Download Notes Set 1

Electrochemistry
In the last chapter, we looked at what happened in a reaction where a
proton was transferred from one chemical species to another. A proton is not,
however, the only part of an atom that can be moved. An electron can also be
moved from one chemical species to another.
Demonstration : Aluminum foil in CuCl2 solution.
- aluminum foil was turned into aluminum ions.
- In order to do this, it had to lose three electrons :
Al (s) ----> Al3+(aq) + 3eWhere did those electrons go ?
- elemental copper produced
- this must have come from the ions in the solution as follows:
Cu2+(aq) + 2e- ----> Cu(s)
In order for the aluminum to lose electrons, there must have been something to
gain the electrons. (If this were not the case, the aluminum foil would disintegrate as soon
as it was made.) Conversely, in order for the copper ions to turn into copper solid, that is
to gain electrons, something must lose the electrons. Clearly, these reactions are
dependant on each other. One doesn't happen without the other.
Definitions :
The loss of electrons (as the aluminum did above) is known as
OXIDATION. It is called oxidation because this is what happens when a metal
oxidizes.
The gaining of electrons (as the copper ions did above) is known as
REDUCTION. It is called reduction because when ores are mined and refined, a
large amount of ore is reduced to a small amount of metal.
The overall reaction that occurred is known as an OXIDATIONREDUCTION reaction, sometimes shortened to a REDOX equation.
An OXIDIZING AGENT is a chemical which causes another to
undergo oxidation. In other words, an oxidizing agent undergoes reduction.
A REDUCING AGENT is a chemical which causes another to
undergo reduction. In other words, a reducing agent undergoes oxidation.
Looking at the above reaction once again, this time in terms of the
HALF-REACTIONS, we get :
Al (s) ----> Al3+(aq) + 3e- oxidation
and
Cu2+(aq) + 2e- ----> Cu(s) reduction
One other thing about the reaction, if we closely examined the solution, we
would find that there are NO free electrons floating around. Nature detests a
solution that has an electrical charge (a POLARIZED solution) therefore we must
look at the reaction once again. If these are the half reactions, one would think
that if we simply added them together,
We would get the overall reaction. When we do this:
Al (s) ----> Al3+(aq) + 3eCu2+(aq) + 2e- ----> Cu(s)
-----------------------------------------------------------Al (s) + Cu2+(aq) ----> Al3+(aq) + Cu(s) + eWe get an excess of electrons produced. This will not happen. We will
not have polarization occurring. In order for us to account for the extra electron,
we must add the two equations together mathematically to eliminate the
electrons .
Balancing REDOX reactions:
a) Write the two half -cell reactions:
Al (s) ----> Al3+(aq) + 3eCu2+(aq) + 2e- ----> Cu(s)
b) The least common multiple of the electrons in the two half reactions is 6,
therefore we multiply each equation in order to make 6 electrons in each :
2 Al (s) ----> 2 Al3+(aq) + 6e3 Cu2+(aq) + 6e- ----> 3Cu(s)
c) Now we can add together the two equations and eliminate the electrons :
2 Al (s) + 3 Cu2+(aq) ----> 2 Al3+(aq) + 3 Cu(s)
The equation is now balanced.
NOTE :
-The total number of atoms of each element is balanced.
- All the charges are balanced.
- Aluminum brings about reduction of Cu 2+ ion and is called
the REDUCING AGENT.
- Cu 2+ ions causes the aluminum to oxidize and they are
called the OXIDIZING AGENT.
Example: The electrical reaction of molten salt
at anode:
Cl- -----> 1/2 Cl2 + eChlorine ion ( anion ) loses an electron to the anode.
- this is oxidation
at cathode: Na+ + e- ------> Na
Sodium ion ( cation ) gains an electron.
- this is reduction
If we add the two equations:
Cl- -----> 1/2 Cl2 + eNa+ + e- ------> Na
.
