Representing electrochemical cells
The electrochemical cell established by the following half
Zn(s) --> Zn2+(aq) + 2 e-
Cu2+(aq) + 2 e- --> Cu(s)
san be represented as:
Anode is shown on the left, the cathode on the right, the
single line separates phases, and the double dotted line
indicates the salt bridge
The electrode need not always be involved in the redox
reaction; it could be an inert substrate through which
electrons flow into or out off.
A cell represented as:
Pt|Fe2+(aq), Fe3+(aq) Cl2(g)|Cl-(aq)|Pt
involves the following half cell reactions:
2Fe2+(aq) --> 2Fe3+ + 2e-
Cl2(g) +2e- --> 2Cl-(aq)
with two Pt strips acting as electrodes
What is the driving force for electron flow in the cell?
The affinity of the two electrodes for electrons is different;
this sets up an electric potential difference DE between the
The potential energy of the electrons is higher at the anode
than at the cathode and the electrons flow spontaneously
from anode to cathode.
The difference in potential energy, DE, per electrical charge
between the two electrodes is measured in units of volts.
One volt (V) is defined so that a charge of one coulomb
(1C) falling through a potential difference of one volt (1 V)
releases one joule (1 J) or energy.
The potential difference between the two electrodes of the
galvanic cell provides the driving force that pushes
electrons through the external circuit.
The potential difference of the cell, DE, is also called the
electromotive force, emf, or cell potential and has units of
A galvanic or voltaic cell, takes advantage of the electron flow
between the two electrodes due to the spontaneous redox
reaction, converting chemical energy into electrical energy.
Under standard conditions (1 bar for gases, or 1M
concentration for solutions) the emf is called the standard
emf or standard cell potential, DEo.
For the Cu/Zn system the standard cell potential across the
Cu and Zn electrodes is 1.10V; DEo = +1.10V
Differences in electric potential drives the flow of current.
This flow of current can do work - electrical work.
welec = - Q DE
Q: amount of charge (in coulomb)
DE : electric potential (in V)
Maximum amount of work done by the system during a
wmax = DG (at constant T and P)
DG = - Q DE
If DE is > 0, DG < 0, cell reaction is spontaneous
It is also possible to drive an electrochemical cell in the
reverse direction of the spontaneous reaction.
This can be done by applying an external voltage of
magnitude greater than the potential difference established
by the spontaneous redox reaction, and in the opposite
For example in the Cu/Zn system applying an external voltage
>1.10 V forces the reverse reaction to occur, converting the
applied electrical energy into chemical energy.
Such cells are called electrolytic cells, and force reactions to
take place that do not occur spontaneously.
In a battery a redox reaction occurs in the self-contained
packaging of the battery.
The reaction produces a cell emf that can be used to drive
electrons through an external circuit, producing electricity.
As the battery operates, reactants are consumed, the cell emf
(which depends on the concentration of the reactants) drops
till eventually it is zero.
Primary Cells: Galvanic cells; reactants sealed in the
packaging. Cannot be recharged.
Anode: powdered Zn immobilized in a gel in contact with a
concentrated solution of potassium hydroxide, KOH.
Cathode: mixture of MnO2(s) and graphite separated from the
anode by a porous separator.
Cell Voltage: 1.5 V
Anode: Zn(s) + 2OH-(aq) --> ZnO(s) + H2O(l) + 2e-
Cathode: MnO2 (s) + 2 H2O(l) + 2e- -> Mn(OH)2(s) + 2 OH-(aq)
Secondary Cells: Galvanic cells that must be charged before
they can be used; rechargeable.
In the charging process, an external source of electricity
reverses the spontaneous cell reaction and creates a non-
After charging, the cell can again produce electricity as the
reaction moves spontaneously in the direction of the
Applications: batteries in laptops and automobiles.
Lead-Acid Battery - used in automobiles (12V battery,
consists of six 2V batteries in series)
The cathode consists of lead dioxide, PbO2, packed on a
The anode consists of Pb.
Both electrodes are immersed in sulfuric acid, H2SO4.
Anode: Pb(s) + HSO4-(aq) -> PbSO4(s) + H+(aq) + 2e-
Cathode: PbO2(s) + 3 H+(aq) + HSO4-(aq) + 2e- ->
PbSO4(s) + 2 H2O(l)
Cell Voltage: 2 V
Fuel Cell: Designed for continuous operation; reactants
supplied, and products removed continuously.
