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Redox Reactions. Reduction Oxidation.

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Presentation on theme: "Redox Reactions. Reduction Oxidation."— Presentation transcript:

1 Redox Reactions. Reduction Oxidation

2 Oxidation: Reduction: Gain of oxygen Loss of electrons Loss of oxygen
Gain of electrons Increase in oxidation number Decrease in oxidation number

3 Oxidized – gains oxygen
2Mg(s) + O2(g)  2MgO(s) Must be a redox! Mg  Mg2+ Oxidized – loss of e- +2e- Put the e- in. O  O2- Reduced – gain of e- +2e-

4 Cu(s) + 2AgNO3(aq)  Cu(NO3 )2(aq) + 2Ag(s)
Complete the half-equations Oxidized? Reduced? Cu  Cu2+ Oxidized – loss of e- +2e- Ag  Ag Reduced – gain of e- +e-

5 Summation of the Oxidation Rules
The oxidation number of a free element is zero Examples of free elements Gold - Au Zinc –Zn The oxidation number of a monatomic ion is equal to its charge Examples of monatomic ions Sodium ion - Na+1 Chloride ion - Cl-1

6 Examples of polyatomic ions Phosphate - PO4-3 Carbonate - CO3-2
3. The sum of the oxidation numbers in the formula of a polyatomic ion is equal to its charge. Examples of polyatomic ions Phosphate - PO4-3 Carbonate - CO3-2 4. The sum of the oxidation number in the formula of a compound is equal to zero Examples of compounds Water - H2O Glucose - C6H12O6

7 Example of a peroxide Hydrogen peroxide - H2O2
5. In compounds, Hydrogen has an oxidation number of +1 6. In compounds, Oxygen has an oxidation number of -2 7. In peroxides, oxygen has an oxidation number of -1 Example of a peroxide Hydrogen peroxide - H2O2

8 Oxidation Numbers Oxidation state of C in CO2? ? – 4 = 0 ? = +4
The oxidation numbers of atoms in a compound add up to zero. Oxidation state of C in CO2? ? – 4 = 0 ? = +4 Put the +!

9 +2 Oxidation Numbers Oxidation state of Mg in MgCl2?
The oxidation numbers of atoms in a compound add up to zero. Oxidation state of Mg in MgCl2? +2

10 -2 Oxidation Numbers Oxidation state of S in S2-?
The oxidation numbers of atoms in an ion add up to the charge on the ion. Oxidation state of S in S2-? -2

11 -3 Oxidation Numbers Oxidation state of N in NH3?
The oxidation numbers of atoms in a compound add up to zero. Oxidation state of N in NH3? -3

12 Oxidation Numbers Oxidation state of S in SO42-? ? – 8 = -2 ? = +6
The oxidation numbers of atoms in an ion add up to the charge on the ion. Oxidation state of S in SO42-? ? – 8 = -2 ? = +6

13 -3 Oxidation Numbers Oxidation state of N in NH4+?
The oxidation numbers of atoms in an ion add up to the charge on the ion. Oxidation state of N in NH4+? -3

14 Oxidation: Reduction: Gain of oxygen Loss of electrons Loss of oxygen
Gain of electrons Increase in oxidation number Decrease in oxidation number

15 Lets Practice H2SO3 + I2 + H2O → H2SO4 + HI
Oxid. # H2SO3 + I2 + H2O → H2SO4 + HI Total FeCl3 + SnCl2 → FeCl2 + SnCl4

16 Lets Practice AgNO3 + NaClO + AgCl → AgClO3 + NaNO3
Oxid. # AgNO3 + NaClO + AgCl → AgClO3 + NaNO3 Total MnO2 + FeSO4 + H2SO4 → MnSO4 + Fe2(SO4)3 + H2O

17 Lets Practice Mn(NO3)2 + PbO2 + HNO3+ → HMnO4 + Pb(NO3)2 + H2O
Oxid. # Mn(NO3)2 + PbO2 + HNO3+ → HMnO4 + Pb(NO3)2 + H2O Total HMnO4 + HNO3 + Mn(NO3)2→ MnSO4 + H2O +O2

18 Answers to Practice quest
11. R= Au O= I R. A.= KI O. A.= AuCl3 12. R= N O= Ag R. A.= Ag O. A.= HNO3 13. R= Fe O= Sn R. A.= SnCl2 O. A.= FeCl3 14. R= N O= Fe, S R. A.= FeS O.A.= HNO3 15. R= Cu O= I R. A.= KI O. A.= CuSO4 16. R= I O= H R.A.= H2 O.A.= KIO3 17. R= Cl O= Cl R.A.= NaClO O. A.= NaClO 18. R= O O= Mn R. A.= MnO2 O. A.= O2 19. R= Mo O= I R. A.= KI O. A.= MoO3 20. R= Br O= Mn R. A.= MnCl2 O. A.= Br2


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