1. Topic: Redox – Half reactions
Do Now: Quiz assign oxidation numbers

2. 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
What was oxidized? Reduced? -3 +1

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

4. 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

5. 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

6. 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

7. Oxidation = electrons lost (becomes more positive) -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-
OIL
But something is wrong! =
and -1 equals 0

8. Something is still wrong! Charge is off now H2  2H+1 + 2e-
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

9. Reduction = electrons gained (becomes more negative) -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
RIG 2 =
2 X -3

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

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

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

13. Zn + 2HCl  H2 + ZnCl2 What is oxidized? Reduced?+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

14. Zn + 2HCl  H2 + ZnCl2 Zn  Zn+2 + 2e- 2H+1 + 2e-  H2 +1 -1 +2 -1+1 -1 +2 -1
Write Half Reactions
Zn  Zn+2 + 2e-
2H+1 + 2e-  H2

15. Taking it one step further

16. The steps… Assign oxidation numbers to all atoms in equation
Determine elements changed oxidation number
Identify element oxidized & element reduced
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
Add half-reactions; Transfer coefficients to skeleton equation
Balance rest of equation by counting atoms

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

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

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