Electrochemistry

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Transcript Electrochemistry

Electrochemistry
Chapter 4
4.4 and 4.8
Chapter 19
19.1-19.5 and 19.8
Oxidation-Reduction Reactions
Electron Transfer Reactions
 Many redox reactions take place in water.
 Half-reactions
 OILRIG

Oxidation-Reduction
Reactions
2Mg (s) + O2 (g)
2MgO (s)
2Mg2+ + 4e- Oxidation half-reaction (lose e-)
2Mg
O2 + 4e-
2O2- Reduction half-reaction (gain e-)
2Mg + O2 + 4e2Mg + O2
2Mg2+ + 2O2- + 4e2MgO
Oxidation-Reduction Reactions
Oxidizing Agent- a substance that
accepts electrons from another substance,
causing the other substance to be
oxidized.
 Reducing Agent- a substance that
donates electrons to another substance,
causing it to become reduced.

Oxidation-Reduction
Reactions
Zn (s) + CuSO4 (aq)
ZnSO4 (aq) + Cu (s)
Oxidation-Reduction
Reactions
Zn (s) + CuSO4 (aq)
Zn
Zn2+ + 2e-
ZnSO4 (aq) + Cu (s)
Zn is oxidized
Zn is the reducing agent
Cu2+ + 2e-
Cu
Cu2+ is reduced
Cu2+ is the oxidizing agent
Oxidation-Reduction
Reactions
Copper wire reacts with silver nitrate to form silver metal.
What is the oxidizing agent in the reaction?
Cu (s) + 2AgNO3 (aq)
Cu
Cu(NO3)2 (aq) + 2Ag (s)
Cu2+ + 2e-
Ag+ + 1e-
Ag
Ag+ is reduced
Ag+ is the oxidizing agent
Oxidation Number
Oxidation Number- the number of
charges the atom would have in a
molecule (or an ionic compound) if
electrons were transferred completely.
 H2(g) + Cl2(g) → 2HCl(g)


0
0
+1 -1
Assigning Oxidation Numbers
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.
Assigning Oxidation Numbers
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.
Assigning Oxidation Numbers
Oxidation numbers of all
the elements in HCO3- ?
HCO3O = -2
H = +1
3x(-2) + 1 + ? = -1
C = +4
Assigning Oxidation Numbers
Redox Titrations
Oxidizing agent/Reducing Agent
 Equivalence point reached when reducing
agent is completely oxidized by the
oxidizing agent.
 Indicator is needed

Redox Titrations
Electrochemistry
Branch of Chemistry that deals with
interconversion of electrical and chemical
energy.
 Importance

 Use
of electricity in everyday life
 Interconversion of energy
 Study of electrochemical processes
Electrochemistry
Electrochemical processes are oxidation-reduction 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
Balancing Redox Equations
Simple reactions easy to balance
 Complex reactions

 Usually
encountered in laboratory
 Can include CrO42-, Cr2O72-, MnO4-, NO3- and
SO42 Follow balancing guidelines
Balancing Redox Reactions

Guidelines





Step 1: Write the unbalanced equation for the reaction in ionic
form.
Step 2: Separate the equation into two half-reactions
Step 3: Balance each half-reaction for number and type of atoms
and charges. (Look at medium)
Step 4: Add the two half-reactions together and balance the final
equation by inspection. The electrons on both sides must cancel.
(Be sure they are equal)
Step 5: Verify that the equation contains the same type and
numbers of atoms and the same charges on both sides of the
equation.
Balancing Redox Equations
The oxidation of Fe2+ to Fe3+ by Cr2O72- in acid solution?
1. Write the unbalanced equation for the reaction ion ionic form.
Fe2+ + Cr2O72-
Fe3+ + Cr3+
2. Separate the equation into two half-reactions.
+2
Fe2+
Oxidation:
+6
+3
Fe3+
+3
Cr3+
Reduction: Cr2O723. Balance the atoms other than O and H in each half-reaction.
Cr2O722Cr3+
Balancing Redox Reactions
4. For reactions in acid, add H2O to balance O atoms and H+ to
balance H atoms.
Cr2O722Cr3+ + 7H2O
14H+ + Cr2O722Cr3+ + 7H2O
5. Add electrons to one side of each half-reaction to balance the
charges on the half-reaction.
Fe2+
Fe3+ + 1e6e- + 14H+ + Cr2O722Cr3+ + 7H2O
6. If necessary, equalize the number of electrons in the two halfreactions by multiplying the half-reactions by appropriate
coefficients.
6Fe2+
6Fe3+ + 6e6e- + 14H+ + Cr2O722Cr3+ + 7H2O
Balancing Redox Reactions
7. Add the two half-reactions together and balance the final
equation by inspection. The number of electrons on both
sides must cancel.
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.
Galvanic Cells




