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CHEMISTRY 161
Chapter 4
REVISION
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
H+(aq) + OH-(aq) → H2O(l)
(Arrhenius)
HCl(aq) + NH3(aq) → NH4Cl(aq)
H+(aq) + NH3(aq) → NH4+(aq)
(Bronsted)
EXAMPLE 1
KOH(aq) + HF(aq) → ???
1. KOH(s) → K+(aq) + OH-(aq)
2. HF(g) →
H+(aq)
+
F-(aq)
reactants
1.+2. K+(aq) + OH-(aq) + H+(aq) + F-(aq)
→
H2O(l) + K+(aq) + F-(aq)
KOH(aq) + HF(aq) → H2O(l) + KF(aq)
EXAMPLE 2
Mg(OH)2(aq) + HCl(aq) → ???
Mg(OH)2(aq) + 2 HCl(aq) → ???
1. Mg(OH)2(s) → Mg2+(aq) + 2 OH-(aq)
2. 2 HCl(g) → 2
H+(aq)
+2
Cl-(aq)
reactants
1.+2. Mg2+(aq) + 2 OH-(aq) + 2 H+(aq) + 2 Cl-(aq)
→
2 H2O(l) + Mg2+(aq) + 2 Cl-(aq)
Mg(OH)2(aq) + 2 HCl(aq) → 2 H2O(l) + MgCl2(aq)
EXAMPLE 3
Ba(OH)2(aq) + H2SO4(aq) → ???
1. Ba(OH)2(s) → Ba2+(aq) + 2 OH-(aq)
2. H2SO4(l) → 2 H+(aq) + SO42-(aq)
reactants
1.+2. Ba2+(aq) + 2 OH-(aq) + 2 H+(aq) + SO42-(aq)
→
2 H2O(l) + Ba2+(aq) + SO42-(aq)
Ba(OH)2(aq) + H2SO4(aq) → 2 H2O(l) + BaSO4(s)
CHEMICAL REACTIONS
a) precipitation reactions
b) acid-base reactions (proton transfer)
c)redox reactions (electron transfer)
H→
+
H
+
e
REDOX REACTIONS
reactions involving transfer of electrons
(solid state reaction of potassium with sulfur)
(1) K → K+ + e-
/×2
oxidation
(2) S + 2 e- → S-2 / × 1
reduction
(1) 2 K → 2 K+ + 2 e-
(2) S + 2 e- → S-2
(1)+(2) 2K + S + 2e-→ 2 K+ + S2- +2e(1)+(2) 2K + S → 2 K+ + S2-
2K(s) + S(s) → K2S(s)
KEY CONCEPTS
1. oxidation
loss/donation of electrons
(minus)
2. reduction
gain/acceptance of electrons
(plus)
o
m
r
p
m
o
p
r
3. balance electrons
4. add oxidation half reaction and reduction half reaction
EXAMPLE
reaction of calcium with molecular oxygen
(1) Ca → Ca2+ + 2e-
oxidation
(2) ½ O2 + 2 e- → O2-
reduction
(1)+(2) Ca + ½ O2 + 2e-→ Ca2+ + O2- +2e-
(1)+(2) Ca + ½ O2 → Ca2+ + O2Ca(s) + ½ O2(g) → CaO(s)
2 Ca(s) + O2(g) → 2 CaO(s)
/ ×2
OXIDATION NUMBER
ionic compounds ↔ molecular compounds
NaCl
HF, H2
Na+Cl-
?
electrons are fully transferred
covalent bond
‘oxidation number’
charges an atom would have if electrons are
transferred completely
EXAMPLE 1
H+ + F-
HF
molecular compound
ionic compound
H+
oxidation state +1
F-
oxidation state -1
EXAMPLE 2
H2O
2 H+ + O2-
molecular compound
ionic compound
H+
oxidation state +1
O2-
oxidation state -2
EXAMPLE 3
H2
molecular compound
H+ + Hionic compound
OXIDATION NUMBER OF FREE ELEMENTS IS ZERO
RULE 1
OXIDATION NUMBER OF FREE ELEMENTS IS ZERO
H2, O2, F2, Cl2, K, Ca
RULE 2
monoatomic ions
oxidation number equals the charge of the ion
group I
M+
group II
M2+
group III
M3+ (Tl: also +1)
group VII
F-
RULE 3
oxidation number of hydrogen
+1 in most compounds
(H2O, HF, HCl, NH3)
-1 binary compounds with metals (hydrides)
(LiH, NaH, CaH2, AlH3)
RULE 4
oxidation number of oxygen
-2 in most compounds
(H2O, MgO, Al2O3)
-1 in peroxide ion (O22-) (H2O2, K2O2, CaO2)
-1/2 in superoxide ion (O2-) (LiO2)
+1 in FO
RULE 5
oxidation numbers of halogens
F: -1 (KF)
Cl, Br, I: -1 (halides) (NaCl, KBr)
Cl, Br, I: positive oxidation numbers if combined
with oxygen (ClO4-)
RULE 6
neutral molecule: sum of oxidation numbers must be zero
NH3
H: +1 N: -3
3 × (+1) + 1 × (-3) = 0
polyatomic ion: sum of oxidation numbers equals charge of ion
NH4+ H: +1 N: -3
4 × (+1) + 1 × (-3) = +1
RULE 7
charges of polyatomic molecules must be integers
(NO3-, SO42-)
oxidation numbers do not have to be integers
-1/2 in superoxide ion (O2-)
MENUE
1.oxidation states of group I – III metals
2.oxidation state of hydrogen (+1, -1)
3. oxidation states of oxygen (-2, -1, -1/2, +1)
4.oxidation state of halogens
5.remaining atoms
K2O
PO43-
NO+
SO42-
KO2
SO3
NO3NO2
BrO-
NO
SO2
NO2-
NO-
SUMMARY
1.redox reactions
2. oxidation versus reduction
3. oxidation numbers versus charges
4. calculation of oxidation numbers
Homework
Chapter 4, p. 116-121
problems