+
1
Na + Cl -----> Na + /2 Cl2
Easy way to remember what reactions are taking place:
OIL
RIG
Translation: oxidation is loss (of electrons) and reduction
is gain (of electrons)
Another easy way to remember what is going on:
GAIN A RED CAT; LOSE AN ODD OX
Translation: gain electrons, reduction cathode;
lose electrons, anode, oxidation.
Oxidation Numbers
Another easy way to tell if oxidation or reduction has occurred is to
calculate the oxidation number of the element in the compound. The oxidation
number is something like the valence of the element. It is assigned to atoms in
order to keep track of the redistribution of electrons during a chemical reaction.
By keeping account of the oxidation numbers of the reactants and products, it is
possible to determine how many electrons are gained or lost in each atom. The
oxidation number of an atom in a compound is assigned according to the
following rules:
1. The oxidation of any substance in its elemental form is ZERO.
examples:
2. The common oxidation number of hydrogen is +1 except in compounds
called hydrides ( here it is -1 )
examples:
3. The common oxidation number of oxygen is -2
known as peroxides ( here it is -1)
examples:
except in compounds
4. The oxidation of Alkali metals ( Group I A ) in a compound is always +1.
examples:
5. The oxidation of Alkali Earth ( Group IIA ) in a compound is always +2.
examples:
6. The oxidation number for halogens is usually -1, however, there are
exceptions. It can range from -7 to + 7.
examples:
7. The sum of the oxidation numbers of all the atoms in a neutral
compound is zero. The sum of the oxidation numbers of the atoms present
in a polyatomic ion is equal to the charge of the ion.
examples:
** To determine the oxidation nymber, we use the above rules and the **
electrical charge on the substance
example 1: Determine the oxidation number for Cr in Cr2O7 2-
example 2: Determine the oxidation number for C in C3H2O2.
example 3: Determine the oxidation number for Mn in KMnO4.
To determine whether reduction or oxidation has occurred, determine the
change in oxidation numbers. If the change is negative, reduction has
occurred. If the change is positive, oxidation has occurred.
example:
MnO4 - -------> MnO2
example:
As ------> HAsO2
Assignment:
Do Review and Practice questions # 1 to 4, on pp. 627
CHEMISTRY 12
OXIDATION NUMBERS
Determine the oxidation numbers of the element underlined in the following
formula :
1. SO3
16. UO3
31. SO22-
46. CH3OH
2. PF3
17. U3O8
32. SrSiO3
47. C4H4O42-
3. PCl5
18. U2O5
33. S2OCl4
48. C5H10O
4. Na3P
19. K2UO4
34. S2O3Cl4
49. C2H3O2-
5. S2F10
20. MgU2O7
35. H2SO5
50. C7H5O2-
6. S2O7
21. MnSO4
36. SO2ClF
7. NO3-
22. MnO2
37. NH2OH
8. NO2
23. KMnO4
38. C2O42-
9. N2O3
24. MnO3+
39. WO42-
10. NO
25. MnCl4-
40. NO2-
11. N2O
26. H2SO3
41. ClO4-
12. N2
27. H2S2O3
42. HIO64-
13. N3-
28. H2S2O7
43. P2O74-
14. N2H5+
29. KHSO4
44. CH4
15. NH4+
30. S4
45. CH3Cl
ANSWERS :
1. +6
11. +1
21. +2
31. +2
41. +7
2. +3
12. 0
22. +4
32. +4
42. +7
3. +5
13. -1/3
23. +7
33. +3
43. +5
4. -3
14. -2
24. +7
34. +5
44. -4
5. +5
15. -3
25. +3
35. +8
45. -2
6. +7
16. +6
26. +4
36. +6
46. -2
7. +5
17. +16/3
27. +2
37. -1
47.+1/2
8. +4
18. +5
28. +6
38. +3
48. 8/5
9. +3
19. +6
29. +6
39. +6
49. 0
10. +2
20. +6
30. 0
40. +3
50. 2/7
Determining if REDOX reactions will proceed:
See appendix D: Table of Standard Reduction of Half Cell.
