General Chemistry II
Chapter 19 Lecture Notes
Electrochemistry
Introduction
Electrochemistry deals with interconversion of electrical and chemical energy. Many chemical
changes can be clearly related to the electrons that move from one species to another. Often, this
electron exchange can be captured to do electrical work external to the chemical system (storage
battery, fuel cell). Other times, electrical energy can be used to bring about chemical change
(electrolysis, battery charging, etc.).
Redox Reactions
In redox reactions, electrons are transferred from one species to another. A species losing electrons
is said to be oxidized; one gaining electrons is said to be reduced. The two processes together are
called redox. One can never occur without the other although they can be studied separately.
In redox reactions, it is useful to write the oxidation numbers of each species above the formula to
track oxidation and reduction.
0 0 +2 -2
Ca(s) + S(s) → CaS(s)
0 +1 -1 +3 -1 0
Fe(s) + 3HCl(aq) → AlCl
3
(aq) + 3/2 H
2
(g)
Balancing Redox Reactions
The above redox reactions are easy to balance by inspection. Others are more difficult, such as
those involving the interaction of the oxyanions CrO
4
-2
, Cr
2
O
7
-2
, NO
3
-1
, MnO
4
-1
with water,
hydroxide ion or hydrogen ion. Note that in the reaction
NO
3
-1
(aq) + 2H
+1
(aq) → NO
2
(g) + H
2
O(l)
the atoms are balanced but electrons are not. Both mass and charge must be balanced in any
chemical reaction. How to ensure that both are balanced without spending a lot of time? How to
gain insight into the process? Use the ion-electron method which divides the reaction into two half
reactions, one for oxidation and one for reduction.
Balance each half reaction separately, leaving out water, hydrogen ion and hydroxide ion until the
end and then combine the equations so that electrons cancel out. Then the redox reaction is
balanced.
2
Ex: Balance the reaction between tin(II) and chromate ion in acid solution to give tin(IV) and
chromium(III) as chromic ion. Note that in many of the stated problems, acid is not shown and the
products water, H
+1
and OH
-1
are left out initially.
Sn
+2
(aq) + Cr
2
O
7
-2
(aq) → Sn
+4
(aq) + Cr
+3
(aq)
Write the oxidation and the reduction equations separately, with the oxidation numbers written
above each species. Write the electrons lost or gained by each. Balance all atoms except H and O.
+2 +4
Sn
+2
→ Sn
+4
Oxidation
+6 +3
CrO
4
-2
→ 2 Cr
+3
Reduction
The reaction takes place in acid, but this has no effect on the tin atoms. In the half-reaction
involving chromium, the only way protons can be accommodated is to use them to make water from
the dichromate oxygen atoms. Once the necessary H and O atoms have been balanced, finish
balancing the half-reactions by inserting electrons as reactants in the reduction equation and as
products in the oxidation equation.
+2 +4
Sn
+2
→ Sn
+4
+ 2e
+6 +3
6e + Cr
2
O
7
-2
+ 14H
+1
→ 2 Cr
+3
+ 7H
2
O
Then multiply each equation by the coefficient that will make the number of electrons lost the same
as the number gained, add the equations together to cancel the electrons. Check for mass and
charge balance.
3Sn
+2
→ 3Sn
+4
+ 6e Oxidation
6e + Cr
2
O
7
-2
+ 14H
+1
→ 2 Cr
+3
+ 7H
2
O Reduction
3Sn
+2
+ Cr
2
O
7
-2
+ 14H
+1
→ 3Sn
+4
+ 2Cr
+3
+ 7H
2
O
Equation is balanced in atoms and charge
Some redox reaction rules:
• If an oxyanion loses its oxygens, they are converted to water in an acidic medium or
converted to hydroxide ion in a basic medium.
• In an acidic medium, hydrogen ion must be consumed or created.
• In a basic medium, hydroxide ion must be created or consumed.
