Balancing Redox Equations

Analyze Mg + S  MgS +2 -2 What is oxidized? What is reduced? +2 -2 Analyze Mg + S  MgS What is oxidized? What is reduced? Assign Oxidation Numbers. Figure out change in oxidation numbers. Mg goes from 0 to +2: Oxidation S goes from 0 to -2: Reduction

Mg + S  MgS 2 electrons 2 electrons Identify what species is oxidized & what species is reduced. Figure out the 2 half-reactions.

Half-Reactions Mg + S  MgS +2 -2 Mg is oxidized: S is reduced: +2 -2 Mg + S  MgS Mg is oxidized: Mg  Mg+2 + 2e- S is reduced: S + 2e-  S-2

Adjust half-reactions so electrons lost = electrons gained. Add half-reactions. Mg  Mg+2 + 2e- S + 2e-  S-2 __________________________________ Mg + S + 2e-  Mg+2 +2e- + S-2 Balance everything else by counting atoms.

Zn + HCl  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 + HCl  H2 + ZnCl2 Zn  Zn+2 + 2e- 2H+1 + 2e-  H2 2 electrons Zn + HCl  H2 + ZnCl2 1 electron per H Zn  Zn+2 + 2e- 2H+1 + 2e-  H2 ______________________________________ Zn + 2H+1 +2e-  Zn+2 +2e- + H2

Transfer the coefficients! Zn + 2H+1  H2 + Zn+2 This is what you’ve got from adding the 2 half-reactions. Zn + HCl  H2 + ZnCl2 This is the skeleton equation you started with. Transfer the coefficients. Zn + 2HCl  H2 + ZnCl2

Balancing Redox Equations Assign oxidation numbers to all atoms in equation. See which elements have changes in oxidation number. Identify atoms that are oxidized & atoms that are reduced. Write the half-reactions. Diatomics have to be written as diatomics. Make the number of electrons lost & gained equal in magnitude by multiplying half-reactions as needed. Add the half-reactions. Transfer coefficients to skeleton equation. Balance remainder of equation by counting up 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 2Ag+1 + 2e-  2Ag Multiply by 2 ______________________ Cu + 2Ag+1 + 2e-  2Ag + Cu+2 + 2e-

Transfer Coefficients Compare skeleton equation & sum of half-reactions: Cu + AgNO3  Ag + Cu(NO3)2 Vs. Cu + 2Ag+1 + 2e-  2Ag + Cu+2 + 2e- Transfer the coefficients! Cu + 2AgNO3  2Ag + Cu(NO3)2) Exception: Do NOT insert the coefficient of any item that appears in more than one place in the equation.

Cu + HNO3  Cu(NO3)2 + NO2 + H2O Assign Oxidation Numbers. +1 +5 -2 +2 +5 -2 +1 -2 +4 -2 Cu + HNO3  Cu(NO3)2 + NO2 + H2O Assign Oxidation Numbers. Identify which species are oxidized & which reduced. Cu from 0 to +2 = oxidized H from +1 to +1 so no change O from -2 to -2 so no change N all starts as +5. Some ends as +5, some as +4 = reduction

Cu + HNO3  Cu(NO3)2 + NO2 + H2O Find change in oxidation number. Change of +2 Cu + HNO3  Cu(NO3)2 + NO2 + H2O 2 2 2( ) = -2 Change of -1 Find change in oxidation number. Write 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 the N in the HNO3. Or the HNO3 or N+5. N+5 Cu

Multiply half-reactions as necessary. Cu  Cu+2 + 2e- 2N+5 + 2e-  2N+4 Now the # of electrons lost = # gained. Cu + 2HNO3  Cu(NO3)2 + 2NO2 + H2O Add half-reactions. Transfer coefficients. Balance remaining atoms by inspection. Cu + 4HNO3  Cu(NO3)2 + 2NO2 + 2H2O