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E° cell and D G. E° cell and work. A galvanic electrochemical cell is capable of generating a flow of electrons. • This flow of electrons (current) can be used to perform work on the surroundings. E° cell and work (cont.). From the definition of electromotive force (emf):.

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e cell and work
E°cell and work
  • A galvanic electrochemical cell is capable of generating a flow of electrons.

• This flow of electrons (current) can be used to perform

work on the surroundings.

e cell and work cont
E°cell and work (cont.)
  • From the definition of electromotive force (emf):

Volt = work (J)/charge (C)

• In English: 1 J of work is done when

1 C of charge is transferred between a potential

difference of 1 V.

e cell and work cont1
E°cell and work (cont.)
  • Recall, we have a system-based perspective such that work done by the system is negative:

E°cell = -w/q

q = charge (not heat!)

w = work

  • Rearranging:

qE°cell = -w

e cell and work cont2
E°cell and work (cont.)
  • An example:

What is the amount of work that is done by a galvanic cell in which 1.5 moles of electrons are passed between a potential of 2 V?

qE°cell = -w

(1.5 mol e-)(2 V) = -w

What do we do with this?

e cell and work cont3
E°cell and work (cont.)
  • Define the “Faraday” (F):

1 F = the amount of charge on 1 mol of e-

1 F = 96,485 C/mol e-

  • Then:
  • q = (1.5 mol e-) x (1 F = 96,485 C/mol e-)
  • = 144,728 C
  • Finally:
  • -w = qE°cell = (144,728 C)(2 J/C) = 298.5 kJ
e cell and d g1
E°cell and DG
  • Since there is a relationship between DG and work, there is also a relationship between DG and E°cell.

DG° = -nFE°cell

  • The above relationship states that there is a direct relationship between free energy and cell potential.
  • For a galvanic cell:

E°cell > 0

Therefore, DG° < 0 (spontaneous)

e cell and d g cont
E°cell and DG (cont.)

• For the following reaction, determine the overall standard cell potential and determine DG°

Cd+2(aq) + Cu(s) Cd(s) + Cu+2(aq)

E°1/2 = -0.40 V

Cd+2 + 2e- Cd

E°1/2 = +0.34 V

Cu+2 + 2e- Cu

E°1/2 = -0.34 V

Cu Cu+2 + 2e-

e cell and d g cont1
E°cell and DG (cont.)

E°1/2 = -0.40 V

Cd+2 + 2e- Cd

E°1/2 = -0.34 V

Cu Cu+2 + 2e-

Cd+2(aq) + Cu(s) Cd(s) + Cu+2(aq)

E°cell = -0.74 V

2

DG° = -nFE°cell

= (-2 mol e-)(96485 C/mol e-)(-0.74 J/C)

= 142.8 kJ

(NOT spontaneous, NOT galvanic)

e cell and d g cont2
E°cell and DG (cont.)

• We now have two relationships for DG°:

DG° = -nFE°cell

= -RTln(K)

-nFE°cell = -RTln(K)

E°cell = (RT/nF) ln(K)

E°cell = (0.0257 V) ln(K)

n

e cell and d g cont3
E°cell and DG (cont.)

E°cell = (0.0257 V) ln(K) = (0.0591) log(K)

n

n

• The above relationship states that by measuring

E°cell, we can determine K.

• The above relationship illustrates that electrochemical

cells are a venue in which thermodynamics is readily

evident

e cell and d g cont4
E°cell and DG (cont.)

• Developing the “big picture”

DG° = -RTln(K)

E°cell = (0.0591 V) log(K)

DG° = -nFE°cell

n

• It is important to see how all of these ideas

interrelate.

an example
An Example
  • Balance, determine E°cell and K for the following:

S4O62- (aq) + Cr2+(aq) Cr3+(aq) + S2O32-(aq)

2e- +

S4O62- S2O32-

2

Cr2+ Cr3+

+ e-

x 2

S4O62- + 2Cr2+ 2Cr3+ + 2S2O32-

an example cont
An Example (cont.)
  • Determining E°cell

2e- +

S4O62- S2O32-

2

E°1/2 = 0.17 V

E°1/2 = 0.50 V

2Cr2+ 2Cr3+

+ 2e-

S4O62- + 2Cr2+ 2Cr3+ + 2S2O32-

E°cell = 0.67 V

an example cont1
An Example (cont.)
  • Determining K

S4O62- + 2Cr2+ 2Cr3+ + 2S2O32-

E°cell = 0.67 V

E°cell = (0.0257 V) ln(K)

= (0.059 V) log K

n

n

n(E°cell)

2 (0.67 V)

=

22.7 = log K

=

(0.059 V)

(0.059 V)

K = 1022.7 = 5 x 1022

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