Electrochemistry
I. Terms and Definitions.
A. Electrochemistry.
1. Study of the interconversion of electrical and chemical energy. Two processes, use
electrical energy to bring about a chemical transformation, or can use a chemical
reaction to do electrical work.
2. Cell = device in which the interconversions take place.
a. In an electrolysis cell one uses electrical energy, from a battery or generator, to bring
about a nonspontaneous, usually endothermic, redox reaction.
b. In a voltaic or galvanic cell one uses a spontaneous redox reaction to generate a voltage.
B. Electrical Terms.
1. Charge (q) = coulomb (C)
a. The magnitude of the electron charge = 1.6022x10–19C.
b. The charge associated with 1 mole of electrons = Faraday (F)
1 F = (6.02x1023)(1.60x10-19) = 96,500 coulombs.
2. Voltage (V,ε) = the difference in electrical potential energy between two points.
joule
a. Basic unit is the volt (= coulomb ).
b.
If one moves a charge of 1 coulomb across a potential energy difference of 1 volt,
will gain or expend 1 joule of energy.
3. Current (I) = measure of the rate of flow of charge through a body.
a. 1 Ampere (Amp) = 1 coulomb of charge flowing past a point per second.
coulomb
b. Dimension = second
c. Number of coulombs = (current in amps)x(time in seconds).
Number of Faradays of charge = Number of coulombs
96500C/F
4. Resistance (R) = measure of the resistance to charge flow.
a. A voltage difference will!generate a current. If a substance has a high resistance, the
current will be small for a given voltage.
Ohm’s Law : V = IR
1
b. Conductance = R
1
c. A good conductor will have a low resistance, while a good insulator will have a high
resistance
C. Types of conductors to be considered.
1. Metallic conductor - metals
a. Charge carriers = electrons.
b. Very low resistances. Resistance increases as temperature increases.
2. Electrolytic conductors - molten salts and solutions of electrolytes.
a. Charge carriers = cations and anions.
b. Resistances higher than metals. Resistance decreases as temperature increases.
II. Voltaic Cells.
A. Introduction.
1. Consider the reaction Cu2+ + Zn(s) -----> Zn2+ + Cu(s)
a. This is a spontaneous redox reaction whose ∆G° = -212.3 kJ/mol
b. Half reactions:
Oxidation : Zn(s) -----> Zn2+ + 2eReduction : Cu2+ + 2e- -----> Cu(s)
c. If the two half reactions could be physically separated so that the transfer of the two
electrons from Zn to Cu2+ takes place through a wire, then a current will flow and
electrical work can be done. This is the principle behind all voltaic cells.
A working voltaic cell is a battery.
2. The maximum electrical work that can be done by such a cell is –∆G.
a. –∆G can be expressed in volts by the equation
–∆G = n Fε
–∆G is in J/mol of the balanced equation; n = the number of electrons transferred per
balanced equation; F = Faraday constant (96500 Coulombs/mol electrons);
ε = Voltage (J/Coulomb)
b. Note that (nFε) is just a unit conversion factor—from joules/mol to volts.
c. The voltage of the Zn/Cu2+ cell under standard conditions is
-∆G°= 212,300 J = (2 mol elect.)(96,500 coul/mol elect.) ε°
212300
ε° = (2)(96500)
J/C = 1.10 volts
2
3. There are several ways to separate the two half reactions.
a. The simplest way is to use a porous partition to divide the half reactions.
b. A salt bridge (a concentrated solution of a salt, such as KCl or NH4NO3, in a gelatin)
can also be used.
e-
-
e-
+
VTVM
Zn
Cu
Zn
2+
2-
SO 4
Cu2+
1M CuSO 4
1 M ZnSO 4
Zn Half Cell
Cu Half Cell
e-
-
VTVM
e-
+
Salt Bridge
Cu
Zn
1M ZnSO4
K+
Cl -
Zn Half Cell
1) Electrode reaction
Cu Half Cell
Zn(s) ----> Zn2+ + 2e–
2) Type of reaction
Oxidation
3) Electrode label
Anode (Electrode at which
4) Electrode charge
5) Ion Migration
1M CuSO4
Cu2+ + 2e– ---> Cu(s)
Reduction
Cathode ( Electrode at which
oxidation takes place)
reduction takes place)
Negative
Positive
Anions move in
3
Cations move in (anions move out)
4. Cell notation: the above cells are written as follows
Zn(s) | Zn2+ (1 M) | | Cu2+(1 M) | Cu(s)
a. Write the anode half cell first, then the cathode half cell.
b. | | separates the two half cells
c. In writing the half cell show only the electrode and those species involved in the
half reaction (do not show spectator ions. Give the concentrations of solutes or partial
pressures of gases (in atm).
d. | separates different phases in a half cell.
