Transcript Slide 1

Exp. 16 – video
(time: 1 hr and 23:08 minutes)
Exp. 16: Volumetric Analysis: Redox Titration
Normality = eq wt of solute
L solution
Acid/bases: #eq = # H+ or OH- ionized
Redox reactions – transfer of ereduction – oxidation reactions
Redox reaction
Equivalent wt - one equivalent of any oxidizing agent
reacts with one equivalent of any reducing agent.
This means #eq/mol is equal to the number of etransferred.
MnO4-(aq) + 8H+(aq) + 5e-  Mn2+(aq) + 4H2O(l)
MnO4- : 5eq
mol MnO4-
Fe2+(aq)  Fe3+(aq) + 1e-
same for KMnO4
1eq
mol Fe2+
(net)
N  M or M  N
N (eq) = M (mol) x #eq
L
L
mol
Note: N equal to or greater than M
0.1 M KMnO4  N?
Goal: eq KMnO4
L soln
MnO4-(aq) + 8H+(aq) + 5e-  Mn2+(aq) + 4H2O(l)
Calc:
Solubility Rules for Ionic Compounds (Dissociates 100%)
1.) All compounds containing alkali metal cations and the ammonium ion
are soluble.
2.) All compounds containing NO3-, ClO4-, ClO3-, and C2H3O2- anions are
soluble.
3.) All chlorides, bromides, and iodides are soluble except those containing
Ag+, Pb2+, or Hg22+.
4.) All sulfates are soluble except those containing Hg22+, Pb2+, Ba2+, Sr2+,
or Ca2+. Ag2SO4 is slightly soluble.
5.) All hydroxides are insoluble except compounds of the alkali metals and
Ca2+, Sr2+, and Ba2+ are slightly soluble.
6.) All other compounds containing PO43-, S2-, CO32-, CrO42-, SO32- and
most other anions are insoluble except those that also contain alkali
metals or NH4+.
Generally, compound dissolves
Hg2Cl2 (s)
KI (aq)
Pb(NO3)2 (aq)
insoluble
soluble
soluble
> 0.10 M - soluble (aq)
< 0.01 M - insoluble (s)
in between - slightly soluble
(this class we will assume slightly soluble as soluble)
4
Strong Acids (Ionizes 100%)
HCl, HBr, HI, HClO4, HNO3, H2SO4
Strong Bases (Dissociates 100%)
NaOH, KOH, LiOH, Ba(OH)2, Ca(OH)2,
Sr(OH)2
5
Ions in Aqueous Solution
Molecular and Ionic Equations
• A molecular/formula unit equation is one in which the
reactants and products are written as if they were
molecules/formula units, even though they may actually
exist in solution as ions.
Calcium hydroxide + sodium carbonate
M.E.
Ca(OH)2 (aq) +
Na2CO3 (aq)  CaCO3 (s) + 2 NaOH (aq)
strong base
soluble salt
insoluble salt
strong base
s solid
l liquid
aq aqueous (acid/bases and soluble salts dissolve in water)
g gases
6
Ions in Aqueous Solution
Molecular and Ionic Equations
• An total ionic equation, however, represents strong electrolytes as
separate independent ions. This is a more accurate representation of the
way electrolytes behave in solution.
– A complete ionic equation is a chemical equation in which strong
electrolytes (such as soluble ionic compounds, strong acids/bases) are
written as separate ions in solution. (note: g, l, insoluble salts (s), weak
acid/bases do not break up into ions)
M.E.
Ca(OH)2 (aq) +
strong base
Na2CO3 (aq)  CaCO3 (s) + 2 NaOH (aq)
soluble salt
insoluble salt
strong base
Total ionic
Ca2+ (aq) + 2OH- (aq) + 2Na+ (aq) + CO32- (aq) 
CaCO3 (s)
+ 2Na+ (aq) + 2OH- (aq)
7
Net ionic equations.
– A net ionic equation is a chemical equation from
which the spectator ions have been removed.
– A spectator ion is an ion in an ionic equation that
M.E. does not take part in the reaction.
Ca(OH)2 (aq) +
Na2CO3 (aq)  CaCO3 (s) + 2 NaOH (aq)
Total Ionic
Ca2+ (aq) + 2OH- (aq) + 2Na+ (aq) + CO32- (aq)  CaCO3 (s) + 2Na+ (aq) + 2OH- (aq)
Net
Ca2+ (aq) + CO32- (aq)
 CaCO3 (s)
8
Types of Chemical Reactions
• Oxidation-Reduction Reactions (Redox rxn)
– Oxidation-reduction reactions involve the
transfer of electrons from one species to another.
– Oxidation is defined as the loss of electrons.
– Reduction is defined as the gain of electrons.
– Oxidation and reduction always occur
simultaneously.
9
27.1 Reduction and Oxidation
Redox reactions – transfer of ereduction – oxidation reactions
Reduction – gain of e- / gain of H / lost of O
Fe3+ + 1e-  Fe2+
(lower ox state)
note: must balance atoms and charges
10
Oxidation - loss of e- / loss of H / gain of O
Fe2+
 Fe3+ + 1e-
(higher ox state)
Br + 4(-2) = -1
Br = -1 +8 = +7
H2O + BrO3-  BrO4- + 2H+ + 2eBr + 3(-2) = -1
(Br oxidized: charge 5+  7+)
Br = -1 +6 = +5
2H+ + 2e-  H2
(H reduced: charge 1+  0)
Oxidizing agent is species that undergoes reduction.
Reducing agent is species that undergoes oxidation.
Note: need both for reaction to happen; can’t have
something being reduced unless something else is being
oxidized.
