Understanding Redox Half Reactions and Oxidation Numbers

Topic: Redox Half reactions Do Now: Quiz assign oxidation numbers

Now we know how to assign oxidation number we can look at redox rxns Haber Process N2(g)+ 3H2(g) 2NH3(g) Start by assigning oxidation numbers 0 0 -3 +1 N2(g)+ 3H2(g) 2NH3(g) What was oxidized? Reduced?

0 0 +1 -3 N2(g) + 3H2(g) 2NH3(g) Oxidation Is Losing electrons O I L 0 -1 -2 -3 -4 4 3 2 1 Began Ended N 0 -3 Reduction Is Gaining electrons R I G Gained 3 electrons = Reduced H 0 +1 Lost 1 electron = Oxidized

Why use the word reduced when electrons are gained? Electrons are Negative! Look how oxidation number changes Ex: Cl gains an electron Cl-1 oxidation # from 0 to -1; the # was reduced

Half Reactions Even though oxidation & reduction reactions occur together we write separate equations for each process and include # of e- gained/lost known as Half-Reactions

Half Reactions Half-reactions must demonstrate: conservation of mass & conservation of charge # atoms on left must = # atoms on right total charge on left must = total charge on right

0 0 -3 +1 N2(g) + 3H2(g) 2NH3(g) Oxidation = electrons lost (becomes more positive) So electrons are on the product side H lost 1 electrons, but since its NH3, there are 3 H so a total of 3 e- are lost H2 H+1 + 1e- 0 +1 1 and -1 equals 0 = OIL But something is wrong!

Remember H2 H+1 + 1e- Total Charge on left = Total Charge on right # atoms on left = # atoms on right H2 2H+1 + 1e- Something is still wrong! Charge is off now H2 2H+1 + 2e- 2 x +1 = +2 and -2 equals 0

0 0 -3 +1 N2(g) + 3H2(g) 2NH3(g) Reduction = electrons gained (becomes more negative) So electrons are on the reactant side Each N gained 3 electrons, but 2 N so N2 + 6e- N-3 0 + -6 2 X -3 = RIG 2

Oxidizing Agent: Substance being reduced Accepts electrons aids oxidation for another species Nitrogen is the Oxidizing Agent Reducing Agent: Substance being oxidized Loses electrons aids reduction for another species Hydrogen is the Reducing Agent

0 +2 -2 0 You Try: Mg + S MgS What is oxidized? What is reduced? Assign Oxidation Numbers Figure out change in oxidation numbers Mg: 0 to +2 = Oxidation S: 0 to -2 = Reduction

Now write the Half-Reactions 0 0 +2 -2 Mg + S MgS S is reduced: Mg is oxidized: Mg Mg+2 + 2e- S + 2e- S-2

0 0 +1 -1 +2 -1 Zn + 2HCl What is oxidized? Reduced? H2 + ZnCl2 Zn goes from 0 to +2 = oxidation H goes from +1 to 0 = reduction Cl goes from -1 to -1; No change

0 0 +1 -1 +2 -1 Zn + 2HCl H2 + ZnCl2 Write Half Reactions Zn 2H+1 + 2e- Zn+2 + 2e- H2

Taking it one step further

The steps 1. Assign oxidation numbers to all atoms in equation 2. Determine elements changed oxidation number 3. Identify element oxidized & element reduced 4. Write half-reactions (diatomics must stay as is, everyone you can write with their oxidation number only NH3 = N-3 but Cl2 stays Cl2 not Cl) Number electrons lost & gained must be equal; multiply half-reactions if necessary 5. 6. Add half-reactions; Transfer coefficients to skeleton equation 7. Balance rest of equation by counting atoms

+2 +5 -2 +1 +5-2 0 0 Cu + AgNO3 Cu(NO3)2 + Ag 1. Assign oxidation numbers to all atoms in equation 2. Determine elements changed oxidation number Cu 0 +2 Ag +1 0 N +5 +5 unchanged O -2 -2 unchanged 3. Identify element oxidized & element reduced Oxidation (OIL) =Cu Reduction (RIG) = Ag

+2 +5 -2 +1 +5-2 0 0 Cu + AgNO3 Cu(NO3)2 + Ag 4. Write half-reactions Cu Cu+2 + 2e- Ag+1 + 1e- 5. Number electrons lost & gained must be equal; multiply half- reactions if necessary 2(Ag+1 + 1e- Ag) 2Ag+1 + 2e- 2Ag Ag

+2 +5 -2 +1 +5-2 0 0 Cu + AgNO3 Cu(NO3)2 + Ag 6. Add half-reactions; Transfer coefficients to skeleton equation Cu Cu+2 + 2e- 2Ag+1 + 2e- +______________________ Cu + 2Ag+1 + 2e- 2Ag + Cu+2 + 2e- Cu + 2Ag+1 2Ag + Cu+2 2Ag Cu + AgNO3 2 original reaction: Cu(NO3)2 + Ag 7. Balance rest of equation by counting atoms 2