Galvanic cell and
Nernst equation
Galvanic cell
• Some times called Voltaic cell
• Spontaneous reaction redox reaction is
used to provide a voltage and an electron
flow through some electrical circuit
• When metallic zinc is dipped into a
solution of copper sulfate: dark brown ,
spongy layer of metallic copper forms on
the zinc.
Galvanic cell
• The reaction can be written as:
Cu2+ (aq) + Zn(s)  Cu(s) + Zn2+ (aq)
The blue color of copper sulfate gradually
disappeared and replaced by the colorless
zinc sulfate.
• This reaction will not produce any flow of
electrons as long as it occurs at the
surface of the zinc.
Galvanic cell
• The flow of the
electrons can be
achieved by making
the oxidation and
reduction halfreactions occur in
separate
compartments of a
galvanic cell.
Galvanic cell
• Each compartment called half-cell.
• Complete isolation of the two species
would lead to an electrical
imbalance…Why?
Galvanic cell
Explanation:
• on the left side Zn ions entering the solution give
the sloution overall positive charge. This would
prevent additional Zn ions to enter the solution.
• On the right side, Cu ions leave the solution, the
SO4 ions left behind would give the solution
negative charge and the electrode becomes
positively charge. This would cause the
electrode to repel Cu ions.
continue
Galvanic cell
• The flow of current accompanied by continues
electrical activity therefore the solution around
the electrodes must be kept electrically neutral
(even Zn leave the electrode compartment or
anions enter to it; and cations enter the
compartment of Cu to balance the charge of
SO4 or SO4 ions must leave .
How this can be achieved???
continue
Galvanic cell
• The salt bridge and the porous partition allow the slow
mixing of the ions in the two solution.
Galvanic cell
• Salt bridge usually a tube filled with an
electrolyte such as KNO3 or KCl in gelatin.
• Cations from the salt bridge can move into
one compartment to compensate for the
excess of negative charge while anions
from the salt bridge diffuse into the other
compartment to neutralize the excess of
positive charge.
The signs of the electrodes in
galvanic cells
• Oxidation occurs at the anode and
reduction occurs at the cathode in both
galvanic cell and electrolyte cell.
What about the sign of the electrode at the
anode and cathode??????
The signs of the electrodes in
galvanic cells
• In the galvanic cell: anode has a negative
charge and cathode has positive charge
this is the opposite to the electrolyte
solution.
Cell potentials
• The force that moves the electrons
through the cell is called electromotive
force (emf) and measured as volts (V).
• Emf is the passage of 1 coulomb is able to
accomplish 1 joule of work.
• (1 volt = 1joule / coulomb)
• The emf in the galvanic cell is called cell
potential (Ecell)
Cell potentials
• The potential of the cell depends on the
concentrations of the ions, the temperature, and
the partial pressure of any gases that might be
involved.
• Standard cell potential (Eocell) is measured when
the concentrations of the ions are 1 M, the
temperature is 25 C and the partial pressures of
the gases are 1 atm.
• The cell potential is measured by a
potentiomenter
Cell diagram
This allow us the describe the galvaic cell to give
informations:
1. The nature of the electrode materials
2. The nature of the solution in contact with the electrodes
(including the concentrations of the ions
3. which of the half cell is anode and cathode
4. The reactants and the products in each of the half cells
In the previous fig suppose the concentrations of both Zn
and cu ions are 1 M
Zn(s)|Zn2+(1M) ||Cu2+(1M)|Cu(s)
Reduction potential
What is the origin of the cell potential??
• In the cell we have two solution; one contains Zn
ion and the other Cu ions.
• Each of these ions (zn and Cu) has a certain
tendency to acquire electrons from its respective
electrode and become reduced.
Zn2+(aq) + 2e- Zn(s)
Cu2+(aq) + 2e- Cu(s)
Reduction potential
• The larger reduction potential the larger
tendency to undergo reduction
• When cell reaction takes place there is “tug of
war” as each species attempts to pull electrons
from its electrode so as to become reduced.
• The species with greater tendency to acquire
electrons (greater reduction potential) wins the
war.
