electrochemistry - philadelphia university · 2015. 4. 9. · 17.2 standard reduction potentials, e...
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Electrochemistry
Chapter 17
Contents Galvanic cells
Standard reduction potentials
Cell potential, electrical work, and free energy
Dependence of cell potential on concentration
Batteries
Corrosion
Electrolysis
Commercial electrolytic processes
Oxidation reduction reactions
Oxidation reduction reactions
involve a transfer of electrons.
Oxidation Involves Loss of electrons
– Increase in the oxidation number
Reduction Involves Gain electrons
– Decrease in the oxidation number
17.1 Galvanic Cells
8H++MnO4-+ 5Fe+2
Mn+2 + 5Fe+3 +4H2O
If we break the reactions into half reactions.
8H++MnO4-+5e- Mn+2 +4H2O (Red)
5(Fe+2 Fe+3 + e- ) (Ox)
Electrons are transferred directly.
This process takes place without doing useful work
Example
H+ MnO4
- Fe+2
When the compartments of the two beakers are connected as shown the reaction starts
Current flows for an instant then stops
No flow of electrons in the wire, Why?
Current stops immediately because charge builds
up.
Oxidant
reductant
H+
MnO4-
Fe+2
Galvanic Cell
Salt Bridge allows
ions to flow without
extensive mixing in
order to keep net
charge zero.
Electrons flow
through the wire
from reductant to
oxidant
Solutions must be connected so ions can flow to
keep the net charge in each compartment zero
H+
MnO4- Fe+2
Porous
Disk
Fe2+
Reducing
Agent
Oxidizing
Agent
MnO4-
e-
e-
e- e-
e-
e-
Anode Cathode
Electrochemical Cells
19.2
Spontaneous redox reaction
_______
__________ _______
__________
Thus a Galvanic cell is a device in which a chemical energy is changed to electrical energy
The electrochemical reactions occur at the interface between electrode and solution where the electron transfer occurs
Anode: the electrode compartment at which oxidation occurs
Cathode: the electrode compartment at which reduction occurs
Cell Potential Oxidizing agent pulls the electrons
Reducing agent pushes the electrons
The total push or pull (“driving force”) is called the cell potential, E
cell
Also called the electromotive force (emf)
Unit is the volt(V)
= 1 joule of work/coulomb of charge
Measured with a voltmeter
Measuring the cell potential
Can we measure the total cell potential??
A galvanic cell is made where one of the two electrodes is a reference electrode whose potential is known.
Standard hydrogen electrode (H+ = 1M
and the H2 (g) is at 1 atm) is used as a
reference electrode and its potential was assigned to be zero at 25
0C.
1 M HCl
H+
Cl-
H2
Standard Hydrogen Electrode
This is the reference all other oxidations are compared to
Eº = 0
(º) indicates standard states of 25ºC, 1 atm, 1 M
solutions.
1 atm
Zn+2
SO4-2
1 M HCl
Anode
0.76
1 M ZnSO4
H+
Cl-
H2
Cathode
Standard Electrode Potentials
Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)
2e- + 2H+ (1 M) 2H2 (1 atm)
Zn (s) Zn2+ (1 M) + 2e- Anode (oxidation):
Cathode (reduction):
17.2 Standard Reduction Potentials, E
The E values corresponding to reduction half-
reactions with all solutes at 1M and all gases at 1
atm.
E can be measured by making a galvanic cell in
which one of the two electrodes is the Standard
Hydrogen electrode, SHE, whose E = 0 V
The total potential of this cell can be measured
experimentally
However, the individual electrode potential can
not be measured experimentally.
Why?