Ex: Balance the equation MnO
4
-1
(aq) + Br
-1
(aq) → MnO
2
(s) + Br
2
(aq)
MnO
4
-1
→ MnO
2
Oxidation
Br
-1
→ Br
2
Reduction
3
+7 +4
2H
2
O
+ MnO
4
-1
→ MnO
2
(s) + 4OH
-1
-1 0
2Br
-1
→ Br
2
3e + 2H
2
O
+ MnO
4
-1
→ MnO
2
(s) + 4OH
-1
2Br
-1
→ Br
2
+ 2e
6e + 4H
2
O
+ 2MnO
4
-1
→ 2MnO
2
(s) + 8OH
-1
6Br
-1
→ 3Br
2
+ 6e
4H
2
O(aq)
+ 2MnO
4
-1
(aq) + 6Br
-1
(aq) → 2MnO
2
(s) + 8OH
-1
(aq) + 3Br
2
(aq)
Electrochemical Cells
An electrochemical or voltaic or galvanic cell is an experimental apparatus for generating an
electric current with a spontaneous redox reaction. Practical applications include the many kinds of
batteries, which generate electric current for practical purposes. Electrochemical cells have various
refinements so they will generate reproducible voltages under standard conditions.
(Contributions of Allesandro Volta and Luigi Galvani)
The (historic) Daniell cell: (Use in old west telegraph offices)
Define (Zn) Anode, (Cu) Cathode, Need to keep the reactants separate (Why?)
External connection for electrons, CuSO
4
(aq) and ZnSO
4
(aq) electrolytes.
The inert sulfate (or chloride, nitrate, etc.) counterions do not often affect emf.
Detailed description of Daniell cell, oxidation and reduction processes and external circuitry
The difference between the anode and cathode emfs is the measured emf (electromotive force or
voltage) of the cell and is characteristic of that cell. Emf of the Daniell Cell is 1.10 volt at 25
o
C.
Cell diagram convention with anode written to the left: Zn(s) |Zn
+2
(1M)||Cu
+2
(1M)| Cu(s)
Standard Reduction Potentials
Standard refers to 25
o
C, 1 atm pressure and 1M electrolyte solutions. (Oxidation potentials were
once tabulated but the convention now is to use reduction reactions.) To find the corresponding
oxidation cell potential, simply reverse the direction of the reaction and change the emf sign.
(Make use of Table 19.1)
4
• The cell voltage (cell potential electromotive force, emf or E) of the Daniell Cell or any
other cell can be calculated from the tabulated standard electrode potentials of the anode and
cathode materials.
• However, there can be no oxidation without a simultaneous reduction so it is impossible to
know the absolute emf for the reduction of any one substance.
• Only the difference between two half-cell emfs can be determined experimentally, that is by
measuring the emf of a real cell.
• It is useful to have a table of relative emfs. These are tabulated as Standard Reduction
Potentials for Half Reactions of the form M
+n
(aq) (1M) + ne → M(s).
• The cell potentials of real cells are calculated by adding the appropriate reduction cell
potential to the appropriate oxidation cell potential.
• One half-reaction, the reduction of 1 M hydrogen ion on a (catalytic) platinum electrode to
hydrogen gas at 1 atm is arbitrarily assigned a standard reduction potential of zero volts.
2H
+1
(aq) (1M) + 2e → H
2
(g) (1atm) E
o
= 0.00 volt by definition
• All other half-cell potentials are calculated relative to the hydrogen half-cell voltage. Any
other arbitrarily chosen hydrogen half-cell voltage would give the same values (differences)
for the calculated cell potentials of real cells. If the reduction cell potential for hydrogen
were given as 3.00 V instead of 0.00 V, all other reduction cell potentials would be 3.00 V
higher and the differences between them would stay the same.
Page 775 of the Chang text tabulates 49 reduction cell potentials under standard conditions which
can be combined to calculate the E
o
values of many real or hypothetical cells.
• The most positive E
o
values represent reactions with the greatest forward tendency and the
most negative E
o
values
represent reactions with the greatest backward tendency.
• The equations are all reversible and any electrode can be used as an anode or as a cathode.
