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