18.4 electromotive force and gibbs free...

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


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


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


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).


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°


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


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


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


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


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


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

18.4 electromotive force and gibbs free...

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