• In principal, under standard-state conditions, any combination of oxidation half-cell and
reduction half-cell reactions where the reactants are physically separated, will proceed
spontaneously if E
o
is greater than zero. If E
o
is negative, the reaction will not proceed
spontaneously. Its reverse reaction, having a positive E
o
will be spontaneous.
• How are reactants kept physically separated (and why) by the KCl bridge.
Ex: Calculate the standard half cell reduction potential of lead metal if the measured E
o
of the cell
Pb(s) | Pb
+2
(1M) || H
+1
(1 M) | H
2
(1 atm)| Pt(s) is + 0.13 V.
The anode where oxidation occurs is listed first. The double bar refers to the salt bridge. The right
side of the cell diagram refers to the cathode where reduction occurs.
Anode: Pb(s) → Pb
+2
(aq) + 2e E
o
anode
= ?
Cathode: 2e + 2H
+1
→ H
2
(g) E
o
cathode
= 0.00 V (by definition)
E
o
cell
= E
o
cathode
- E
o
anode
= 0.00 V - E
o
anode
= +0.13 V
E
o
anode
= 0.00 V - 0.13 V = - 0.13 V
Ex: What is the emf of the (hypothetical) cell I
2
(s) | I
-1
(1M) || Au
+3
(1M) | Au(s) |? Is the reaction
of the cell as written spontaneous? Write the spontaneous chemical reaction that occurs in the cell.
5
The anode reaction (oxidation) comes first (left) in the cell diagram so the direction of the equation
and the sign of the E
o
value in the table must changed.
I
-1
(aq) → I
2
(s) + 2e E
o
cathode
= – 0.53V
The cathode reaction is on the right where reduction occurs as it is written in table 19.1.
Au
+3
(aq) + 3e → Au(s) E
o
anode
= + 1.50 V as tabulated.
Au(s) → Au
+3
(aq) + 3e E
o
rxn
= - 1.50 V
E
o
cell
= E
o
cathode
- E
o
anode
= + 0.53 V - (+ 1.50 V) = - 0.97 V
The cell reaction, 3I
2
(s) + 2Au(s) → 6I
-1
+ 2Au
+3
The spontaneous reaction is the reverse, where iodide is reduced by auric ion to elemental iodine.
6I
-1
+ 2Au
+3
→ 3I
2
(s) + 2Au(s)
Spontaneity of Redox Reactions
How does E
o
cell
relate to ∆G, and K
eq
?
The Electrical Work done by a voltaic cell or done on an electrolytic cell is the same as the Free
(useful) Energy, ∆G.
How to calculate Electrical Work? (Contrast actual work obtained with W
max
)
Electrical work done is the movement of charge (electrons) across an electrical gradient, the
voltage.
Electrical Energy(J) = emf(volts or J/cou) x charge(cou)
Electric charge(cou) = no of electrons x charge per electron (inconvenient)
= moles of electrons x electric charge per mole
= n(mol) x F(coulombs/mol)
= n(mol) x 96,485.3 coulombs/mol
Electrical Energy = - emf
(volts or J/cou) x electric charge(cou)
= - E
o
cell
(volts) x n(mole) x F(cou/mol)
W
max
= -nFE
cell
or ∆G = -nFE
cell
Earlier, ∆G = -RTlnK ∴nFE
cell
= RTlnK and
E
cell
= (RT/nF) lnK
eq
R = 8.314 J/K and, under standard conditions, (T = 298K)
E
o
cell
(V) = (0.0257/n)ln K
eq
or since ln K
eq
= 2.302 log K
eq
, then
6
E
o
cell
(V) = (0.0592/n)log K
eq
(This equation will be supplied for tests. Know how to solve for both E
cell
and K
eq
.
Now can calculate K
eq
values from measured cell potentials or E
cell
values from measured K
eq
Ex: Calculate the (difficult to measure) K
eq
of any cell reaction using half-cell reduction potentials.
Under standard conditions, calculate K
eq
for the reaction Ag
+1
(aq) + Al(s) → 3Ag(s) + Al
+3
(aq)
To get E
o
cell
values,
add together the actual half cell potentials for the two reactions as they occur:
or Subtract the listed reduction potential for the anode (oxidation) from the listed reduction
potential for the cathode (reduction).
