1 © Prof. Zvi C. Koren 20.07.2010

Electrochemistry & Electrochemical Cells (Batteries)

Zn

Cu2+

Cu2+

Cu2+ Cu2+

Cu2+

Zn2+ Cu2+

Zn2+Zn2+

Zn

Cu

Cu

Reduction Half-Rxn.:

Oxidation Half-Rxn.:

Cu2+ + 2e– Cu(s)

Zn(s) Zn2+ + 2e–

Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq)Net-Ionic Redox Reaction:

Introduction: Direct Reaction(no useful work is being done; E = q + wEnergy released by the rxn. Is dissipated as heat, q)

Spontaneous Rxn.: G < 0

Zn

Electricity ↔ Chemistry(A Chicken-and-Egg Question: “Who Came First?”

2 © Prof. Zvi C. Koren 20.07.2010

Galvanic Cell (also Voltaic or Self-Driven Cell)Indirect Reaction: Daniell Cell with Porous Semi-Permeable Membrane

Cathode: Reduction Half-Rxn.:

Anode: Oxidation Half-Rxn.:Cu2+ + 2e– Cu(s)Zn(s) Zn2+ + 2e–

Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq)Cell Reaction:

Cu2+

e– e–

Anode Half-Cell Cathode Half-Cell

SO42–

external circuitvoltmeter

External circuit:Electronic

conduction by free electrons

Internal circuit:Electrolytic

conduction by electrolytes

(ions)

poroussemi-permeable

membrane

Red CatAn Ox “ חים-א "

An

od

e

Cat

ho

de– +

Electrode:Object that conducts

(releases/collects) electrons.

Two Electrodes:Oxidation always

occurs at the Anode.Reduction always

occurs at the Cathode

Spontaneous Rxn.: G < 0

Cu

Questions: Why is the Anode (–)? Why do the ions diffuse, migrate, to the electrodes?

“Anion”

Zn2+

Zn

“Cation”

Copper-plating

(Energy released by the rxn.can perform useful work!)

3 © Prof. Zvi C. Koren 20.07.2010

Cu2+

e– e–

Anode Half-Cell Cathode Half-Cell

SO42–

An

od

e

Cat

ho

de– +

Cu

Zn2+

Zn

Daniell Cell with Salt Bridge

Na+

Note: Salt ions must not themselves undergo reduction or oxidation

poroussemi-permeable

membrane

Indirect Reaction

NO3–

4 © Prof. Zvi C. Koren 20.07.2010

Line Notation(Shorthand Notation for Electrochemical Cells)

electrode sp.1(aq), sp.2(aq) … sol. liq. gas sp.1(aq), … sol. liq. gas electrode

Anode half-cell Cathode half-cell

Notes:1. Only the species appearing in each net-ionic half rxn. are listed.2. Species in same solution are separated by a comma.3. Species in different phases are separated by a vertical line.4. The first half-cell listed is always for the Anode.5. The electrodes are listed at the extremes.6. A double line is used if there is a physical separation between the two half-

cell compartments.

Example for the Daniell cell (with and without the salt bridge cells before):

Zn Zn2+ Cu2+ Cu

Question: How would you design an electrochemical cell for which the net-ionic rxn. is:

Ni + 2Ag+ Ni2+ + 2Ag

e–

5 © Prof. Zvi C. Koren 20.07.2010

Cell Potential (ε), units in Voltse–

e–

voltmeter

–+

An

od

e

Cat

ho

de

HIGHelectric

potential lowelectric

potential

The existence of apotential difference

between the two electrodes is the

“driving force” ,also known as the

electromotive force (emf),for the electron flow, and

thus for the rxn.

Electrical Work (Electrical “Energy”), wel

1 J of energy is released when 1 C of charge passes through a potential drop of 1 V:1 J 1 VC C Coulomb, קולון

In general: Q charge (in C) = neFne # of moles of electronsF Faraday constant = 96,485 C/mol e

wel = – εQ

Thermodynamics, ΔGIf wel is the maximum work that the system can perform, then it also represents the “available energy”: ΔG = wel = – neFε

For a spontaneous rxn. (“self-driven” cell, battery): ε > 0, ΔG < 0

6 © Prof. Zvi C. Koren 20.07.2010

Measuring Standard Half-Cell Potentials Standard Cell Potential.

