Chapter 17, Electrochemistry

1) Electrochemistry has tremendous commercialimportance.

2) You will rely heavily on your understanding of redoxrxns. to describe electrochemical cells (2 types):

a. Galvanic (a.k.a.: voltaic; rxn. produces energy)b. Electrolytic (a.k.a.: rxn. requires/consumes energy)

3) The chemical (redox) rxns involve e! transfer; we canrelate these rxns to a common scale, EE, standardreduction potential.

4) Relationship of EE to ÄGE?Background, Fig. 17.1, p. 698.

1

1) This clearly energetically downhill. (Rxn done in lab?)

Cu2+ + Zn ÿ Cu + Zn2+

2

2) Is there a way to use/capture/store some of the energythat is released when this rxn. occurs? Comparisons tobiological systems?

3) Can we force a normally non-spontaneous rxn. to occurby adding energy? (See electrolytic cells.) Comparisons to biological systems?

3

I. Galvanic Cells

A. The rxn in a galvanic cell is spontaneous.

1. Example: Fig. 17.1, p. 698:

2. Net ionic equation for the above:

Zn(s) + Cu2+(aq) ÿ Zn2+(

aq) + Cu(s)

3. Half-rxns for the above are:

Oxidation: ÿ +

Reduction: + ÿ

4

Summed half-reactions give the overall balanced rxn.

4. This is not a homogeneous system with respect tophases of the reactants & products. How does thatinfluence the way you think about this rxn?

B. It’s possible to get this rxn to go even when thereactants are not in direct physical contact if:

1. a wire that conducts e! is placed in direct contact withthe appropriate electrodes and

2. an ion conducting channel is present so redox inertions can move to minimize charge accumulation. SeeFig. 17.2 a) p. 699.

5

3. A design variation is the Daniell cell Fig. 17.2 b).

6

C. Nomenclature

1. Electrode where the oxidation half-rxn occurs iscalled the anode. Anions move toward the anode.

2. Electrode where the reduction half-rxn occurs is thecathode. Cations move toward the cathode.

Read the paragraph at the bottom of p. 699 (-700) to geta better feel for the charge designations assigned to theanode and cathode. From the perspective of the liquidphase, they may seem illogical. View the Zn2+ ions givenoff by the anode as attracting the anions.

Try Prob. 17.1, p. 702.

7

II. Shorthand Notation for Galvanic Cells

A. As in many other areas of humancommunication, efficiency is useful inrepresenting a galvanic cell.

1. The cell that we looked at above employing the rxn:

Zn(s) + Cu2+(aq) ÿ Zn2+(aq) + Cu(s)

8

can be described by:

Zn(s)* Zn2+(aq) 2 Cu2+(aq)*Cu(s) where:

a) * represents a phase boundaryb) 2 represents a salt bridgec) anode half-cell on left, cathode on rightd) electrodes are on extreme left and right (s)e) reactants within each half-cell come first

9

2. So

anode half-cell cathode half-cell bridge

Zn(s)* Zn2+(aq) 2 Cu2+(aq)*Cu(s) _ a

phase boundary e! flow ÷ phase boundary

Try prob. 17.2, p. 704.

10

III. Cell Potentials & ÄGE? for Cell Reactions

A. What makes the e! move through the wire?

1. This force is an electrical potential called theelectromotive force (emf), a.k.a. cell potential (E). Galvanic E defined to be +. Unit = volt.

2. More on units:a) 1 J = 1 C × 1 V joules, coulombs, voltsb) 1 C = 1 amp for 1 secc) a) means 1 C moving across 1 V yields 1 J work

11

B. Relationship between free energy (G) and cellpotential (E):

ÄG = !nFE note the !sign

1. n = number of moles of e! transferred2. F = Faraday constant (96,486 C/mol e!) You will

confirm this value in the last lab.3. E = cell potential

C. As before, it is useful in comparing reactions todefine a specific set of standard conditions:

ÄGE = !nFEE Try prob. 17.5, p. 706.

12

IV. Standard Reduction Potentials (EE)

A. Sum of the standard half-cell potentials foranode and cathode is equal to the standardpotential for the cell:

EEcell = EEox + EEred

1. Example: Cu2+/H2 cell (comment re. H2 rxn)

EEcell = EEH2ÿH+ + EECu2+2ÿCu = 0.00... V + 0.34 V = 0.34 V

Why is EEH2ÿH+ expressed as 0.00...?

13

B. Think back to previous quantitative treatment ofG & H.

1. We can’t define these in absolute terms (contrast with ).

2. Therefore, we approached quantitative treatment of Gand H by examining changes: ÄG and ÄH.

3. The approach is similar for quantitative problems withE. In this case we define one of the half-reactions ashaving a numerical value of 0.000...V, and thenanalyze all other rxns relative to that one.

14

4. Reference rxn is the standard H electrode (S.H.E.):

2 H+ (aq, 1 M) + 2 e! ÿ H2 (g, 1 atm) EE = 0 V

H2 (g, 1 atm) ÿ 2 H+ (aq, 1 M) + 2 e! EE = 0 V

Try prob. 17.6, p. 709.

Al(s)*Al3+(aq) 2 Cr3+(aq)*Cr(s)

If EE = 0.92 V, look up EE for the Al3+'Al half-cell inTable 17.1 and then calculate EE for the Cr3+'Cr half-cell.

Scan table a bit.

15

16

V. Using EE Values

A. Because redox rxns all use e! (commoncurrency) we can relate these types of rxnsthrough a common scale.

1. For a redox table with 100 half-cell entries, we cancalculate

100 × 99 = 9,900 cell potentials

17

2. Example: Zn(s) + 2 Ag+(aq)

ÿ Zn2+(aq) + 2 Ag(s)

EEcell = EEox + EEred

EE

Oxidation half-cell:

Reduction half-cell:

Cell:

18

B. Remember: EE is an intensive property (doesn’tdepend on how many electrons), therefore, youdon’t multiply the Ag EE value by 2.

Derive logic from ÄG = !nFE

!ÄG'nF = E

ÄG (J/mol)'[n (mol) × F(C'mol)] = E

(J/mol)'C × [(V×C'J)] = V'mol = E

You can see that the potential E is in terms of volts per mole. Itdoes not matter if you have X# or Y# of moles, because it is thequantity per mole.

Try prob. 17.8, p. 711.

19

VI. Ecell & Composition of Rxn Mixture:Nernst

A. From Chap 16: amount of energy derived from asystem is related both to the energies of thereactants/products & their concentrations:

ÄG = ÄGE + RT ln QRecall that:

1. ÄGE free energy change under standard conditions.2. Q is the rxn quotient: [products]'[reactants], which

you can clearly control by how you set it up.3. ÄG therefore can vary quite a bit depending on the

way the system’s initial concentrations are set.

20

B. By direct substitution from V. B. above (ÄG = !nFE):

1. !nFE = !nFEE + RT ln Q divide by !nF

2. E = EE ! (RT'nF) ln Q the Nernst Equation

3. Because pH is based on the common log (base 10),the following version is often used:

E = EE ! (0.0592 V'n) log Q applies at 25E C

21

Try prob. 17.10, p. 713.

Cu(s) + 2 Fe3+(aq) ÿ Cu2+(aq) + 2 Fe2+(aq)

What is E for a cell at 25E C that has the [following]:

[Fe3+] = 1.0 × 10!4 M [Cu2+] = 0.25 M [Fe2+] = 0.20 M

22

Try Conceptual prob. 17.11, p. 714.

23

VII. Electrochemical Determination of pH What a pH meter does.

A. Cell description:

Pt*H2 (1 atm)* H+ (?M) 2 ref cathode

1. Overall cell potential: Ecell = EH2ÿH+ + Eref

2. Calculate E for the hydrogen half-cell rxn via Nernst:

H2(g) ÿ 2 H+(aq) + 2 e!

EH2ÿH+ = EEH2ÿH+ ! (0.0592 V'n) (log [H+]2'PH2)

24

Because the standard hydrogen cell potential isdefined to be zero and PH2 = 1 atm:

EH2ÿH+ = ! (0.0592 V'2) (log [H+]2)

Extracting the 2 terms & using the definition of pH:

EH2ÿH+ = ! (0.0592 V'2) 2 (log [H+])

EH2ÿH+ = (0.0592 V) (pH)

3. Now substitute this term back into:

Ecell = EH2ÿH+ + Eref

25

Ecell = ! (0.0592 V) (pH) + Eref

solve for pH: pH = ( Ecell ! Eref )' 0.0592 V

B. In practice, a glass electrode is used in place ofthe standard hydrogen electrode & the referenceelectrode used is the calomel electrode. Half-cellrxns are:

1. glass: 2 [Ag(s) + Cl!(aq) ÿ AgCl(s) + e!] EE =!0.22 V

Glass electrode dips into a dilute HCl soln. inside glassmembrane separating pH electrode from soln whosepH is to be measured. See Fig. 17.7, p. 715, below.

26

2. calomel: Hg2Cl2(s) + 2 e! ÿ 2Hg(l) + 2Cl!(aq) EE = 0.28 V

D. Based on the above two half-cell rxns, youmight (should?) ask, “How does pH fit in to theEcell?”

H+ outside of the thin glass membrane develops apotential relative to the H+ of the dilute HCl solution.

Try prob. 17.12, p. 716.

