1. ELECTROCHEMISTRY Chapter 9
Electric automobile
ELECTROCHEMISTRY Chapter 9

2. Why Study Electrochemistry?
Batteries
Corrosion
Industrial production of chemicals such as Cl2, NaOH, F2 and l
Biological redox reactions
The heme group

3. TRANSFER REACTIONS Atom transfer Electron transfer
HCl (g) + H2O (l)  Cl- (aq) H3O+ (aq)
Electron transfer
Cu(s) Ag+(aq)  Cu2+(aq) Ag(s)
- 2 e-
2 x +1 e-

4. Electron Transfer Reactions
Electron transfer reactions are oxidation-reduction or redox reactions.
Redox reactions can result in :
generation of an electric current, or
be caused by imposing an electric current.
When external electric current is involved, this field of chemistry is called ELECTROCHEMISTRY.

5. Terminology for Redox Reactions
OXIDATION—loss of electron(s) by a
species; increase in oxidation number.
REDUCTION—gain of electron(s);
decrease in oxidation number.
OXIDIZING AGENT—electron acceptor;
species is reduced.
REDUCING AGENT—electron donor; species is oxidized.

6. Demo: Balanced Chemical Equation Total Net Ionic Equation
Zn(s) + CuSO4(aq)  _______________
ACTIVITY SERIES OF METALS
Balanced Chemical Equation
Total Net Ionic Equation
Net Ionic Equation

7. Let’s Examine Reaction…
Zn (s) + Cu+2 + SO4-2  Zn+2 + SO4-2 + Cu(s)
Zn (s) + Cu+2  Zn+2 + Cu(s)
LEO GER
Lost 2 e
Gained 2 e
Zn – lost 2 e  OXIDAZED
Cu – gained 2 e  REDUCED

8. DEFINITION
Reactions in which one reactant is oxidized (lose e-) and the other reactant is reduced (gains e-) are called oxidation-reduction reaction, or REDOX
NOTE: e- gained by one reactant are the e- lost by the other one

9. To understand electrochemistry, we need to be able to keep track of electrons, we do so by assigning oxidation numbers.

11. SUMMARY: Redox Reactions
The charge on each atom in any uncombined element is always ZERO [Cu(s); Br2(g)…]
In a redox reaction, electrons are transferred from one reactant to another reactant
Oxidation – lose e-  Reducing agent
Reduction – gain e-  Oxidizing agent

12. Rules for assigning Oxidation Numbers  pg.658 Table 1
Examples:
Cl2
Na2O
Fe2O3
PbSO4
KNO2
Fe(NO3)3
Supply the oxidation number of the underlined element in the following formulas;
a) Zn3(PO4)2, b) NaNO2, c) SnBr2, d) HSbO2, e) Mg(MnO4)2, f) NH4NO3

13. Identify the reactant oxidized and the reactant reduced in each of the following equations. Which of the following equations represent redox reactions?
Pg. 662 (#19, 20)

14. Direct Redox Reactions
Oxidizing and reducing agents in direct contact.
Cu(s) Ag+(aq)  Cu2+(aq) Ag(s)
2Al (s) + 3Cu2+ 
2 Al Cu (s)

15. Electroplating

16. Indirect Redox Reactions
A battery functions by transferring electrons through an external wire from the reducing agent to the oxidizing agent.
Electron transfer
Oxidation
Reduction
11_battry.mov
21mo2an2.mov
Ions

17. How to balance for both charge and mass ?
Balancing Equations
Cu (s) + Ag+ (aq)  Cu2+ (aq) + Ag (s)
How to balance for both charge and mass ?
Step 1: Identify the oxidation and reduction HALF-REACTIONS:
OX: Cu  Cu e-
RED: Ag+ + e-  Ag
Step 2: Balance each HALF-REACTION for charge and mass (done)
Step 3: Multiply each half-reaction by a factor that makes the reducing agent supply as many electrons as the oxidizing agent requires - ELECTRON TRANSFER NUMBER (2 here)
Oxidation (Reducing agent) Cu  Cu e-
Reduction (Oxidizing agent) Ag e-  2 Ag
Step 4: Add half-reactions to give the overall equation.
Cu (s) + 2 Ag+ (aq)  Cu2+ (aq) + 2Ag (s)

18. Balancing Equations (2)
Balance the following in acid solution—
VO Zn  VO Zn2+
Step 1: Write the half-reactions
Ox Zn  Zn2+
Red VO2+  VO2+
Step 2: Balance each half-reaction for mass.
Red 2 H+ + VO2+  VO H2O
Add H2O on O-deficient side and add H+ on other side for H-balance.

