18.4 electromotive force and gibbs free...
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1 January 13 Electromotive Force and Gibbs’ Free Energy
18.4 Electromotive Force and Gibbs’ Free Energy
The Spontaneity of Electron Transfer
Dr. Fred Omega Garces Chemistry 201 Miramar College
2 January 13 Electromotive Force and Gibbs’ Free Energy
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• F
W max = - n F E°max
3 January 13 Electromotive Force and Gibbs’ Free Energy
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 e- therefore; q = n•F (moles of electron with Coulomb charge)
W max = - q Emax = - n•F •Emax
4 January 13 Electromotive Force and Gibbs’ Free Energy
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).
5 January 13 Electromotive Force and Gibbs’ Free Energy
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°
6 January 13 Electromotive Force and Gibbs’ Free Energy
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 e- C
7 January 13 Electromotive Force and Gibbs’ Free Energy
Concentration and EMF When a voltaic cell is discharge E = 0 V (dead cell) EMF is a function of Concentration [Reactant] > large, EMF c large [Product] > large, EMF c small
ΔG = ΔG° + Corr = ΔG° + RT lnQ but, 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 = E° - 2.303 RT log Q (Nerst Eqn) nF n F
at 298 K E = E°(V) - 0.0592 (V) log Q n
8 January 13 Electromotive Force and Gibbs’ Free Energy
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) ,
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
10 January 13 Electromotive Force and Gibbs’ Free Energy
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) 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: Zn s) + 2 H+
(aq) g Zn+2 (aq)
+ H2 (g) E°cell = 0.76 V
12 January 13 Electromotive Force and Gibbs’ Free Energy
Summary Free Energy, Equilibrium and Electromotive Force
The signs of ΔG°, E°cell and Keq determine the reaction direction at standard-state conditions.
The interrelationship of ΔG°, E°, and K eq . Any one of these three central thermodynamic variables can be used to find the other two.
ΔG°
E°cell K
ΔG° =
-nF
E°ce
ll
ΔG° = -RT ln K
E°cell = RT ln KnF
ΔG° K E°Cell
Reaction at standard state conditions
<0 0 >0
>1 Q= Keq
<1
>0 0 <0
Spontaneous At Equilibrium
NonSpontaneous
13 January 13 Electromotive Force and Gibbs’ Free Energy
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
€
2Fe3+(aq) + 2H2SO3 (aq) → Fe2+
(aq) + S2O6
2-(aq) + 4H+
(aq)
E°cell = 0.17V
ΔG° = −nF E°cell = - 2mol (96485 Cmol
) (0.17 JC
)
ΔG° = - 32.805 kJ
Keq = e-ΔG°
RT
% & '
( '
) * '
+ ' = e
−-32805J
8.314 JK∗298K
%
&
' '
(
' '
)
*
' '
+
' ' = 5.62•105
i) Calculate Ecell, ΔG and Keq for the reaction at 298 K for the following conditions-
ii) [Fe+3] = 1.50 M, [Fe+2] = .75 M [S2O6
-2] = 1.25 M, [H2SO3] =0.50 M, pH = 3.00
€
2Fe3+(aq) + 2H2SO3 (aq) → 2Fe2+
(aq) + S2O6
2-(aq) + 4H+
(aq) ΔG = ΔG° + RT lnQ
ΔG = - 32804 J + (8.324 Jmol •K
) (298 K) ln [0.75]2 [1.25] [1e-3]4
[1.50]2 [0.50]2
ΔG = - 32804 J + (8.324 Jmol •K
) (298 K) ln(1.25•10-12)
ΔG = - 32804 J + (8.324 Jmol •K
) (298 K) (-27.4) = - 32804 J - 67905 J
ΔG = -100.71 kJ
E = ΔG-nF
= - (-100709 J)
(2mol) (96485 Cmol
) = 0.522 V
Keq = e-ΔG°
RT
$ % &
' &
( ) &
* & = e
−-100709J
8.314 JK∗298K
$
%
& &
'
& &
(
)
& &
*
& & = 4.50 •1017
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