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 What was oxidized? Reduced? -3 +1

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 0 -3 H 0 +1 -1 -2 -3 -4 Gained 3 electrons = Reduced 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

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! = +1 1 and -1 equals 0

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

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 0 + -6 = 2 X -3

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

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: 0 to +2 = Oxidation S: 0 to -2 = Reduction

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

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

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

Taking it one step further

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

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+2 Ag +1 0 N +5  +5 unchanged O -2  -2 unchanged Identify element oxidized & element reduced Oxidation (OIL) =Cu Reduction (RIG) = Ag

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

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)2 + 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: