZnCl2 Zn  Zn+2 + 2e- 2H+1 + 2e-  H2 0 +1 -1 0 +2 -1 Zn  Zn+2 + 2e- 2H+1 + 2e-  H2 ______________________________________ Zn + 2H+1 +2e-  Zn+2 +2e- + H2

Balancing Redox Equations assign oxidation # to all elements determine elements that changed oxidation number identify element oxidized & element reduced write half-reactions (diatomics must stay as is) # electrons lost/gained must be equal; (multiply half-reactions if necessary) add half-reactions; transfer co-efficients to skeleton equation balance rest of equation by counting atoms

Cu + AgNO3  Cu(NO3)2 + Ag +1 +5 -2 +2 +5 -2 +1 +5 -2 +2 +5 -2 Cu + AgNO3  Cu(NO3)2 + Ag Cu goes from 0 to +2 = oxidation Ag goes from +1 to 0 = reduction N goes from +5 to +5; no change O goes from -2 to -2; no change

Half-Reactions +______________________ Cu  Cu+2 + 2e- Ag+1 + 1e-  Ag multiply by 2 Cu  Cu+2 + 2e- 2Ag+1 + 2e-  2Ag +______________________ Cu + 2Ag+1 + 2e-  2Ag + Cu+2 + 2e-

Transfer Co-efficients compare skeleton equation & sum of ½ rxns: Cu + AgNO3  Ag + Cu(NO3)2 vs. Cu + 2Ag+1 + 2e-  2Ag + Cu+2 + 2e- transfer co-efficients! Cu + 2AgNO3  2Ag + Cu(NO3)2

Cu + HNO3  Cu(NO3)2 + NO2 + H2O Cu: 0 to +2 = oxidized +1 +5 -2 +2 +5 -2 +1 -2 +4 -2 Cu + HNO3  Cu(NO3)2 + NO2 + H2O Cu: 0 to +2 = oxidized H: +1 to +1; no change O: -2 to -2; no change * N: starts as +5 ends as +5 [no change – Cu(NO3)2] ends as +4 = reduction [NO2] can’t just say N! say:N in HNO3 or N+5 half-reactions: Cu  Cu+2 + 2e- N+5 + 1e-  N+4

multiply half-reactions as necessary: # of electrons lost = # gained Cu  Cu+2 + 2e- N+5 + 1e-  N+4 multiply half-reactions as necessary: # of electrons lost = # gained Cu  Cu+2 + 2e- 2 (N+5 + e-  N+4) balance remaining atoms by inspection balanced half-reaction co-efficients might not completely balance the equation Cu + 4HNO3  Cu(NO3)2 + 2NO2 + 2H2O