If the cathode compartment of the cell is SHE, then the half reaction would be
2H+ + 2e H2 (g); Eo = 0V
And the anode compartment is Zn metal in Zn2+, (1 M) then the half reaction would be
Zn Zn2+ + 2e
The total cell potential measured experimentally was found to be + 0.76 V
Thus, +0.76 V was obtained as a result of this calculation:
Eº cell = EºZn Zn2+ + Eº H
+ H2
0.76 V 0.76 V 0 V
Standard Reduction Potentials
The E values corresponding to reduction half-
reactions with all solutes at 1M and all gases at
1 atm. can be determined by making them half
cells where the other half is the SHE.
E0 values for all species were determined as
reduction half potentials and tabulated. For
example:
– Cu2+ + 2e Cu E = 0.34 V
– SO42 + 4H+ + 2e H2SO3 + H2O E = 0.20 V
– Li+ + e- Li E = -3.05 V
Some Standard Reduction Potentials
Li+ + e- ---> Li -3.045 v
Zn+2 + 2 e- ---> Zn -0.763v
Fe+2 + 2 e- ---> Fe -0.44v
2 H+(aq) + 2 e- ---> H2(g) 0.00v
Cu+2 + 2 e- ---> Cu +0.337v
O2(g) + 4 H+(aq) + 4 e- ---> 2 H2O(l) +1.229v
F2 + 2e- ---> 2 F- +2.87v
Standard Reduction Potentials at 25°C
• E0 is for the reaction as
written
• The more positive E0 the
greater the tendency for the
substance to be reduced
• The more negative E0 the
greater the tendency for the
substance to be oxidized
• Under standard-state
conditions, any species on
the left of a given half-
reaction will react
spontaneously with a
species that appears on the
right of any half-reaction
located below it in the table
• 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
Can Sn reduce Zn2+ under standard-state conditions?
•How do we find the answer?
•Look up the Eº values in in the table of reduction
potentials
•\Which reactions in the table will reduce Zn2+(aq)?
Zn+2 + 2 e- ---> Zn(s) -0.763v
Sn+2 + 2 e- ---> Sn -0.143v
Look up the Eº values in in the table of reduction potentials
Standard cell potential
Zn(s) + Cu+2 (aq) Zn+2(aq) + Cu(s)
The total standard cell potential is the sum of the potential at each electrode.
Eº cell = EºZn Zn2+ + Eº Cu
+2 Cu
We can look up reduction potentials in a table.
One of the reactions must be reversed, in order to change its sign.
Standard Cell Potential
Determine the cell potential for a galvanic cell based on the redox reaction.
Cu(s) + Fe+3(aq) Cu+2(aq) + Fe+2(aq)
Fe+3(aq) + e- Fe+2(aq) Eº = 0.77 V
Cu+2(aq)+2e- Cu(s) Eº = 0.34 V
Cu(s) Cu+2(aq)+2e- Eº = -0.34 V
2Fe+3(aq) + 2e- 2Fe+2(aq) Eº = 0.77 V
Eo cell = EoFe
3+ Eo Fe
2+ + EoCuEo
Cu2+
Eo cell = 0.77 + (-0.34) = o.43 V
The total reaction:
Cu(s) Cu+2(aq)+2e- Eº = -0.34 V
2Fe+3(aq) + 2e- 2Fe+2(aq) Eº = 0.77 V
Cu(s) + 2Fe+3(aq) Cu2+ + 2Fe2+
Eºcell = +0.43 V
Line Notation
SolidAqueousAqueoussolid
Anode on the leftCathode on the right
Single line different phases.
Double line porous disk or salt bridge.
Zn(s)Zn2+(aq)Cu2+Cu
If all the substances on one side are aqueous, a platinum electrode is indicated.
For the last reaction
Cu(s)Cu+2(aq)Fe+2(aq),Fe+3(aq)Pt(s)
Complete description of a Galvanic Cell
The reaction always runs spontaneously in the direction that produces a positive cell potential.
Four parameters are needed for a complete description:
1. Cell Potential
2. Direction of flow
3. Designation of anode and cathode
4. Nature of all the components- electrodes and ions
Exercise
Describe completely the galvanic cell based on the following half-reactions under standard conditions.
