1. Copyright Sautter 2003

2. ELECTROCHEMISTRY All electrochemical reactions involve oxidation and reduction. Oxidation means the loss of electrons (it does not always involve oxygen). Reduction means the gain of electrons (gaining of negatives, that is electrons, reduces the oxidation number of an atom. We will discuss oxidation number latter in this program). Whenever electrons are lost by one substance they must be gained by another. The substance that loses electrons is referred to as a reducing agent (it lets another substance be reduced, that is, gain the electrons). The substance that gains electrons is referred to as an oxidizing agent (it lets another substance be oxidized, that is, lose the electrons). These terms are important in electrochemistry!!

3. OXIDATION & REDUCTION OXIDATION: Zn (s)  Zn +2 (aq) + 2 e - Metallic zinc is oxidized to zinc ion. Metallic zinc is serving as a reducing agent.(electron loser) REDUCTION: Cu +2 (aq) + 2 e -  Cu (s) Copper ion is reduced to copper metal. Copper ion is serving as an oxidizing agent (electron gainer) In the overall reaction two electrons are transferred from the zinc metal to the copper ion. Zn (s) + Cu +2 (aq)  Zn +2 (aq) + Cu (s)

4. OXIDATION NUMBERS In the previous reactions the loss and gain of electrons was obvious. Zn (s) must lose two negative electrons in order to form a Zn +2 (aq) ion. And Cu +2 (aq) must gain two negative electrons to form a Cu (s) atom. What about more complicated reactions? How do we tell which substance loses and which gains electrons and how many? For example: H + + MnO 4 - + Cl -  Mn +2 + Cl 2 + H 2 O Here we need oxidation numbers!

5. OXIDATION NUMBERS Before discussing the mechanics of oxidation numbers it is important to realize that: (1) oxidation numbers have no physical significance. They are merely a way to tell which substances gain or lose electrons and how many. (2) oxidation numbers are assigned to atoms never molecules. Molecules contain atoms with oxidation numbers but they themselves cannot be assigned oxidation numbers. (3) there a several rules used to assign oxidation numbers. They must be observed carefully.

6. OXIDATION NUMBERS RULES (1) Elements have oxidation numbers of zero. For example Cu has an oxidation number of zero. In Cl 2 each atom of chlorine has an oxidation number of zero. (2) The oxidation number of a monatomic ion (consisting of one atom) is the charge on the ion. In Cu +2 the oxidation number is +2. In Cl - the oxidation number is –1. (3) The oxidation number of combined oxygen is always –2 except in peroxides such as hydrogen peroxide, H 2 O 2. In CO 2 each oxygen atom has an oxidation number of –2. In SO 3 each oxygen is –2.

7. OXIDATION NUMBERS RULES (4) Combined hydrogen is always +1 except in hydrides (metal atoms and hydrogen like NaH). In water H 2 O each hydrogen has an oxidation number of +1. In methane, CH 4, each hydrogen is +1. (5) The sum of all of the oxidation numbers of the atoms in a molecule equal its charge. If no charge is shown the sum must equal zero. This allows us to determine the oxidation numbers of elements not discussed in the first four oxidation number rules! Oxidation numbers of other elemental groups: Elements of Column I in a compound = +1 Elements of Column II in a compound = +2 It is important to realize that the oxidation number of an atom can change when it combines differently during reaction. This is the basis for deciding how electrons are transferred during a chemical reaction!

8. FINDING UNKNOWN OXIDATION NUMBERS Problem: What is the oxidation number of Mn in MnO 4 -, the permanganate ion? Solution: Mn is unknown Each O is –2 (rule 3) All of the oxidation numbers must add up to –1 (rule 5) Mn + 4(-2) = -1, Mn = +7 in permanganate Problem: What is the oxidation number of Cr in K 2 Cr 2 O 7, potassium dichromate? Solution: Cr is unknown Each K from Column I is +1 Each O is –2 (rule 3) All the oxidation numbers add up to 0 (no charge is shown for K 2 Cr 2 O 7 ) 2(+1) + 2 Cr + 7(-2) = 0, 2 Cr = 12, Cr = 12 / 2 = +6 Cr = +6 in potassium dichromate