Oxidation-Reduction Reaction
Oxidizing agent and Reducing agent are not in
contact
External conducting medium
Galvanic Cell/Volatic Cell- the experimental
apparatus for generating electricity through the
use of a spontaneous reaction.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Voltmeter
e–
Zinc
anode
e–
Copper
cathode
Cl– K +
Salt
bridge
SO 2–
4
Zn 2+
Zn2+
ZnSO4
solution
Cotton
plugs
Cu2+
SO 2–
4
CuSO4
solution
Cu2+
2e–
Cu
Zn is oxidized to
Zn2+ at anode.
Zn(s) Zn2+ (aq) + 2 e–
Cu2+ is reduced
to Cu at cathode.
2e– + Cu2+ (aq)
Net reaction
Zn2+ (aq) + Cu(s)
Zn(s) + Cu2+ (aq)
Cu(s)
Galvanic Cells



Cell Voltage- the difference in electrical
potential between the anode and the cathode.
Measured by a voltmeter.
Electromotive force (emf) and Cell Potential
Voltage of a cell depends on:
 Composition
of electrode and ions
 Concentration of ions
 Temperature
Galvanic Cells
Cell Diagram
Zn (s) + Cu2+ (aq)
Cu (s) + Zn2+ (aq)
Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s)
anode
cathode
Standard Reduction Potentials
Standard reduction potential (E0) is the voltage associated
with a reduction reaction at an electrode when all solutes
are 1 M and all gases are at 1 atm.
Used to calculate the emf of a galvanic
cell.
 Use the values from anode and cathode

0 )
Standard emf (Ecell
0
0 = E0
Ecell
cathode - Eanode
Standard Reduction Potentials
Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)
Anode (oxidation):
Zn (s)
Cathode (reduction): 2e- + 2H+ (1 M)
Zn (s) + 2H+ (1 M)
Zn2+ (1 M) + 2eH2 (1 atm)
Zn2+ + H2 (1 atm)
Standard Reduction Potentials
0 = 0.76 V
Ecell
0 )
Standard emf (Ecell
0
0 = E0
Ecell
cathode - Eanode
Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)
0 = E 0 + - E 0 2+
Ecell
H /H2
Zn /Zn
0 2+
0.76 V = 0 - EZn
/Zn
0 2+
EZn
/Zn = -0.76 V
Zn2+ (1 M) + 2e-
Zn
E0 = -0.76 V
Standard Reduction Potentials
0 = 0.34 V
Ecell
0
0 = E0
Ecell
cathode - Eanode
0 = E 0 2+
0
Ecell
Cu /Cu – EH +/H 2
0 2+
0.34 = ECu
/Cu - 0
0 2+
ECu
/Cu = 0.34 V
Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s)
Anode (oxidation):
H2 (1 atm)
Cathode (reduction): 2e- + Cu2+ (1 M)
H2 (1 atm) + Cu2+ (1 M)
2H+ (1 M) + 2eCu (s)
Cu (s) + 2H+ (1 M)
19.3
Tips





E0 is for the reaction as written
The more positive E0 the greater the tendency
for the substance to be reduced
The half-cell reactions are reversible
The sign of E0 changes when the reaction is
reversed
Changing the stoichiometric coefficients of a
half-cell reaction does not change the value of
E0
Spontaneity of Redox Reactions

Relationship between Eºcell, ΔGº and K.
0
DG0 = -nFEcell
n = number of moles of electrons in reaction
J
F = 96,500
= 96,500 C/mol
V • mol
0
Ecell
0.0257 V
ln K
=
n
0
Ecell
0.0592 V
log K
=
n
Spontaneity of Redox Reactions
Relationships between Eºcell, ΔGº
and K
The Effect of Concentration on Cell
Emf
Not all reactions in galvanic cells can
occur under standard state conditions.
 Nernst Equation- equation used to
calculate cell potential under nonstandard
state conditions.

E = E0 -
0.0257 V
ln Q
n
E = E0 -
0.0592 V
log Q
n
Will the following reaction occur spontaneously at 250C if
[Fe2+] = 0.60 M and [Cd2+] = 0.010 M?
Fe2+ (aq) + Cd (s)
Fe (s) + Cd2+ (aq)
Oxidation:
Cd
Reduction: 2e- + Fe2+
0
0
E0 = EFe
2+/Fe – ECd2+/Cd
E0 = -0.44 – (-0.40)
E0 = -0.04 V
Cd2+ + 2en=2
2Fe
0.0257 V
ln Q
n
0.010
0.0257 V
ln
E = -0.04 V 2
0.60
E = 0.013
E = E0 -
E>0
Spontaneous
19.5
Concentration Cells





Galvanic cells consisting of the same type of
electrodes in the same solution. Each solution is
a different concentration.
Zn(s) | Zn2+(0.10M) || Zn2+(1.0M) | Zn(s)
E = Eº - 0.0257V/ 2 x ln [Zn2+] dil / [Zn2+]conc
E = 0- 0.0257V/2 ln (0.10/1.0)
E = 0.0296 V
Concentration Cells

Importance
 Unequal
ion concentrations
 Membrane potential
 Propagation
Electrolysis
Electrolysis- is the process in which
electrical energy is used to cause a
nonspontaneous chemical reaction to
occur.
 Electrolytic Cell- an apparatus for
carrying out electrolysis.