H2(g) + 1/2O2(g) -------> H2O(l) DH = -286 kJ/mol
Anode: H2(g) -> 2 H+(aq) + 2e-
Cathode: 2 H+(aq) + 1/2 O2(g) + 2e- -> H2O(l)
Cell voltage: 0.615 V
Ecell ~ 0.8 V
Individual fuel cells can be stacked; the number of fuel
cells determines the total voltage, the surface area of
each determines the total current.
Voltage x total current = Electrical Power
200 kW UTC Power Plant in Central Park
Corrosion: an undesirable application of spontaneous redox
When iron is exposed to damp air, with both O2 and H2O
present, the half reaction
O2(g) + 4H+(aq) + 4 e- -> 2H2O(l)
Fe(s) is first oxidized by O2 to Fe2+
The Fe2+ is then further oxidized to Fe3+
Fe3+ then forms Fe2O3.H2O - rust
Electrical energy to drive a non-spontaneous redox reaction.
For example, electricity can be used to decompose molten
NaCl to Na and Cl2, a non-spontaneous process
2NaCl(l) --> 2Na(l) + Cl2(g)
DGo > 0
(for the reaction of 2NaCl(s) -> 2 Na(s) + Cl2(g)
DGo ~ 770 kJ/mol )
Electrolysis is a process driven by an external source of
electrical energy and takes place in an electrolytic cell.
Cathode: 2Na+(l) + 2e- --> 2Na(l) Reduction
Anode: 2Cl-(l) --> Cl2(g) + 2e- Oxidation
In an electrolytic cell oxidation occurs at the anode and
reduction at the cathode, just as in a voltaic cell, but the
signs are reversed.
Quantitative Electrochemistry - Faraday’s laws
In redox reactions charge must be balanced; the number of
electrons given up in the oxidation process must equal the
number of electrons involved in the reduction process.
The amount of charge transferred between the two
electrodes must be related to the amount of each species
reacting at each electrode.
The stoichiometry of a half reaction indicates how many
electrons are needed to achieve an electrolytic process.
For example, in the reduction of Na+ to Na(s), one mole of
electrons reduces one mole of Na+ resulting in the
deposition of one mole of Na(s) on the cathode
Similarly two moles of electrons produce one mole of Cu(s)
from one mole of Cu2+
Cu2+ + 2e- --> Cu
Or 3 moles of electrons produce one mole of Al(s)
Al3+ + 3e- --> Al
Hence, for any half reaction, the amount of substance
reduced or oxidized is directly proportional to the number of
electrons passed into the cell.
This is the basis of Faraday’s laws
1) the quantities of substance produced and consumed at the
electrodes are directly proportional to the amount of electric
charge passing through the cell.
2) When a given amount of electric charge passes through
the cell, the quantity of substance produced or consumed at
an electrode is proportional to its molar mass divided by the
number of moles of electrons required to produce or
consume one mole of the substance.
A coulomb (C) is the quantity of charge passing a point in a
circuit in 1 s when the current is 1 ampere (A)
If a current of I amperes flows for a period of t secs, the
amount of charge transferred = I amperes x t sec
(coulomb = amperes x seconds)
Current is the rate of flow of charge
1 A = 1C/sec
The charge on a single electron = 1.6022 x 10-19 C
The charge associated with one mole of electrons
= 1.6022 x 10-19 C/e- x 6.0221 x 1023 e-/mole = 9.6485x104 C/mol
The charge associated with one mole of electrons, 9.6485x104
C/mol, is the FARADAY CONSTANT, F
To use Faraday’s law to calculate the amount of substance
oxidized or reduced:
1) Need to know the amount of current passed through the
cell and for how long
2) Calculate the quantity of charge transferred- coulombs
3) Using the Faraday constant to calculate the number of
moles of electrons that corresponds to the quantity of charge
determined in step 2
4) Relate the number of moles of electrons to the number of
moles of substance oxidized or reduced
5) Convert moles of substance to grams of substance
A galvanic cell generates an average current of 0.121 A for
15.6 min. The cathode half-reaction in the cell is
Pb2+(aq) + 2e- -->Pb(s)
Determine the amount of lead, in grams, deposited at the
The amount of charge passed through the cell
= (0.121 A) (15.6 min x 60sec/min) = 113 C
Amount of electrons in moles = 113 C/(9.6485x104 C/mol) =
1.17 x 10-3 mol
Amount of Pb deposited in moles = 5.85x10-4 mol
Amount of Pb deposited in grams = 0.122 g