MnO4- + 8 H+ +5e- Mn+2 + 4H2O Eº=1.51
Fe+3 +3e- Fe(s) Eº=0.036V
1. Write the total cell reaction
2. Calculate Eo cell
3. Define the cathode and anode
4. Draw the line notation for this cell
17.3 Cell potential, electrical work and free energy
The work accomplished when electrons are transferred through a wire depends on the “push” (thermodynamic driving force) behind the electrons
The driving force (emf) is defined in terms of potential difference (in volts) between two points in the circuit
emf = potential difference (V)
= work (J) / Charge(C) =
q
w
The work done by the system has a
–ve sign
Potential produced as a result of doing a work should have a +ve sign
The cell potential, E, and the work, w, have opposite signs.
Relationship between E and w can be expressed as follows:
E = work done by system / charge
( )
q
wE
=
Charge is measured in coulombs. Thus,
-w = qE
Faraday = 96,485 C/mol e-
q = nF = moles of e- x charge/mole e-
w = -qE = -nFE = DG
Thus, DG = -nFE and
DGo = -nFEo
q
wE
=
Potential, Work, DG and spontaneity
DGº = -nFE º
if E º > 0, then DGº < 0 spontaneous
if E º < 0, then DGº > 0 nonspontaneous
In fact, the reverse process is spontaneous.
Spontaneity of Redox Reactions
DG = -nFEcell
DG0 = -nFEcell 0
n = number of moles of electrons in reaction
F = 96,500 J
V • mol = 96,500 C/mol
DG0 = -RT ln K = -nFEcell 0
Ecell 0 =
RT
nF ln K
(8.314 J/K•mol)(298 K)
n (96,500 J/V•mol) ln K =
= 0.0257 V
n ln K Ecell
0
= 0.0592 V
n log K Ecell
0
Spontaneity of Redox Reactions
If you know one, you can
calculate the other…
If you know K, you can
calculate DEº and DGº
If you know DEº, you can
calculate DGº
Spontaneity of Redox Reactions
Relationships among DG º, K, and Eºcell
2(3e- + Al3+ Al)
3 (Mg Mg2+ + 2e-) Oxidation:
Reduction:
Calculate DG0 for the following reaction at 250C.
2Al3+(aq) + 3Mg(s) 2Al(s) + 3Mg+2(aq)
n = ?
DG0 = -nFEcell 0
E0 = Ered + Eox cell 0 0
DG0 = -nFEcell 0 = ___ X (96,500 J/V mol) X ___ V
DG0 = _______ kJ/mol
17.4 Dependence of Cell Potential on Concentration
Qualitatively: we can predict direction of change in E from LeChâtelier pinciple
2Al(s) + 3Mn+2(aq) 2Al+3(aq) + 3Mn(s); Eo cell = 0.48 V
Predict if Ecell will be greater or less than Eºcell for the following cases:
if [Al+3] = 1.5 M and [Mn+2] = 1.0 M
if [Al+3] = 1.0 M and [Mn+2] = 1.5M
An increase in conc. of reactants would favor forward reaction thus increasing the driving force
for electrons; i.e. Ecell becomes > Eo cell
Concentration Cell: both compartments contain same
components but at different concentrations
Half cell potential are not
identical
Because the Ag+ Conc.
On both sides are not
same
Eright > Eleft
• To make them equal, [Ag+]
On both sides should same
• Electrons move from left to
right
The Nernst Equation
Effect of Concentration on Cell Emf
DG = DG0 + RT ln Q DG = -nFE DG0 = -nFE 0
-nFE = -nFE0 + RT ln Q
E = E0 - ln Q RT
nF Nernst equation
At 298K
- 0.0257 V
n ln Q E 0 E = -
0.0592 V n
log Q E 0 E =
The Nernst Equation
As reactions proceed concentrations of products increase and reactants decrease.