9. USING OXIDATION NUMBERS TO DETERMINE ELECTRON TRANSFERS Using oxidation numbers we will find the oxidizing agent, the reducing agent, the number of electrons lost and gained, the oxidation half-reaction, the reduction half-reaction and the final balanced equation for: H + + MnO 4 - + Cl -  Mn +2 + Cl 2 + H 2 O Step I – Find the oxidation number of each atom. H + + MnO 4 - + Cl -  Mn +2 + Cl 2 + H 2 O +1 (+7, -2) -1 +2 0 (+1, -2) Step II – Determine which oxidation numbers change MnO 4 -  Mn +2 2 Cl -  Cl 2 +7  +2 2(-1)  0 5 electrons gained 2 electrons lost (MnO 4 - is the oxidizing agent. It gained electrons from Cl - ) (Cl - is the reducing agent. It lost electrons to MnO 4 - ) If oxidization # s decrease, Reduction occurs If oxidization # s increase, Oxidation occurs

10. USING OXIDATION NUMBERS TO DETERMINE ELECTRON TRANSFERS Step III - Electrons lost by a reducing agent must always equal electrons gained by the oxidizing agent! Therefore: 2 (MnO 4 - + 5e -  Mn +2 ) 2MnO 4 - + 10e -  2Mn +2 (reduction half-reaction*) And 5(2Cl -  Cl 2 + 2e - ) 10 Cl -  5Cl 2 + 10e - (oxidation half-reaction) Step IV -Completing the first half-reaction with H + ions and H 2 O molecules: 16 H + + 2MnO 4 - + 10e -  2Mn +2 + 8 H 2 O Step V - Adding the oxidation half-reaction and reduction half- reaction the 10 electrons gained and lost cancel to give the overall reaction: 16 H + + 2MnO 4 - + 10 Cl -  2 Mn +2 + 5 Cl 2 + 8 H 2 O * Half-reaction refers to the reaction showing either the electron gain (reduction) or the electron loss (oxidation) step of the reaction.

11. ELECTRICAL POTENTIAL, CHARGE, ELECTRON FLOW For electrons to transfer from one substance to another the oxidizing agent (electron gainer) must have a greater attraction for those electrons than the reducing agent (electron loser). This where electrical potentials come into play. Electrical potential is measured in volts. Electrical charge is measured in coulombs. The flow of electrons from one substance to another is measured in amperes. 1 coulomb is the charge on 6.25 x 10 18 electrons.(The charge on 1 electron is –1.6 x 10 -19 coulombs). 1 ampere (amp) is the flow rate of 1 coulomb per second 1 volt is 1 joule of energy moving 1 coulomb of charge between two substances.

12. 6.25 x 10 18 electrons 1 coulomb 1 second 1 amp = 1 coul / sec 1 joule 1 coulomb of charge 1 volt = 1 joule / coul ELECTRICAL QUANTITIES

13. ELECTRICAL POTENTIAL Electrical potential is measured in volts and indicates the tendency of electrons to move from one substance to another. Potential depends on a variety of factors such as the concentration of reactant materials, temperature, gas pressures and the nature of the materials involved. Standard Reduction Potentials (measuring the tendency of a substance to gain electrons) are determined at 25 0 C, 1 molar concentrations, 1 atm pressures and are compared to hydrogen ion’s tendency to be reduced (gain electrons) to hydrogen gas (2H + (aq) + 2e -  H 2(g) ) which is assigned a potential of 0.00 volts. Substances which gain electrons better than H + ion are assigned positive potentials. Those which gain electrons more poorly than H + ion are assigned negative potentials.

14. ELECTRICAL POTENTIAL Some Standard Reduction Potentials E 0 (volts)

15. ELECTRICAL POTENTIALS The very best oxidizing agent (electron gainer) available is flourine gas, F 2. Its reduction potential is +2.87 volts. The very best reducing agent (electron loser) is lithium metal, Li. Its oxidization potential is +3.05 volts. When Li is combined with F 2, the reaction is highly spontaneous. Electrons move readily from the Li to the F 2. The standard potential of this cell (redox reaction) is +5.92 volts. By contrast, combining Li + ion with F - ion results in a nonspontaneous process.(Li + has already lost its electrons and F - has already gained its electrons). The standard potential for this cell is –5.92 volts

16. ELECTRICAL POTENTIALS From the Standard Reduction Potential Table: (1) F 2(g) + 2e -  2 F - (aq) 2.87 volts (2) Li + (aq) + 1 e -  Li (s) -3.05 volts Since Li metal serves as a reducing agent, it must be shown to lose electrons, and the half-reaction must be reversed. This requires that the E 0 sign be changed. (3) Li (s)  Li + (aq) + 1e - +3.05 volts and since 2 electrons must be transferred to F 2 (4) 2 Li (s)  2 Li + (aq) + 2e - +3.05 volts (note that the E 0 is not multiplied by two.) Combining half-reaction (1) and (4) we get Li (s) + F 2(g)  Li + (aq) + 2 F - (aq) + 5.92 volts For the reverse reaction which is nonspontaneous, the potential is –5.92 volts.