Strongest
Oxidizing agents
Strongest
Reducing
Agents
To determine if two species in the table will proceed spontaneously, note
the relative placement of each specie.
*** Any oxidizing agent will react with any reducing agent on the right
which is LOWER on the list***
Experiment 21-A
Some notes on Expt. 21-A
Part 1 - if a reaction occurs, the metal will turn
black.
Part 2 - reaction need to be done in the fume hood.
The colours of the halogens in heptane are as
follows : chlorine : yellow
bromine : orange
iodine : purple
The reason why we use heptane as a solvent rather than water is that all the
halogens are non-polar in nature and will dissolve more readily in a non-polar liquid (such as
heptane) rather than water, a polar molecule. Besides, heptane makes it easier to identify bromine
from iodine.
If a reaction occurs, the colour of the heptane layer will indicate the halogen
present.
Discussion of lab 21-A
Part 1
When a metal undergoes reduction, it gains electrons. In order to tell which of two metals
is the stronger oxidizing agent, we must do one of two experiments as we did in the lab :
A (s) + B+(aq)
- If a reaction occurs then the ion of element B, B+(aq) , has a greater affinity for
electrons than the ion of element A. If it were not the case, the B+(aq) ions could not pull the
electrons away from the metal and a reaction would not occur.
- The other reaction we could do would be exactly the opposite : A+(aq) + B (s). As we
can see in the experiment, the exact opposite result will happen. In the lab, this can be used as a
check of your results.
Example from the lab :
Ag (s) + Pb2+(aq) ----- no reaction therefore Ag+ is a
stronger oxidizing agent than Pb2+
Ag+(aq) + Pb (s) ----- reaction therefore Ag+ is a
stronger oxidizing agent than Pb2+
NOTE : The results are different but the logic is the same.
Part 2
In this section of the experiment, it is not the ion that is gaining the electrons, but it is the
halogen itself. The halide ion can not accept another electron because its outermost electron shell
is full. The results therefore will be opposite that of the metals.
Example - Cl2 (aq) + Br-(aq) - reaction therefore the
- Cl2 is a stronger oxidizing agent than Br2
(because the C l2 was reduced).
Br2 (aq) + Cl-(aq) - no reaction therefore
- the Cl2 is a stronger oxidizing agent than Br2
(because the Cl2 was not reduced).
Questions :
In question #3, the information is not quite explained as good as it could be. The Ag+ ion are
indeed in between the two halogens listed but it should be noted that the rest of the metals used
in this lab are all below the iodine molecule.
Follow- up Questions :
In order to see if there will be a reaction between two chemical species, we must ask ourselves
the following questions :
1. Is there one species gaining electrons and one
losing ? If not, there will not be a reaction.
2. Is the one gaining the electrons a better oxidizing
agent than the one giving the electrons ? If not, a
spontaneous reaction will not occur. This is not to
say that a reaction between the two can not happen,
it is just that the reaction will not happen all by
itself.
- Discussion of the lab Follow up Questions 1. a) A reaction will occur Pb + Br2 ---> Pb2+ + 2Brb) no reaction because I2 is a stronger oxidizing
agent than Zn2+.
- continue through the rest of the questions using the same logical technique.
Assignment :
- Complete the lab write-up. All questions and all follow-up questions are to be
done for tomorrow.
Standard Reduction Potential Chart
The table you have in front of you is a much more complete list than the one you made in
the lab. It will be provided for you on the final exam. It is important that you know how to use this
table.
As you can see, there are several reactions that simply involve the transfer of electrons
from an ion to an element (as we did in the lab). Also, there are a number of other reactions that
have either H+ ions and water in them or OH- ions and water in them. These are the half
reactions that will take place in either acidic or basic solutions.
It should be stressed that These are only half of the reaction that will take place. They are
all reduction reactions. We have seen that reduction can only take place if oxidation is also
occurring. To get an oxidation half reaction all we have to do is to write the reaction in the reverse
order. Notice that the arrows are bidirectional - this does not mean that they are in equilibrium but
that they can proceed in either direction.
