Download (redox ) reaction

Exercise 4
Oxidation-reduction (redox) reaction – oxidimetry.
Theoretical part
In oxidation-reduction (or redox) reactions, the key chemical event is the net movement of
electrons from one reactant to the other. Redox reactions constitute some of the most
important of all chemical processes, including the formation of a compound from its elements,
all combustion reactions, the reactions in batteries that generate electricity, and the production
of biochemical energy.
1. Redox terminology
Oxidation is the loss of electrons, reduction is the gain of electrons.
During the formation of magnesium oxide (for example) Mg undergoes oxidation (electron
loss) and O2 undergoes reduction (electron gain). The loss and gain are simultaneous, but we
can imagine them occurring in separate steps:
Oxidation: Mg → Mg 2+ + 2e- (electron lost by Mg)
Reduction: ½ O2 → O2- (electrons gained by O2)
Since O2 gained the electrons that Mg lost when Mg was oxidized, we say that O2
oxidized Mg, or that O2 is the oxidizing agent, the species doing the oxidizing. Similarly,
since Mg gave up the electrons that O2 gained when O2was reduced, Mg reduced O2. Mg is the
reducing agent, the species doing the reducing.
It is extremely important to realize that the oxidation and the reduction occur
simultaneously. There is no such chemical change as an oxidation reaction or a reduction
reaction; only an oxidation-reduction reaction can occur.
e-
X
Transfer
or shift of
electrons
Y
X loses electron (s)
Y gains electron (s)
X is oxidized
Y is reduced
X is the reducing agent
Y is the oxidizing agent
X increases in oxidation
Y decreases in oxidation number
number
1
Each atoms is assigned an oxidation number (O.N., or oxidation state), which is
determined by the set of rules.
General rules
1. For an atom in its elemental form (Na, O2, Cl2, and so forth): O.N. = 0
2. For a monatomic ion: O.N. = ion charge
3. The sum of O.N. values for the atoms in a compound equals zero. The sum of O.N.
values for the atoms in a polyatomic ion equals the ion charge.
Rules for specific atoms or periodic table groups
1. For Group 1A:
O.N. = +1 in all compounds
2. For Group 2A:
O.N. = +2 in all compounds
3. For hydrogen: O.N. = +1 in combination with nonmetals
O.N. = -1 in combination with metals and boron
4. For fluorine:
O.N. = -1 in all compounds
5. For oxygen:
O.N. = -1 in peroxides
O.N. = -2 in all other compounds 9exept with F)
6. For Group 7A:
O.N. = -1 in combination with metals, nonmetals (except O), and other
halogens lower in the group
Balancing redox reactions
Since oxidation and reduction occur simultaneously, the transferred electrons are never free
and must be accounted for in the overall process. Therefore, the total number of electrons lost
by the oxidized reactant must equal the total number of electrons gained by the reduced
reactant. By keeping track of the changes in oxidation numbers and the number of electrons
transferred, we can balance redox reactions.
Two methods used to balance redox reactions are the oxidation number method and
the half-reaction method.
The oxidation number method. The oxidation number method consists of five steps that use
the changes in oxidation numbers to generate balancing coefficients.
Step 1. Assign oxidation numbers to all elements in the equation.
0
+5
+1
-2
Cu + HNO3
+2 +5 -2
+4 -2
+1 -2
Cu(NO3)2 + NO2 + H2O
Step 2. From the changes in oxidation numbers, identify the oxidized and reduced species.
2
Step 3. Compute the number of electrons lost in the oxidation and gained in the reduction
from the oxidation number changes. Draw tie-lines between these atoms to show electron
changes.
-2e Cu + HNO3
Cu(NO3)2 + NO2 + H2O
+1e -
Step 4. Multiply one or both of these numbers by appropriate factors to make the electrons lost
equal the electrons gained, and use the factors as balancing coefficients.
Cu + 2HNO3
Cu(NO3)2 + 2NO2 + H2O
Step 5. Complete the balancing by inspection, adding states of matter.
Cu + 4HNO3
Cu(NO3)2 + 2NO2 + 2H2O
The half-reaction method. The other method for balancing redox reactions is called the halfreaction method because the overall oxidation-reduction process is viewed as the sum of
oxidation “half-reaction” and a reduction “half-reaction”, each of which involves either the
loss or the gain of electrons.
Consider the reaction between zinc metal and hydrochloric acid:
0
+1 -1
+2 -1
Zn + HCl
0
ZnCl2 + H2
We can break up this equation into two half-reactions.