17. ELECTRICAL POTENTIALS From the previous frames we can conclude: (1) Standard redox potentials for a reaction can be obtained from standard reduction half-reactions by adding the half-reactions. (2) When a substance is to act as a reducing agent, the half-reaction from the standard reduction potentials table must be reversed and the sign of the E 0 value changed. (3) When half-reactions are added, electrons lost and gained must be balanced. When balancing the electrons, although the half-reactions are multiplied, the E 0 values are not. (4) Reactions with positive E 0 values are spontaneous. Those with negative E 0 values are nonspontaneous in the forward direction

18. USING ELECTRICAL POTENTIALS Problem: What is the electrical potential (E 0 ) for a standard cell (redox reaction) composed of zinc metal and copper II ions? The products are zinc ion and copper. Is the reaction spontaneous? Identify the cathode (reduction site) and anode (oxidation site). Solution: Zn (s) + Cu +2 (aq)  Zn +2 (aq) + Cu (s) 0 +2 +2 0 (oxidization #s) Zn is the reducing agent, loses electrons (0  +2) Cu +2 is the oxidizing agent, gains electrons (+2  0) Zn (s)  Zn +2 (aq) + 2e - +0.76 volts (anode reaction) Cu +2 (aq) + 2e -  Cu (s) +0.34 volts (cathode reaction) Overall reaction: Zn (s) + Cu +2 (aq)  Cu (s) + Zn +2 (aq) E 0 = +1.10 volts The reaction is spontaneous (E 0 net is positive)

19. Cu +2 Cu Electrons move through the external circuit from anode to cathode. Copper is deposited at the cathode

20. Zn +2 Zn Zn +2 Zn Electrons move through the external circuit from anode to cathode. Zinc is dissolved at the anode.

21. ELECTROCHEMICAL CELL CONVENTIONS Electrochemical cell that release energy spontaneously are call voltaic or galvanic cells. Those which require energy input in order to function are call electrolytic cells. Voltaic cells are often represented as follows for the reaction: Zn (s) + Cu +2 (aq)  Zn +2 (aq) + Cu (s) Zn / Zn +2 (1.0 M) // Cu +2 (1.0 M) / Cu Anode Compartment components Cathode Compartment components Salt bridge Solution concentration Solution concentration

22. AN ELECTROCHEMICAL CELL EMF 1.100 VOLTS ANODE (OXIDATION OCCURS) CATHODE (REDUCTION OCCURS) Zn (S)  Zn +2 (aq) + 2 e - E 0 = 0.763 volts Cu +2 (aq) + 2 e -  Cu (s) E 0 = 0.337 volts SALT BRIDGE

23. USING NONSTANDARD POTENTIALS Reactions do not always involve Standard Electrical Potentials (that is 25 0 C, 1molar, 1atm). Calculating potential at other concentrations and temperatures requires the use of the Nernst Equation. E cell = E 0 – (RT / n F) ln Q where: E cell = the voltage of the non standard cell E 0 = standard cell potential R = 8.31 joules / mole K, T = Kelvin temperature n = the number of electrons transferred in the balanced equation F = Faraday’s constant (96,487 coulombs / mole) Q = reaction quotient (product concentrations / reactant concentrations) At equilibrium E cell = 0 and Q = K e therefore: E 0 = (RT / n F) ln K e At 25 o C (the most common temperature used), substituting 298 K, 8.31 joules / mole K, 96,487 coulombs / mole and 2.203 (the conversion factor from base e to base 10 logs) we get: E cell = E 0 – (0.0592 / n) log Q and E o = (0.0592 / n) log K e

24. USING THE NERNST EQUATION Problem: Calculate the K e value for the zinc / copper cell shown in the previous diagram. Solution: The standard potential for the cell is 1.10 volts and 2 electrons are transferred from the zinc anode to the copper cathode as shown. E o = (0.0592 / n) log K e, K e = log -1( (E 0 x n) / 0.0592) K e = log -1 (37.16) = 1.45 x 10 37 (the reaction is very spontaneous)

25. USING THE NERNST EQUATION Problem: What is the potential of the zinc / copper cell if the zinc ion concentration is 0.10 M and the copper ion concentration is 0.00010 M? (non standard cells) Zn (s) + Cu +2 (aq)  Zn +2 (aq) + Cu (s), E 0 net = 1.10 volts E cell = E 0 – (0.0592 / n) log Q Q = [Zn +2 ] / [Cu +2 ] = 0.10 / 0.00010 = 1000 E cell = 1.10 – (0.0592 / 2 ) log 1000 = 1.01 volts Based on LeChatelier’s Principle, the higher concentration of zinc ion on the right side of the reaction should drive the system to the left and reduce it spontaneity. The reduced voltage of 1.01 volts as compared to the standard state voltage of 1.10 volts confirms that observation.