E
o
cell
= E
o
Ag
(actual, reduction)
+ E
o
Al
(actual, oxidation)
= +0.80V + +1.66V = +2.46V
or
E
o
cell
= E
o
Ag
(actual, reduction)
− E
o
Al
(as a reduction) = +0.80V − −1.66V = +2.46V
Now calculate K
eq
from E
cell
(V) = (0.0592/n)log K
eq
log K
eq
= E
cell
x (3/0.0592) = 2.46 x (3/0.0592) = 125
K
eq
= 1 x 10
125
(This value could not have been obtained from reactant concentrations)
The Effect of Concentration on Cell Emf
Just as the direction of equilibrium is affected by changing reactant and product concentrations, so
is the Emf of a cell can be affected by changing concentrations of cell components.
The Nernst Equation:
Remember the effect of reactant and product concentration of the direction of equilibrium.
Similarly, cell component concentrations affect the cell Emf (E
cell
).
Use intuitive reasoning in the cell:
Pt | H
2
(g) | H
+1
(aq) || Cu
+2
(aq) | Cu(s) with anode and cathode reactions
H
2
(g) → 2H
+1
(aq) + 2e and Cu
+2
(aq) + 2e → Cu(s)
The overall reaction is H
2
(g) + Cu
+2
(aq) → 2H
+1
(aq) + Cu(s)
The forward reaction is driven by increased pH
2
, and [Cu
+2
].
The backward reaction is favored by increased [H
+1
].
How to quantify these facts?
7
E under conditions other than standard (1M, 1 atm) is given (without derivation) by
The Nernst Equation:
E = E
o
- (0.0592/n)log Q
Under standard conditions when all concentrations are 1M, Q = 1, log Q = 0 and E = E
o
.
Ex: Calculate E for the cell: Zn(s) | Zn
+2
(aq) || Br
-1
(aq) | Br
2
(l) | under standard conditions and
when Zn
+2
(aq) = 3.00 M and Br
-1
(aq) = 2.00 M.
The overall cell reaction is: Zn(s) + Br
2
(l) ¥ Zn
+2
(aq) + 2Br
-1
(aq)
Zinc metal (written first in the cell diagram) is the anode and Br
2
(l) acts as the cathode.
E
o
cell
= E
o
cathode
- E
o
anode
= + 1.07V − − 0.76V = + 1.83V
Raising Zn
+2
(aq) and Br
-1
(aq) both shift the equilibrium to the left and will lower E
cell
E
cell
= E
o
cell
− (0.0592/n)log Q = + 1.83V − (0.0592/2)log 6.00 = + 1.83V −
(0.0592/2)(0.778)
= + 1.83V − 0.0230 = + 1.81V
Ex: At what [Fe
+2
]/[Cd
+2
] ratio will the voltaic cell Fe(s) | Fe
+2
(aq) || Cd
+2
(aq) | Cd(s) have zero
Emf?
The anode (oxidation) reaction is Fe(s) → Fe
+2
(aq) + 2e E = − 0.44 V
The cathode (reduction) reaction is Cd
+2
+ 2e → Cd(s) E = − 0.40V
The allover reaction is Fe(s) + Cd
+2
(aq) → Fe
+2
(aq) + Cd(s)
E
o
cell
= E
o
cathode
- E
o
anode
= − 0.40V − − 0.44V = + 0.04V (a very small positive Emf)
Zero Emf means E
cell
= 0.00V
E
cell
= E
o
cell
− (0.0592/n)log Q, so when E
cell
= 0.00V,
E
o
cell
= +(0.0592/n)log Q
+ 0.04V = +(0.0592/2)log Q
log Q = +0.04V · (2/0.0592) = 1.35 (good to one significant figure)
Q = [Fe
+2
]/[Cd
+2
] = 22 ≅ 2 x 10
1
(Does this make sense intuitively? Look at the overall reaction.)
Concentration Cells
Since the concentrations of cell components can influence cell Emf, a cell can be constructed
having the same metal for anode and cathode and the same metal ions, but at different
concentrations in the two parts of the cell.