Zn → Zn2+ + 2e- [oxidation half-reaction]
e- + H+ → ½ H2 [reduction half-reaction]
In each half-reaction, moles of atoms and amount of charge are balanced. To obtain integer
coefficients and to make the number of electrons lost equal the number gained, we double the
reduction half-reaction:
Zn → Zn2+ + 2e- [oxidation half-reaction]
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2e- + 2H+ → H2 [reduction half-reaction doubled]
Next, we add the half-reactions together, and the electrons cancel:
Zn + 2H+ → Zn2+ + H2
Adding the appropriate number of spectator gives the balanced total equation:
Zn + 2HCl → ZnCl2 + H2
Voltaic Cells.
Electrochemistry is the study of chemical reactions by use of electrical circuits. This is a
natural way to investigate oxidation/reduction reactions since there is an electron transfer
occurring. By forcing the transferred electrons to move through a wire, we can measure
reaction
progress
(kinetics),
composition
(equilibrium
constants),
energy changes
(thermodynamics), and add or subtract energy from the system.
A typical electrochemical cell:
The oxidation and reduction half–reactions are physically separated and connected by an
electrical circuit.
The electrical circuit is completed by using two different types of connections.
The wire: this is the path that electrons flow from the oxidation to the reduction.
The salt bridge: this completes the electrical circuit but does not allow electron flow. Rather,
the charge is carried by ions (cations or anions). The salt bridge also prevents the two reacting
solutions from mixing.
The electrode where oxidation occurs is called the anode.
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The electrode where reduction occurs is called the cathode.
If the net oxidation/reduction reaction is spontaneous, the cell is called a voltaic cell.
Useful work can be done by voltaic cells (these are batteries).
If the net oxidation/reduction reaction is nonspontaneous, the cell is called an electrolytic cell.
Energy must be supplied to an electrolytic cell in order to drive the oxidation/reduction
reaction.
Shorthand notation used to describe electrochemical cells:
anode reaction||cathode reaction
The double bars (||) represent the salt bridge
The essential components of each half–reaction are described and each piece of information is
separated by a single bar (|). This may include the nature of the electrode, phase information,
concentrations or partial pressures, or temperature if the cell is not at standard conditions.
Examples
Write the half–reactions and net reaction for the cell Mg(s)|Mg2+(aq)||Fe3+(aq)|Fe(s)
Anode (oxidation) reaction:
Mg(s)
Mg2+(aq) + 2e–
The solid magnesium serves as a physical electrode.
Cathode (reduction) reaction:
Fe3+(aq) + 3e–
Fe(s)
The solid iron serves as a physical electrode.
Net reaction:
3Mg(s) + 2Fe3+(aq)
3Mg2+(aq) + 2Fe(s)
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Write the half–reactions and net reaction for the cell Al(s)|Al3+(aq)||Hg2+(aq)|Hg(l)|Pt(s)
Anode (oxidation) reaction:
Al(s)
Al3+(aq) + 3e–
The solid aluminum serves as the physical electrode.
Cathode (reduction) reaction:
Hg2+(aq) + 2e–
Hg(l)
The solid platinum serves as the physical electrode.
Net reaction:
2Al(s) + 3Hg2+(aq)
2Al3+(aq) + 3Hg(l)
Voltaic cells are used to convert chemical energy to electrical energy that can be used to
produce useful work
Calculation of the Cell Potential of Standard Voltaic Cells:
Whenever two standard half-cells are joined to create a voltaic cell as in figure 1, the one with
the more negative Eo will function as the anode since it is the metal that is most easily
oxidized. The value for the standard potential of the cell, Eocell, is the difference between the
standard potential of the cathode and anode:
Eocell = Eocathode - Eoanode
The use of standard reduction potentials to predict the potential of the voltaic cell
illustrated below.
6
is
Reduction Reactions
Eo
Zn2+(aq) + 2 e– Æ Zn(s)
-0.76 (more negative potential, therefore the anode)
Cu2+(aq) + 2 e– Æ Cu(s)
+0.34 (more positive potential, therefore the cathode)
Cathode:
Anode:
Cu2+(aq) + 2 e– Æ Cu(s)
Zn(s) Æ Zn2+(aq) + 2 e–
Eocathode = +0.34 V
Cell Reaction: Cu2+(aq) + Zn(s)Æ Zn2+(aq) + Cu(s)
- Eoanode = -(-0.76 V)
Eocell = 1.10 V
Calculation of the Cell Potential of a Voltaic Cell NOT at Standard Conditions:
If the potential that is measured is not a standard cell, then the Nernst equation can be
used to calculate the standard cell potential.