When equilibrium is reached
Q = K ; Ecell = 0
and DG = 0 (the cell no longer has
the ability to do work)
Qualitatively: we can predict the direction of change in E from Lechatelier principle
Find Q
Calculate E
E > 0; the reaction is spontaneous to the
right
E < 0; the reaction is spontaneous to the
left
Predicting spontaneity using Nernst equation
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)
2e- + Fe2+ 2Fe
Cd Cd2+ + 2e- Oxidation:
Reduction: n = 2
E0 = -0.44 + (+0.40)
E0 = -0.04 V
E0 = EFe /Fe + ECd /Cd 2+ 0 0
2+
- 0.0257 V
n ln Q E 0 E =
- 0.0257 V
2 ln -0.04 V E =
0.010
0.60
E = ____________
E ___ 0 ________________
Exercise- p. 843
Determine the cell potential at 25oC for the following cell, given that
2Al(s) + 3Mn+2(aq) 2Al+3(aq) + 3Mn(s)
[Mn2+] = 0.50 M; [Al3+]=1.50 M; E0 cell = 0.4
Always we have to figure out n from the balanced equation
2(Al(s)+ Al+3(aq) + 3e-)
3(Mn+2(aq) + 2e- Mn(s))
n = 6
- 0.0592 V
n log Q E 0 E =
Calculation of Equilibrium Constants
for redox reactions
At equilibrium, Ecell = 0 and Q = K.
Qn
EE o log059.0
=Then,
Kn
Eo log0591.0
0 =
0591.0log
onEK = at 25 oC
2e- + Fe2+ Fe
2Ag 2Ag+ + 2e- Oxidation:
Reduction:
What is the equilibrium constant for the following reaction
at 250C? Fe2+ (aq) + 2Ag (s) Fe (s) + 2Ag+ (aq)
= 0.0257 V
n ln K Ecell
0
E0 = -0.44 –0.80= -1.24 V
E0 = -1.24 V 0.0257 V
x n E0 cell exp K =
n = ___
0.0257 V
x 2 -1.24 V = exp
K = ________________
E0 = EFe /Fe+ EAg /Ag 0 0
2+ +
17.5 Batteries
Lead-Storage Battery
A 12 V car battery consists of 6 cathode/anode
pairs each producing 2 V.
Cathode: PbO2 on a metal grid in sulfuric acid:
PbO2(s) + SO42-(aq) + 4H+(aq) + 2e-
PbSO4(s) + 2H2O(l)
Anode: Pb:
Pb(s) + SO42-(aq) PbSO4(s) + 2e-
Batteries are Galvanic Cells
Anode:
Cathode:
Lead storage battery
PbO2 (s) + 4H+ (aq) + SO2- (aq) + 2e- PbSO4 (s) + 2H2O (l) 4
Pb (s) + SO2- (aq) PbSO4 (s) + 2e- 4
Pb (s) + PbO2 (s) + 4H+ (aq) + 2SO2- (aq) 2PbSO4 (s) + 2H2O (l) 4
Lead-Storage Battery
The overall electrochemical reaction is
PbO2(s) + Pb(s) + 2SO42-(aq) + 4H+(aq)
2PbSO4(s) + 2H2O(l)
for which
Ecell = Ered(cathode) - Ered(anode)
= (+1.685 V) - (-0.356 V)
= +2.041 V.
H2SO4 is consumed while the battery is
discharging
H2SO4 is 1.28g/ml and must be kept
Water is depleted thus the battery should be
topped off always
Dry cell Batteries
Zn (s) Zn2+ (aq) + 2e- Anode:
Cathode: 2NH4 (aq) + 2MnO2 (s) + 2e- Mn2O3 (s) + 2NH3 (aq) + H2O (l) +
Zn (s) + 2NH4 (aq) + 2MnO2 (s) Zn2+ (aq) + 2NH3 (aq) + H2O (l) + Mn2O3 (s)
Dry Cell Battery
Anode: Zn cap:
Zn(s) Zn2+(aq) + 2e-
Cathode: MnO2, NH4Cl and C paste:
2NH4+(aq) + 2MnO2(s) + 2e- Mn2O3(s) + 2NH3(aq)
+ 2H2O(l)
Total reaction:
Zn + NH4+ +MnO2 Zn2+ + NH3 + H2O
This cell produces a potential of about 1.5 V.