A Concentration Probe can be constructed from a metal anode and its (for convenience) 1M cation
solution with a salt bridge which is inserted in a solution of the same metal of unknown
8
concentration. A small Emf will be generated that depends on the concentration of the metal ion
solution being tested (and on the number of electrons transferred).
From the Nernst Equation, E = E
o
- (0.0592/n)log Q
E
o
is zero since both anode and cathode are the same material.
Therefore, E = - (0.0592/n)log Q
log Q = – nE /(0.0592)
(Note that since E
o
is zero, the Emf generated by such a cell depends on the ratio of the anode and
cathode cell concentrations, and not on the nature of the reacting species. Moreover, it does not
matter if the concentration ratio or its reciprocal is used. The sign will change but not the
magnitude of the ratio. A test solution concentration less than 1M will give a positive E; one with a
higher concentration will have a negative E value.
Ex: What is the concentration of a Zn
+2
(aq) solution when a 1M Zn
+2
probe gives a voltage of
+0.014V?
log Q = – nE /(0.0592) = – 2 x (+0.014V)/(0.0592) = – 0.47
Q = [Zn
+2
]/[1M Zn
+2
] = 0.34
The unknown solution has [Zn
+2
] = 0.34M
Batteries
The tem battery comes form a series or pile of cells that can be used as a source of direct current at
a constant voltage. Modern commercial batteries are constructed to deliver a constant voltage for
most of their lifetime, be self-contained, leak-proof and with convenient external electrodes.
The Dry Cell Battery
(Almost) dry as opposed to the wet Daniell cell. Used in flashlights and cheap electronics, the
Leclanché cell delivers 1.5 V using a zinc can as the anode that contains an ammonium chloride-
zinc chloride-starch paste designed (almost) to prevent leakage. Higher voltages are attained by
wiring several cells in series. The anode is a conductive (unreactive) graphite rod in contact with
solid manganese dioxide. Anode and cathode reactions are respectively
Zn(s) → Zn
+2
(aq) + 2e
2NH
4
+1
(aq) = 2MnO
2
(s) + 2e → Mn
2
O
3
(s) + 2NH
3
(aq) + H
2
O(l)
9
Battery life ends when much of the MnO
2
is
used up or extensive polarization of the electrolyte
occurs. (A rest from service is helpful) Dead dry-cell batteries leak in time and ruin equipment.
The Mercury Battery
Consists of amalgamated zinc anode and a stainless steel casing which supports the mercuric oxide
cathodic material. Only the solid components change, not the electrolyte, keeping volume and
internal pressure constant. Used in high-end applications where removing battery is inconvenient.
Anode: Zn(Hg) + 2OH
-1
(aq) → ZnO(s) + H
2
O(l) + 2e (plus unchanged Hg)
Cathode: HgO(s) + H
2
O + 2e → Hg(l) + 2OH
-1
(aq)
The Lead Storage Battery
The 2 V lead –acid battery is rechargeable. Must be kept upright to prevent spillage of sulfuric acid
electrolyte. Automotive 6V and 12V (soon 42V?) batteries are built from multiple cells. Loss of
power in cold weather is due to increased viscosity of cold electrolyte which hinders movement of
ions.
Anode: Pb(s) + SO
4
-2
(aq) → PbSO
4
(s) + 2e
Cathode: PbO(s) + 4H
+1
+ SO
4
-2
(aq) + 2e → PbSO
4
(s) + 2H
2
O(l)
The sulfuric acid electrolyte changes to water as the battery is discharged, with a corresponding
decrease in viscosity which can be measures with a hydrometer.
Charging a battery requires the input of electrical energy from the generator which reverses the
reactions and replenishes the battery’s “charge”. Battery engineering involves building longer-
lasting batteries that sustain little electrode damage in repeated charge-discharge cycles. Batteries
are recycled to recover the expensive and environmentally harmful lead metal.
The Solid State Lithium Battery
Nontoxic lithium has the largest reduction potential on the table and has the lowest density, making
a battery that has a high power/weight ratio.
