1. CH 104: Voltaic Cells
Oxidation is a loss of an electron or electrons by an atom or group of atoms.
Zn(s) → Zn2+(aq) + 2e–
Zn(s) is oxidized to Zn2+(aq) in this half-reaction.
Reduction is a gain of an electron or electrons by an atom or group of atoms.
Cu2+(aq) + 2e– → Cu(s)
Cu2+(aq) is reduced to Cu(s) in this half-reaction.
These 2 half-reactions make an oxidation-reduction reaction. That is, electrons are transferred from 1 reactant to another reactant.
Oxidation: Zn(s) → Zn2+(aq) + 2e–
Reduction: Cu2+(aq) + 2e– → Cu(s)___________
Oxidation-reduction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

2. Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
MAKING A BATTERY
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Reactants
Products

3. MAKING A BATTERY
An electrochemical cell (or battery) is made of 2 half-cells with electrodes joined by a wire and solutions joined by a salt bridge.
The salt bridge lets ions move from 1 half-cell to the other half-cell; however, it does not let the bulk solutions move.
A potentiometer (or voltmeter) measures the difference in electric potential (or voltage) between these 2 electrodes. In this battery the difference is volts.

4. Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
MAKING A BATTERY
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Oxidation occurs at the anode. Which electrode is the anode?
Hint: oxidation and anode both start with vowels.
Reduction occurs at the cathode. Which electrode is the cathode?
Hint: reduction and cathode both start with consonants.

5. Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
MAKING A BATTERY
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
What is the oxidation half-reaction?
What is the reduction half-reaction?
Therefore, electrons flow from the to the .
Electrons are produced at the anode and are consumed at the cathode.
The mass of Zn(s) at the anode will increase, decrease, or stay the same?
The mass of Cu(s) at the cathode will increase, decrease, or stay the same?
Zn(s) → Zn2+(aq) + 2e–
Cu2+(aq) + 2e– → Cu(s)
anode
cathode
Decrease, Zn(s) is converted into Zn2+(aq).
Increase, Cu2+(aq) is converted into Cu(s).

6. Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
MAKING A BATTERY
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Therefore, anions (NO3–) flow from the salt bridge to the
That is, anions from the salt bridge are attracted to the Zn2+(aq) produced at the anode.
Therefore, cations (K+) flow from the salt bridge to the
That is, cations from the salt bridge are attracted to the deficit of Cu2+(aq) at the cathode.
What is the voltage of this battery?
1.097 volts (V).
anode
cathode

7. MAKING A SECOND BATTERY
Now let’s examine a different oxidation-reduction reaction:
Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
______ is being oxidized to _______.
______ is being reduced to _______.
Cu(s)
Cu2+(aq)
Ag+(aq)
Ag(s)
Reactants
Products

8. Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
MAKING A SECOND BATTERY
Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
Notice that replacing Zn(s) and 1.00 M Zn(NO3)2(aq) with Ag(s) and 1.00 M AgNO3(aq) decreases the voltage from V to V.
Which electrode is the anode?
Hint: oxidation and anode both start with vowels.
Which electrode is the cathode?
Hint: reduction and cathode both start with consonants.
Notice the first battery had a Cu(s)|Cu2+(aq) cathode. This second battery has a Cu(s)|Cu2+(aq) anode.

9. Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
MAKING A SECOND BATTERY
Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
What is the oxidation half-reaction?
What is the reduction half-reaction?
Therefore, electrons flow from the to the .
Electrons are produced at the anode and are consumed at the cathode.
The mass of Cu(s) at the anode will increase, decrease, or stay the same?
The mass of Ag(s) at the cathode will increase, decrease, or stay the same?
Cu(s) → Cu2+(aq) + 2e–
Ag+(aq) + e– → Ag(s)
anode
cathode
Decrease, Cu(s) is converted into Cu2+(aq).
Increase, Ag+(aq) is converted into Ag(s).

10. Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
MAKING A SECOND BATTERY
Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
Therefore, anions (NO3–) flow from the salt bridge to the
That is, anions from the salt bridge are attracted to the Cu2+(aq) produced at the anode.
Therefore, cations (K+) flow from the salt bridge to the
That is, cations from the salt bridge are attracted to the deficit of Ag+(aq) at the cathode.
anode
cathode

11. STANDARD REDUCTION POTENTIALS, E°red
Standard reduction potentials (E°red) are measured with all ions at 1.0 M concentration, all gases at 1.0 atmosphere pressure, all solids in their most thermodynamically stable forms, and the temperature at 298 K.
Therefore, the standard hydrogen electrode has pure H2(g) at 1.0 atm bubbling into a solution of 1.0 M H+(aq) at 298 K. This electrode is assigned a standard reduction potential of zero volts.
2 H+(aq) + 2e– → H2(g), E°red = 0 volts
All standard reduction potentials are measured relative to the standard hydrogen electrode.

12. STANDARD REDUCTION POTENTIALS, E°red
What is the voltage of the anode?
The anode is a standard hydrogen electrode. By definition it has 0 volts.
What is the voltage of the cathode?
The cathode is a standard Cu2+(aq)|Cu(s) electrode. It has 0.34 volts, according to the meter. This voltage agrees with the following table of standard reduction potentials (E°red).

