Transcript Must

Electrochemistry
Chapter 19
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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
Redox = Reduction - Oxidation
Redox Reaction - A reaction where one species gains
electrons and one species loses electrons.
Must have both processes occurring.
Remember Bronsted acid-base reactions where protions are
transferred. One has to give, the other has to take.
0
0
2+ 2-
2Mg (s) + O2 (g)
2Mg
O2 + 4e-
2MgO (s)
2Mg2+ + 4e- Oxidation half-reaction (lose e-)
2O2-
Reduction half-reaction (gain e-)
Reduction Involves Gain of electrons.
(oxidation number )
 OIL RIG
Oxidation Involves Loss of electrons.
(oxidation number )
Oxidizing agent - reactant of reduction
O 0
-2 O2-
Reducing agent - reactant of oxidation
Mg 0
+2 Mg2+
Oxidation number
The charge the atom would have in a molecule (or an
ionic compound) if electrons were completely transferred.
Number assigned to each atom to estimate its excess
charge or deficiency.
There is ALWAYS a change in oxidation numbers
in a redox reaction.
Rules for Determination of Oxidation Number
1. Free elements (uncombined state) and purely
covalent compounds all have atoms with 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.
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.
•
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.
What are the oxidation numbers of all the
atoms in HCO3- ?
HCO3From rules we know:
O = -2
H = +1
Can calculate C:
3x(-2) + 1 + ? = -1
C = +4
4.4
What are the oxidation numbers of Mn in the
following compounds and ions ?
Mn - element  oxidation # = 0
Mn2+ - ion  oxidation # = +2
metal
colorless
MnO2 - compound  oxidation # = +4
brown
+4 + (2x(-2)) = 0
MnO4- - ion  oxidation # = +7
purple
+7 + (4x(-2)) = -1
MnO4-2 - ion  oxidation # = +6
green
+6 + (4x(-2)) = -2
 Mn can have at least 5 oxidation states.
Balancing Redox Reactions
Reaction can be balanced in acidic, neutral or basic solutions.
MnO4- + HSO3-
Mn2+ + SO4-2
acid
MnO4- + HSO3-
MnO2 + SO4-2
neutral
MnO4- + HSO3-
• Balance mass.
MnO42- + SO4-2
base
• Balance charge.
• Balance number of electrons exchanged
(treat electrons as a reactant).
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
+3
Fe2+
Oxidation:
+6
Reduction:
Cr2O7
Fe3+
+3
2-
Cr3+
Oxidation and reduction must take place simultaneously,
but we can consider them separately for simplicity.
Can look in tables of standard reduction potentials for
half reaction or build the half reaction ourselves.
6e- + 14H+ + Cr2O72Fe2+
2Cr3+ + 7H2O
Fe3+ + 1e-
NOTE: We reversed the
direction of the oxidation
reaction (table is of reduction
potentials.).
3. Balance the atoms other than O and H in each half-reaction.
Cr2O72-
2Cr3+
4. For reactions in acid, add H2O to balance O atoms and H+ to
balance H atoms.
Cr2O7214H+ + Cr2O72-
2Cr3+ + 7H2O
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
19.1
Balancing Redox Equations
6. If necessary, equalize the number of electrons in the two halfreactions by multiplying the half-reactions by appropriate
coefficients.
6Fe2+
6Fe3+ + 6e6e- + 14H+ + Cr2O72-
2Cr3+ + 7H2O
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+
6Fe3+ + 6eReduction: 6e- + 14H+ + Cr2O72-
14H+ + Cr2O72- + 6Fe2+
2Cr3+ + 7H2O
6Fe3+ + 2Cr3+ + 7H2O
Overall reaction should be balanced if halfreactions are properly balanced.
Balancing Redox Equations
8. Verify that the number of atoms and the charges are balanced.
14x1 – 2 + 6x2 = 24 = 6x3 + 2x3
14H+ + Cr2O72- + 6Fe2+
6Fe3+ + 2Cr3+ + 7H2O
9. For reactions in basic solutions, add OH- to both sides of the
equation for every H+ that appears in the final equation.
19.1
Balancing Redox Reactions
Balance reaction in acidic solution.
