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Topic: Redox – Half reactions

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 N 2(g) + 3H 2(g)  2NH 3(g) Start by assigning oxidation numbers N 2(g) + 3H 2(g)  2NH 3(g) What was oxidized? Reduced?. 0. 0.

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Topic: Redox – Half reactions

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  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 • N2(g) + 3H2(g) 2NH3(g) • What was oxidized? Reduced? 0 0 -3 +1

  3. 0 0 +1 -3 N2(g) + 3H2(g) 2NH3(g) • Oxidation • Is • Losing electrons • Reduction • Is • Gaining electrons • O • I • L • R • I • G 4 3 2 1 BeganEnded N 0 -3 H 0 +1 0 -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. 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- OIL But something is wrong! +1 1 and -1 equals 0 0 =

  8. 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. 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 RIG 2 0 + -6 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. 0 +2 0 -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

  12. 0 0 +2 -2 Mg is oxidized: Mg  Mg+2 + 2e- Now write the Half-Reactions Mg + S  MgS S is reduced: S + 2e- S-2

  13. 0 0 +1 -1 +2 -1 Zn + 2HCl  H2 + ZnCl2What is oxidized? Reduced? • Zn goes from 0 to +2 = oxidation • H goes from +1 to 0 = reduction • Cl goes from -1 to -1; No change

  14. 0 0 +1 -1 +2 -1 Zn + 2HCl  H2 + ZnCl2 • 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 • Balancerest of equation by counting atoms

  17. +2 +1 -2 +5 0 +5 -2 0 Cu + AgNO3 Cu(NO3)2 + Ag • 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

  18. +2 +1 -2 +5 0 +5 -2 0 Cu + AgNO3 Cu(NO3)2 + Ag • 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. +2 +1 -2 +5 0 +5 -2 0 Cu + AgNO3 Cu(NO3)2 + Ag • Add half-reactions; Transfer coefficients to skeleton equation Cu  Cu+2 + 2e- 2Ag+1 + 2e- 2Ag Cu + AgNO3 Cu(NO3)2 + Ag • Balancerest of equation by counting atoms +______________________ Cu + 2Ag+1 + 2e- 2Ag + Cu+2 + 2e- Cu + 2Ag+12Ag + Cu+2 2 2 original reaction:

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