18.4 Electromotive Force and
Gibbs’ Free Energy
The Spontaneity of Electron Transfer
Dr. Fred Omega Garces
Chemistry 201
Miramar College
1
Electromotive Force and Gibbs’ Free Energy
January 13
Electrochemical Cell and Electrical Work
Electromotive Force (E° )
E° - Potential difference (in volts) between two point in
the circuit.
emf
= Potential difference (V)
E
= work (J) / charge (C )
E
= - W / q ( relative to system)
Relative to system (neg. sign)
Therefore:
W
max
= - q Emax
Max work related to max cell potential.
charge due to flow of electron = n•
W
2
max
F
= - n F E°max
Electromotive Force and Gibbs’ Free Energy
January 13
Charge; Electrons moving through a wire.
Electrical Work - Flow of current
W max = - q Emax where q - Charge in Coulombs
but charge of 1 mole of electrons = faraday
F = 96,485 C / mol e- or 96,500 C/mol etherefore;
q = n•F (moles of electron with Coulomb charge)
W
3
max
= - q Emax
= - n•F •Emax
Electromotive Force and Gibbs’ Free Energy
January 13
Free Energy and Electron Transport
The potential of Voltaic (Galvanic) Cell is
related to Free Energy by:
ΔG = W max = - n•F •E max
therefore;
ΔG = - n•F•E max = - n•F •E
ΔG° = - n•F•E° (Std State conditions)
• This equation states that maximum cell potential is directly
related to the free energy difference between the reactants
and the products in the cell.
(Recall E° is the potential difference between anode and
cathode c E°cell = Ered(red) - Ered(ox).
4
Electromotive Force and Gibbs’ Free Energy
January 13
In Class Exercise
Example 20.45 (B&L, 8th ed)
Given: the following reduction half-reaction
Half Rxn (red): (1)
Fe3+ (aq) + e- D
Fe2+ (aq)
E°red = 0.77 V
Half Rxn (red): (2)
S2O6-2 (aq) + 4H+ (aq) + 2e- D 2 H2SO3 (aq) E°red = 0.60 V
i) Calculate E°, ΔG° and Keq for the reaction at 298 K
Find the combination (coupled) reaction between the two half-reaction above that is
the most spontaneous?
Hint first determine the reaction of interest.
Determine the combine half reactions that gives largest positive E°
5
Electromotive Force and Gibbs’ Free Energy
January 13
Free Energy & EMF Calculations
In Class exercise
Given: the following reduction half-reaction
Half Rxn (red): (1)
Fe3+ (aq) + e- D
Fe2+ (aq)
E°red = 0.77 V
Half Rxn (red): (2)
S2O6-2 (aq) + 4H+ (aq) + 2e- D 2 H2SO3 (aq) E°red = 0.60 V
i) Write the balanced chemical eqn for the oxidation of Fe2+(aq) by S2O6-2 (aq).
i) 2 Fe3+ + 2 H2SO3 D
2Fe2+ + S2O6-2 + 4H+
ii) Add (-1) to (2) E°Cell = 0.17 V
ΔG° = -n F E = - 2 mol • (96,500 C ) • (0.17 V)
Note: Volt (V) = 1 J / C or 1 C = 1J / V
ΔG° = - 2 mol • 96,500 C• 0.17 J = - 32.8 KJ ...... Keq = 1.73 •1013
mol e6
C
Electromotive Force and Gibbs’ Free Energy
January 13
Concentration and EMF
When a voltaic cell is discharge E = 0 V (dead cell)
EMF is a function of Concentration
[Reactant] > large,
[Product] > large,
ΔG
but,
EMF c large
EMF c small
= ΔG° + Corr = ΔG° + RT lnQ
ΔG = -nFE
-n F E = -nFE° + (RT) lnQ ,
R = 8.314 J/mol•K
with some algebra and conversion of ln to log.