• The loser must supply the electrons (oxidized)
Reduction potential
• The potential that we measure arises from
the differences in the tendency of the two
ions to become reduced and is equal to:
Eocell = Eosubstances reduced - Eosubstances oxidized
There fore for the previous reaction
Eocell = EoCu2+|Cu(s) - EoZn2+|Zn
Reduction potential
• Experimentally we can measure the overall cell
potentials but can be calculated.
• We use standard hydrogen electrode which
(arbitrarily) sets its reduction potential of zero
volts.
• It consists of a platinum wire encased in a glass
sleeve through which hydrogen gas is passing at
pressure of 1 atm.
• The platinum wire is attached to a platinum foil
that is coated with a black velvety layer of finely
divided platinum.
Reduction potential
Standard Hydrogen Electrode
The convention is to select a particular electrode and assign its
standard reduction potential the value of 0.0000V. This electrode is
the Standard Hydrogen Electrode.
2H+(aq) + 2e–  H2(g)
H2
Pt
H+
The “standard” aspect to this cell is that the
activity of H2(g) and that of H+(aq) are both 1.
This means that the pressure of H2 is 1 atm
and the concentration of H+ is 1M, given that
these are our standard reference states.
Reduction potential
• Eo H+|H2 =0.00V, means any substance that is more
easily reduced than H+ has a positive value of Eo.
• What could happened if hydrogen electrode is
paired with another half-cell?
• If the reduction potential of the species is greater
than that for hydrogen electrode oxidation at
hydrogen electrode
H2 (g) 2H+(aq) + 2e-
Reduction potential
• If the reduction potential of the species is
less than that for hydrogen electrode 
reduction at hydrogen electrode
2H+(aq) + 2e-  H2 (g)
Reduction potential
when connecting a voltmeter, connect the positive
terminal to the positive electrode. If it reads a
positive potential, you have correctly identified
all the terminals. If you read a negative
potential, then you have misidentified the
reactions in the cells, and you have hooked it up
backwards. Reverse your assignment of anode
and cathode.
• in a galvanic cell the cathode is +ive
• in an electrolytic cell the cathode is –ive.
Reduction potential
Suppose we connect both Cu and H electrode:
to obtain proper reading (+) terminal should
connected to Cu electrode and (-) terminal to H
electrode.
Cu2+(aq) + 2e- Cu(s)
(cathode)
H2 (g) 2H+(aq) + 2e(anode)
Eocell = EoCu2+|Cu - EoH+|H2
0.34 V = EoCu2+|Cu – 0.00 V
o
ٍE Cu2+|Cu = +0.34 V
Reduction potential
If both Zn and H electrode are connected then:
Positive terminal is for H electrode
Negative terminal is for Zn electrode
2H+(aq) + 2e-  H2 (g) (cathode)
Zn(s) Zn2+(aq) + 2e(anode)
Eocell = EoH+|H2 - Eozn2+|zn
0.76 V = 0.000 V - Eozn2+|zn V
o
ٍE zn2+|zn = -0.76 V
Eo negative means Zn is more difficult to reduce than H+
Reduction potential
• To calculate Eocell For Cu and Zn cell
Eocell= EoCu2+|Cu - EoZn2+|Zn
Eocell=+0.34 V– (-0.76 V)
Eocell= +1.10 V
• This value is precisely the value that we
observe experimentally
• Bear in mind, Eocell positive means
spontaneous reaction
More Example
Fe2+ + 2e–  Fe
V2+ + 2e–  V
-0.44
-1.19
Sn2+ + 2e–  Sn
-0.14
Ag+ + e–  Ag
+0.80
To get a final positive cell
potential, the more negative
half-reaction (V) must act as
the anode.
More negative potential reaction is
the anode.
Multiply the Ag reaction by 2, but
don’t modify the cell potential.
Fe2+ + V  Fe + V2+
2 Ag+ + Sn  2 Ag + Sn2+
Ecell = -0.44 - (-1.19) = +0.75 V
Ecell = +0.80 - (-0.14) = +0.94 V
Standard Potential Tables
All of the equilibrium electrochemical data is cast in Standard
Reduction Potential tables.