As we said before, in order for a reaction to occur spontaneously, both oxidation and
reduction must occur and the reduction reaction must be above the oxidation reaction on
the chart.
Assignment: Standard reduction potential ws.
Chemistry 12
Standard Reduction Potential WS
1. Calculate the Eo cell for each reaction and state whether the reaction is expected to be
spontaneous.
a) Cr + 3 Ag+ --- Cr 3+ + 3 Ag
b) Cu + Fe 3+ -- Cu 2+ + Fe 2+
c) Mn 2+ + 2 H2O + I2 -- MnO2 + 4 H+ + 2 I –
d) 3 Cu + 2 NO3- + 8 H+ --- 3 Cu 2+ + 4 H2O + 2 NO
e) 2 Cr 3 + + 7 H2O + 3 Pb 2+ --- Cr2O7 2- + 14 H+ + 3 Pb
2. Can aluminum foil immersed in an electrolyte solution be used to restore the lustre of
silverware? Show the balanced equations and the cell potential for the overall reaction.
( assume 1M solutions for all reactants).
3. When silver metal ia placed in 1 M HCl solution under standard conditions, there is no
observable reaction. However, when magnesium metal is placed in the same HCl solution,
the metal oxidizes and hydrogen gas is produced. Explain these two phenomena.
4. Will silver metal in a solution of chloride ions produce silver ions and chlorine gas?
5. What will happen if an aluminum spoon is used to stir a solution of Fe(NO3)3?
6. Can a 1M solution of Iron III sulfate be stored in a container made of nickel metal?
Explain your answer.
7. Predict the overall reaction:
a) a mixture of Br2(aq) and Cl2(aq) is added to a beaker containing
CuSO4 (aq) and copper rod.
b) A mixture of powdered aluminum and iron is added to a beaker of Cr 3+ solution.
c) A tin strip is immersed in HNO3(aq)
d) A copper strip is immersed in HNO3(aq)
8. A series of experiments is performed to measure the Eo produced by various
combinations of metals in 1.00 M solution of salts.
Anode
Cathode
Eo (V)
Be
Cd
1.297
Be
Ga
1.180
Ti
Be
0.050
Based on the above,
a) List the metals in order of their activity (strongest reducing agent first)
b) Predict Eo of a Ti/Cd cell.
9. The voltage of the cell
Pb 2+ + Zn(s) -- Pb(s) + Zn 2+ changes depending on the
conditions used. Arrange the following three sets of conditions in order of the voltages
they produce, from highest to lowest. Each pair of half-cells is connected to form an
electrochemical cell using a salt bridge.
a) A thin zinc wire is immersed in 0.5 M Zn 2+ and a thick lead strip is immersed in 1M
Pb 2+.
b) A thick zinc strip is immersed in 1M Zn 2+ and a thin lead wire is immersed in 0.5 M
Pb 2+.
c)A thick zinc strip is immersed in 1M Zn 2+ and a thick lead wire is immersed in 1M Pb
2+.
10. a) Which is a stronger oxidizing agent: 1M Pb 2+ or 2M Pb 2+?
b) Which is stronger reducing agent: 1M I- or 2M I -?
11. A battery is simply an electrochemical cell. If a battery is used continuously in a
flashlight, why does the light from the flashlight grow dimmer and dimmer, rather than
staying at the same intensity until the battery dies?
Balancing Half-Reactions The chart we have is good to use if we have a reaction that is on it (or if we actually have
it in our possession at the time we need it). If it isn't on the chart, or if we don't have a chart
handy, there is a straight forward, simple way to balance all half reactions :
Rules for balancing half reactions :
1. Balance all elements except hydrogen and oxygen. Do this the same way we
would normally balance equations, by changing coefficients.
2. Balance the oxygen atoms by adding water molecules.
3. Balance the hydrogen atoms by adding H+ ions.
4. Balance the electrical charge by adding electrons.
**5. To make the solution basic (if asked for) add the exact same number of OH- ions
as H+ i ons to both sides of the equation. This will eliminate H+ ions by producing water.