B. Cell Voltages and Spontaneity.
1. Since ε°∝ -∆G°, the cell voltage reflects how spontaneous the cell reaction is.
ε° = 1.10 V.
a. For the cell reaction Zn(s) | Zn2+ (1 M) | | Cu2+(1 M) | Cu(s)
b. If Zn is replaced by a better reducing agent, for example, Mg, the reaction would be
more spontaneous, ∆G would be more negative, and the cell voltage would be higher.
Mg(s) | Mg2+(1M) | | Cu2+(1M) | Cu(s)
ε° = 2.71
c. Since the cathode half cell was the same, the increased voltage can be attributed to an
increased in the “contribution” of the anode half cell.
2. Just as ∆G can be thought of as a measure of the tendencies of the two half reactions to
proceed in the direction indicated by the overall reaction,
ε
ell
can be thought of as being the
sum of an anode half cell potential and a cathode half cell potential.
a. ∆Gcell = ∆G(anode half reaction) + ∆G(cathode half reaction)
b.
εcell = εanode + εcathode
εanode = anode half cell potential.
εcathode =
cathode half cell potential.
One cannot directly measure the absolute values of the half cell potentials but one can develop
tables of relative values by using a common half cell as a standard and comparing all other
half cells to this standard.
4
C. Standard Half Cell Potentials.
1. The standard reference electrode is the Standard Hydrogen Electrode (S.H.E.).
a. S.H.E.
Consists of a glass hood in which a platinum electrode is sealed. The platinum electrode is
immersed in a 1M H3O+ solution and H2(g), at 1 atm pressure, is bubbled over the
electrode.
A sketch of the S.H.E. is given below.
Salt Bridge
H 2 (g) 1atm
Glass Hood
[H3 O+ ] = 1.0M
Pt
b. If the S.H.E. acts as the anode of a cell:
Half cell reaction is
H2(g)(1 atm) ---> 2H+ + 2e-
(H+ is used for H3O+)
Half cell notation is Pt | H2(1 atm) | H+(1M) | |
c. If the S.H.E. acts as the cathode of a cell,
Half cell reaction is
2H+(1M) + 2e- ---> H2(1 atm)
Half cell notation is | | H+(1M) | H2(1 atm) | Pt
d. Since the S.H.E. is used as a reference, its half cell potential, ε°S.H.E. = 0.0000 V.
2. Consider: Zn | Zn2+(1M) | | S.H.E.
ε = 0.76 V.
a. The relative tendency for the half reaction Zn(s) ---> Zn2+(1M) + 2e– is given by 0.76 V.
0.76 V = Standard Oxidation Potential of Zn. It reflects -∆G for the half reaction.
This is a measure of the tendency of Zn to be oxidized
b. The relative tendency for the reverse half reaction, Zn2+(1M) + 2e- ---> Zn(s), is given
by -0.76 V = Standard Reduction Potential of Zn2+
This is a measure of the tendency of Zn2+ to be reduced.
5
3. Consider: Mg | Mg2+(1M) | | S.H.E.
ε° = 2.37 V
2.37 V = standard oxidation potential of Mg = reflects relative tendency for half reaction
Mg(s)
Mg2+ + 2e- to take place.
–2.37 V = the standard reduction potential of Mg2+ = measure of the relative
tendency of the half-reaction Mg2+ + 2e4. Consider: S.H.E. | | F2(g)(1 atm) | F-(1M) | Pt
Mg(s) to take place.
ε°= 2.87 V.
a. 2.87 V = Standard reduction potential of F2 = measure of the relative tendency of F2
to be reduced = tendency for the half reaction F2 (1 atm) + 2e- ---> 2F-(1M)
to take place.
b. -2.87 V = standard oxidation potential of F- = measure of the relative tendency of the
half reaction 2F-(1M) ---> 2e- + F2(1 atm) to take place.
That is, the tendency of F- to be oxidized.
5. A large number of standard electrode potentials have been measured at 25 °C and have
been tabulated. By convention, standard reduction potentials (ε°) are usually listed.
The following table is a list of the standard reduction potentials of some common
substances.
6
a. The half reactions are of the form:
Ox + ne- -----> Rd
1) Ox is an oxidizing agent. Rd is a reducing agent.
2) If ε° is large and positive, then Ox is a good oxidizing agent, and Rd is a poor
reducing agent.
3) If ε° is large and negative, then Rd is a good reducing agent and Ox is a poor
oxidizing agent.
b. In the list on the following page, F2(g) is the best oxidizing agent and Li(s) is the
best reducing agent.
Conversely, F- is the poorest reducing agent and Li+ is the poorest oxidizing agent.