19. Balancing Equations (3)
Step 3: Balance half-reactions for charge.
Ox Zn  Zn e-
Red e H+ + VO2+  VO H2O
Step 4: Multiply by an appropriate factor to balance the electron transfer in OX. and RED.
Ox Zn  Zn e-
Red 2e H VO2+  2 VO H2O
Step 5: Add half-reactions
Zn H VO2+  Zn VO H2O

20. Tips on Balancing Equations
Never add O2, O atoms, or O2- to balance oxygen.
Never add H2 or H atoms to balance hydrogen.
Be sure to write the correct charges on all the ions.
Check your work at the end to make sure mass and charge are balanced.

21. Electrochemical Cells
An apparatus in which a redox reaction occurs by transferring electrons through an external connector.
VOLTAIC CELL
Product favored reaction
chemical reaction  electric current
Batteries are
voltaic cells
ELECTROLYTIC CELL
Reactant favored reaction
electric current  chemical reaction

22. CHEMICAL CHANGE  ELECTRIC CURRENT
With time, Cu plates out onto Zn metal strip, and Zn strip “disappears.”
Zn is oxidized and is the reducing agent Zn(s)  Zn2+(aq) + 2e-
Cu2+ is reduced and is the oxidizing agent Cu2+(aq) + 2e-  Cu(s)

23. CHEMICAL CHANGE  ELECTRIC CURRENT (2)
Oxidation: Zn(s)  Zn2+(aq) + 2e-
Reduction: Cu2+(aq) + 2e-  Cu(s)
Cu2+(aq) + Zn(s)  Zn2+(aq) + Cu(s)
Electrons are transferred from Zn to Cu2+, but there is no useful electric current.

24. CHEMICAL CHANGE  ELECTRIC CURRENT (2)
To obtain a useful current, we separate the oxidizing and reducing agents so that electron transfer occurs thru an external wire.
This is accomplished in a VOLTAIC cell.
(also called GALVANIC cell)
A group of such cells is called a battery.

25. ANODE
OXIDATION
CATHODE
REDUCTION
• Electrons travel thru external wire.
Salt bridge allows anions and cations to move between electrode compartments.
This maintains electrical neutrality.

26. Electrochemical Cell Electrons move from anode to cathode in the wire.
Anions & cations move through the salt bridge.
Electrochemical Cell

27. Standard Notation for Electrochemical Cells
Phase boundary
Salt bridge
Phase boundary
ANODE Zn / Zn2+ // Cu2+ / Cu CATHODE
Cathode electrode
Anode electrode
Active electrolyte in
reduction half-reaction
Active electrolyte in
oxidation half-reaction
OXIDATION
REDUCTION

28. CELL POTENTIAL, E Zn  Zn2+ + 2e- 2e- + Cu2+ Cu ANODE CATHODE
Electrons are “driven” from anode to cathode by an electromotive force or emf.
For Zn/Cu cell, this is indicated by a voltage of 1.10 V at 25°C and when [Zn2+] and [Cu2+] = 1.0 M.

29. CELL POTENTIAL, Eo STANDARD CELL POTENTIAL, Eo
For Zn/Cu, voltage is 1.10 V at 25°C and when [Zn2+] and [Cu2+] = 1.0 M.
This is the
STANDARD CELL POTENTIAL, Eo
Eo is a quantitative measure of the tendency of reactants to proceed to products when all are in their standard states at 25 °C.

30. Eo and DGo
Michael Faraday
Eo is related to DGo, the free energy change for the reaction.
DGo = - n F Eo
F = Faraday constant = x 104 J/V•mol
n = the number of moles of electrons transferred.
Discoverer of
electrolysis
magnetic props. of matter
electromagnetic induction
benzene and other organic chemicals
Zn / Zn2+ // Cu2+ / Cu
n = 2
n for Zn/Cu cell ?