Cancel the excess water.
Major OHExamples :
Mn3+ -----> MnO4- acidic conditions
done - to check, count all the individual atoms on both sides of the equation. As well, the
TOTAL CHARGE on both sides of the equation MUST be EQUAL.
example:
I2 -----> IO3- acidic
Example:
MnO4- ----> MnO2 basic
How do we tell whether or not oxidation or reduction has occurred ?
- Assignment :
- Balancing Half-Cell Reactions
Balancing Redox equations using oxidation numbers.
The overall change in oxidation numbers must be zero.
1) Assign the oxidation numbers to all atoms involved in a change of oxidation number.
2) Since the change in oxidation number must be zero, cross-multiply each half reaction
by the oxidation numbers involved.
3) Add the two half-reactions.
4) Balance the “O’s” by adding H2O
5) Balance the “H’s” by adding H+
Example:
Balance the following equation:
1) U4+ + MnO4 - ---- Mn2+ + UO2
2+
( in acidic solution)
2) Zn + As2O3 ----- AsH3 + Zn 2+ ( in basic solution)
3) P4 ---- H2PO2
-
+ PH3 ( acidic solution)
CHEMISTRY 12
BALANCING HALF-CELL REACTIONS
Balance the following half-cells.
1.
Ce4+ ---> Ce2+
2.
I2 ---> I-
3.
Mn2+ ---> MnO2 (acidic solution)
4.
O2 ---> H2O2 (acidic solution)
5.
S2O82- ---> HSO4- (acidic solution)
6.
MnO4- ---> MnO2 (acidic solution)
7.
NO3- ---> NO (acidic solution)
8.
ClO3- ---> Cl- (acidic solution)
9.
H3AsO4 ---> HAsO2 (acidic solution)
10.
H2SeO3 ---> Se (acidic solution)
11.
HO2- ---> O2 (basic solution)
12.
N2H4 ---> N2 (basic solution)
13.
Cr2O72- ---> Cr3+ (acidic solution)
14.
HXeO4- ---> HXeO63- (basic solution)
15.
Cr(OH)3 ---> CrO42- (basic solution)
16.
FeS ---> Fe3+ + SO42- (acidic solution)
17.
Cu2S ---> Cu2+ + H2SO3 (acidic solution)
18.
FeHPO3 ---> PO43- + Fe(OH)3 (basic solution)
ANSWERS :
1.
Ce4+ + 2e- ---> Ce2+
2.
I2 + 2e- ---> 2I-
3.
Mn2+ + 2H2O ---> MnO2 + 4H+ + 2e-
4.
O2 + 2H+ + 2e- ---> H2O2
5.
S2O82- + 2H+ + 2e- ---> 2HSO4-
6.
MnO4- + 4H+ + 3e- ---> MnO2 + 2H2O
7.
NO3- + 4H+ + 3e- ---> NO + 2H2O
8.
ClO3- + 6H+ + 6e- ---> Cl- + 3H2O
9.
H3AsO4 + 2H+ + 2e- ---> HAsO2 + 2H2O
10.
H2SeO3 + 4H+ + 4e- ---> Se + 3H2O
11.
HO2- + OH- ---> O2 + H2O + 2e-
12.
N2H4 + 4OH- ---> N2 + 4H2O + 4e-
13.
Cr2O72- + 14H+ + 6e- ---> 2Cr3+ + 7H2O
14.
HXeO4- +4OH- ---> HXeO63- + 2H2O + 2e-
15.
Cr(OH)3 + 5OH- ---> CrO42- + 4H2O + 3e-
16.
FeS + 4H2O ---> Fe3+ + SO42- 8H+ + 9e-
17.
Cu2S + 3H2O ---> 2Cu2+ + H2SO3 + 4H+ + 8e-
18.
