Electrochemistry

1
Electrochemistry

• Chapter 19

2
Electrochemistry

• The oxidation state is the charge on the atom if
  it were an ion in solution.
• e.g. KBr K1 Br1-
• Lose Electrons Oxidation
• Gain Electrons Reduction

3
Lets Consider a Reaction

• Cu (s) HNO3 (aq) Cu(NO3)2 (aq) H2 (g)

4
Lets Consider a Reaction

• Cu (s) 2 HNO3 (aq) Cu(NO3)2 (aq) H2 (g)

5
Lets Consider a Reaction

• Cu (s) 2 HNO3 (aq) Cu(NO3)2 (aq) H2 (g)

1 1-
0
Assign oxidation states
6
Assigning Oxidation States

• Cu (s) 2 HNO3 (aq) Cu(NO3)2 (aq) H2 (g)

0 1-
2 2(1-)
0
0
Change in oxidation state Electron Transfer
Occurs!
7
Half Reactions

• Cu (s) 2 HNO3 (aq) Cu(NO3)2 (aq) H2 (g)

0 1-
2 2(1-)
0
0
Electron Transfer Occurs!
Any Redox Reaction can be broken
into two half reactions.
8
Half Reactions

• Cu (s) 2 HNO3 (aq) Cu(NO3)2 (aq) H2 (g)

0 1-
2 2(1-)
0
0
Electron Transfer Occurs!
Any Redox Reaction can be broken
into two half reactions.
Cu (s) Cu2 (aq) 2 e- 2H (aq) 2 e-
H2 (g)
9
Balancing Redox Equations

• Assign Oxidation States to all atoms
• Decide which element is oxidized and determine
  the number of electrons
• Decide which element is reduced and determine the
  number of electrons
• Choose coefficients so that the of electrons
  gained equals the of electrons lost.
• Balance the rest of the atoms.

10
Voltaic (galvanic) Cells

• We can connect half reactions in separate
  containers through an electrical circuit. This
  will produce a current (electron flow) and
  voltage according to the spontanaiety of the
  reactions.

(Figure 19.2 Here)
11
The Reaction

• Zn (s) CuSO4 (aq) ZnSO4 (aq) Cu (s)
• Half Reactions

12
The Reaction

• Zn (s) CuSO4 (aq) ZnSO4 (aq) Cu (s)
• Half Reactions
• Zn (s) Zn2 2 e-
• Cu2 2 e- Cu (s)

13
The Reaction

• Zn (s) CuSO4 (aq) ZnSO4 (aq) Cu (s)
• Half Reactions
• Zn (s) Zn2 2 e-
• Cu2 2 e- Cu (s)

oxidation reduction
14
The Reaction

• Zn (s) CuSO4 (aq) ZnSO4 (aq) Cu (s)
• Half Reactions
• Zn (s) Zn2 2 e-
• Cu2 2 e- Cu (s)

oxidation reduction
anode cathode
15
Electromotive Force-EMF

• The voltage difference between two solutions
  provides a measure of the driving force of the
  electron transfer reaction.

16
Then We Can Compare

• Eocell Eoox - Eored
• A positive cell potential, Eocell, indicates a
  spontaneous reaction.

17
Cell EMFs

• Calculate the standard emf of the following cell
  at 25 oC. Cr(s)Cr3(aq)Hg22(aq)Hg(l)

18
Cell EMFs

• Calculate the standard emf of the following cell
  at 25 oC. Cr(s)Cr3(aq)Hg22(aq)Hg(l)

Determine the half reactions,
19
Cell EMFs

• Calculate the standard emf of the following cell
  at 25 oC. Cr(s)Cr3(aq)Hg22(aq)Hg(l)

Balance the two half reactions,
Cr3 3 e- Cr(s) Eo -0.74 Hg22 2 e-
2Hg(l) Eo 0.80
20
Cell EMFs

• Calculate the standard emf of the following cell
  at 25 oC. Cr(s)Cr3(aq)Hg22(aq)Hg(l)

Balance the two half reactions,
2x 3x
Cr3 Cr(s) 3 e- Eo 0.74 Hg22 2 e-
2Hg(l) Eo 0.80
21
Cell EMFs