D. Uses of ε°‘s.
1. Calculate cell voltages under standard conditions.
a. Calculate
ε
o
cell
Zn(s) | Zn2+(1M) | | Cu2+(1M) | Cu(s)
for:
1) From the list of standard reduction potentials find
ε°
Half Reaction
Zn2+ + 2e- ----> Zn(s)
-0.76 V
Cu2+ + 2e- ----> Cu(s)
0.34 V
2) Write the anode, cathode and cell reactions
Electrode Reactions
Tendency
Anode: Zn(s) ----> Zn2+ + 2e-
0.76 V
Cathode: Cu2+ + 2e- ---> Cu(s)
0.34 V
o
Cell: Zn(s) + Cu2+ ----> Zn2+ + Cu(s) ε cell = 1.10 V
b. Calculate
ε ocell
for: Fe(s) | Fe2+(1M) | | MnO4 (1M) , H+(1M), Mn2+ (1M) | Pt(s)
Given:
ε°
Half Reaction
MnO4 + 8H+ + 5e- ----> Mn2+ + 4H2O
Fe2+ + 2e- ----> Fe(s)
1.51 V
-0.74 V
7
Electrode Reactions
Tendency
Anode: 5(Fe(s) ----> Fe2+ + 2e–)
0.74 V
Cathode: 2(MnO4 + 8H+ + 5e- -----> Mn2+ + 4H2O)
1.51 V
Cell: 5Fe(s) + 2MnO4 + 16H+ ----> 5Fe2+ + 2Mn2+ + 8H2O
ε ocell
= 0.74 + 1.51 = 2.25 V
c. Note:
1) The standard oxidation potential (minus the standard reduction potential) is used
for the anode tendency .
2) Even when the half reactions must be multiplied by factors so that the electrons
cancel, the tendencies are not multiplied by those factors.
3) Any spontaneous reaction can be used as the basis of a voltaic cell that will
generate a positive voltage.
2. Prediction of the spontaneity of redox reactions under standard conditions.
a. A reaction is spontaneous if a positive
ε ocell
is obtained when the reaction is used
in a voltaic cell.
b. Will Au3+ oxidize H2O to give O2 and Au(s) under standard conditions?
1) Calculate
ε ocell
for the cell reaction Au3+ + H2O ----> Au(s) + O2 + H+
Reduction potentials are:
εo
Half Reaction
Au3+ + 3e- ---> Au(s)
1.50 V
O2 + 4H+ + 4e- ---> 2H2O 1.23 V
2)
Electrode Reactions
Tendency
Anode: 3(H2O ----> O2 + 4H+ + 4e-)
-1.23 V
Cathode: 4(Au3+ + 3e- ---> Au(s))
Cell: 3H2O + 4Au3+ ----> 3O2 + 12H+ + 4Au(s)
1.50 V
ε ocell
o
Since ε cell is positive, the reaction is spontaneous.
8
= 0.27 V
c. Will Cu+ be stable under standard conditions or will it disproportionate to give
Cu2+ and Cu(s)? Is the reaction 2Cu+
Cu2+ + Cu(s) spontaneous?
1) Standard reduction potentials.
ε°
Half Reaction
Cu+ + 1e- ----> Cu(s)
0.52 V
Cu2+ + 1e- ----> Cu+
0.15 V
2) Electrode Reactions
Tendency
Anode: Cu+ ----> Cu2+ + 1e-
-0.15 V
+
Cathode: Cu + 1e- ----> Cu(s)
0.52 V
o
Cell: 2Cu+ ----> Cu2+ + Cu(s) ε cell = 0.37 V
o
Since ε cell is positive, the disproportionation reaction is spontaneous.
A solution of Cu+ is not stable.
d. Note that any oxidizing agent will spontaneously react with any reducing agent below it
in a list of Standard Reduction Potentials.
II. Cell Voltages Under Nonstandard Conditions. The Nernst Equation.
A. Thermodynamic Origin.
1. ∆G = ∆G° + RT ln Q
a. Since –∆G = –nFε and ∆G° = –nFε°, –nFε= –nFεo+ RT ln Q
ε = ε °- RT
nF
ln Q
This is the Nernst Equation
b. The working form of the Nernst equation at 25 °C that is usually used is
ε = ε° - 0.0591
n
0.0591 =
log Q
log = base 10 logarithm
(8.314)(2.303)(298)
96500
2. Note that at equilibrium ∆G = 0, Q = K, and, therefore, ε = 0
The Nernst equation becomes
ε = 0 = ε° - 0.0591
log K or
n
n! "
