Transcript Electrochemistry - Professor Monzir Abdel

Electrochemistry
Chapter 19
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Electron Transfer reactions
• Electron transfer reactions are oxidation-
reduction or redox reactions.
• Electron transfer reactions result in the
generation of an electric current
(electricity) or be caused by imposing an
electric current.
• Therefore, this field of chemistry is called
ELECTROCHEMISTRY.
2
Electrochemical processes are oxidationreduction reactions in which:
The energy released by a spontaneous reaction is
converted to electricity or
Electrical energy is used to cause a nonspontaneous
reaction to occur
0
0
2Mg (s) + O2 (g)
2Mg
O2 + 4e3
2+ 2-
2Mg O (s)
2Mg2+ + 4e- Oxidation half-reaction (lose e-)
2O2-
Reduction half-reaction (gain e-)
Important Concepts
OXIDATION—loss of electron(s) by a species;
increase in oxidation number; increase in oxygen.
REDUCTION—gain of electron(s); decrease in
oxidation number; decrease in oxygen; increase
in hydrogen.
OXIDIZING AGENT—electron acceptor; species is
reduced.
REDUCING AGENT—electron donor; species is
oxidized.
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Review of Oxidation numbers
Numbers associated with atoms and ions (can also
be fractions), where:
1. Free elements (uncombined state) have an
oxidation number of zero.
Na, Be, K, Pb, H2, O2, P4 = 0
2. In monatomic ions, the oxidation number is equal
to the charge on the ion.
Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2
3. The oxidation number of oxygen is usually –2. In
H2O2 and O22- it is –1.
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4. The oxidation number of hydrogen is +1 except when
it is bonded to metals in binary compounds. In these
cases, its oxidation number is –1.
5. Group IA metals are +1, IIA metals are +2 and fluorine is
always –1.
6. The sum of the oxidation numbers of all the atoms in
a molecule or ion is equal to the charge on the
molecule or ion.
HCO3Find the oxidation number
of carbon in HCO3- ?
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O = -2
H = +1
3x(-2) + 1 + ? = -1
C = +4
4.4
Redox Reaction Example
A redox reaction:
Cu(s) + 2Ag+(aq)
ON:
0
+1
oxidation
Cu2+(aq) + 2Ag(s)
+2
0
reduction
Cu(s) is oxidized, therefore it is a reducing agent.
Ag+ is reduced, therefore it is an oxidizing agent
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We can envision breaking up the full redox
reaction into two half reactions:
Cu(s) + 2Ag+(aq)
Cu2+(aq) + 2Ag(s)
reduction
oxidation
Cu(s)
Ag+(aq) + e-
Cu2+(aq) + 2eAg(s)
“half-reactions”
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Note that the half reactions are combined to
make a full reaction:
Cu(s) + 2Ag+(aq)
oxidation
Cu2+(aq) + 2Ag(s)
reduction
The important thing to remember: electrons
are neither created nor destroyed during a
redox reaction. They are transferred from
the species being oxidized to that being
reduced.
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Identify the species being oxidized and reduced in
the following (unbalanced) reactions:
reduction
+5
ClO3- + I-
-1
I2 + Cl-
-1
0
oxidation
NO3- + Sb
+5
10
Sb4O6 + NO
0 oxidation +3
reduction
+2
Balancing Redox Equations
The oxidation of Fe2+ to Fe3+ by Cr2O72- in acid solution?
1. Write the unbalanced equation for the reaction in ionic
form.
Fe2+ + Cr2O72Fe3+ + Cr3+
2. Separate the equation into two half-reactions.
+2
+3
Fe2+
Oxidation:
+6
Reduction:
Cr2O7
Fe3+
+3
2-
Cr3+
3. Balance the atoms other than O and H in each halfreaction.
Cr2O722Cr3+
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4. For reactions in acid, add H2O to balance O atoms and H+
to balance H atoms.
Cr2O72-
2Cr3+ + 7H2O
14H+ + Cr2O72-
2Cr3+ + 7H2O
5. Add electrons to one side of each half-reaction to
balance the charges on the half-reaction.