The graphite rod in the center is an inert
cathode.
For an alkaline battery, NH4Cl is replaced with
KOH.
Anode: oxidation of Zn
Zn(s) + 2OH- ZnO + H2O + 2e-
Cathode: reduction of MnO2.
2MnO2 + H2O + 2e- Mn2O3 + 2OH-
• Total reaction
Zn(s) + 2 MnO2(s) ---> ZnO(s) + Mn2O3(s)
It lasts longer because Zn anode corrodes
less rapidly than under acidic conditions.
Alkaline Cell Battery
Alkaline Battery
Nickel-Cadmium (Ni-Cad) Battery
Anode: Cd(s) + 2OH- Cd(OH)2 + 2e-
Cathode: NiO2 + 2H2O + 2e- ` Ni(OH)2 + 2OH-
NiO2 + Cd + 2H2O Cd(OH)2 +Ni(OH)2
NiCad 1.25 v/cell
The products adhere to the
electrodes thus the battery can
be recharged indefinite number
of times.
Fuel Cells
A fuel cell is a galvanic cell that requires a continuous
supply of reactants to keep functioning
Anode:
Cathode: O2 (g) + 2H2O (l) + 4e- 4OH- (aq)
2H2 (g) + 4OH- (aq) 4H2O (l) + 4e-
2H2 (g) + O2 (g) 2H2O (l)
17.6 Corrosion
Rusting - spontaneous oxidation of metals.
Most metals used for structural purposes have reduction potentials that are less positive than O2 . (They are readily oxidized by O2)
Fe+2 +2e- Fe Eº= - 0.44 V
O2 + 2H2O + 4e- 4OH- Eº= 0.40 V
When a cell is formed from these two half reactions a cell with +ve potential will be obtained
Au, Pt, Cu, Ag are difficult to be oxidized (noble metals)
Most metals are readily oxidized by O2 however, this process develops a thin oxide coating that protect the internal atoms from being further oxidized.
Al that has Eo = -1,7V is easily oxidized. Thus, it is used for making the body of the airplane.
Water
Rust
Iron dissolves forming a pit e-
Salt speeds up process by increasing conductivity
Anodic area
Cathodic area
Fe Fe+2 + 2e-
Anodic reaction
O2 + 2H2O + 4e- 4OH-
cathodic reaction
Electrochemical corrosion of iron
Fe2+ (aq) + O2(g) + (4-2n) H2O(l)
2F2O3(s).nH2O (s)+ 8H+(aq)
Fe on the steel surface is oxidized (anodic regions)
Fe Fe+2 +2e- Eº=- 0.44 V
e-’s released flow through the steel to the areas that have O2 and moisture (cathodic regions). Oxygen is reduced
O2 + 2H2O + 4e- 4OH- Eº= 0.40 V
Thus, in the cathodic region Fe+2 will react with O2
The total reaction is:
Fe2+ (aq) + O2(g) + (4-2n) H2O(l)
2F2O3(s).nH2O (s)+ 8H+(aq)
Thus, iron is dissolved to form pits in steel
Moisture must be present to act as the salt bridge
Steel does not rust in the dry air
Salts accelerates the process due to the increase in conductivity on the surface
Preventing of Corrosion
Coating to keep out air and water.
Galvanizing - Putting on a zinc coat
Fe Fe2+ + 2e- Eoox = 0.44V
Zn Zn2+ + 2e- Eoox = 0.76 V
Zn has a more positive oxidation potential than Fe, so it is more easily oxidized.