Fuel Cells
Many fuels are burned in internal or external combustion engines to produce power for locomotion
or electric generation. These engines are notoriously inefficient. In a fuel cell the electrons could
be captured during the (redox) burning, electric generation would be more efficient. Oxygen and
10
fuels such as hydrogen or simple hydrocarbons are bubbled over catalytic metal electrodes where
oxidation and reduction takes place under standard, not flame conditions. .
Anode: 2H
2
(g) + 4OH
-1
(aq) → 4H
2
O(l) + 4e
Cathode: O
2
(g) + 4H
2
O(l) + 4e → 4OH
-1
(aq)
E
o
cell
= E
o
cathode
- E
o
anode
= + 0.40 V - (-0.83V) = 1.23 V
Corrosion
Corrosion is defined as the deterioration of metal by an electrochemical process. Rusting of iron
metal is the prime example. Other examples are pitting of aluminum and the patina on outdoors
copper and brass.
Iron rusts in the presence of moisture and air which supplies oxygen and CO
2
which gives H
+1
via
H
2
CO
3
.
Anode: (iron surface) Fe(s) → Fe
+2
(aq) + 2e
Cathode: (also the iron surface) O
2
(g) + 4H
+1
(aq) + 4e → 2H
2
O(l)
The overall redox reaction is
2Fe(s) + O
2
(g) + 4H
+1
(aq) → 2Fe
+2
(aq) + 2H
2
O(l)
E
o
cell
= E
o
cathode
- E
o
anode
= 1.23 V - (-0.44V) = + 1.67 V
Iton(II) is then oxidized by atmospheric O
2
to Fe(III), forming the familiar red-brown flaky solid
hydrated iron(III) oxide Fe
2
O
3
· xH
2
O that we call rust.
Coating iron with paint helps as long as the paint film stays intact.
Alloys of iron with carbon and chromium give “stainless steel”, expensive but durable. (Building
codes require stainless steel nails and screws in certain seaside areas.)
Aluminum corrodes very quickly in air but is protected by the adherent aluminum oxide film. (Iron
rust is flakes off and does not protect.)
Cathodic Protection relies on the preferential corrosion of magnesium metal electrically connected
to an underground iron structure. The flow of electrons to the iron resulting from Mg(s) → Mg
+2
+ 2e opposes the analogous oxidation of iron,
Fe(s) → Fe
+2
(aq) + 2e
11
Electrolysis
Electrolysis means literally taking apart with electricity. Electrolysis is a nonspontaneous chemical
change produced by electrical energy. It is used commercially to prepare many reactive metals in
their elemental form and to prepare fluorine and chlorine gas.
Electrolysis of Molten Sodium Fluoride
Elemental chlorine can be used to oxidize bromide ion to elemental bromine. How would one
prepare elemental fluorine, already the strongest elemental oxidizing agent. There is no other
halogen with a higher electronegativity that can take electrons away from F
-1
to make F
2
. The
solution is to supply the energy from an outside source, a battery, which can supply any voltage.
Typically NaF or KF is melted to mobilize the ions. The anode must be an inert material such as
graphite to prevent attack by fluorine. The cathode must be a metal which will not amalgamate
with molten sodium and the products must be kept separate to prevent the highly exothermic
reaction between fluorine gas and molten Na.
(Extreme reactivity of F
2
gas with water, glass and asbestos. Fatal consequences of inhaling F
2
.)
Anode: 2F
-1
(l) → F
2
(g) + 2e
Cathode: 2Na
+1
+ 2e → 2Na(l)
E
o
= E
o
anode
- E
o
cathode
= - 2.87 V - (+ 2.71 V) = - 5.58 V
The large negative E
o
shows this is a highly endothermic process requiring a large energy input to
drive it. There are no existing cells with E
o
approaching 5.58 V but any number of cells of lower E
o
can be wired in series to get a voltage of the desired value.
This is the commercial method typically used to prepare pure alkali metals which otherwise could
not be prepared from their salts or solutions.
Electrolysis of Water
Electrolysis of water is a common laboratory demonstration. Do not expect water to decompose
spontaneously to hydrogen and oxygen.
