
The potential difference between two electrodes in a cell is called the
electromotive force, or __________________________

The EMF of a voltaic cell is called the ________________________________

The cell voltage of a voltaic cell will be a _______________
Note: We are used to spontaneous processes having negative values for energy
terms. However, since batteries were historically measured looking at the
change in energy of the surroundings rather than for the reaction, the opposite
sign convention applies.
Work
• We can now write an expression for the maximum work attainable by a
voltaic cell.
– Let n be the number of mole electrons transferred in the overall
balanced cell reaction.
– The maximum work for molar amounts of reactants is:
LP#5. The EMF of a voltaic cell with the following reaction is 0.650 V.
Hg22+(aq) + H2(g) → 2 Hg(l) + 2H+(aq)
Calculate the max work that can be done by this cell when 0.500g H2 is consumed.
Step 1: Identify the number of electrons transferred during the overall reaction.
Oxid ½ Rxn:
Red. ½ Rxn:
Step 2: Compute the maximum work per mole.
Wmax = -(n)(F)(Ecell) =
=
Step 3: Determine the total energy based on the total moles of e- transferred.
19-13
Standard Cell EMF’s and Standard Electrode Potentials
• A cell emf is a measure of the driving force of the cell reaction.
– A reduction potential is a measure of the
_______________________________
in the reduction half-reaction.
– You can look at the oxidation half-reaction as the reverse of a
corresponding reduction half-reaction.
– The oxidation potential for an oxidation half-reaction (the reverse
reaction) is the __________of the reduction potential.
• By convention, the Table of Standard Electrode Potentials are tabulated
as reduction potentials.
Standard Reduction Potentials
• E0cell = standard cell potential under standard conditions,
• Voltage listed at standard conditions are at
temperature:
concentration:
19-14
The actual cell voltage will depend upon
a) _____________________
b) _____________________
c) _____________________

Individual potentials cannot be measured, only differences in potential.
By convention, all others must be compared to a standard reference reaction

The reference reaction is the reduction of H+(aq) ions to produce H2(g):

2H+(aq, 1M) + 2e-  H2(g, 1 atm)
Eºred =
– This is known as the: ___________________________________
Standard electrode potentials are measured relative to this hydrogen reference.

A cell will always include one oxidation reaction (at the anode) and
one reduction reaction (at the cathode).
When we reverse a reaction to get an oxidation half reaction,
the sign of the potential must be changed.
We can then sum the energy potentials the same way we have treated other
energy variable (e.g. ∆G & ∆H) this semester.
E°cell =

Note: The book uses the convention E°cell = E°cathode - E°anode
Where all potentials are reduction potentials!

Both processes will give the same correct answer.
19-15
Lets look at a section of an Activity Series for metals and hydrogen ion:

Standard
Potential, (Eºred) in
Volts
Reduction HalfReaction
-3.04
Li+(aq) + e-  Li(s)
-0.76
Zn2+(aq) + 2e-  Zn(s)
0
2H+(aq) + 2e-  H2(g)
0.34
Cu2+(aq) + 2e-  Cu(s)
0.80
Ag+(aq) + e-  Ag(s)

The more positive the value of Eºred the greater the driving force for
reduction

Reactions with positive E values
- want to undergo: ____________ .
- are good: __________________

Reactions with negative E values
- prefer to undergo: ___________
- are good: __________________
The net difference between the standard reduction potentials of the two reactions,
is the excess potential that can be used to drive electrons through the cell, Eºcell
 The more active they are, the greater the oxidation potential for the metal.
 Since tables usually list reduction potentials,
the more active they are, the more negative the reduction potential for the
ion.
 Active metals are _____________________ .
19-16
When comparing Zinc with Copper in the activity series, which reaction would
we expect to happen?
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
OR
Zn (aq) + Cu(s)  Zn(s) + Cu2+(aq)
2+
Zinc is higher on the activity series.
Zinc is more easily oxidized than copper.
We would expect the ________ reaction to happen.
Example: Consider the zinc-copper cell described earlier.
–
Zn( s ) | Zn 2 (aq ) || Cu 2 (aq) | Cu ( s )
– The two half-reactions are:
Oxidation ½ Rxn:
____________________________________
Reduction ½ Rxn:
____________________________________
– Find the oxidation potential for this half reaction.
Zn(s)  Zn2+(aq) + 2e-
Eoox =
– Find the oxidation potential for this half reaction.
Cu2+(aq) + 2e-  Cu(s)
Eored =
– Find the overall cell potential.
Eºcell = ___________________________
The electrode potential is an intensive property whose value is independent of
the amount of species in the reaction.
–
Thus, the electrode potential for the half-reaction would be:
2Cu2+(aq) + 4e-  2Cu(s)
19-17
Eo =
LP#6. Consider a cell constructed of the following two half-reactions. What
would the overall reaction be that would create a voltaic cell and what would the
voltage of that cell be?
Cd
2
( aq )  2 e   Cd ( s ); E o   0.40 V
Ag  (aq )  1e   Ag ( s ); E o  0.80 V
– Which reaction should be reversed?___
– Half reactions and associated voltages therefore are:
Cd(s) → Cd2+(aq) + 2e- ; E =
Ag+(aq) + 1e- → Ag(s) ; E = 0.80V
– We must double the silver half-reaction so that the electrons cancel.
Cd(s) → Cd2+(aq) + 2e- ; E = 0.40 V
2Ag+(aq) + 2e- → 2Ag(s) ; E =
– Now we can add the two half-reactions.
– The corresponding cell notation would be:
Spontaneity of Redox Reactions
Eºcell = Eºred + Eºoxid