E = E° - RT ln Q
nF
= E° - 2.303 RT log Q
n F
(Nerst Eqn)
at 298 K
E
= E°(V) - 0.0592 (V) log Q
n
7
Electromotive Force and Gibbs’ Free Energy
January 13
Variable Concentration EMF Calculations
Example 20.51 (B&L, 8th ed)
Given: Redox reaction:
Zn + Cd2+ g
Zn+2 + Cd
i) Calculate the E° for : Zn(s)|Zn+2 (aq, 1.0M)||Cd2+ (aq,1.0M) |Cd(s) ,
ii) Calculate the emf for Zn(s)|Zn+2 (aq, 0.150M)||Cd2+ (aq,1.5M) |Cd(s) ,
8
Half Rxn (oxidation): (1)
Zn+2 + 2e- D Zn(s)
E°red = - 0.76 V
Half Rxn (reduction): (2)
Cd+2 + 2e- D Cd(s)
E°red = - 0.403 V
Electromotive Force and Gibbs’ Free Energy
January 13
Variable Concentration EMF Calculations
Example 20.55 (#49, B&L, 7th ed (Zn|Zn
+2
(0.10M)||H
+
( ? )
,H2 (.90atm) ||Pt , Ecell = 0.72V )
Given: Redox reaction Zn|Zn+2 (0.10M) || H+ ( ? ) ,H2 (1atm) |Pt , Ecell = 0.72 V,
calculate E°Cell and pH at 298 K.
Half Rxn (oxid): (1)
10
Zn(s) D Zn2+ (aq) + 2e-
E°red = 0.76 V
Half Rxn (red): (2) 2H+ (aq) + 2e- D H2 (g)
E°red = 0.00 V
Net Rxn:
E°cell = 0.76 V
Zn s) + 2 H+ (aq) g Zn+2 (aq) + H2 (g)
Electromotive Force and Gibbs’ Free Energy
January 13
Summary
Free Energy, Equilibrium and Electromotive Force
ΔG°
-R
-nF
E°
ce
ll
°=
ΔG
ΔG
°=
nK
Tl
E°cell
12
E°cell = RT ln K
nF
ΔG°
K
E
°Cell
Reaction at standard state
conditions
<0
0
>0
Q= Keq
>1
>0
0
<0
Spontaneous
At Equilibrium
NonSpontaneous
<1
The interrelationship
of ΔG°, E°, and K eq .
Any one of these
three central
thermodynamic
variables can be used
to find the other two.
K
The signs of ΔG°, E°cell and
Keq determine the reaction
direction at standard-state
conditions.
Electromotive Force and Gibbs’ Free Energy
January 13
Free Energy & EMF Calculations
In Class exercise
Given: the following reduction half-reaction
Half Rxn (red): (1)
Fe3+ (aq) + e- D
Half Rxn (red): (2)
S2O6-2 (aq) + 4H+ (aq) + 2e- D 2 H2SO3 (aq)
i)  Calculate Ecell, ΔG and Keq for the
reaction at 298 K for the following
conditions-
2Fe3+(aq)
+ 2H2SO3 (aq)
→
Fe2+(aq)
+
S 2 O6
2(aq)
+
Fe2+ (aq)
E°red = 0.77 V
E°red = 0.60 V
ii) [Fe+3] = 1.50 M, [Fe+2] = .75 M
[S2O6-2] = 1.25 M, [H2SO3] =0.50 M, pH = 3.00
4H+(aq)
2Fe3+(aq)
ΔG°
= −nF E°cell
%
' -ΔG° )
'
&
*
' RT +
'
= - 2mol (96485
%
)
'
'
'
'
-32805J
−&
*
'
J
'
' 8.314 ∗298K '
(
+
K
=e
C
mol
) (0.17
J
C
ΔG = - 32804 J + (8.324
)
ΔG = - 32804 J + (8.324
ΔG =
E =
- 100.71 kJ
ΔG
-nF
= 5.62 •105
K eq = e
13
2Fe2+(aq)
Electromotive Force and Gibbs’ Free Energy
€
J
mol • K
+
S 2 O6
J
mol • K
J
mol • K
) (298 K) ln
2(aq)
+
4H+(aq)
(2mol) (96485
$
(
&
&
& -100709J &
−%
)
&
J
&
∗298K &
& 8.314
'
*
K
=e
[0.75]2 [1.25] [1e-3]4
[1.50]2 [0.50]2
) (298 K) ln(1.25•10-12 )
) (298 K) (-27.4) = - 32804 J - 67905 J
(-100709 J)
= -
$
& -ΔG° (
&
%
)
& RT *
&
'
€
→
ΔG = - 32804 J + (8.324
ΔG° = - 32.805 kJ
K eq = e(
+ 2H2SO3 (aq)
ΔG = ΔG° + RT lnQ
E°cell = 0.17V
C
mol
= 0.522 V
)
= 4.50 • 1017
January 13