MnO4- + HSO3Mn2+ + SO4-2
Reduction: MnO4Mn2+
add H2O to balance O atoms.
MnO4MnO4- + 8 H+
add H+ to balance H atoms.
add electrons to balance charge.
Mn2+ + 4 H2O
Mn2+ + 4 H2O
-1 + 8  +2 + 0 need 5 e-’s
MnO4- + 8 H+ + 5 eMn2+ + 4 H2O
Oxidation:
HSO3SO4-2
SO4-2
add H2O to balance O atoms. HSO3- + H2O
add H+ to balance H atoms.
HSO3- + H2O
add electrons to balance charge.
HSO3- + H2O
SO4-2 + 3 H+
-1 + 0  -2 + 3 need 2 e-’s
SO4-2 + 3 H+ + 2 e-
equalize the number of electrons in the two half-reactions by multiplying
the half-reactions by appropriate coefficients.
2x(MnO4- + 8 H+ + 5 eMn2+ + 4 H2O)
5x(HSO3- + H2O
SO4-2 + 3 H+ + 2 e- )
2 MnO4- + 16 H+ + 10 e- + 5 HSO3- + 5 H2O
2 Mn2+ +
1
8 H2O + 5 SO4-2 + 15 H+ + 10 e3
2 MnO4- + H+ + 5 HSO3-
Check balance:
H
Mn
1+5=3x2
2=2
O 8 + 15 = 3 + 20
2 Mn2+ + 3 H2O + 5 SO4-2
S
5=5
Charge - 2 + 1 - 5 = 4 - 10
-6 = -6
Balance reaction in neutral solution.
MnO4- + HSO3MnO2 + SO4-2
Reduction: MnO4-
MnO2
Choose half-reaction with neutral product from table.
MnO4- + 4 H+ + 3 eOxidation:
HSO3-
MnO2 + 2 H2O
SO4-2
Use same half-reaction as before.
HSO3- + H2O
SO4-2 + 3 H+ + 2 e-
equalize the number of electrons in the two half-reactions by multiplying
the half-reactions by appropriate coefficients.
2x(MnO4- + 4 H+ + 3 e3x(HSO3- + H2O
MnO2 + 2 H2O)
SO4-2 + 3 H+ + 2 e- )
2 MnO4- + 8 H+ + 6 e- + 3 HSO3- + 3 H2O
2 MnO2 +
4 H2O + 3 SO4-2 + 9 H+ + 6 e1
1
2 MnO4- + 3 HSO32 MnO2
Check balance:
S
H
3=2+1
Charge -2
Mn
2=2
O 8 + 9 = 4 + 1 + 12
+ H2O + 3 SO4-2 + H+
3=3
-3=0+0-6+1
-5 = -5
Balance reaction in basic solution.
MnO4- + HSO3MnO42- + SO4-2
Reduction: MnO4MnO42add electrons to balance charge.
MnO4- + eOxidation:
HSO3-
MnO42SO4-2
Use same half-reaction as before.
HSO3- + H2O
SO4-2 + 3 H+ + 2 e-
equalize the number of electrons in the two half-reactions by multiplying
the half-reactions by appropriate coefficients.
2x(MnO4- + e1x(HSO3- + H2O
2 MnO4- + 2 e- + HSO3- + H2O
MnO42-)
SO4-2 + 3 H+ + 2 e- )
2 MnO42- + SO4-2 + 3 H+ +
+ 2 e-
2 MnO4- + HSO3- + H2O
Check balance:
H 1+2=3
Mn
2=2
O 8+3+1=8+4
2 MnO42- + SO4-2 + 3 H+
S
1=1
Charge -2 - 1 = - 4 - 2 + 3
-3 = -3
But, are we done ??? Solution needs to be basic and
we are generating 3 H+
2 MnO4- + HSO3- + H2O
3x(OH- + H+
2 MnO4- + HSO3- + 3 OH-
Check balance:
H
1+3=4
Mn
2=2
O 8+3+3=8+4+2
2 MnO42- + SO4-2 + 3 H+
H2O)
Neutralize H+
2 MnO42- + SO4-2 + 2 H2O
S
1=1
Charge -2 - 1- 3 = -4 - 2 + 0
-6 = -6
Redox Chemistry and Electrical Energy
Electrical Conduction  moving electrical charges
In Metals:
electric charges that move are electrons in the metal
electric potential difference is what causes electrons to
move (similar to gravitational forces causing mass to move
or chemical potential driving reactions)
resistance interferes with the flow of electrons - usually
atomic motion
circuit - electrons move in circles, must have a place to go
and other electrons to replace them
electric current is the rate charges move - # charges / time
Symbol
Units
Charge
Q
Coulombs (c)
Faradays (F)
Potential Difference
Resistance
E
R
Volts (V)
Ohms ()
Current
I
Amperes (A)
I=
1 A = 1 c / 1 sec
Q
time
c = A . sec
1 Faraday of charge = charge of 1 mole of electrons = 96,500 c
Energy = 1 J = (1 V) . (1 c)
E=IR
Ohm’s Law
Conduction is good if potential is high
and resistance is low.