Fe2+
6e- + 14H+ + Cr2O72-
Fe3+ + 1e2Cr3+ + 7H2O
6. If necessary, equalize the number of electrons in the two
half-reactions by multiplying the half-reactions by
appropriate coefficients.
6Fe2+
6Fe3+ + 6e12
6e- + 14H+ + Cr2O72-
2Cr3+ + 7H2O
7. Add the two half-reactions together and construct the
final equation. The number of electrons on both sides
must cancel. You should also cancel like species.
Oxidation:
6Fe2+
Reduction:6e- + 14H+ + Cr2O7214H+ + Cr2O72- + 6Fe2+
6Fe3+ + 6e2Cr3+ + 7H2O
6Fe3+ + 2Cr3+ + 7H2O
8. Verify that the number of atoms and the charges are
balanced.
14x1 – 2 + 6x2 = 24 = 6x3 + 2x3
9. For reactions in basic solutions, add OH- to both sides of
the equation for every H+ that appears in the final
equation. You should combine H+ and OH- to make H2O.
13
Balance the following reaction:
CuS(s) + NO3-(aq) g Cu2+(aq) + SO42-(aq) + NO(g)
Step 1: Identify and write down the unbalanced
half reactions.
oxidation
-2
CuS(s) + NO3-(aq)
Cu2+(aq) + SO42-(aq) + NO(g)
+5
14
+6
reduction
+2
CuS(s)
Cu2+(aq) + SO42-(aq) oxidation
NO3-(aq)
NO(g)
reduction
Step 2: Balance atoms and charges in each half
reaction. Use H2O to balance O, and H+
to balance H (assume acidic media). Use
e- to balance charge.
4H2O + CuS(s)
3e- + 4H+ + NO3-(aq)
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Cu2+(aq) + SO42-(aq)+ 8H+ + 8eNO(g) + 2H2O
Step 3: Multiply each half reaction by an integer
such that electrons cancel.
4H2O + CuS(s)
Cu2+(aq) + SO42-(aq)+ 8H+ + 8e- x 3
3e- + 4H+ + NO3-(aq)
3CuS + 8 NO3- + 8H+
NO(g) + 2H2O x 8
8NO + 3Cu2+ + 3SO42- + 4H2O
Step 4: Add the two half reactions and cancel
like species.
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Step 5. For reactions that occur in basic
solution, proceed as above. At the end,
add OH- to both sides for every H+ present,
combine to yield water on the H+ side.
3CuS + 8 NO3- + 8H+
8NO + 3Cu2+ + 3SO42- + 4H2O
+ 8OH-
+ 8OH-
3CuS + 8 NO3- + 8H2O
4
8NO + 3Cu2+ + 3SO42- + 4H2O
+ 8OH-
The final equation in basic medium is
3CuS + 8 NO3- + 4H2O g 8NO + 3Cu2+ + 3SO42- + 8OH17
Balance the following redox reaction
occurring in basic media:
BH4- + ClO3BH4-5
ClO3-
+5
18
H2BO3- + ClH2BO3-
oxidation
+3
Cl-
-1
reduction
3H2O + BH46e- + 6H+ + ClO3-
H2BO3- + 8H+ + 8eCl- + 3H2O
x3
x4
3
3 BH4- + 4 ClO3-
4 Cl- + 3H2BO3- + 3H2O
Here, the net balanced equation contains no H+ or
OH-, and thus it can be used as is for acidic or
basic solutions.
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Galvanic Cells
In redox reactions, electrons are transferred
from the oxidized species to the reduced
species.
Imagine separating the two half cells
physically, then providing a circuit
through which the electrons travel from
one half cell to the other.
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Ionic Conduction: Salt Bridge
Salt bridge/porous disk: allows for ion migration
such that the solutions will remain neutral.
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Calculate your score in the second exam:
Suppose your score on the exam paper was x:
The final score = ([x+2]/42)*40
If the points on your paper were 38, then your
score in the exam is: ([38+2]/42)*40 = 38.
If your score on the exam paper is 28, your final
score is: ([28+2]/42)*40 = 28.57 = 29
And so on.
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