FeHPO3 + 6OH- ---> PO43- + Fe(OH)3 + 2H2O + 3e-
Balancing Full Reactions
In order to balance overall equations, it is important to do the following:
1. Isolate the two half reactions
2. Balance each half reaction separately
3. Find Least Common Multiple for the electrons
4. Add equations together to get rid of free
electrons
Example: Mn3+ + Au 3+
------->
MnO4-
+
Au
Note: in the overall balanced equation the numbers of atoms of each type are the same on
each side of the equation. Also, the electrical charges are equal. Both of these rules are
the same as with half reactions but, most importantly, in an overall reaction, there are NO
ELECTRONS FREE TO ROAM AROUND!!!
Example: Cr2O7 2-
+ HXeO4-
---->
Cr3+
+
HXeO63-
example:
Example:
MnO4-(aq) + Fe2+(aq) ---------- > Fe3+(aq) + Mn2+(aq)
Ag(s) + CN-(aq) + O2(aq) -------------- > Ag(CN)2-(aq)
Further Notes on Balancing Redox Eqations
1.
Sometimes, there are bizarre circumstances where the reactant
undergoes BOTH oxidation and reduction.
In order to balance
these equations, we must make two equations -- one oxidation and
one reduction. From here, we can go through the procedures we did
in the previous section.
example:
2.
Br2
----> HBr
+ HBrO3
Sometimes there are more than one species undergoing either
oxidation or reduction ( an even more bizarre case ). In situations
such as these, we should try and isolate one element in its own half
reaction and place the others in a separate half reaction.
example:
Mn3+ + FeSO4
---->
MnO4- +
Fe + SO2 ( basic )
example: Gold metal will not dissolve in either concentrated nitric acid or
concentrated hydrochloric acid. It will dissolve, however, in aqua regia, a mixture
of the two concentrated acids. The products of the reaction are the AuCl 4- and
gaseous NO. Write a balanced equation for the dissolution of gold in aqua
regia.
example: Chlorine was first prepared in 1774 by C.W. Scheele by oxidizing
sodium choloride with manganese (IV) oxide. The reaction is:
NaCl(aq) + H2SO4(aq) + MnO2(s) ---------> Na2SO4(aq) + MnCl2(aq) + H2O(l) + Cl2(aq)
Balance this equation:
CHEMISTRY 12
BALANCING OXIDATION-REDUCTION EQUATIONS
1.
Cl2 + SO2 ---> Cl- + SO42- (acidic)
2.
Cu + NO3- ---> Cu2+ + NO (acidic)
3.
S2- + ClO3- ---> Cl- + S (basic)
4.
Zn + As2O3 ---> AsH3 + Zn2+ (acidic)
5.
Fe2+ + Cr2O72- ---> Cr3+ + Fe3+ (acidic)
6.
U4+ + MnO4- ---> Mn2+ + UO22+ (acidic conditions)
7.
CN- + IO3- ---> I- + CNO-
8.
Mn2+ + HBiO3 ---> Bi3+ + MnO4- (acidic)
9.
HSO3- + IO3- ---> I2 + SO42- (acidic)
The following equations involve a species which undergoes BOTH oxidation and reduction at the
same time. Use two separate half cells with the same reactant.
10.
OCl- ---> Cl- + ClO3-
11.
HNO2 ---> HNO3 + NO
12.
Br2 ---> Br- + BrO3- (basic)
The following equations involve three changes in oxidation numbers. HINT : One of the species
can be isolated. Use this in a half cell by itself. The other two can not be separated. Use a half cell
involving both.
13.
Sb2S3 + NO3- ---> NO2 + SO42- + Sb2O5 (acidic)
14.
As2S3 + NO3- ---> NO + SO42- + H3AsO4 (acidic)
15.
FeS + NO3- ---> NO + SO42- + Fe3+ (acidic)
16.
FeHPO3 + Cr2O72- ---> Cr3+ + H3PO4 + Fe3+ (acidic)
17.