K = 10 0.0591
B. Uses of the Nernst Equation.
9
1. Calculate ε for the cell:
Pt(s) | Fe2+(0.01M), Fe3+(0.5M) | | MnO4– (1.0x10-3M), H+(2.0M), Mn2+(0.03M) | Pt(s)
The standard reduction potentials are:
MnO4
Half Reaction
ε°
+
2+
+ 8H + 5e ----> Mn + 4H2O 1.51 V
–
Fe3+ + 1e- ----> Fe2+
0.77
Electrode Reactions
Tendency
5(Fe2+ ----> Fe3+ + 1e-)
Anode:
–0.77 V
Cathode: MnO4– + 8H+ + 5e- ----> Mn2+ + 4H+
Cell: 5Fe2+ + MnO4– + 8H+ ----> 5Fe3+ + Mn2+ + 4H2O
3+ 5
2+ 5
ε ocell
= 0.74 V
2+
[Fe ] [Mn ]
Q=
1.51 V
+ 8
where [ ] = molar concentration
-
[Fe ] [H ] [MnO4]
Q=
(0.50)5 (0.03)
= 3.66x107
(0.01)5 (2.0)8 (1.0x10 !3 )
b. The Nernst equation
ε
cell
o
0.0591
= ε cell - n
log Q
ε
cell
= 0.74 -
0.0591
log (3,66x107)
5
ε
cell
= 0.74 -
0.0591
0.0591
log 3.66x107 = 0.74 (7.564)
5
5
ε
= 0.74 - 0.09 = 0.65 V
c. K for the cell reaction.
cell
n! o
K = 10
0.0591
(5)(0.74 )
= 10
0.0591
= 10 62.61 = 4.0x10 62
For all practical purposes the reaction goes 100% to completion.
2. Concentration Cells.
Calculate the voltage of the cell: Cu(s) | Cu2+(0.001M) | | Cu2+(2M) | Cu(s)
a. Cell reactions.
Anode: Cu(s) ----> Cu2+ (0.001M) + 2eCathode: Cu2+(2M) + 2e- ----> Cu(s)
Cell: Cu2+(2M) -----> Cu2+(0.001M)
10
b. Nernst equation.
o
[Cu2 + ]anode 0.001M
ε cell = 0.000; Q = [Cu2+ ]cathode = 2M = 5.0x10-4
ε
cell
= 0.00 -
0.0591
0.0591
log 5.0x10-4 = - 2 (-3.30) = 0.098 V
2
3. Calc ulate the voltage of the cell Pt | H+ (1.1x10–3 M) | H2(g) (1atm) || SHE
H+ (1.0x10–3M)
Anode: 1/2H2(g)
Cathode: H+ (1M)
1/2H2(g)
H+ (1M)
! 0 = 0.00 "
H+(1.0x10–3 M)
0.0591
log(1.0x10 -3 ) = 0.0591( "log(1.0x10 "3 ) = 0.0591(3.0)
1
In general for a M molar acid solution in the anode, ε cell = 0.0591(-log M) = 0.0591 pH
∴pH =
o
"cell
0.0591
3. Measurement of Ksp’s. Calculate Ksp for AgCl
!
a. Half cell potentials.
ε
Half Reaction
AgCl(s) + e- ----> Ag(s) + Cl–
0.22 V
Ag+ + e- -----> Ag(s)
0.80 V
b. Cell: Electrode Reactions
Tendency
Anode: Ag(s) ----> Ag+ + e-
-0.80
Cathode: AgCl(s) + e– ----> Ag(s) + Cl–
Cell: AgCl(s) ----> Ag+ + Cl-
K
sp
= 10
ε ocell
(1)(-0.58)
0.0591
0.22
= -0.58 V
-9.81
= 10
11
-10
= 1.5x10
III. Electrolysis.
A. Electrolysis of Molten Salts Using Inert Electrodes.
1. Consider the following cell for the electrolysis of molten NaBr using inert electrodes.
e-
e-
-
generator
+
Inert Cathode
Inert Anode
-
Na +
+
Br Molten NaBr
-
a. Inert electrodes: ones that do not chemically react during electrolysis. The
most common are C (graphite), Pt, Au.
b. The generator is any source of direct current. It moves electrons through
the circuit from the negative pole to the positive pole.
c. Outer circuit (generator, connecting wires, electrodes) current is carried by electrons
moving from the negative to the positive poles of the generator.
d. In the body of the melt current is carried by the motion of the Na+ ions towards the
negative electrode and Br– ions towards the positive electrode.
2. In order for current to flow, negative charge must move continuously from the negative
to the positive pole of the generator and, at any instant in time, electrical neutrality must
be maintained at all points in the cell. Therefore, electrons must be simultaneously
transferred at the two electrodes.
a. Negative electrode.
1) Na+ ions approach. Electrons must be transferred from the electrode to the melt to
complete the circuit and the positive charge in the melt around the electrode must
be reduced.
2) Reduction must take place at this electrode. Electrode is the cathode.
3) Na+ ions are reduced.
12
Na+ + e- -----> Na
This electrode reaction allows for the transfer of electrons and also helps maintain
electrical neutrality in the melt around the electrode.
b. Positive electrode.
1) Br– ions approach. Electrons must be transferred from the melt to the electrode and
the negative charge in the melt around the electrode must be reduced.
2) Oxidation must take place at this electrode. Electrode is the anode.
3) Br- is oxidized.
