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Redox (electron transfer) reactions are unique in that
they do not necessarily require the reactants to come
into physical contact with one another. They must be
in electrical contact.
Cu(s) + 2Ag+(aq)
Cu2+ + 2Ag(s)
If beaker on the left initially contains 0.020 M CuSO4 with
a piece of copper metal immersed in it and the beaker on
the right initially contains 0.020 M AgNO3 with a piece of
silver metal in it, additional silver metal will be plated on
the silver electrode if measures are made to allow for a
complete electrical circuit (a metal wire and the salt bridge)
The two half reactions:
Cu(s)
2Ag+ +2e-
Cu2+ + 2e-
oxidation
2Ag(s) reduction
proceed simultaneously in the separate beakers.
Overall current flow takes place via mobile ions (solution)
and electrons (metal). Electron flow in the metal is
balanced by ion flow in the salt bridge and the solutions in
the two beakers.
Electrochemical cells:
In the preceding schematic diagram, it is apparent that
positive ions (called cations) are attracted to the silver
electrode. This electrode is referred to as the cathode
(electrode to which cations are attracted). The copper
electrode is attracting negative ions (called anions). In
this circumstance, the copper electrode is referred to as the
anode (electrode to which anions are attracted).
reduction takes place at the cathode
oxidation takes place at the anode
The salt bridge is a concentrated salt solution contained
within a container with porous regions that allow ions to
pass. The salt bridge serves to complete the electrical
circuit without allowing direct mixing of the solutions of
the two cells.
If the two beakers are connected to a voltmeter (very
high resistance to current flow such that almost no
current flows in the two cells), a voltage or potential
is apparent. The magnitude of the potential is
determined by the identities of the reactants and their
concentrations. This fact makes the phenomenon
useful as a quantitative analytical technique, as we
shall see.
As shown, the cell will spontaneously conduct
current such that silver cations will be reduced at the
silver electrode and copper metal will be oxidized to
copper II ions. When a cell operates in the direction
of the spontaneous reaction, it is referred to as a
galvanic cell.
A cell can be driven in the opposite direction by the
application of a potential greater than that generated
by the spontaneous reaction. In this case, copper II
ions will be reduced to copper metal and silver metal
will be oxidized to silver ions. The copper electrode
will serve as the cathode and the silver electrode will
serve as the anode. A cell operated to drive the
reaction opposite to the spontaneous direction is
called an electrolytic cell.
Cu(s) + 2Ag+(aq)
Cu2+ + 2Ag(s)
Before electrical connection of the two half-cells:
After connection:
As current flows, the concentration of Cu2+ in the beaker on
the left increases and the concentration of Ag+ in the beaker
on the right decreases until the overall reaction reaches
equilibrium. At this point the potential difference between
the electrodes is zero and no net current flows:
Cell conventions and definitions:
Cu(s) + 2Ag+(aq)
Cu2+ + 2Ag(s)
short-hand description of this cell:
Cu|Cu2+ (0.020 M) || Ag+ (0.020 M)|Ag
where | represents a phase boundary or interface
where a potential develops and || represents two
phase boundaries associated with the salt bridge.
plus right rule: convention that the right hand
electrode is connected to the +ve voltmeter lead and
the –ve or ground lead is connected to the left
electrode.
The overall cell potential therefore indicates the
voltage associated with the reaction going from left
to right.
The relationship between cell potential and
thermodynamics:
Cu(s) + 2Ag+(aq)
Cu2+ + 2Ag(s)
ΔG = − nFEcell
where n is the number of electrons transferred and F
is Faraday’s constant (=96,485 C/mole e-)
When reactants and products are in their standard
states,
o
ΔG o = − nFEcell
= − RT ln K eq
Note that when Ecell is positive, the free energy is
negative (a negative free energy is associated with a
spontaneous reaction).
The overall cell potential is determined by the
difference between the potentials of the electrodes:
Ecell = E right − Eleft
and
o
o
o
Ecell
= Eright
− Eleft
The overall cell potentials of a wide variety of cells
can be calculated from known half-cell potentials.
We can only measure differences in potentials.
Therefore, half-cell potentials are reported relative to
that of the Standard Hydrogen Electrode, :
2H+(aq) + 2e-
H2(g)
Pt,H2(p=1.00 atm)|(aH+ = 1.00 M)|| or SHE||
2Ag+ + H2(g)
2Ag(s) + 2H+
Pt,H2(p=1.00 atm)|(aH+ = 1.00 M)||Ag+(aAg=1.00 M)|Ag
SHE||Ag+(aAg=1.00 M)|Ag
o
o
o
Ecell
= Eright
− Eleft
= 0.799V − 0.0000V
Ag+ + e-
Ag
E0Ag+/Ag = +0.799 V
Cd2+ + H2(g)
Cd(s) + 2H+
Pt,H2(p=1.00 atm)|(aH+ = 1.00 M)||Cd2+(aAg=1.00 M)|Cd
SHE||Cd2+(aCd=1.00 M)|Cd
o
o
o
Ecell
= Eright
− Eleft
= −0.403V − 0.0000V
Cd2+ + 2e-
Cd E0Cd2+/Cd = -0.403 V
Note that this reaction is not spontaneous. Rather,
Cd(s) + 2H+
Cd2+ + H2(g)
E0cell = +0.403 V
Potentials for half-cell reactions are referred to as
Electrode Potentials. All standard electrode
potentials are reported as reduction potentials and are
relative to the SHE:
Strong oxidizing agents are found at the top of the list
(large positive reduction potentials) whereas strong
reducing agents are found at the bottom of the list
(large negative reduction potentials).
See also Appendix 5
+ve SEPs: electrode will serve as cathode in a cell with a SHE
-ve SEPs: electrode will serve as anode in a cell with a SHE
How to determine electrode potentials when reactants
and products are not present in their standard states:
aA + bB +…+ne-
cC + dD + …
o
ΔG o = −nFEcell
= − RT ln K eq
E
o
cell
RT
=
ln K eq
nF
At equilibrium, E=0.
E = Eo −
RT
ln K = 0
nF
c
d
RT
[
C
]
[
D
]
...
E = Eo −
ln
nF [ A]a [ B]b ... Nernst equation
Hence, E is a measure of the difference of the
concentrations of reactants and products from
equilibrium…
0.0592
[C ]c [ D]d ...
E=E −
log
n
[ A]a [ B]b ... at 25 °C
o
(note that when the reactants and products are in their
standard states, the second term =0 and E=Eo)