26. RELATING FREE ENERGY AND CELL POTENTIALS Since both free energy and cell potential predict the spontaneity of a reaction they logically must be related. The relationship is given by:  G = - n F E cell and  G 0 = - n F E 0 cell where n = number electrons transferred in the balanced equation F = Faraday’s constant (96,487 coulombs / mole) E = the cells potential in volts Problem: What is the  G 0 for the zinc / copper cell? Solution:  G 0 = - n F E 0 cell  G 0 = - (2) (96,487) (1.10) = - 2.12 x 10 5 joules / mole or - 212 KJ / mole A large positive E 0 value correlates well with a large negative  G 0 value. Both indicate a very spontaneous reaction!

27. ELECTROLYSIS Electrolysis involves the input of energy into an electrochemical cell. Electrolytic cells are used to decompose substances, deposit metals and determine the charges on ions in solution. The exact number of electrons passing into an electrolytic cell can be determined by monitoring the current flow in amperes and the time elapsed. One ampere is equal to a flow rate of one coulomb (6.25 x 10 18 electrons) per second. Coulombs = amps x time (in seconds) Since chemistry measurements are most always done in moles a conversion factor between coulombs and moles is called Faraday’s constant is needed. One mole (6.02 x 10 23 ) divided by one coulomb (6.25 x 10 18 ) gives 96,487 coulombs per mole (Faraday’s constant - we have used this number before in this program)

28. ELECTROLYTIC SILVER PLATING Ag DIRECT CURRENT SOURCE Ag + solution cathode anode electrons Ag + + 1 e -  Ag

29. ELECTROLYSIS Problem: How many grams of copper can be deposited from a copper sulfate solution by a current of 2 amperes for 20 minutes? Solution: Cu +2 (aq) + 2e -  Cu (s) Amperes x time in seconds = coulombs (2.00 coul / sec) x (20 x 60) sec = 2400 coul. charge in coulombs / Faraday’s constant = moles of e - 2400 coul x (1 moles / 96,487coul) = 0.0249 moles of e - 1 mole of copper atom requires 2 moles of electrons 0.0249 moles e - / 2 = 0.0125 moles of Cu atoms 0.0125 moles Cu x 63.3 grams / mole = 0.791 g of Cu

30. ELECTROLYSIS Problem: What current is needed to deposit 0.500 grams of Cr +3 ion from solution in 1.00 hour? Solution: Cr +3 + 3 e -  Cr grams Cr / (grams per mole) = grams 0.500 / 52.0 = 0.00962 moles 1 mole of Cr +3 needs 3 moles of electrons 0.00962 moles Cr +3 x 3 = 0.0288 moles of e - current is measured in amperes (coulombs per second) 0.0288 moles of e - x 96487 coul per mole = 2780 coul One hour = 60 x 60 = 3600 seconds Amperes = 2780 coulombs / 3600 seconds = 0.773 amps

31. SUMMARY OF KEY IDEAS (1) Oxidation number rules: (i) Elements = 0 (ii) Monatomic ion = the ionic charge (iii) Combined oxygen = -2 (iv) Combined hydrogen = +1 (v) The sum of all oxidation numbers = the charge on the ion or molecule (2) Oxidation is electron loss (oxidation number increases. Reduction is electron gain (oxidation decreases) (3) Cathode is the site of reduction. Anode is the site of oxidation.

32. SUMMARY OF KEY IDEAS (4) Potential is measured in volts (1 volt = 1 joule / 1 coul). If the net E 0 is + the reaction is spontaneous. If net E 0 is – the reaction is non spontaneous. (5) Standard state is 25 C 0, 1 molar and 1 atm (6) Faraday’s constant = 96,487 coulombs per mole (7) 1 ampere = 1 coulomb per second (8) 1 coulomb = 6.25 x 10 18 electrons (9) E cell = E 0 – (RT / n F) ln Q (10) E 0 = (RT / n F) ln K e (11)E cell = E 0 – (0.0592 / n) log Q (12) E o = (0.0592 / n) log K e (13)  G = - n F E cell and  G 0 = - n F E 0 cell (14) Large K e values, + E values and -  G values mean spontaneous reactions