1
Br-– -----> 2 Br2 + e–
This electrode reaction allows electron transfer and helps maintain electrical neutrality
in the melt around the electrode.
c. Cell reactions
1
Anode: Br– -----> 2 Br2 + e–
Cathode: Na+ + e- ----> Na
1
Cell: Na+ + Br – ------> Na + 2 Br2
Na reacts spontaneously and exothermically with Br2 to form NaBr. Therefore, have
used electrical energy from the generator to bring about a nonspontaneous,
endothermic reaction.
B. Electrolysis of Aqueous Solutions using Inert Electrodes.
1. Consider the electrolysis of an aqueous solution of NaBr.
a. The cell diagram is basically the same as that found for the molten NaBr
electrolysis.
1) Current is carried in the outer circuit by the movement of electrons from the
negative to the positive pole of the generator.
2) Current in the body of the melt is carried by the movement of the Na+ ions
towards the negative electrode and the Br– ions towards the positive electrode.
Since H2O is a neutral molecule, its motion is not important in determining
the current.
b. Negative electrode (cathode). Two possibilities for reduction
Na+ + e– ----> Na
1
H2O + e– ----> 2 H2 + OH–
13
1) Both allow the transfer of electrons from the electrode to the solution and
help maintain electrical neutrality, either by removing positive ions from the
solution around the electrode (the Na+ reduction) or by generating negative ions
(the H2O reduction).
2) The electrode reaction requiring less energy will occur. Sodium is a very
reactive metal (ε° = -2.71 V), therefore, Na+ is difficult to reduce. It takes
much less energy to reduce H+ in water than it does Na+; H2O is reduced
and H2(g) is evolved at the cathode.
c. Positive electrode (Anode). Two possibilities
1
Br– ----> 2 Br2 + e–
1
H2O -----> 2H+ + 2 O2 + 2e–
It takes less energy to oxidize Br – than it does to oxidize H2O. Therefore, Br2
is produced at the anode.
d. Cell reactions.
1
Anode: Br– -----> 2 Br2 + e1
Cathode: H2O + e- ----> 2 H2 + OH–
Cell:
H2O + Br— ----->
1
1
H2 + 2 Br2 + OH—
2
1) H2 reacts spontaneously with Br2. Therefore, have used energy from the
generator to bring about a nonspontaneous reaction.
2) In solution, the net effect of electrolysis is to replace Br— by OH—.
As electrolysis continues, NaBr is converted into NaOH.
2. Overvoltage.
a. In theory one should be able to use the Nernst equation to predict what electrode
reactions will occur. According to the Nernst equation, the reduction potential ε of H+
as a function of [H+] is
ε
= 0.000 - 0.0591 log
1
= -0.0591pH
[H + ]
at pH = 7.0, ε = -0.414 V.
b. Therefore, only metals whose reduction potentials are less negative than -0.41 V
14
should be reduced and plate out at the cathode in the electrolysis of aqueous
solutions of electrolytes.. This means that it should not be possible to reduce
metal ions such as Zn2+ (ε° = -0.76 V), Cr3+(ε° = -0.74 V) and
Fe2+(ε° = -0.44 V) from aqueous solutions. In fact, these metals do plate out of
solution in electrolysis.
c. The problem is that the Nernst equation is a thermodynamic equation that says
nothing about kinetics. Some reactions, such as the evolution of H2, are too slow
to occur at the potentials given by the Nernst equation and only take place at higher
voltages.
d. The overvoltage = the potential at which an electrode reaction takes place - the
potential calculated from the Nernst equation.
e. The overvoltage depends on the electrode reaction, the current passing through the
cell, and the electrode material. Hydrogen overvoltages in excess of 1 V are not
uncommon.
f. Overvoltage can also cause the actual voltages measured in voltaic cells to be less
than expected from the Nernst equation, especially when H2(g) or O2(g) are involved.
C. Rules for Predicting Aqueous Electrolysis Reactions.
1. Inert Electrodes.
a. Cathode.
Either the metal cation is reduced or H2 is evolved.
+
H2 is produced if the metals are in Groups 1, 2, Al3+, or when the cation is NH4 .
Otherwise, the metal will plate out on the electrode.
b. Anode.
Either the anion is oxidized or O2 is evolved.
There are two types of anions commonly encountered.
1) Monatomic anions (halides). O2 is evolved from all F- solutions and from
dilute Cl- solutions, otherwise the halogen is produced.
n2) Oxyanions, XOy . In general , O2 is produced if X is in its highest
oxidation state, otherwise X is oxidized to that state.
The normal maximum oxidation states for the main group elements are:
Group 13, +3; Group 14, +4; Group 15, +5; Group16, +6; Group 17, +7.
2. Active Electrodes (metals other than Pt or Au).
a. Cathode. No Change from inert electrodes.
15
b. Anode. Irrespective of the anion, the electrode metal will be oxidized.
3. Examples.
a. Table of some electrolysis reactions.