Eº will be positive for the case where the reaction is: __________________

Eº will be zero for a redox reaction that is: ___________________________

Eº will be negative for the case where the reaction is:
_______________________
______________
LP#7. Can copper be dissolved (oxidized) by acid (i.e., H+)?
 Really asking if the following reaction is spontaneous:
Cu(s) + 2H+(aq)  Cu2+(aq) + H2(g)

This reaction can be broken down into two half-reactions:
Oxidation: Cu(s)  Cu2+(aq) + 2e-
E0ox =
Reduction: 2H+(aq) + 2e-  H2(g)
E0red =
19-18

The standard cell potential for this reaction is:
Eºcell =
Since this value is __________________ ,
this redox reaction is ________________
and ______________________________ .
What if we only had the activity series without any numerical voltages?
(Remember the most active at the top is most easily
Since H is higher than Cu, it is more easily oxidized.
Cu cannot replace it as the oxidized species.
Li
Zn
H
Cu
Ag
EMF, Free Energy Changes, and Equilibrium
Free energy change, Gº, Kc, and Eºcell, all measure spontaneity of a reaction.
What is the relationship between these variables?
Gº and Eºcell
– Previously, we saw that G is the free energy available which equals the
maximum useful work of a reaction.
– Remember, for a voltaic cell, work = -nFEcell,
so when reactants are in their standard states
– The Gibbs Free Energy (G) can be related to the EMF of the cell.
Where n= number of moles of e- transferred
F = Faraday’s constant = 96,485 C/mole e
C = Coulombs
Our typical unit of energy, Joules can be related to the cell EMF:
Since 1 V = 1J/C
19-19
Gº and Kc
The measurement of cell EMF’s gives you yet another way of calculating
equilibrium constants.
– Combining the previous equation, Go = -nFEocell,
with the equation Go = -RT lnK, (from our previous chapter) we get
– Or, rearranging, we get
LP#8. The standard EMF for the following cell is 1.10 V.
Zn( s ) | Zn 2 (aq) || Cu 2 (aq) | Cu ( s )
Calculate the equilibrium constant Kc at 25oC for the reaction:
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
– Note that n=__. Substituting into the equation relating Eocell and K gives
logK=
– Solving for log Kc, you find:
Now take the antilog of both sides:
–
Kc= 
Effect of Concentration on Cell EMF
The EMF of a voltaic cell is determined by
a) the identity of the redox reaction and
b) the concentrations of the reactants and products.

The EMF of the cell will fall as the reactants are used up and products
increase in concentration
19-20
The Nernst Equation
This is an equation that related the EMF of a redox reaction on the concentration
of reactants and products.
Developed by Walther Hermann Nernst (1864 - 1941), a German chemist.
RT
ln Q = _______________________
nF
o
Ecell  Ecell


At equilibrium concentrations of reactants and products, the EMF = __.
Electrons flow spontaneously in a redox reaction because the system is
attempting to achieve equilibrium. When equilibrium is achieved, net
electron flow is zero.
o
0  Ecell

0
Ecell 

RT
ln K
nF
at equilibrium
A common form of the equation does away with the natural log and puts
the equation in the form of log10:
o

Ecell  Ecell

RT
ln K
nF
2.303RT
log Q Nernst Equation
nF
At 25 °C and using base 10 logs this becomes: E = E° - 0.0592 log Q
n
The Nernst Equation.

Allows us to determine cell potentials at
____________________________
19-21
LP#9. A voltaic cell that utilizes the oxidation of zinc by copper ion is set up with
an initial concentration of 5.0M copper ion and 0.050M zinc ion. Calculate the
cell potential at 20C.
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
E0cell = 1.10V
n=
Determination of pH
A pH electrode works by relating cell EMF (i.e., potential) to concentration.
As the concentration of H+ in the solution changes, the amount of voltage that can
be measured changes. We can then calibrate the voltage to actual pH values.
19-22
CONCENTRATION CELLS
Since cell potentials cepend not only on the half reactions but on the
concentrations, it is possible to create a cell where the two half reactions are the
same, but only th e concentrations are different.

The standard cell potential would be zero:
Oxidation: Cu(s)  Cu2+(aq) + 2e-
E0ox =
Reduction: Cu2+(aq) + 2e-  Cu(s)
E0red =
Using the Nernst Equation the actual cell potential is:
19-23
4. Electrolysis
Voltaic cells are based upon the spontaneous redox reactions
However, we have seen that some commercial voltaic cells can be "recharged"

This application of current in the opposite direction will cause the redox
reaction to proceed in the non-spontaneous direction
Sodium metal and chlorine gas spontaneously react to form the ionic compound
Sodium Chloride:
2Na(s) + Cl2(g)  2NaCl(s)

This reaction is a redox reaction:
oxidation: _________________________
reduction: _________________________

The reverse reaction, can be achieved by the application of an external
current to molten sodium chloride. (This is not the same as aqueous!)
2NaCl(l)  2Na(l) + Cl2(g)
Redox reactions that are driven by an external EMF are called
_________________ reactions and take place in
_________________ cells
19-24