Standard conditions: [solute] = 1.0 M, Pgas = 1.0 atm

εcell = εanode + εcathode

εoxidation + εreduction

For example, for the Daniell cell:

Anode (Oxidation): Zn Zn2+ + 2e–, εA,ox

Cathode (Reduction): Cu2+ + 2e– Cu, εC,red

Cell Reaction: Zn + Cu2+ Zn2+ + Cu, εcell,redox

Note: ε is an intensive property; it depends on concentration and not on amount.(Other examples are: d, T, P, M)

Thus, if a half-rxn is multiplied by a number, its half-cell potential is not multiplied by that number.

For example: A: Zn Zn2+ + 2e–, εox

C: 2(Ag+ + e– Ag), εred (and not 2 εred )

Cell: Zn + 2Ag+ Zn2+ + 2Ag, εcell = εox + εred

7 © Prof. Zvi C. Koren 20.07.2010

Standard Hydrogen Electrode (“SHE”)

[H+] = 1.0 MPH2

= 1.0 atm

If SHE is the Cathode:2H+ + 2e– H2(g)H+(1 M)H2(1 atm)Pt

If SHE is the Anode:H2(g) 2H+ + 2e–

PtH2(1 atm) H+(1 M)

V 0.0 εo

SHE

Standard Conditions:

SHE Unknown

M

H+

Pt

Pt

Mm+

0H2(g)1 atm

“Platinum black” is a Pt film plated onto Pt metal:Advantages: Pt is inert (does not corrode);

Pt film increases surface area and serves as a catalyst for the rxn.;adsorbs H2(g) onto it.

Problem: Given the following cell: Zn Zn2+ H+ H2 Pt, V 0.76 εo

cell Determine the half-rxns at the electrodes, the cell rxn, wel, , .

o

Znεo

rxn ΔG

8 © Prof. Zvi C. Koren 20.07.2010

Saturated Calomel Electrode (SCE)

Replaces the technically difficult SHE as a reference for electrode potential measurements of other half-cells.

4.5 M

Hg2Cl2(s) + 2e– 2Hg(l) + 2Cl–, εSCE,25 oC = 0.24 V (Note: this is not εo)

Crystal structure of Hg2Cl2

“calomel” is an old name for mercury(I) chloride

9 © Prof. Zvi C. Koren 20.07.2010

Half-Reaction ε0 (V)Li+(aq) + e- -----> Li(s) –3.05K+

(aq) + e- -----> K(s) –2.93Ba2+

(aq) + 2 e- -----> Ba(s) –2.9Sr2+

(aq) + 2 e- -----> Sr(s) –2.89Ca2+

(aq) + 2 e- -----> Ca(s) 2.87–Na+

(aq) + e- -----> Na(s) –2.71Mg2+

(aq) + 2 e- -----> Mg(s) –2.37Be2+

(aq) + 2 e- -----> Be(s) –1.85Al3+

(aq) + 3 e- -----> Al(s) –1.66Mn2+

(aq) + 2 e- -----> Mn(s) –1.182 H2O + 2 e- -----> H2(g) + 2 OH-

(aq) –0.83Zn2+

(aq) + 2 e- -----> Zn(s) –0.76Cr3+

(aq) + 3 e- -----> Cr(s) –0.74Fe2+

(aq) + 2 e- -----> Fe(s) –0.44Cd2+

(aq) + 2 e- -----> Cd(s) –0.4PbSO4(s) + 2 e- -----> Pb(s) + SO4

2-(aq) –0.31

Co2+(aq) + 2 e- -----> Co(s) –0.28

Ni2+(aq) + 2 e- -----> Ni(s) –0.25

Sn2+(aq) + 2 e- -----> Sn(s) –0.14

Pb2+(aq) + 2 e- -----> Pb(s) –0.13

2 H+(aq) + 2 e- -----> H2(g) 0

Sn4+(aq) + 2 e- -----> Sn2+

(aq) 0.13

Cu2+(aq) + e- -----> Cu+

(aq) 0.13

SO42-

(aq) + 4 H+(aq) + 2 e- -----> SO2(g) + 2 H2O 0.2

AgCl(s) + e- -----> Ag(s) + Cl-(aq) 0.22

Cu2+(aq) + 2 e- -----> Cu(s) 0.34

O2(g) + 2 H2 + 4 e- -----> 4 OH-(aq) 0.4

I2(s) + 2 e- -----> 2 I-(aq) 0.53

MnO4-(aq) + 2 H2O + 3 e- -----> MnO2(s) + 4 OH-

(aq) 0.59

O2(g) + 2 H+(aq) + 2 e- -----> H2O2(aq) 0.68

Fe3+(aq) + e- -----> Fe2+

(aq) 0.77

Ag+(aq) + e- -----> Ag(s) 0.8

Hg22+

(aq) + 2 e- -----> 2 Hg(l) 0.85

2 Hg2+(aq) + 2 e- -----> Hg2

2+(aq) 0.92

NO3-(aq) + 4 H+

(aq) + 3 e- -----> NO(g) + 2 H2O 0.96

Br2(l) + 2 e- -----> 2 Br-(aq) 1.07

O2(g) + 4 H+(aq) + 4 e- -----> 2 H2O 1.23

MnO2(s) + 4 H+(aq) + 2 e- -----> Mn2+

(aq) + 2 H2O 1.23

Cr2O72-

(aq) + 14 H+(aq) + 6 e- -----> 2 Cr3+

(aq) + 7 H2O 1.33

Cl2(g) + 2 e- -----> 2 Cl-(aq) 1.36

Au3+(aq) + 3 e- -----> Au(s) 1.5

MnO4-(aq) + 8 H+

(aq) + 5 e- -----> Mn2+(aq) + 4 H2O 1.51

Ce4+(aq) + e- -----> Ce3+

(aq) 1.61

PbO2(s) + 4 H+(aq) + SO4

2-(aq) + 2 e- -----> PbSO4(s) + 2 H2O 1.7

H2O2(aq) + 2 H+(aq) + 2 e- -----> 2 H2O 1.77

Co3+(aq) + e- -----> Co2+

(aq) 1.82

O3(g) + 2 H+(aq) + 2 e- -----> O2(g) + H2O 2.07

F2(g) + 2 e- -----> F-(aq) 2.87

(strong oxidizing agent)

Standard Reduction Potentials at 25 oC

o

redε

(strong reducing agent)Great interactive table in Wikipedia data:http://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)

10 © Prof. Zvi C. Koren 20.07.2010

Alphabetical List

11 © Prof. Zvi C. Koren 20.07.2010

Problems (consult Table of Standard Half-Cell Reduction Potentials):

Question: Can Cu2+ oxidize H2 with all species in their standard state?(Show all reactions.) Answer: Yes; εo

rxn = +0.34 V.

Question: Can Ag reduce Cu2+ with all species in their standard state?(Show all reactions.) Answer: No; εo

rxn = –0.46 V

What happens to cell with time?

time

time

or

12 © Prof. Zvi C. Koren 20.07.2010

Inert Electrodes:Pt, Au, C(graphite)

Problem: Show a diagram of the following cell and all the relevant reactions,and calculate the standard cell potential:

Pt Fe2+, Fe3+ MnO4–, Mn2+, H+ C

Mn2+

e– e–

MnO4–

An

od

e

Cat

ho

de–

+

C

Fe2+

Pt

NO3–

Na+

Fe3+

H+

5(Fe2+ Fe3+ + e–), oAn = –0.77 V

5e– + 8H+ + MnO4– Mn2+ + 4H2O, o

Cath = 1.52 V

5Fe2+ + 8H+ + MnO4– 5Fe3+ + Mn2+ + 4H2O, o

cell = 0.75 V

13 © Prof. Zvi C. Koren 20.07.2010

Nernst Equation: Non-Standard Cell PotentialsThe calculation of and G from o and Go, respectively.Consider the following overall rxn.: aA + bB cC + dD

From Lewis:G = Go + RTlnQ (will be proved in the Thermodynamics course)

Reaction Quotient for solutes and gases: Q = [C]c[D]d/[A]a[B]b

For example, for the previous rxn.: 5Fe2+ + 8H+ + MnO4– 5Fe3+ + Mn2+ + 4H2O

Q = [Fe3+]5[Mn2+]/[Fe2+]5[H+]8[MnO4–]

ΔG = wel = – neFεFrom before: ΔGo = – neFεo

ε = εo – (RT/neF)lnQ Nernst Eqn.: ε conc.

At equilibrium, ε = 0: εo = (RT/neF)lnKeq

Could also prove this from Geq = 0: Go = –RTlnK

Problem: Calculate εcell and Grxn for the following cell at 25 oC (and show all rxns.):

Au AuCl4–(0.50 M), Cl–(2.0 M) ClO3

–(0.60 M), ClO4–(0.80 M), H+(3.0 M) Pt

Answer (partial): Q = [AuCl4–]2 [ClO3

–]3 / [Cl–]8 [ClO4–]3 [H+]6 = 5.65 x 10–7.

εocell = 0.19 V; ne = 6. εcell = 0.25 V (spontaneous rxn.)

(Note: εocell could be –, but the actual εcell could be +.)

Q has the same expression as K, but Q is for any time, whereas K is only for equilibrium.

ε = (RT/neF)ln(Keq /Q)

14 © Prof. Zvi C. Koren 20.07.2010

The permanganate ion is a powerful oxidizing agent that oxidizes water to oxygen under acidic conditions, while it gets reduced to Mn2+.

What is the potential of a permanganate-oxygen cell operating at pH = 7.00, with oxygen at 0.200 atm, 0.100 M MnO4

–, and 0.100 M Mn2+?

Another Nernst Eqn. Problem (Note : mixed concentrations and pressures):

Measuring Ksp Potentiometrically

Recall those very low values for Ksp, even 10–50 ? How were they measured?

For example,PbCl2(s) Pb2+ + 2Cl–, Ksp = [Pb2+][Cl–]2 From before: εo = (RT/neF)lnKeq

We can “create” this chemical equation from the following two half-rxns.:

Pb(s) Pb2+ + 2e–, εoox =

PbCl2(s) + 2e– Pb(s) + 2Cl–, εored =

PbCl2(s) Pb2+ + 2Cl–, εocell =

Thus: εocell = (RT/neF)lnKsp

Note: A non-redox rxn. was “created” from a reduction and oxidation half-rxns.

e– e–

Pb2+

PbAn

+–

PbCl2

PbCatNa+NO3

–

Cl–

15 © Prof. Zvi C. Koren 20.07.2010

ElectrolysisGalvanic cell, Voltaic cell, Self-Driven cell (“battery”): Chemistry ElectricityElectrolytic cell (in an electrolysis process): Electricity Chemistry

Electrolysis of H2Oe–

–

+

“Anion”Na+

Pt Pt

SO42–

An

od

e

Ca

tho

de

+

–

“Cation”

An. (Ox.): 2H2O(l) O2(g) + 4H+ + 4e–, εo = –1.23 V

Cat. (Red.): 2(2H2O(l) + 2e– H2(g) + 2OH–), εo = –0.83 V

Cell: 6H2O(l) O2(g) + 2 H2(g) + 4H+ + 4OH–, ocell = –2.06 V

2H2O(l) O2(g) + 2 H2(g)O2(g)

H2(g)

In an electrolytic cell, a power source – a battery – is used.The battery acts like a “bully” with “two strong arms”:

takes e’s with one and gives with the other

e–

battery

Note the different signs at the electrodes.But Oxidation always occurs at the Anode, no matter what the cell type.

16 © Prof. Zvi C. Koren 20.07.2010

Electrolysis of Pure Molten NaCl

NaCl(s) Na+(l) + Cl–(l)

e– e–

battery

–

+

“Anion”Na+

Pt Pt

Cl–

An

od

e

Ca

tho

de

+

–

“Cation”

An (ox): 2Cl–(l) Cl2(g) + 2e–

2(Na+(l) + e– Na(l))

This is a pure molten liquid, a “melt”, not in an aqueous solution.

2Na+(l) + 2Cl–(l) 2Na(l) + Cl2(g), ocell = –4.07 V

Need to keep the products separated to prevent them reacting (explosively) with each other.

SchematicLab

Diagram

SchematicIndustrialDiagram:

Downs Cell

Na(l) floats through the molten NaCl(l) and is collected

17 © Prof. Zvi C. Koren 20.07.2010

If there are competing half-rxns., the one with the higher will be the most favored.

Electrolysis of Aqueous NaCl

What are the possible reduction rxns.?

Na+(aq) + e– Na(s), εo = –2.71 V

2H2O(l) + 2e– H2(g) + 2OH–, εo = –0.83 V

NaCl(aq) Na+(aq) + Cl–(aq)

What are the possible oxidation rxns.?

2Cl–(aq) Cl2(g) + 2e–, εo = –1.36 V

2H2O(l) O2(g) + 4H+ + 4e–, εo = –1.23 V (So, is it really a bit easier to oxidize H2O)

But , note the following:1. Non-Standard State effects favor Cl– oxidation: The cell is never allowed to reach

standard-state conditions. The solution is typically 25% NaCl by mass, which significantly decreases the potential required to oxidize the Cl – ion.

2. pH Effects do not favor oxidation of H2O: The pH of the cell is also kept very high (see H2O reduction above), which decreases the oxidation potential for water.

3. Overvoltage: This is the deciding factor. It is the extra voltage that must be applied to a reaction to get it to occur at the rate at which it would occur in an ideal system.

18 © Prof. Zvi C. Koren 20.07.2010

So the actual rxns. are: Cat (Red): 2H2O(l) + 2e– H2(g) + 2OH–(aq) An (ox): 2Cl–(aq) Cl2(g) + 2e–

Cell: 2Cl–(aq) + 2H2O(l) Cl2(g) + H2(g) + 2OH –(aq)

Electrolysis of Aqueous NaCl (continued)

Compare with the electrolysis of molten NaCl: 2Na+(l) + 2Cl–(l) 2Na(l) + Cl2(g)

Because the demand for chlorine is much larger than the demand for sodium, electrolysis of aqueous sodium chloride is a more important process commercially. Electrolysis of an aqueous NaCl solution has two other advantages. It produces H2

gas at the cathode, which can be collected and sold. It also produces NaOH, which can be drained from the bottom of the electrolytic cell and sold.

If the porous barrier is removed, the products of the electrolysis of aqueous sodium chloride react to form sodium hypochlorite, which is the first step in the preparation of hypochlorite bleaches, such as Chlorox.

Cl2(g) + 2OH–(aq) Cl–(aq) + OCl–(aq) + H2O(l)

OH–

19 © Prof. Zvi C. Koren 20.07.2010

Electroplating

• The process of depositing one metal on top of another.

• Done usually for cosmetic reasons.

20 © Prof. Zvi C. Koren 20.07.2010

Faraday’s Laws: Electrochemical Stoichiometry

If 96,485 C of charge is passed through a cell,equivalent amounts react and are produced

The charge on 1 mol of e’s is 96,485 C and an “equivalent” amount of any substance is that quantity that reacts with or is produced from 1 mol of e’s in a redox rxn.:

96,485 C ↔ 1 mol e’s ↔ 1 eq. of substanceIn short, use stoichiometry on the half-rxns. or the overall rxn. (example to follow)

Some Relevant Electrical PropertiesEquationUnitSymbolProperty

s, secondtTime

V, Volt, EPotential

C, CoulombQCharge

I = Q/tA, AmpereICurrent

= IR, ohmRResistance

P = wel/tW, WattPPower

wel = QJ, Joule

welEnergy, work wel = Pt(1000 W)(3600 s)

1 kWh = 3.6 MJkilowatt-hour, kWh

21 © Prof. Zvi C. Koren 20.07.2010

Faraday Problems

Question:If 1.50 A of current flows through an aqueous solution containing CuCl2 for 15.0 min, how many grams of Cu(s) are deposited at the _____ode?

Answer:Write the relevant net-ionic half-rxn.: Cu2+ + 2e– Cu(s)Use the Unit-Factor Method for the Electrical Stoichiometry:

Cu g _________ Cu mol 1

Cu g 546.63

e mol 2

Cu mol 1

C 96,485

e mol 1s .900

s

C 5.1

Q=It

Question:In the electrolysis of CuCl2(aq) above, the _____tion of Cl– to Cl2(g) occurs at the _____ode. ? Assume that standard state conditions apply with the current and time from the problem above, how much electrical energy (in J and in kWh) and power (in W) were needed?

Answer:

ocell wel = Q P = wel/t

22 © Prof. Zvi C. Koren 20.07.2010

Problem:

If the electrolytic cell for the electrolysis of water draws a current of 0.775 A for 45.0 minutes, calculate the volumes of H2 and O2 produced if each gas is collected at 25 oCand P = 1.00 atm.

Problem:

An electroplating apparatus is used to coat jewelry with gold. What mass of gold can be deposited from a solution that contains Au(CN4)– ions if a current of 5.0 A flows for 30.0 min?

23 © Prof. Zvi C. Koren 20.07.2010

Corrosion (שיתוך)

The oxide of a metal is more stable than the metal itself:M(s) + O2(g) ---> MxOy(s), G < 0 (spontaneous)

So all metals naturally “want” to be oxidized.