31. For a product-favored reaction Reactants  Products
Eo and DGo (2)
DGo = - n F Eo
For a product-favored reaction
battery or voltaic cell: Chemistry  electric current
Reactants  Products
DGo < 0 and so Eo > 0 (Eo is positive)
For a reactant-favored reaction
- electrolysis cell: Electric current  chemistry
Reactants  Products
DGo > 0 and so Eo < 0 (Eo is negative)

32. Calculating Cell Voltage
Balanced half-reactions can be added together
to get the overall, balanced equation.
Anode: 2 I-  I e-
Cathode: 2 H2O e-  2 OH H2
Net rxn: 2 I H2O  I OH H2
If we know Eo for each half-reaction, we can calculate Eo for the net reaction.

33. STANDARD CELL POTENTIALS, Eo
Can’t measure half- reaction Eo directly. Therefore, measure it relative to a standard HALF CELL:
the Standard Hydrogen Electrode (SHE).
2 H+(aq, 1 M) e H2(g, 1 atm)
Eo = 0.0 V

34. Zn/Zn2+ versus H+/H2 Zn/Zn2+ half-cell combined with a SHE.
Eo for the cell is V

35. Eo for Zn/Zn2+ half-cell
Overall reaction is reduction of H+ by Zn metal.
Zn(s) + 2 H+ (aq)  Zn2+ + H2(g) Eo = V
Therefore,
Eo for Zn  Zn2+ (aq) + 2e- is ??
+0.76 V.

36. Standard REDUCTION potentials
Zn  Zn2+ (aq) + 2e Eo = V
Q. Relative to H2 is Zn a (better/worse) reducing agent ?
A. Zn is a better reducing agent than H2.
What is Eo for the reverse reaction ?
Zn2+ + 2e-  Zn
The value for the REDUCTION 1/2-cell is the negative of
that for the OXIDATION 1/2-cell:
Zn  Zn2+ (aq) + 2e- Eo = V
THUS
Zn2+ + 2e-  Zn Eo = V

37. Cu/Cu2+ and H2/H+ Cell
Eo = V

38. Overall reaction is reduction of Cu2+ by H2 gas.
Cu/Cu2+ half cell Eo
Overall reaction is reduction of Cu2+ by H2 gas.
Cu2+ (aq) + H2(g)  Cu(s) H+(aq)
Measured Eo = V
Therefore, Eo for Cu e-  Cu is ??
+0.34 V

39. Zn/Cu Electrochemical Cell
What is Eo for the Zn/Cu cell (Daniel’s cell) ??
Anode, negative, source of electrons
Cathode, positive, sink for electrons
Anode: Zn(s)  Zn2+(aq) + 2e- Eo = V
Cathode: Cu2+(aq) + 2e-  Cu(s) Eo = V
Net: Cu2+(aq) + Zn(s)  Zn2+(aq) + Cu(s)
Eo = Eo(anode) + Eo(cathode) = = V

40. Uses of Eo Values This shows we can
a) decide on relative ability of elements to act as reducing agents (or oxidizing agents)
b) assign a voltage to a half-reaction that reflects this ability.

41. STANDARD REDUCTION POTENTIALS
Oxidizing ability of ion
Half-Reaction Eo (Volts)
Reducing ability
of element
Cu e-  Cu
2 H+ + 2e-  H
Zn e-  Zn
BEST Oxidizing agent ? ?
Cu2+
BEST Reducing agent ? ? Zn

42. Standard Redox Potentials, Eo
Any substance on the right will reduce any substance higher than it on the left.
Zn can reduce H+ and Cu2+.
H2 can reduce Cu2+ but not Zn2+
Cu cannot reduce H+ or Zn2+.
Use tabulated reduction potentials to analyse
spontaneity of ANY REDOX REACTION

43. Determining Eo for a Voltaic Cell
Cd  Cd2+ + 2e- or
Cd2+ + 2e-  Cd
Fe  Fe2+ + 2e- or
Fe2+ + 2e-  Fe

44. Eo for Fe/Cd Cell • Fe is a better reducing agent than Cd
LHS species
is better oxidizing agent
RHS species
is better reducing agent
Cd e-  Cd
Fe e-  Fe
• Fe is a better reducing agent than Cd
Cd 2+ is a better oxidizing agent than Fe 2+
Overall reaction as written is spontaneous:
Fe + Cd 2+  Cd + Fe Eo = V
The reverse reaction is not spontaneous:
Cd + Fe 2+  Fe + Cd Eo = V