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Nernst Equation
Where…
n = number of electrons transferred in the cell reaction
o
Ecell = Ecell
-
0.0591
log Q
n
Qc = Reaction quotient
Eo = standard cell potential (1 M solutions at 298K and 1 atm)
E = Cell potential at nonstandard conditions
(both potentials are measured in volts)
Sample calculation of the standard cell potential for a voltaic cell made of a Sn electrode in
0.10 M Sn2+ in one half-cell and Al in Al3+ 0.10 M in the other.
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Reduction Reactions
Eo
Sn2+(aq) + 2 e– Æ Sn(s)
-0.14 V (more positive potential, therefore the cathode)
Al3+(aq) + 3 e– Æ Al(s)
-1.66 V(more negative potential, therefore the anode)
Cathode:
(Sn2+(aq) + 2 e– Æ Sn(s)) x 3
(Al(s) Æ Al3+(aq) + 3 e–) x 2
Anode:
Eocathode = -0.14 V
Cell Reaction: 3Sn2+(aq) + 2Al(s)Æ 2Al3+(aq) + 3Sn(s)
- Eoanode = -(-1.66 V)
Eocell = 1.52 V
o
Ecell = Ecell
-
0.0591
0.0591
[Al 3 + ] 2
log Q = 1.52V log
n
6
[Sn 2+ ] 3
= 1 .52 V -
0.0591
[ 0.10] 2
log
= 1.52 V − 0.0099 V
6
[0.10]3
Ecell = 1.51 V (nearly the same as the standard cell potential!)
Redox Titrations
Just as a base is used to determine the concentration of an acid (or vice versa) in an acid-base
titration, as known concentration of oxidizing agent can be used to determine the unknown
concentration of a reducing agent (or vice versa) in redox titration. This indispensable
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analytical tool is used in a wide range of situations, including measuring the iron content in
drinking water, the calcium in blood and the vitamin C content in fruits and vegetable.
A common oxidizing agent in redox titrations is the permanganate ion, MnO4-.
Because it is strongly colored, the ion also serves as an indicator for monitoring the reaction.
In the following example, MnO4- is used to determine the concentration of oxalate ion, C2O42-.
As long as C2O42- is present, the deep purple MnO4- is reduced to the very faint pink (nearly
colorless) Mn2+ ion. As soon as all the available C2O42- ions have been oxidized to CO2, the
next drop of MnO4- turns the solution light purple. This color change indicates the point at
which the total number of electrons lost by the oxidized species (C2O42-) equals the total
number of electrons gained by the reduced species (MnO4-). We calculate the concentration of
the C2O42- solution from its known volume, the known concentration and volume of the MnO4solution.
+7
+2
+3
2MnO4-(aq) + 5C2O4 2-(aq) + 16H+(aq)
+4
2Mn2+(aq) + 10CO2(g) + 8H2 O(l)
Experimental
1. Determination of Fe2+
Reaction:
MnO42+ + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O
Procedure:
Complete the tested solution with distilled water up to 100mL in given flask. Mix well and
transfer 10mL of the tested solution to the beaker (100mL) using a pipette, next add 10mL
15% H2SO. Stir all precisely and titrate with 0.02 M KmnO4 solution until the mixture turns
into faint pink.
Record the amount of mL of KmnO4.
Calculate the mass of Fe2+ (in terms of mg)
m = V (KMnO4) • C (KMnO4) • 55.8 • 5
2. Determination of K2Cr2O7 using iodometric method
Reactions:
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K2Cr2O7 + 6KJ + 7H2SO4 → 4 K2SO4 + Cr2(SO4)3 + 3J2 + 7H2O
Received iodine is titrated according to the reaction:
J2 + 2 Na2S2O3 → 2NaJ + Na2S4O6
Summation:
Cr2O72- → 3J2 → 6S2O32Procedure:
Complete the tested solution using distilled water to 100mL in measured flask. Mix well and
transfer 10mL of the tested solution to the beaker (100mL) using a pipette and add small
amount of KI and 10mL 15% H2SO. Stir all precisely. Cover the mixture with glass and leave
for 5 minutes. After this time, titrate with 0.1 M Na2S2O3 to observe brown-yellow color, add
the starch solution and titrate until the blue color disappears (at the end-point solution should
be colourless or faint blue).
Record the amount of mL of Na2S2O3.
Calculate the mass of K2Cr2O7 (in terms of mg)
m = V (Na2S2O3) • C (Na2S2O3) • 294 • 6
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