In Solutions (or liquids ionic compounds at temperatures
high enough to melt):
electric charges that move are ions
cations (positively charged) move towards anything
negatively charged
anions (negatively charged) move towards anything
positively charged
Electrochemical Cell - A circuit that includes electrolytic
and metallic conductors
Electrode - connects metal conductor to electrolytic
conductor
Electrochemical Cells
anode
oxidation
cathode
reduction
spontaneous
redox reaction
Attracts anions
Attracts cations 19.2
Electrochemical Cells
The difference in electrical
potential between the anode
and cathode is called:
• cell voltage
• electromotive force (emf)
• cell potential
Cell Diagram
Zn (s) + Cu2+ (aq)
Cu (s) + Zn2+ (aq)
[Cu2+] = 1 M & [Zn2+] = 1 M
anode
cathode
Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s)
Use || to separate half cells
Use | to separate reactants/phases in each half cell
19.2
Salt Bridge
Electrolytic connection between two half cells
to complete the circuit
Tube containing a solution of inert salt
(usually KNO3)
Two Types of Cells
Electrochemical Cell - need “direct current source” = an
electron pump (ex: battery)
•
Electrons are forced in one direction, independent of
sponteneity
•
Electrical energy is used to cause a nonspontaneous
reaction to occur - drive the desired chemistry
•
Electrons are added to the cathode by electron pump
causing reduction.
Two Types of Cells
Voltaic Cell or Galvanic Cell - passively electric
(no need of “direct current source” = an electron pump)
•
Electrons move because of spontaneous reaction
•
Use chemistry to get desired energy
•
Electrons are taken from the cathode by reduction, causing
electrons to flow to it (from outside it appears )
•
Can use as a direct current source for electrolytic cell.
How do we know if reaction is spontaneous?
What is the electric potential?
How do concentrations effect the process?
Electromotive Force (emf) is the electric potential of cell
E (emf)  units = volts (V)
emf is the potential difference between the anode and the
cathode
Standard Electrode 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)
19.3
Standard Electrode 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.
Reduction Reaction
2e- + 2H+ (1 M)
H2 (1 atm)
Defines E0 = 0 V
Use as a reference to measure
all other potentials. 
Standard hydrogen electrode (SHE)
19.3
Standard Electrode Potentials
0 = 0.76 V
Ecell
0 )
Standard emf (Ecell
0
0 = E0
Ecell
cathode - Eanode
reduction oxidation
Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)
0 = E 0 + - E 0 2+
Ecell
H /H2
Zn /Zn
Convention:
0
0.76 V = 0 - EZn2+/Zn
E° > 0 spontaneous reaction
0
EZn 2+/Zn = -0.76 V
Zn2+ (1 M) + 2e-
Zn
E0 = -0.76 V
19.3
Standard Electrode 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
•
E0 is for the reaction as
written
•
The half-cell reactions are
reversible
•
The sign of E0 changes
when the reaction is
reversed (E° red = -E°oxid)
•
Changing the stoichiometric
coefficients of a half-cell
reaction does not change
the value of E0
19.3
Strongest oxidizing agent
•
The more positive E0 the
greater the tendency for the
substance to be reduced
Zero reference point
Strongest reducing agent
0 = 0.76 V
Ecell
0 = 0.34 V
Ecell
Combine
Zn (s)
Zn2+ (1 M) + 2e-
2e- + Cu2+ (1 M)
Zn (s) + Cu2+ (1 M)
Cu (s)
Cu (s) + Zn2+ (1 M)
0 = 0.76 V + 0.34 V = 1.10 V
Ecell
What is the standard emf of an electrochemical cell made
of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr
electrode in a 1.0 M Cr(NO3)3 solution?