Any oxidation dissolves Zn rather than Fe
Alloying is also used to prevent corrosion. stainless steel contains Cr and Ni that make make steel as a noble metal
Cathodic Protection - Attaching large pieces of an active metal like magnesium by wire to the pipeline that get oxidized instead. By time Mg must be replaced since it dissolves by time
Cathodic Protection of an Underground Pipe
Cathodic Protection of an Iron Storage Tank
Running a galvanic cell backwards.
Put a voltage bigger than the galvanic potential and reverse the direction of the redox reaction.
Electrolysis: Forcing a current through a cell to produce a chemical change for which the cell potential is negative.
That is causing a nonspontaneous reaction to occur
It is used for electroplating.
17.7 Electrolysis
1.0 M
Zn+2
e- e-
Anode Cathode
1.10
Zn Cu 1.0 M
Cu+2
Galvanic cell based on spontaneous reaction:
Zn + Cu2+ Zn2+ + Cu
1.0 M
Zn+2
e- e-
Anode Cathode
A battery
>1.10V
Zn Cu 1.0 M
Cu+2
Electrolytic cell
Zn2+ + Cu Zn + Cu2+
Galvanic Cell Electrolytic Cell
Calculating plating
How much chemical change occurs with the flow of a given current for a specified time?
Determine quantity of electrical charge in coulombs
Measure current, I (in amperes) per a period of time
1 amp = 1 coulomb of charge per second
coulomb of charge = amps X seconds =
q = I x t
q/nF = moles of metal
Mass of plated metal can then be calculated
x ss
C
Stoichiometry of Electrolysis
• How much chemical change occurs with the flow of a given current for a given time?
• Current and time quantity of charge
moles of electrons moles of
analyte grams of analyte
Exercise
Calculate mass of Cu that is plated out when a current of 10.0 amps is passed for 30.0 min through a solution of Cu2+
Excercise
How long must 5.00 amp current be applied to produce 10.5 g of Ag from Ag+?
Electroplating
How many grams of chromium can be plated from a Cr+6 solution in 45 minutes at a 25 amp current?
How many grams of chromium can be plated from a Cr+6 solution in 45 minutes at a 25 amp current?
(45 min) #g Cr = ------------
(45 min)(60 sec) #g Cr = --------------------- (1 min)
How many grams of chromium can be plated from a Cr+6 solution in 45 minutes at a 25 amp current?
(45) (60 sec) (25 amp) #g Cr = --------------------------- (1)
(45)(60 sec)(25 amp)(1 C) #g Cr = ----------------------------- (1) (1 amp sec)
Faraday’s constant
(45)(25)(60)(1 C)(1 mol e-) #g Cr = ---------------------------------- (1)(1)(96,500 C)
(45)(60)(25)(1)(1 mol e-)(52 g Cr) #g Cr = ------------------------------------------- (1)(1)(96,500) (6 mol e-)
Electroplating
How many grams of chromium can be plated from a Cr+6 solution in 45 minutes at a 25 amp current?
(45)(60)(25)(1)(1 mol e-)(52 g Cr) #g Cr = ------------------------------------------- (1)(1)(96,500)(6 mol e-)
= 58 g Cr
Michael Faraday Lecturing at the Royal Institution Before Prince Albert and Others (1855)
Electrolysis of water
Electrolysis of Water
The Electrolysis of Water Produces Hydrogen Gas at the Cathode (on the Right) and Oxygen Gas at the Anode (on the Left)
Other uses of electrolysis
Separating mixtures of ions.
More positive reduction potential means the reaction proceeds forward.
We want the reverse.
Most negative reduction potential is easiest to plate out of solution.
17.8 Commercial electrolytic processes
A Schematic Diagram of an Electrolytic Cell for
Producing Aluminum by the Hall-Heroult Process.
The Downs Cell for the Electrolysis of Molten Sodium Chloride
The Mercury Cell for Production of Chlorine and
Sodium Hydroxide