13. STANDARD REDUCTION POTENTIALS, E°red
This table of standard reduction potentials (E°red) is used to calculate standard cell potentials (E°cell).
E°cell = E°ox + E°red

14. STANDARD CELL POTENTIALS, E°cell
For example, use this table of standard reduction potentials (E°red) to calculate the standard cell potential (E°cell) of the following battery.
Oxidation: Cu(s) → Cu2+(aq) + 2e–, E°ox = –0.34 volts
Reduction: 2Ag+(aq) + 2e– → 2Ag(s), E°red = 0.80 volts
Net reaction: Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
Important: Oxidation is the opposite of reduction; therefore, the sign of E°ox for a half-reaction is always the opposite of E°red. For example, the volts for the standard reduction potential of this oxidation half-reaction was changed to –0.34 volts.
Important: Standard reduction potential is an intensive property; therefore, the values of E°ox and E°red are not changed if the stoichiometric coefficients are multiplied by a constant. For example, the 0.80 volts for Ag+(aq) + e– → Ag(s) was not changed when these stoichiometric coefficients were multiplied by 2 to balance the net reaction.
Important: If E°cell is +, then the cell reaction is spontaneous as written.
Important: If E°cell is –, then the cell reaction is non-spontaneous as written.
E°cell = E°ox + E°red
E°cell = –0.34 volts volts = 0.46 volts

15. STANDARD CELL POTENTIALS, E°cell
Calculate the standard cell potential (E°cell) of the following battery.
Oxidation: 4Ag(s) → 4Ag+(aq) + 4e–, E°ox = –0.80 volts
Reduction: O2(g) + 4H+(aq) + 4e– → 2H2O(l), E°red = 1.23 volts
Net reaction: 4Ag(s) + O2(g) + 4H+(aq) → 4Ag+(aq) + 2H2O(l)
E°cell = E°ox + E°red
E°cell = –0.80 volts volts = 0.43 volts
Is this reaction spontaneous?
Yes, E°cell is +.

16. THE NERNST EQUATION
The Nernst equation is used to calculate cell potentials (Ecell) at non-standard conditions.
For the cell reaction: aA + bB = cC + dD
Where
Ecell = the cell potential at non-standard conditions
E°cell = the standard cell potential
R = the gas constant = JK–1mol–1
T = the temperature in Kelvin
n = the total number of electrons transferred in the equation as written
F = the Faraday constant = 96,500 Jvolt–1mol–1

17. THE NERNST EQUATION
Use the Nernst equation to calculate Ecell for this battery at 25° C.
Oxidation: Fe2+(aq) → Fe3+(aq) + e–, E°ox = –0.77 volts
Reduction: Ag+(aq) + e– → Ag(s), E°red = 0.80 volts
Net reaction: Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s)
E°cell = E°ox + E°red =
–0.77 volts volts = 0.03 volts

18. ELECTROCHEMISTRY AND CORROSION
Oxidation-reduction causes the corrosion of iron and steel. Steel is refined Fe containing less than 1.7% C.
The oxidation of Fe(s) to Fe2+(aq) occurs at the strained regions of these iron nails: the heads, tips, and bend. This generation of Fe2+(aq) is marked by the formation of a blue precipitate in the presence of K3[Fe(CN)6].
The reduction of O2(aq) to OH–(aq) occurs at the other regions of these nails. This generation of OH–(aq) is marked by the pink color of phenolphthalein.
Use these observations to write a balanced equation for the corrosion of iron or steel.
2Fe(s) + O2(aq) + 2H2O(l) → 2Fe2+(aq) + 4OH–(aq)
Notice that exposure to air and water causes the corrosion of iron and steel.

19. ELECTROCHEMISTRY AND CORROSION
Coatings, such as paints or Zn(s), are used to retard the corrosion of iron and steel by preventing this exposure to air and water.
Galvanized iron or steel is coated with Zn(s). This retards the corrosion of Fe(s).
Coating iron or steel with Cu(s) accelerates the corrosion of Fe(s).

20. ELECTROCHEMISTRY AND CORROSION
Why is stainless steel stainless?
Stainless steel contains at least 10.5% Cr. This Cr reacts with the O2(g) in air to form a complex chrome-oxide surface layer that prevents further corrosion. Unlike the rusting surface of iron, this chrome-oxide layer does not flake away and expose additional atoms to air and water. Higher levels of Cr and the addition of other alloying elements such as Ni and Mo further improve this surface layer and the corrosion resistance of stainless steels.

21. SOURCES
Barnes, D.S., J.A. Chandler Chemistry Workbook and Laboratory Manual. Amherst, MA: University of Massachusetts.
McMurry, J., R.C. Fay Chemistry, 4th ed. Upper Saddle River, NJ: Prentice Hall.
Merriam-Webster, Inc Webster’s 9th New Collegiate Dictionary. Springfield, MA: Merriam-Webster, Inc.
Petrucci, R.H General Chemistry Principles and Modern Applications, 4th ed. New York, NY: Macmillan Publishing Company.
Specialty Steel Industry of North America SSINA: Stainless Steel: About. Available: [accessed 12 October 2006].