Electrochemistry

1
Electrochemistry Utilizing electrons from
oxidation-reduction reactions to do electrical
work Applications Batteries, Electroplating
2
Review
Oxidation Loss of electrons Species which is
oxidized is the reducing agent Reduction
Gain of electrons Species which is reduced is
the oxidizing agent
3
Write a balanced reaction between iron (II)
nitrate and potassium permanganate in acidic
conditions
8 H (aq)
4 H2O (l)
MnO4- (aq) ? Mn2 (aq)
5 e-
Fe2 (aq) ? Fe3 (aq)
5 5 5
e-
5 Fe2 (aq) MnO4- (aq) 8 H (aq) ? Mn2 (aq)
5 Fe3 (aq) 4 H2O (l)
4

• Galvanic Cells spontaneously converting
  chemical to electrical
  energy
• Separate each half reaction into their own
  chamber (cell)
• Connect each half cell with a wire connected to
  an inert electrode (Pt) or a metal that is a part
  of the half-cell.
• Use a porous disk or salt bridge to allow for
  ions to flow
• More positive reduction potential is site of
  reduction (called the cathode)
• More negative reduction potential is site of
  oxidation (called the anode)

5
Line Notation for Electrochemical Cells

• Anode is listed on the left, cathode on the right
• Double vertical lines represent porous disk
• Phase boundary is indicated by a single vertical
  line
• States are indicated with symbols (concentrations
  or pressures may be indicated if known)
• Anode Cathode
• Pt(s) Fe2(aq) , Fe3(aq) MnO4-(aq) ,
  Mn2(aq), H (aq) Pt (s)
• Oxidation
  Reduction

6
Determination of Standard Cell Potentials

• Standard Cell Potential (voltage) is the
  difference of the Standard Reduction Potentials
  of the Cathode and Anode
• ?o cell Eocathode Eoanode
• At standard state condition (1 M, 1 atm, 25oC)

• 5 Fe2 (aq) MnO4- (aq) 8 H (aq) ? Mn2
  (aq) 5 Fe3 (aq) 4 H2O (l)
• ?o cell Eo
  MnO4-/Mn2 Eo Fe2/Fe3
• Eocell 1.51
  V 0.77 V 0.74 V

7
e-
e-
? anions

Cu2 (aq)
Fe2(aq)
Cu
Eocell 0.34 V (-0.44 V) 0.78 V
8
Cell Potential, Electrical Work and Free Energy

• The electrical work that can be done (?Go) at
  standard state conditions is
• ?Go -nFEocell
• n number of electrons transferred in the
  reaction
• F Faradays constant (96,485 Coulombs/mole e-)
• Eo measured in volts (J/Coulomb)
• Eo is the electromotive force or work per charge.

9
The Nernst Equation

• Cell Potential Depends on Concentration
• ?G ?Go RT ln Q (?Go - nFEo)
• - nFE - nFEo RT ln Q
• Or E Eo RT/nF ln Q
• At 25oC
• Ecell Eocell 0.0592/n log Q

10
Concentration Cells

• A cell may be constructed where cathode and anode
  are identical in all aspects except
  concentration.
• The anode will have the smaller concentration of
  ions and the cathode will have the greater
  concentration of ions.
• What is Eo of this cell?
• 0.00V
• What is Ecell when Ag concentrations
• are 0.10 M and 10. M?
• E 0.00 V -0.0592/1 log 0.10 / 10.
• E 0.12 V

11
Practice

• Sketch the Ag/Ag Zn/Zn2 galvanic cell.
  Show the direction of e- flow and the direction
  of ion migration.
• Calculate Eo, ?Go and K for the cell. Give the
  line notation for the cell.
• Determine Ecell when Ag 0.010 M and Zn2 2.0
  M

12
e-
? Anions
Zn2(aq)
Ag(aq)
Zn

Eocell 0.80 V (-0.76 V)
Eocell
1.56 V ?Go -nFEo log K n Eo/0.0592
?Go - (2 mole e-)(96485 C / mol e-)(1.56 J/C)
K 10(2)(1.56) / 0.0592 ?Go -3.01.105
J K 5.04.1052 Ecell Eo 0.0592/n
log Q or o Ecell 1.56 V -
0.0592/2 log Zn2 / Ag2 K e - ?G / RT
Ecell 1.56 V - 0.0592/2 log 2.0 / 0.0102
K e301033 J/(8.3145 J/molK(298K) E 1.56 V
0.127 V 1.43 V K 5.82.1052
Ag
13
Electrolysis

• Utilizing electrical current to produce chemical
  change
• How many grams of Zinc can be produced by
  passing10.0 Amps of current through a Zn2
  solution for 10.0 minutes?
• Amps Coulombs / sec
• g 10.0 min 60 s 10.0 C mole e- mol Zn
  65.38 g Zn 2.03 g
• min s
  96485 C 2 mol e- mol Zn
• How much time will be required to plate 2.00 g of
  Au when 5.00 Amps is passed through a Au3
  solution?
• sec 2.00 g mol Au 3 mole e- 96485 C
  sec 588 s
• 197.0 g Au mol Au
  mol e- 5.00 C

14

• When a mixture of chemicals undergo electrolysis
• 1. The more positive Eo will determine what
  will first be reduced at the cathode .
• 2. The more negative Eo will determine what
  will first be oxidized at the anode.

Example When a solution of KF undergoes
electrolysis, what will be the reactions at the
cathode and anode? the most negative half cell
will be oxidized and the most positive half cell
will be reduced. possible oxidations at
the anode possible reductions at the
cathode 2 F-(aq) ? F2 (g) 2 e- Eo 2.87
V K (aq) e- ? K (s) Eo -2.92 V 2
H2O (l) ? O2 (g) 4 H (aq) 4e- Eo 1.23 V
2H2O(l) 2e- ? H2 (g) 2OH-(aq) Eo
-0.83 V The most negative sight for oxidation
is the electrolysis of water to O2 Anode 2 H2O
(l) ? O2 (g) 4 H (aq) 4e- The most
positive sight for reduction is the electrolysis
of water to hydrogen Cathode 2H2O(l) 2e- ? H2
(g) 2OH-(aq)