Cd2+ (aq) + 2e-
Cd (s) E0 = -0.40 V Cd is the stronger oxidizer
Cr3+ (aq) + 3e-
Cr (s)
Anode (oxidation):
E0 = -0.74 V
Cr3+ (1 M) + 3e- x 2
Cr (s)
Cathode (reduction): 2e- + Cd2+ (1 M)
2Cr (s) + 3Cd2+ (1 M)
Cd will oxidize Cr
Cd (s)
x3
3Cd (s) + 2Cr3+ (1 M)
0
0 = E0
Ecell
cathode - Eanode
0 = -0.40 – (-0.74)
Ecell
0 = 0.34 V  spontaneous
Ecell
19.3
Spontaneity of Redox Reactions
Sponteneity DG < 0
Energy = Q E = -nFEcell
Total charge
DG = -nFEcell
n = number of moles of electrons in reaction
J
=
F = 96,500
= 96,500 C/mol
V • mol
0
DG0 = -RT ln K = -nFEcell
DG0
0
-nFEcell
0
Ecell
(8.314 J/K•mol)(298 K)
RT
ln K =
ln K
=
nF
n (96,500 J/V•mol)
0
Ecell
0.0257 V
ln K
=
n
0
Ecell
=
0.0592 V
log K
n
Spontaneity of Redox Reactions
DG0 = -RT ln K
DG = -nFEcell
19.4
What is the equilibrium constant for the following reaction
at 250C? Fe2+ (aq) + 2Ag (s)
Fe (s) + 2Ag+ (aq)
0
Ecell
=
0.0257 V
ln K
n
Oxidation:
Reduction:
2e-
+
2Ag
2Ag+ + 2e-
Fe2+
Fe
n=2
0
0
E0 = EFe
2+/Fe – EAg + /Ag
E0 = -0.44 – (0.80)
E0 = -1.24 V
0
Ecell
xn
-1.24 V x 2
= exp
K = exp
0.0257 V
0.0257 V
K = 1.23 x 10-42
19.4
The Effect of Concentration on Cell Emf
DG = DG0 + RT ln Q
DG = -nFE
DG0 = -nFE 0
-nFE = -nFE0 + RT ln Q
Nernst equation
E = E0 -
RT
ln Q
nF
Can look at
concentration
effects / nonstandard
conditions
At 298 K
E=
E0
0.0257 V
ln Q
n
E=
E0
0.0592 V
log Q
n
19.5
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:
Reduction:
Cd
2e-
+
Cd2+ + 2e-
Fe2+
2Fe
n=2
0
0
E0 = EFe
2+/Fe – ECd2+/Cd
E0 = -0.44 – (-0.40)
E0 = -0.04 V
0.0257 V
ln Q
n
0.010
0.0257 V
ln
E = -0.04 V 2
0.60
E = 0.013 V
E = E0 -
E>0
Spontaneous
19.5
Concentration Cells have the same half-reaction in each
cell, but at different concentrations.
Will the following reaction occur spontaneously at 250C if [Ag+]
= 0.10 M and [Ag+] = 0.010 M?