Electrode Material
Anode Cathode
Electro__lyte_
Inert
Inert
NiI2
Inert
Cu
K2CO3
Inert
Cu
Cu
Cu
Anode half
____Reaction
1
I- ---> 2 I2 + e1
H2O ---> 2 O2---> 2H++2e-
Ca(NO2)2 NO2 +H2O --> NO3 +2H++2eCuSO4
Cu ---> Cu2+ + 2e-
_
Cathode half
Reaction______
Ni2+ + 2e- ---> Ni
1
H2O + e- ---> 2 H2 + OH1
H2O + e- ---> 2 H2 + OHCu2+ + 2e- ---> Cu
b. The last reactions are used in the electro-refining of Cu. Impure Cu can be obtained by
heating copper(II) sulfide with O2. This Cu, called blister copper, contains two types
of metal impurities, those metals that are more active than Cu, such as Zn, and less
active metals, such as Ag. In electro-refining, the blister copper is made the anode of a
cell in which a solution of CuSO4 is electrolyzed and a thin piece of pure Cu is used as
the cathode. The voltage of the cell is set so that Cu and the more active metal impurities are
oxidized but the less active metal, Ag, is not. The Ag falls to the bottom of the cell as the
electrode is consumed (this is called anode slime). The Cu2+ is then re-reduced at the cathode
but the more active Zn2+ remains in solution. This gives very pure copper metal at the
cathode.
D. Quantitative Aspects of Electrolysis. Faraday’s Laws.
1. Since electrical neutrality must be maintained during electrolysis, for every electron
transferred at the cathode an electron must be simultaneously transferred at the anode.
a. Recall that the equivalent mass was defined as the mass that will react with 1 mole
of electrons. Therefore, if 1 mole of electrons (1 Faraday of charge) is passed
through an electrolysis cell, one equivalent mass of material will be produced at the
anode and one equivalent mass will be produced at the cathode.
b. When a current of I amperes is passed through a cell for t seconds,
Number of coulombs of charge passed through the cell =
(current in amps)x(time in sec) = (I)(t)
Number of Faradays =
No. Of coulombs
96500 coul/F
Grams of material produced = (equivalent mass)x(No. of Faradays)
16
2. Examples.
a. A current of 5.0 amps is passed through an aqueous solution of MgI2 for a period
of 2.0 hours. Calculate the number of grams of material produced at each
electrode.
1) Number of coulombs passed = (5.0 C/s)(3600 s/hr)(2.0 hr) = 36,000 C
36000 C
Number of Faradays = 96500 C/F = 0.373 F
254
Eq. mass I2 = 2 = 127 g/F
2) Anode: 2I- ----> I2 + 2e-
g I2 = (127 g/F )(0.373 F ) = 47.3 g
1
3) Cathode: H2O + e- ---> 2 H2 + OH–
Eq. mass H2 = 1.0 g/F
g H2 = (1.0 g/F )(0.373 F ) = 0.37 g
b. How long must a current of 2.0 amps flow through aqueous Ni(NO3)2 in order
to produce 0.50 g of material at the cathode?
1) Cathode: Ni2+ + 2e- -----> Ni
58.7
Eq. mass Ni = 2 = 29.4 g/F
1F
1s
t = (0.50 g)(29.4 g )(96500 C/F )(2.0 C ) = 821 s or 13.7 min.
2) How many grams of material is simultaneously produced at the anode?
1
Anode: H2O ----> 2 O2 + 2H+ + 2e-
16
Eq. mass O2 = 2 = 8 g/F
1F
g O2 = (821 s)(2.0 C/s)(96500 C )(8 g/F) = 0.14 g
17
IV. Batteries.
A. Primary Cells.
1. Working voltaic cells that cannot be recharged.
2. Dry Cell (Leclanché cell)
a. Zn can separated from a paste of MnO2, NH4Cl and ZnCl2 by a porous partition.
b. The cathode is a graphite rod in the middle of the paste and the anode is the Zn can.
c. The cell reactions are complex and can change with the age of the battery and
the amount of current being drawn from the battery.
1) Anode reaction: Zn(s) ----> Zn2+ + 2e+
2+
The Zn2+ is complexed by NH3 : Zn2+ + 2NH4 ---> Zn(NH3)2 + 2H+
2) Cathode reaction: Complex series of reactions that involve MnO2 being reduced as in
MnO2(s) + H2O(l) + e- ----> MnO(OH) + OH–
+
2+
3) Net: : 2MnO2(s) + Zn(s) + 2NH4 ----> 2MnO(OH) + Zn(NH3)2
d. The cell delivers 1.5 V. Due to the buildup of electrode products and gases in the cell,
the voltage can fall to about half that value if current is rapidly drawn from the cell.
The voltage is restored somewhat on standing.
+
e. Since the acidic NH4 can attack the Zn directly, the batteries will deteriorate over time.