27

VIII. EE & Keq

A. We now want to examine the relationshipbetween standard cell potentials and equilibriumconstants.

28

B. Derivation of relationship:

1. As noted in III above: ÄGE = !nFEE

2. From Chap 17: ÄGE = !RT ln K

3. Combining 1 & 2: !nFEE = !RT ln K

4. As before, we can convert to log and set T = 25E C:

EE = (0.0592 V'n) log K

29

Graphically: See Fig. 17.8, p. 718.

Note that small voltage differences give large Kdifferences.

30

C. We have now added a third way to determine Kvalues. The different approaches are:

1. from measuring [solute] values: K = [prod]'[react]

2. from thermodynamic data: K = e!(ÄGE'RT)

3. from electrochemical data: ln K = nFEE'RT

Try prob. 17.13, p. 719.

31

IX. Batteries (“...most important practical application of galvanic cells.”)

A. Physics element. In a multicell battery, anumber of batteries are wired together in series (asopposed to in parallel). The voltage for theoverall battery is the sum of the voltages for theindividual cells.

Some examples of batteries:

32

B. Pb Storage Battery Anode rxn: oxidation of Pb(s)

1. Single cell has a voltage . 2 V. Six cells yield 12 V.

2. This battery can be readily recharged (run reactions innon-spontaneous direction) because the rxn product,PbSO4(s), adheres to the electrodes.

3. What part of a car normally recharges the battery andwhat is its energy source?

33

C. Dry-Cell Battery Anode rxn: oxidation of Zn(s)

1. A number of types of these are commercially in use.

2. The major differences are in the cathodic rxn.

3. Which type did the bunny (Eveready) peddle(originally)?

D. Ni-Cadmium Anode rxn: oxidation of Cd(s)

1. A.k.a.”Ni-cad”2. Can be recharge efficiently because the rxn products

adhere to the electrodes.

34

E. Lithium Battery Anode rxn: oxidation of Li(s)

1. Rechargeable

2. Relatively high voltage per cell (3 V)

3. Light weight (Have we seen this before in a differentcontext???)

35

F. Fuel cell (rxns below)

1. Anode: 2 H2(g) + 4 OH!(aq) ÿ 4 H2O(l) + 4 e!

2. Cathode: O2(g) + 2 H2O(l) + 4 e! ÿ 4 OH!(aq)

3. Overall: 2 H2(g) + O2(g) ÿ 2 H2O(l)

4. What are advantages and disadvantages for this rxn?

5. Have we seen this before in a different context?

36

X. Corrosion

A. In our society, probably the most importantcorrosion rxn is for Fe. See Fig. 17.12, p. 725:

B. Corrosion half-cell rxns:

1. Anode region: Fe(s) ÿ Fe2+(aq) + 2 e! EE = 0.45 V

2. Cathode region:

O2(g) + 4 H+(aq) + 4 e! ÿ 2 H2O(l) EE = 1.23 V

37

a) Note that under normal environmental conditions, [H+] ismuch less than 1 M.

b) Under normal conditions cathodic EE ~ 0.81 V.c) This means Ecell is still quite positive.d) What does c) say about spontaneity and whether the rxn

shown above goes toward products?

38

C. Why don’t metals such as Al and Ti, which areelectrochemically as vulnerable as Fe corrode?

39

D. Preventing Corrosion

1. Shielding with paint, but what if paint chips?2. Galvanizing with Zn. (See Fig. 17.13, p. 715.)

a) The metal you want to preserve (normally Fe) is coated with Zn. b) From the reduction potentials, you can see that any Fe that is oxidized will be re-

reduced by the Zn, yielding Zn2+, which eventually forms ZnCO3, which adheresrelatively tightly to the metal surface.

3. Cathodic protection (the sacrificial anode). a) Involves long range e! movement in the metal. b) Anode material must eventually be replenished.

40

We won’t cover the Chapter 17 topics shown below during spring 2016.

41

XI. Electrolysis and Electrolytic Cells

A. Electrolysis of molten NaClProducts obtained are? and

B. E l ectrolysis of water.Products obtained are? and

XII. Commercial Applications of Electrolysis

A. Production of Na(s) from NaCl

B. Production of Cl2 and NaOH

C. Production of Al(s)

1. Large current requirement: 1 mol of e! produces only 9 g of Al(s). (Can you derive thatnumber?)

2. Electrolytic production of Al is the largest single process consumer of electricity inU.S.A. today.

3. Comment re. Al recycling.

D. Electrorefining and Electroplating Examples of:

42

1. Chrome plating of auto components2. Ag/Au plating of jewelry, dining implements, musical instruments, etc.

XIII. Quantitative Aspects of Electrolysis

Interesting, but we don’t have enough time to cover.

43