Electrochemistry
It deals with reactions involving a transfer of electrons:
1.
Oxidation-reduction phenomena
2.
Voltaic or galvanic cell
Chemical reactions can be used to produce eletrical
energy:
3.
Electrolytic cells
Electrical energy can be used to bring about
chemical transformations:electrolysis
Oxidation and reduction
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Zn(s)  Zn2+(aq) + 2eOxidation half-reaction
Cu2+(aq) + 2e-  Cu(s)
Reduction half-reaction
Galvanic /voltaic/ cells
Electrode:
e.g. a metal strip which can funcion as a
cathode or an anode
Half-cell:
an electrode immersed in a solution
containing the metal ions. Half of a voltaic
cell in which an oxidation or a reduction
occurs.
Galvanic cell
Electrode conventions
Cathode
Anode
Ions attracted
Cations
Anions
Direction of electron
movement
Into cell
Out of cell
Reduction
Oxidation
Positive
Negative
Negative
Positive
Half-reaction
Sign
Galvanic cell
Electrolytic cell
There is a difference in electrical potentials between the solution
and the electrode
Types of electrodes
1. Electrodes of the first kind:
(metal electrodes) Zn, Cu, Fe, Au, Pt (indifferent)
2. Electrodes of the second kind:
(calomel electrode)
- metal: Hg
- slightly soluble salt: Hg2 Cl2
precipitate
- another soluble salt of its anion: KCl
Hg2Cl2(s)  Hg2(aq)2+ + 2Cl(aq)Ksp = [Hg22+ ][Cl-]2 /= 1.3 x 10-18/
2Hg  Hg22+ + 2e-
3. Gas electrodes: H2 /S.H.E./, Cl2
Membrane electrodes
Glass electrode
Ion-selective electrode: Na+, K+
It can be calculated by means of NERNST-equation
E = E0 +
RT
ln c
nF
0.059
E = E0 +
n
log c
R = 8.314 J/K mol
T = 298.15 K (25°C)
F = 96 485 C/mol
n = number of moles of electrons
transferred
Zn loses electrons, enters the
solution. Cu gains electrons,
deposits as metal.
Electromotive force (E.M.F.) is the
difference in potentials between the
two half cells.
E.M.F. = Ecat.(+) – Ean.(-)
Cell diagrams
Zn(s) I Zn (aq)2+ II Cu (aq)2+ I Cu(s)
half-cell
half-cell
salt bridge
At the left:
At the right:
- oxidation
- reduction
- anode
- cathode
- negative
- positive
Standard electrode potentials
0.059
E = E0 +
n
log c
We can measure the
potential differences only.
Standard hydrogen
electrode: S.H.E.
E 0 = 0.0000 volt
E=E0
if c = 1 mol
Standard reduction potentials
(at 25°C)
Half-reaction
Li+(aq) + e-  Li(s)
Ca2+(aq) + 2e-  Ca(s)
2H+(aq) +2e-  H2(g)
E°(volts)
- 3.05
- 2.76
0.00
Cu2+(aq) + 2e-  Cu(s)
Ag+(aq) + e-  Ag(s)
Cl2(g) + 2e-  2 Cl-(aq)
F2(g) + 2e-  2F -(aq)
+ 0.34
+ 0.80
+ 1.36
+ 2.87
Concentration cells
Zn I 0.01M Zn2+ II 1M Zn2+ I Zn
[Zn2+]1
0.059
2+ ] > [Zn 2+ ]
E.M.F. =
log
;
[Zn
1
2
2
[Zn 2+]2
Measurement of pH
Pt H2(g,1atm) I H+(xM) II H+(1M) I H2(q,1atm) Pt
Ecell = 0.059 log
[1]
[H+]
= 0.059 (-log [H+ ])
Ecell = 0.059 pH
Redox cells
Fe2+  Fe3+ + ereduced
E = E0 +
oxidized
RT
nF
ln
[ox]
[red]
[ox]
0.059
E = E0 +
log
n
[red]
T = 25°C
Fe2+/Fe3+
S.H.E.
Electrode potentials of some reduction oxidation systems
Co2+ /Co3+
+ 1.80V
Pb2+ / Pb4+
+ 1.80V
MnO4- /Mn2+
+ 1.52V
Fe2+ /Fe3+
+ 0.77V
I2/2I -
+ 0.62V
Sn2+ /Sn4+
+ 0.15V
Cytochrom a Fe3+ /Fe2+
0.29V
Cytochrom c Fe3+ /Fe2+
0.22V
Hemoglobin Fe3+ /Fe2+
0.17V
Cytochrom b Fe3+ /Fe2+
0.07V
Vitamin C ox/red
0.06V
oxidizer
reducer
Electrolytic cell
conduction
Metallic
Electrolytic
Molten salt
Aqueous soln.
ClNa+
Electrolytic cell
(Molten sodium chloride)
Electrolysis: the use of electricity to bring
about chemical change
- Primary process: redox reaction,
overvoltage
- Secondary process: …..
Stoichiometry of electrolysis
Faraday’s law
m = kIt
m = mass …
k = electrochem. constant
I = electric current in amperes
t = time
A = 1 C/s
ampere
sec.
coulomb
The quantity of charge equivalent to one mole of
electrons is called the faraday (F)
1F = 96 485C
1F  1 g-equivalent