3. Alkaline Battery.
a. It is similar to the Leclanché cell except that the NH4Cl is replaced by KOH or NaOH.
b. The reactions are:
Anode: Zn(s) + 2OH– ----> ZnO(s) + H2O(l) + 2eCathode: MnO2(s) + H2O(l) + e- ----> MnO(OH) + OH–
c. The cell delivers 1.54 V. It gives a more constant voltage than the Leclanché cell and
has a longer shelf life. It is more expensive than the acid dry cell.
4. Mercury Battery.
a. These batteries consist of Zn anode. The cathode is a steel (inert) electrode container
that holds a paste of HgO, KOH and Zn(OH)2. The cell reaction is the oxidation of Zn by HgO.
Anode: Zn(s) + 2OH- ----> ZnO(s) + H2O(l) + 2eCathode: HgO(s) H2O(l) + 2e- ----> Hg(l) + 2OHNet: HgO(s) + Zn(s) -----> Hg(l) + ZnO(s)
18
b. Each cell delivers 1.35 V. Since the reactants and products are in condensed phases,
the battery maintains a constant voltage through its life. The batteries can also be made
very small and compact and are used in calculators, watches, cameras, etc.
c. Since Hg is a poisonous heavy metal, there are environmental problems associated with
the widespread use of these batteries.
B. Secondary Cells (storage batteries).
Secondary cell can be recharged.
1. Lead Storage Cell.
a. Consists of a lead anode and supported PbO2 anode immersed in a solution of
sulfuric acid as the electrolyte.
2Anode: Pb(s) + SO4 -----> PbSO4(s) + 2e2Cathode: PbO2(s) + 4H+ + SO4 + 2e- ----> PbSO4(s) + 2H2O(l)
Net: Pb(s) + PbO2(s) + 2H2SO4 -----> PbSO4(s) + PbSO4(s) + 2H2O(l)
b. Each cell delivers 2.0 V; a 12 V car battery is six cells connected in series.
The battery is capable of delivering very high current densities.
c. As the cell discharges the Pb(s) is converted to PbSO4(s) on the anode and the
PbO2(s) is converted to PbSO4(s) on the cathode. No material leaves either electrode.
Therefore, the above cell reaction can be reversed by reversing the original anode and
cathode electrodes in an electrolysis cell. The Pb(s) and the PbO2(s) are restored and
the cell can be discharged again.
d. As the cell discharges, H2SO4 is consumed and H2O is produced. Since the density
of H2SO4 is greater than that of H2O, the state of discharge of the battery
can be determinedby measuring the density (or specific gravity) of the electrolyte.
2. Nickel-Cadmium Alkaline Battery. (Ni-Cad battery)
a. Consists of a number of anode and cathode plates separated by layers of porous
material soaked in NaOH or KOH.
Anode: Cd(s) + 2OH- -----> Cd(OH)2(s) + 2eCathode: Ni(OH)3(s) + e- -----> Ni(OH)2(s) + OHNet: 2Ni(OH)3(s) + Cd(s) ----> 2Ni(OH)2(s) + Cd(OH)2(s)
b. Each cell delivers 1.4 V. Since the anode and cathode materials never leave the
electrode surfaces the cell can be recharged.
c. The Ni-Cad batteries are easy to recharge and the cells can be sealed, they are used in
many cordless appliances.
19
C. Fuel Cells.
1. Primary cells in which the cell reaction is a combustion reaction.
a. The combustion reaction of methane gas, CH4(g) + 2O2(g) ----> CO2(g) + 2H2O(l)
is a redox reaction that could be used as the basis of a voltaic cell. The half reactions are:
Anode: CH4(g) + 2H2O -----> CO2(g) + 8H+ + 8eCathode: O2(g) + 4H+ + 4e- ----> 2H2O
Net: CH4(g) + 2O2(g) -----> CO2(g) + 2H2O
b. These have the potential advantage that a high percent of the energy released in the
combustion reaction could be directly converted into electricity.
2. A fuel cell based on the combustion of H2(g) as the fuel is used on the space shuttle.
20
Electrochemistry Problems
(Use ε°’s
in the unit and the text in answering these questions)
1. Calculate εcell for each of the following Voltaic cells.
a. Cr(s) | Cr3+(0.050M) | | Cd2+(3.00M) | Cd(s)
b. Pt(s) | Sn2+(0.020M), Sn4+(0.002M) | | Cl–(0.50M) | AgCl(s) , Ag(s)
c. Pt(s) | Fe2+(2.00M), Fe3+(0.100M) | | Ag+(0.50M) | Ag(s)
2. Calculate the equilibrium constants for the following redox reactions.
a. Fe2+ + Br2(l) ----> Fe3+ + Br –
b. Ce3+ + MnO4 + H+ -----> Ce4+ + Mn2+
c. Al(s) + Mn2+ -----> Al3+ + Mn(s)
3. Calculate the value of the reduction potential for O2(g) + 4H+ + 4e- ---> 2H2O under normal
atmospheric conditions, that is, neutral water and an atmosphere that is 20 mole percent O2.