• Calculate the standard emf of the following cell
  at 25 oC. Cr(s)Cr3(aq)Hg22(aq)Hg(l)

Balance the two half reactions,
2x 3x
Cr3 Cr(s) 3 e- Eo 0.74 Hg22 2 e-
2Hg(l) Eo 0.80
2Cr(s) 3Hg22(aq) 2Cr3(aq) 6Hg(l)
Eocell 1.54 V
22
EMF and Free Energy

• Faraday showed that the work performed by a
  system was proportional to the EMF and
  subsequently the Free Energy.
• ?G -nFE
• Where F is Faradays constant
• 96,485 C/mole of e-

23
The Effect of Concentration on Eo

• What happens if we decrease the concentration of
  our redox reactants?
• ?G ?Go RT ln(Q)
• Thus, we can relate the equilibrium quotient to
  our concentrations
• and ?G - n F E
• -nFE -nFEo RT ln(Q)
• E Eo - ln(Q)
• Eo - 0.0592/n log(Q) under standard conditions

RT
nF
24
A Reaction Under Non-standard Conditions

• Al (s) Mn2 (aq) Al3 (aq) Mn (s)

25
Non-standard Conditions

• 2Al (s) 3Mn2 (aq) 2Al3 (aq) 3Mn (s)

If Mn2 0.50 M and Al3 1.50 M what is
the voltage associated with the cell?
26
Non-standard Conditions

• 2Al (s) 3Mn2 (aq) 2Al3 (aq) 3Mn (s)

If Mn2 0.50 M and Al3 1.50 M what is
the voltage associated with the cell?
The half reactions are
2 Al 2Al3 6 e-
so, n 6
3Mn2 6 e- 3 Mn
27
Non-standard Conditions

• 2Al (s) 3Mn2 (aq) 2Al3 (aq) 3Mn (s)

If Mn2 0.50 M and Al3 1.50 M what is
the voltage associated with the cell?
The half reactions are
1.66 -1.18 0.48
2 Al 2Al3 6 e-
3Mn2 6 e- 3 Mn
Eo
28
Non-standard Conditions

• 2Al (s) 3Mn2 (aq) 2Al3 (aq) 3Mn (s)

If Mn2 0.50 M and Al3 1.50 M what is
the voltage associated with the cell?
so, n 6
Finally, E 0.48 - 0.592/6 log
(1.50)2/(0.50)3 0.48 - .00987 log
(18) 0.47 V
29
Electrolysis

• By applying a potential greater than Ecell, we
  can push electrons in the opposite direction.

30
Electrolysis - A Stoichiometry Problem

• By applying a potential greater than Ecell, we
  can push electrons in the opposite direction.

current Amps coulomb/sec coulomb charge
1 mol of electrons F coulombs
96,500 C
31
Electrolysis

• In the commercial preparation of aluminum,
  aluminum oxide, Al2O3, is electrolyzed at 1000
  oC. (The mineral cryolite is added as a
  solvent.) Assume that the cathode reaction is
  Al3 3 e- Al. How many coulombs of
  electricity are required to prepare 5.12 kg of
  aluminum?

32
Electrolysis

• In the commercial preparation of aluminum,
  aluminum oxide, Al2O3, is electrolyzed at 1000
  oC. (The mineral cryolite is added as a
  solvent.) Assume that the cathode reaction is
  Al3 3 e- Al. How many coulombs of
  electricity are required to prepare 5.12 kg of
  aluminum?

3 mol e- 1 mol Al
9.65 x 104C 1 mol e-
1 mol Al 26.98 g
5,120 g x
x
x
33
Electrolysis

• In the commercial preparation of aluminum,
  aluminum oxide, Al2O3, is electrolyzed at 1000
  oC. (The mineral cryolite is added as a
  solvent.) Assume that the cathode reaction is
  Al3 3 e- Al. How many coulombs of
  electricity are required to prepare 5.12 kg of
  aluminum?

3 mol e- 1 mol Al
9.65 x 104C 1 mol e-
1 mol Al 26.98 g
5,120 g x
x
x
5.49 x 107C