Ag+ (aq) + Ag (s)
Ag (s) + Ag+ (aq)
Ag (s) | Ag+ (0.10 M) || Ag+ (0.010 M) | Ag (s)
Oxidation:
Reduction:
Ag
e- + Ag+
Ag+ + eAg
Agoxid + Ag+red
0
0
E0 = EAg
+ /Ag – EAg+ /Ag
n=1
Agred + Ag+oxid
E0 = -0.7991 V – (-0.7991 V)
E0 = -0.000 V
0.0257 V
ln Q
E = E0 n
E = -0.000 V - 0.0257 V
1
E < 0 Non-spontaneous
+
[Ag
oxid]
ln
0.0257 V
[Ag+red] = -
ln
0.10 = - 0.0592 V
0.010
Batteries
Dry cell
Leclanché cell
Anode:
Cathode:
Zn (s)
2NH+4 (aq) + 2MnO2 (s) + 2e-
Zn (s) + 2NH4 (aq) + 2MnO2 (s)
Zn2+ (aq) + 2eMn2O3 (s) + 2NH3 (aq) + H2O (l)
Zn2+ (aq) + 2NH3 (aq) + H2O (l) + Mn2O3 (s)
19.6
Batteries
Mercury Battery
Anode:
Cathode:
Zn(Hg) + 2OH- (aq)
HgO (s) + H2O (l) + 2eZn(Hg) + HgO (s)
ZnO (s) + H2O (l) + 2eHg (l) + 2OH- (aq)
ZnO (s) + Hg (l)
19.6
Batteries
Lead storage
battery
Anode:
Cathode:
Pb (s) + SO2-4 (aq)
PbSO4 (s) + 2e-
PbO2 (s) + 4H+ (aq) + SO24 (aq) + 2e
Pb (s) + PbO2 (s) + 4H+ (aq) + 2SO2(aq)
4
PbSO4 (s) + 2H2O (l)
2PbSO4 (s) + 2H2O (l)
19.6
Batteries
Solid State Lithium Battery
19.6
Batteries
A fuel cell is an
electrochemical cell
that requires a
continuous supply of
reactants to keep
functioning
Anode:
Cathode:
2H2 (g) + 4OH- (aq)
O2 (g) + 2H2O (l) + 4e2H2 (g) + O2 (g)
4H2O (l) + 4e4OH- (aq)
2H2O (l)
19.6
Corrosion
Dissolved oxygen in
water causes oxidation
E°red = -0.44 V
E°red = 1.23 V
Since E°red (Fe3+) < E°red (O2)
Rust Fe2O3
Fe can be oxidized by oxygen
19.7
Cathodic Protection of an Iron Storage Tank
E°red = -2.37
V
E°red = 1.23 V
Mg reduction potential is too negative to be overcome by oxygen.
19.7
Electrolysis is the process in which electrical energy is used
to cause a nonspontaneous chemical reaction to occur.
19.8
Electrolysis of Water
19.8
Electrolysis and Mass Changes
Quantitative Analysis
How much current?
time ?
product?
charge (C) = current (A) x time (s)
1 mole e- = 96,500 C
19.8
How much Ca will be produced in an electrolytic cell of
molten CaCl2 if a current of 0.452 A is passed through the
cell for 1.5 hours?
Anode:
Cathode:
2Cl- (l)
Ca2+ (l) + 2eCa2+ (l) + 2Cl- (l)
Cl2 (g) + 2eCa (s)
Ca (s) + Cl2 (g)
2 mole e- = 1 mole Ca
C
s 1 mol e- 1 mol Ca
mol Ca = 0.452
x 1.5 hr x 3600 x
x
s
hr 96,500 C 2 mol e= 0.0126 mol Ca
= 0.50 g Ca
19.8
Consider the stoichiometryof the electrolytic cell:
Cathode:
e- + Ag+
Ag
Anode: 2 H2O
O2 (g) + 4 H+ + 4 e-
What current (in amps) is required to convert 0.100 mole
Ag+ to Ag in 10.0 min?
1 mole electrons = 1 F
A = c/sec
Q = nF
I=Q/t
Find Q
Q = 0.10 mol Ag 1 mole electrons
1F
96,500 c =
mole Ag
mole electrons
F
Q = 9,650 c
t = 10 min
60 seconds = 600 seconds
1 minute
I = 9,6500 c / 600 s = 16 c/s = 16 A
Stoichiometry of prodcuts at different electrodes
What is the pH of the anode half cell (assume volume =
0.100 L) after 6.00 g of Ag plates out at the cathode?
Cathode:
e- + Ag+
Ag
Anode: 2 H2O
O2 (g) + 4 H+ + 4 eFind [H+]
6.00 g Ag
1 mol Ag 1 mole electrons
4 mol H+
107.9 g Ag 1 mole Ag 1 mol electrons
= 0.05567 mol H+
[H+] = 0.0556 mol = 0.56 M
0.10 L
pH = 0.25
=