2H
2
O(l) → 2H
2
(g) + O
2
(g) ∆G = + 474.4 kJ
The reaction can be easily driven by an electrical current. Pure water is made conductive with
sulfuric acid, chosen because sulfate anion is not easily oxidized.
Anode: 2H
2
O(l) → O
2
(g) + 4H
+1
(aq) + 4e
Cathode: 4H
+1
(aq) + 4e → 2H
2
(g)
12
If the gases are collected separately in burets, the volume of hydrogen is exactly twice that of
oxygen, illustrating their molar volumes. Why is it important to keep the evolved hydrogen and
oxygen gases separate? (Incidental dangerous electrolysis of water in nuclear reactors.)
Electrolysis of Aqueous Sodium Chloride Solution
The result of electrolyzing this solution is not as predicted. The solution contains the oxidizable
species chloride and water and the reducible species water, hydrogen ion and sodium ion. What
actually is produced in electrolysis of this mixture?
The oxidation of chloride and water have almost the same E
o
.
2Cl
-1
(aq) → Cl
2
(g) + 2e E
o
= -1.36 V
2H
2
O(l) → O
2
(g) + 4H
+1
(aq) + 4e E
o
= -1.23 V
Oxygen should be oxidized preferentially at the anode but it is not. Chlorine gas is formed because
of the overvoltage for water oxidation. (Overvoltage, the difference between the calculated and the
actual voltage needed to produce electrolysis is a common phenomenon.)
Possible cathode reactions are:
2H
+1
(aq) + 2e → H
2
(g) E
o
= 0.00 V
2H
2
O(l) + 2e → H
2
(g) + OH
-1
(aq) E
o
= - 0.83 V
Na
+1
(aq) + e → Na(s) E
o
= - 2.71 V
This time the predicted reaction occurs; hydrogen ion is preferential reduced and hydrogen gas
evolves. Other useful products of the Chlor-alkali process are chlorine gas and sodium hydroxide.
Quantitative Aspects of Electrolysis
Faraday observed that the mass of metal deposited by electrolysis was proportional to the amount of
current that flowed through the cell.
Electric current flow is expressed in amperes, and by definition,
Charge(coulombs) = Current(amperes) x Time(s) and
1 mole of electrons = 96500 coulombs
Ex: A 0.430 amp current flowed through a molten magnesium fluoride electrolysis cell for 1 hour,
35.0 minutes. What mass of magnesium and what mass of fluorine was produced?
The overall reaction is: MgF
2
(l) → Mg(l) + F
2
(g)
Anode (oxidation) 2F
-1
(l) → F
2
(g) + 2e
Cathode (reduction) Mg
+2
(l) + 2e → Mg(l)
First, calculate moles of electrons, then calculate moles of fluorine and magnesium.
13
Charge(coulombs) = Current(amperes) x Time(s)
= 0.430 amp x 5700 s = 1.05 x 10
3
coulomb (Total charge that flowed through the cell.)
1.05 x 10
3
coulomb x 1 mole of electrons/96500 coulombs = 1.09 x 10
-2
mol electrons
1.09 x 10
-2
mol electrons x (1 mol Mg/2 mol electrons) x (24.31 g Mg/mol Mg) =
= 0.133 g Mg
1.09 x 10
-2
mol electrons x (1 mol F
2
/2 mol electrons) x (38.00 g F
2
/mol F
2
) =
= 0.207 g F
2
An electrolytic cell can also be used to determine total charge flow or size of current.
Ex: If 1.456 g of silver was deposited in 1.500 hours from Ag
+1
solution, what was the current
flow?
1.456 g Ag x (1 mol Ag/107.9 g Ag) = 1.349 x 10
-2
mol Ag = 1.349 x 10
-2
mol electrons
1.349 x 10
-2
mol electrons x (96,485.3 coulombs/mol electrons) = 1302 coulombs
Charge(coulombs) = Current(amperes) x Time(s)
Current(amperes) = Charge(cou)/Time(s) = 1302 coulombs/5400. s = 0.2411 amps