SnS2O3 + MnO4- ---> Mn2+ + SO42- + Sn4+ (acidic)
18. FeHPO3 + OCl- ---> Cl- + PO43- + Fe(OH)3 (acidic)
Answers :
1.
Cl2 + SO2 + 2H2O ---> 2Cl- + SO42- + 4H+
2.
3Cu + 2NO3- + 8H+ ---> 3Cu2+ + 2NO + 4H2O
3.
3S2- + ClO3- + 3H2O ---> 3S + Cl- + 6OH-
4.
6Zn + As2O3 + 12H+ ---> 2AsH3 + 6 Zn2+ + 3H2O
5.
6Fe2+ + Cr2O72- + 14H+ ---> 2Cr3+ + 6Fe3+ + 7H2O
6.
2H2O + 5U4+ + 2MnO4- ---> 2 Mn2+ + 5UO22+ + 4H+
7.
3CN- + IO3- ---> I- + 3CNO-
8.
2Mn2+ + 5HBiO3 + 9H+ ---> 5Bi3+ + 2MnO4- + 7H2O
9.
5HSO3- + 2IO3- ---> I2 + 5SO42- + 3H+ + H2O
10.
3OCl- ---> 2 Cl- + ClO3-
11.
3HNO2 ---> HNO3 + 2NO + H2O
12.
3Br2 + 6OH- ---> 5Br- + BrO3- + 3H2O
13.
Sb2S3 + 22H+ + 28NO3- ---> 3SO42- + Sb2O5 + 28NO2 + 11H2O
14.
3As2S3 + 4H2O + 10H+ + 28NO3- ---> 9SO42- + 6H3AsO4 + 28NO
15.
FeS + 3NO3- + 4H+ ---> SO42- + Fe3+ + 3NO + 2H2O
16.
2FeHPO3 + Cr2O72- + 14H+ ---> 2H3PO4 + 2Fe3+ + 2Cr3+ + 5H2O
17.
SnS2O3 + 2MnO4- + 6H+ ---> 2SO42- + Sn4+ + 2Mn2+ + 3H2O
18.
2FeHPO3 + 5H2O + 3OCl- ---> 2PO43- + 2Fe(OH)3 + 6H+ + 3Cl-
19.
6Hg4Fe(CN)6 + 66H+ + 47ClO3- ---> 36NO + 36CO2 + 6Fe3+ +
24Hg2+ + 47Cl- + 33H2O
Stoichiometry and Redox Reactions: Redox Titrations
Similar to acid base titrations.
1. Oxidizing Agents: Acidic Potassium permanganate:
- strong oxidizing agent.
- MnO4- - purple ----- > Mn2+ - colourless
- Other oxidizing agents:
Use standard KMnO4 to determine the concentration of an unknown solution of
Fe2+.
Equation:
MnO4- + Fe2+ ----- Mn2+ + Fe3+
Balanced equation:
Example 1:
Example: A 3.00 g sample of pure iron is dissolved in hydrochloric acid and the
resulting solution treated to produce Fe2+. The final volume of the solution is
500.0 mL. What volume of 0.0500 M KMnO4 is required to titrate 25.0 mL
sample of Fe 2+ solution? (10.8 mL)
2. Reducing Agents: NaI, or KI (for I-)
Thiosulphate (S2O3 2-)
Sulphite (SO3 2-)
Oxalic acid (H2C2O4)
Example:
Analysis of Bleach can be done by reacting the solution with an excess amount
of I-. The reaction:
Titrate a known volume of this sample by reacting the solution with Na 2S2O3 and
using starch solution as indicator. The solution is a dark blue colour because of
the reaction between the starch and the Iodine.
At the endpoint, the iodine reacts completely, forming I-. The endpoint is
reached when the blue colour fades.
Example: A 25 mL sample of bleach is reacted with an excess amount of NaI.
The solution is then titrated with 0.500M solution of sodium thiosulphate. If
35.00 mL of the thiosulphate solution was required to bring the solution to
endpoint, what is the [OCl -] in bleach? (0.35 M)