Rusting of Fe:Anode: Fe Fe 2+ + 2e–. Also: Fe2+ Fe3+ + e–

Cathode (also Fe, but inert): 2H2O + 4e– 4OH–

O2 + 4H+ + 4e– 2H2O (other rxns. are possible)Also: 2Fe + 1½O2(g) Fe2O3(s); Fe3+ + 3OH– Fe(OH)3(s)

Galvanized Iron (Plated with Zn):Zn is the “sacrificial” anode; easier to oxidize Zn than Fe (check it!).Fe is now “forced” to be the cathode, but of course an inert one:Anode: Zn Zn 2+ + 2e–

Cathode: 2H2O + 4e– 4OH–

Aluminum:Al undergoes “passivation”: 2Al + 1½O2(g) Al2O3(s)Once the oxide is formed, it serves as a thin film that prevents any more Al to come into contact with the environment and react.

24 © Prof. Zvi C. Koren 20.07.2010

Different Types of Batteries

Cell: Zn + 2MnO2 ZnO + Mn2O3, ocell = 1.54 V

Alkaline Battery (“Dry Cell”):An(-): Zn + 2OH– ZnO + H2O + 2e–

Cat(+): 2MnO2 + H2O + 2e– Mn2O3 + 2 OH–

25 © Prof. Zvi C. Koren 20.07.2010

Mercury Battery:An: Zn + 2OH– ZnO + H2O + 2e– (as alkaline, above)Cat: HgO + H2O + 2e–

Hg(l) + 2 OH–

26 © Prof. Zvi C. Koren 20.07.2010

Zinc-Air Battery

• Zinc – Air Battery – deteriorates quickly once exposed to air, Zn at the anode and O2 at the cathode.

• Used in watches, calculators, hearing aids and cameras.

Ni-Cad:An: Cd + 2OH– Cd(OH)2 + 2e–

Cat: 2[NiO(OH) + H2O + e– Ni(OH)2 + OH–]

Storage Batteries – Rechargeable:

ocell = 1.35 V

Rechargeable because the nickel and cadmium hydroxides adhere tightly to the electrodes.

27 © Prof. Zvi C. Koren 20.07.2010

Lead Storage Battery:An: Pb + SO4

2– PbSO4 + 2e–

Cat: PbO2 + 4H+ + SO42– + 2e– PbSO4 + 2H2O

28 © Prof. Zvi C. Koren 20.07.2010

Problem:

Automobile headlights typically draw 5.9 A of current. The galvanic cell of a lead storage battery consumes Pb and PbO2 as it operates. A typical electrode contains about 250 g of PbO2. Assuming that the battery can supply 5.9 A of current until all the PbO2 has been consumed, how long will it take for a battery to run down if the lights are left on after the engine is turned off?

Hydrogen Fuel Cell (70 – 140 oC):An: 2H2(g) + 4OH– 4H2O + 4e–

Cat: O2(g) + 2H2O + 4e– 4OH–

Fuel Cell is an electrochemical cell that produces electricity from a fuel tank. The electricity is generated through the reaction, triggered in the presence of an electrolyte, between the fuel (on the anode side) and an oxidant (on the cathode side). The reactants flow into the cell, and the reaction products flow out of it, while the electrolyte remains within it. Fuel cells can operate virtually continuously as long as the necessary flows are maintained.Fuel cells are different from conventional electrochemical cell batteries in that they consume reactant from an external source, which must be replenished – a thermodynamically open system. By contrast, batteries store electrical energy chemically and hence represent a thermodynamically closed system.

29 © Prof. Zvi C. Koren 20.07.2010

The pH Meter Consists of a “Combination” Electrode

pH –logaH+

aH+ = H+ MH+

Kw = aH+aOH– /aH2O

The net effective rxn. is:

H+(inside glass electrode)↕

H+(in the unknown solution)

Nernst:

= o – (RT/neF)ln([H+]outside/[H+]inside)

= o + 2.303(RT/neF)(pHoutside – pHinside)pHinside is constant, so

= o’ + 2.303(RT/neF)pHoutside

pH meter is calibrated with the pH of known

buffer solutions, so o’ is cancelled.

2.303(RT/neF)pHoutside

1.98x10–4TpHoutside

These two electrodes are combined into one “combo” electrode

This combo electrode is inserted into a solution with unknown pH