F2 + 2e–  2F–
+2.87
2H+ + 2e–  H2
0.0000
Co3+ + e–  Co2+
+1.81
Pb2+ + 2e–  Pb
-0.13
Au+ + e–  Au
+1.69
Sn2+ + 2e–  Sn
-0.14
Ce4+ + e–  Ce3+
+1.61
In3+ + 3e–  In
-0.34
Br2 + 2e–  2Br–
+1.09
Fe2+ + 2e–  Fe
-0.44
Ag+ + e–  Ag
+0.80
Zn2+ + 2e–  Zn
-0.76
Cu2+ + 2e–  Cu
+0.34
V2+ + 2e–  V
-1.19
AgCl + e–  Ag + Cl–
+0.22
Cs+ + e–  Cs
-2.92
Sn4+ + 2e–  Sn2+
+0.15
Li+ + e–  Li
-3.05
Oxidative Strength
F2 + 2e–  2F–
+2.87
Co3+ + e–  Co2+
+1.81
Au+ + e–  Au
+1.69
Ce4+ + e–  Ce3+
+1.61
Br2 + 2e–  2Br–
+1.09
Ag+ + e–  Ag
+0.80
Cu2+ + 2e–  Cu
+0.34
AgCl + e–  Ag + Cl–
+0.22
Sn4+ + 2e–  Sn2+
+0.15
Consider a substance on the left of one of
these equations. It will react as a
reactant with something below it and
on the right hand side.
• higher in the table means more likely to
act in a reducing manner.
• when something is reduced, it induces
oxidation in something else.
• it is an oxidizing agent or an oxidant.
• F2 is a stronger oxidant than Ag+.
• Cu2+ is a weaker oxidant than Ce4+.
Reductive Strength
F2 + 2e–  2F–
+2.87
Co3+ + e–  Co2+
+1.81
Au+ + e–  Au
+1.69
Ce4+ + e–  Ce3+
+1.61
Br2 + 2e–  2Br–
+1.09
Ag+ + e–  Ag
+0.80
Cu2+ + 2e–  Cu
+0.34
AgCl + e–  Ag + Cl–
+0.22
Sn4+ + 2e–  Sn2+
+0.15
Substances on the right hand side of
the equations will react so as to be
oxidized.
• LOWER in the table means a
greater tendency to be oxidized.
• when oxidized, it induces reduction
in something else. It is a reducing
agent or reductant.
• Ag is a stronger reductant than Au.
• Co2+ is a weaker reductant than
Sn2+
Gibbs and the Cell Potential
• Here we can easily see how this Gibbs function relates to a potential.
• By convention, we identify work which is negative with work which is
being done by the system on the surroundings. And negative free
energy change is identified as defining a spontaneous process.
GT,P  welectrical  n F E
• Note how a measurement of a cell potential directly calculates the Gibbs
free energy change for the process.
Gibbs and the Cell Potential
• Consider reaction with
n (number of electron) =2
F (number of coulombs per mole
electrons=96.500 C/mole e-)
E (cell potential) = +1.1 V
Then ∆ Go =-212 kJ
Calculation of thermodynamic constants from
standard cell potential
• ∆ Go = -RT ln K
R=8.314 J/mole
• ∆ Go =- 2.303 RT log K
• ∆ Go = -nFEo= =- 2.303 RT log K
• Eo = (2.303 RT/nF) log K
• At T = 25, the quantity 2.303 RT/F is constant
And equal to (2.303x8.314x298)/96.500 =0.0592V
Calculation of thermodynamic constants
from standard cell potential
• Finally we can get
log K = (nEo)/ 0.059
Nernst Equation
• ∆ G = ∆ Go + RT ln Q
Q is the reaction quotient
• ∆ G = ∆ Go + 2.303RT log Q
• -nFE = -nFEo + 2.303RT log Q
Rearrangement
E = Eo – (2.303RT/nF) log Q
E = Eo – (0.059/n) log Q
Nernst Equation
• Consider the reaction of both Cu and Zn ar
discussed earlier :
• [Zn2+] = o.4M and [Cu2+] = 0.02M, what is
E?
E = Eo – (0.059/n) log Q
Q = [Zn2+] / [Cu2+] and Eo = 1.1V
Then
E = 1.1 – (0.059/2) log [Zn2+] / [Cu2+] =1.06