4. Consider the cell Zn(s) | Zn2+(0.002M) | | Cr3+(5.00M) | Cr(s) . What is the cell voltage?
5. Answer the following questions and give suitable
εocell
to justify your answers. Assume
standard conditions.
2a. Will Au+ oxidize Cr3+ to Cr2O7 ?
b. Will H2O2 disproportionate in solution to give O2 and H2O?
c. Will nitric acid oxidize Ag to Ag+?
6. Consider the following cell:
Au(s) | Au3+(1.5x10-4M) | | Cl- (2.0x10-3M) | Cl2(g) (10.0 atm) | Pt
a. Write the anode, cathode, and cell reaction.
o
b. Calculate εcell and ∆Go for the cell reaction.
c. Calculate εcell and ∆G for the cell reaction in the above cell.
d. Is the cell reaction spontaneous under standard conditions? Is the cell reaction
spontaneous under the concentration conditions given in the cell?
e. Calculate the equilibrium constant, K, for the cell reaction.
7. Diagram the cell in which the cell reaction is:
3 Ni(s) + 2 Al3+ ----> 3 Ni2+ + 2 Al(s)
o
Calculate εcell . Is the cell reaction spontaneous under standard conditions?
21
8. Consider the following half reactions under standard conditions:
2+
Fe3+ + e- -----> Fe2+
Hg + 2 e- -----> Hg(l)
Cl2(g) + 2 e- ----> 2 Cl-
Ag+ + e- -----> Ag(s)
Mn2+ + 2 e- -----> Mn(s)
Give the formulas for the best oxidizing agent and the best reducing agent in the above list.
Under standard conditions, will Fe3+ oxidize Cl- to Cl2? Will Cl2 oxidize Mn to Mn2+ under
standard conditions? Give cell voltages or half cell voltages to justify your
answer.
9. Calculate Ksp for PbSO4.
10.
Suppose that a current of 0.50 amps is passed through an aqueous solution of Zn(NO3)2 for a
period of 3.0 hr. Calculate the grams of material produced at each electrode. Assume inert
electrodes.
11.
A lead storage cell is recharged by passing a current of 4.0 amps through the cell for 5.0 hrs.
How many coulombs of charge could this cell deliver?
12.
How long must a current of 2.0 amps be passed through the following cells to produce 6.0 g
of material at the cathode? Assume inert electrodes.
a. Aqueous MnSO4.
b. Molten AlCl3.
c. Aqueous Na2CO3.
13. Consider the following electrolysis cell in which the electrolyte is in an aqueous solution.
In
each case, write the anode, cathode, and cell reactions.
a.
Anode
inert
Cathode
inert
Electrolyte
KNO3
b.
Ni
inert
KNO3
c.
inert
Ni
K2SO4
d. inert
inert
ZnI2
14. In 1(a) above, how long must a current of 1.50 amperes have to flow in order to produce
1.00 g of product at the cathode? In 1(d) above, how many grams of product would be
formed at the cathode if a current of 5.00 amperes flowed for 3.00 hours?
22
15.
Suppose two aqueous electrolysis cells were connected in series as shown below.
+
Pt
Generator
-
Ag
Pt
Ag
1M NaI 2
1M AgNO3
Cell I
Cell II
a. Label the anode and cathode of each cell.
b. After a current had been passed through the cell for 2.00 hrs., it was found that the anode of Cell
II had lost 1.006 g.
1) What current was passed through the cell?
2) Calculate the grams of material produced at the other electrode in Cell II and the two electrodes
in Cell I?
23
Answers to problems
(Electrochemistry)
1. a. 0.38 v b. 0.14 v c. 0.09 v
2. a. 1.4x1010 b. 3.5x10-9 c. 5.4x1048
3. 0.83 v
4. 0.11 volts
5. a. yes, ε° = 0.17 v b. yes,
ε°= 1.10 v
c. yes, ε° = 0.16 V
6.(b) -0.14 v, 81.06 kJ; (c) 0.12 v, -69.5 kJ; (d) no, yes; (e) 6.1x10-15
o
7. Ni(s) | Ni2+ | | Al3+ | Al(s) εcell = -1.41 v, no
8. No; Cl2 is the best oxidizing agent; Mn is the best reducing agent.
9. Ksp = 8.1x10-7
10.
Cathode: 1.83 g Zn; Anode: 0.45 g O2
11.
7.2x104 coulombs.
12 a.2.93 hrs. b. 8.93 hrs. c. 80.4 hrs.
13. Electrode products only: Anode Cathode
a.
O2
H2
2+
b.
Ni
H2
c.
O2
H2
d.
I2
Zn
14 . (a) 17.8 hrs. (b) 18.3 g.
15. b.1) 0.125 amp. 2) Cell II: 1.006 g; Cell I: anode, 1.183 g I2 ; cathode, 9.32x10-3g H2
24