Balancing Redox Equations

Analyze: Mg + S  MgS Mg: 0 to +2 = Oxidation S: 0 to -2 = Reduction -2 +2 Analyze: Mg + S  MgS assign Oxidation Numbers 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 electrons 2 electrons half-reactions Mg + S  MgS +2 -2 +2 -2 Mg is oxidized: Mg  Mg+2 + 2e- S is reduced: 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 2 electrons Zn +2HCl  H2 + ZnCl2 1 electron per H Zn  Zn+2 + 2e- 2H+1 + 2e-  H2 ______________________________________ Zn + 2H+1 +2e-  Zn+2 +2e- + H2

Balancing Redox Equations assign oxidation numbers to all atoms in equation 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 coefficients 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 Coefficients Compare skeleton equation & sum of ½ rxns: Cu + AgNO3  Ag + Cu(NO3)2 vs. Cu + 2Ag+1 + 2e-  2Ag + Cu+2 + 2e- Transfer coefficients! Cu + 2AgNO3  2Ag + Cu(NO3)2

Cu + HNO3  Cu(NO3)2 + NO2 + H2O Cu from 0 to +2 = oxidized +1 +5 -2 +2 +5 -2 +1 -2 +4 -2 Cu + HNO3  Cu(NO3)2 + NO2 + H2O Cu from 0 to +2 = oxidized H from +1 to +1; no change O from -2 to -2; no change N starts as +5; ends as +5 [no change – Cu(NO3)2] and ends as +4 = reduction [NO2] half-reactions Cu  Cu+2 + 2e- N+5 + 1e-  N+4

What’s oxidized? What’s reduced? What is oxidized? What is reduced? What is the oxidizing agent? What is the reducing agent? Cu Can’t just say N! It’s N in the HNO3 or N+5 N+5 Cu

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 coefficients might not completely balance the equation Cu + 4HNO3  Cu(NO3)2 + 2NO2 + 2H2O