11
27.3 Balancing Redox Reactions
- Must know charges (oxidation numbers) of species
including polyatomic ions
- Must know strong/weak acids and bases
- Must know the solubility rules
Oxidation Numbers – hypothetical charge assigned to the
atom in order to track electrons; determined by rules.
12
Rules to balance redox
1.) Convert to net ionic form if equation is originally in molecular form
(eliminate spectator ions).
2.) Write half reactions.
3.) Balance atoms using H+ / OH- / H2O as needed:
– acidic: H+ / H2O put water on side that needs O or H (comes from
solvent)
– basic: OH- / H2O put water on side that needs H but if there is no H
involved then put OH- on the side that needs the O in a 2:1 ratio
2OH- / H2O balance O with OH, double OH, add 1/2 water to
other side.
4.) Balance charges for half rxn using e-.
5.) Balance transfer/accept number of electron in whole reaction.
6.) Convert equation back to molecular form if necessary (re-apply
spectator ions).
13
Zn(s) + AgNO3(aq)  Zn(NO3)2(aq) + Ag(s)
Total ionic:
Zn(s) + Ag+(aq) + NO3-(aq)  Zn2+(aq) + 2NO3-(aq) + Ag(s)
Net ionic:
Zn(s) + Ag+(aq)  Zn2+(aq) + Ag(s)
14
Zn(s) + Ag+(aq)  Zn2+(aq) + Ag(s)
Net:
Oxidation:
Zn(s)  Zn2+(aq) + 2e-
Reduction:
[ 1e- + Ag+(aq)  Ag(s) ] 2
Balanced net:
Zn(s) + 2 Ag+(aq)  Zn2+(aq) + 2 Ag(s)
Balanced eq:
Zn(s) + 2 AgNO3(aq)
 Zn(NO3)2(aq) + 2 Ag(s)
15
MnO4-(aq) + Fe2+(aq)
Net:
[ Fe2+(aq)  Fe3+(aq) + 1e- ] 5
Ox:
Red:
H+
 Mn2+(aq) + Fe3+(aq)
5e- + 8 H+(aq) + MnO4-(aq)  Mn2+(aq) + 4 H2O(l)
Balanced net:
8 H+(aq) + MnO4-(aq) + 5 Fe2+(aq)  Mn2+(aq) + 5 Fe3+(aq) + 4 H2O(l)
16
KMnO4(aq) + NaNO2(aq) + HCl(aq)  NaNO3(aq) + MnCl2(aq) + KCl(aq) + H2O(l)
Net:
MnO4-(aq) + NO2-(aq) + H+(aq)  NO3-(aq) +
Mn2+(aq)
+ H2O(l)
H2O(l) + NO2-(aq)  NO3-(aq) + 2 H+(aq) + 2 e- ] 5
Ox:
[
Red:
[ 5 e- + 8 H+(aq) + MnO4-(aq)  Mn2+(aq) + 4 H2O(l) ] 2
Balanced net:
2 MnO4-(aq) + 5 NO2-(aq) + 16 H+(aq) + 5 H2O(l)  2Mn2+(aq) + 8 H2O(l) + 5 NO3-(aq) +10 H+(aq)
2 MnO4-(aq) + 5 NO2-(aq) + 6 H+(aq)  2Mn2+(aq) + 3 H2O(l) + 5 NO3-(aq)
Balanced eq:
2 KMnO4(aq) + 5 NaNO2(aq) + 6 HCl(aq)  2MnCl2(aq)+ 3 H2O(l)+ 5 NaNO3(aq) + 2 KCl
17
OH-
Net:
CrI3 (s) + Cl2 (g)  CrO42-(aq) + IO4-(aq) + Cl-(aq)
Ox:
[ 32 OH-(aq) + CrI3(s)  CrO42-(aq) + 3 IO4-(aq) + 16 H2O(l) + 27 e- ] 2
Red:
[ 2 e- + Cl2(g)  2 Cl-(aq) ] 27
Balanced net:
64 OH-(aq) + 2 CrI3(s) + 27 Cl2(g)  2 CrO42-(aq) + 6 IO4-(aq) + 54 Cl-(aq) + 32 H2O(l)
18
Exp 16:
S2O32- (aq)
thiosulfate ion
Ox:
Red:
Balanced net:
+
I2

iodine
S4O62-(aq) + I-(aq)
2 S2O32-(aq)  S4O62-(aq) + 2 e-
2 e- + I2(aq)  2 I-(aq)
2 S2O32-(aq) + I2(aq)  S4O62-(aq) + 2 I-(aq)
Outside exercise II page 199 – posted on my website
S2O32I2
2eq
=
2mol S2O32-
1 eq
mol S2O32-
2eq
mol I2
Exp today
First: Standardize thiosulfate against 0.100 N I2 standard
solution.
Changes in sample preparation:
10 mL I2, 30 mL deionized H2O, 1 mL starch (20 drops)
Starch – indicator (add from beginning)
Starch + I2 gives blue color
At end pt (all I2 consumed), solution will be colorless
Since using normality can use
NiodineViodine = NthiosulfateV thiosulfate
minimum 3 runs ± 0.005 N (around ± 0.5 mL)
report
Avg N ± s N thiosulfate ion (S2O32-)
Convert average N to M
Second: Same exact procedure as standardization
except using unknown conc. of I2.
minimum 3 runs ± 0.005 N (around ± 0.5 mL)
report
Avg N ± s N iodine (I2) unknown
Convert average N to M
Amount of chemicals to obtain in small beaker per
group:
Na2S2O3.5H2O – 150 mL
(source of thiosulfate ions)
0.100 N I2 standard solution – 50 mL
Unknown I2 solution – 45 mL