21elecchem3weq

21 Electrochemistry 1

21 ElectrochemistryElectricity is such a mystery till we understand chemistry. Plain facts and sciences of electrochemistry continue to be important in technologies.


ElectricityAncient people noticed electricity

1746 B. Franklin demonstrated lightening as electric effect and performed the kite experiment in 1751. Two people tried to repeat his kite flying experiment were killed by thunder.

1767 L. Galvani inserted two different metals in frog fluid and constructed a electric cell

1800 A. Volta substituted frog fluid; made batteries, consisted of several cells.

1802 G. Romagnosi noticed magnetism related to electricity

Michael Faraday 1791-1867 discovered many theories of electricity and magnetism


Galvani

Luigi Galvani (1737-1798): erroneously concluded that the frog's nervous system generated an electrical charge, his work stimulated much research into the electrochemistry.

The depiction of his laboratory


Electrochemistry

A. Volta (1745-1827) experimented with different materials, and made voltaic piles (batteries)

William Nicholson (1753-1815) observed bubbles forming on the surfaces of metals submerged in water when they are connected to a voltaic pile

Humphry Davy (1778-1829) observe electrolysis of water and metal salts. Following that, … Michael Faraday (1791-1860) studied electrolysis, and discovered the relationship between charges and chemical stoichiometry


Electrons

Voltaic piles (batteries) made the following study possible

W. Crookes (1832-1919) observed cathode rays in low-pressure tubes.

1897: J.J. Thomson determined the charge to mass ratio (e– / me) of cathode rays (electrons).

1916 R. Millikan (1868-1953) measured the amount of charge of e–.

qe = –1.60217733e-19 C F = 96485 C me = 0.00054856 amu = 9.1093897e-31 kgspin = ½ (two state) magnetic moment = 9.284770e–24 J/tesla


Redox reactions and electronsEnergy drives chemical reactions.

Redox reactions involve the transfer of electrons.

Loss of electron (increase oxidation state) is oxidation, (Leo).Gain of electron (decrease oxidation state) is reduction, (ger).

Zn Zn2+ + 2 e– leo Cu2+ + 2 e – Cu ger

net: Zn + Cu2+ Cu + Zn2+ redox

Chemical energy in redox reactions may be convert to electric energy by applying electrochemistry.

In the meantime, we should learn to balance the redox reaction equations.


Galvanic CellA galvanic cell consists of two different metals inserted into a solution of an electrolyte (salt, acid or base), simulated

Representation:Zn | Zn2+ || Cu2+ | Cu


Assign oxidation states–3 NH3

–2 N2H4

–1 NH2OH

0 N2

+1 N2O

+2 NO

+3 NO2–

+4 NO2

+5 NO3–

0 for any element

1 for H in compounds, but –1 for LiH, NaH, etc

– 2 for O in compounds, but –1 for H2O2, Na2O2

+1 for alkali metals, +2 for alkaline earth metals

The oxidation states of other elements are then assigned to make the algebraic sum of the oxidation states equal to the net charge on the molecule or ion.

–1 Cl–

0 Cl2

+1 ClO–

+3 ClO2–

+4 ClO2

+5 ClO3–

+7 ClO4–


Half reaction equationsOxidation and reduction can be written as half-reaction equations such as

Zn Zn2+ + 2 e– leo

Cu2+ + 2 e – Cu ger

net: Zn + Cu2+ Cu + Zn2+ redox

Demonstrate how to balance these

Fe2+ Fe3+ + __ e–

C2O42- 2 CO2 + __ e –

MnO4 – + __ e– Mn2+

Cr2O72– + __ e– + __ H+ 2 Cr3+ + __ H2O

Steps to balance half reaction

Assign oxidation number

Figure out what is oxidized or reduced.

Add electrons according to oxidation number change

Balance charge with H+ (acid) or OH – (base)

Balance atoms with H2O


More half-reaction equations 2 I– I2 + __ e–

ClO2 + __ OH– ClO3- + __ e– + H2O (in basic solution)

2 S2O32– S4O6

2- + __ e–

HS(=S)O3– S + __ e- + HSO4

–

__ H3O+ + __ e – H2(g) + __ H2O

H2O2 + __ e – 2 H2O

ClO2 + __ e – ClO2–

NO3– + __ e – NH4

+


Electrochemical SeriesAn electrochemical series is a list of metals in order of decreasing strength as reductant, or increasing strength as oxidant.An example of an activity series of metals based on the Standard Potentials given would be:

K > Ba > Ca > Na > Mg > Al > Mn > Zn > Fe > Ni > Sn > Pb > Cu > Ag

In this series the most active metal is potassium (K) and the least active metal is silver (Ag)

Reactions and cells are illustrated in 21-1 of Text (PHH).


Constructing half cellsA half cell consists of an oxidizing and its oxidized species

Zn | Zn2+

Cu | Cu2+

Pt | H2 | H+ (Pt as conductor)Pt |Fe2+ , Fe3+

Cl– | Cl2 | Pt

Student cell set from School Master Science $30

Explain the cell convention and reactions of cells.


Galvanic cellsUsing corns for a galvanic cells is illustrated by an Internet site: (schoolnet.ca/general/electric-club/e/page9.html)

This picture illustrates a way to make a pact of battery using coins of different metals.

Apply the principles you have learned regarding electrochemical series and electrolytes to make such a pile will be an interesting exercise.


Cell convention Oxidation takes place always at the anode

Zn (s) Zn2+ (aq) + __ e– Zn(s) | Zn2+(aq)

Fe2+ (aq) Fe3+ (aq) + e– Pt | Fe2+ , Fe3+

H2 (g) 2 H+ (aq) + __ e– Pt | H2(g) | H+(aq)

Reduction takes place at the cathodeCu2+ (aq) + __ e– Cu (s) Cu2+ | Cu(s)

Cl2 (g) + __ e– Cl– (aq) Cl2(g) | Cl –(aq) | Pt

Fe3+ (aq) + __ e – Fe2+ (aq) Fe3+(aq), Fe2+(aq)| Pt

2 H+ (aq) + __ e– H2 (g) H+(aq) | H2(g)| Pt

Concentration (1.1 M) and pressure (0.9 atm) are also included in cell notations.


Electric energy and workElectric energy or electric work = charge * potential difference

W = q * V (1 J = 1 Coulomb Volt, C V)compare W = m g h

The Faraday constant F is the charge for one mole of electrons,F = 96485 C; F / NA = 96485 C / 6.022e23 = 1.602177e-19 C, charge per e–

The maximum chemical energy of the cell that can be converted to electric work is the Gibbs free energy change, G

G = – n F E (n F = q is the charge) electromotive force (emf, or V)number of electrons in the reaction equationn F is the charge qSee slides in

16 Equilibria


Standard cell emf’s and electrode potentials

The standard cell emf is the emf of a voltaic cell operating under standard conditions (1 M, 1 atm, 25oC etc). E.g.

Zn (s) | Zn2+ (aq, 1 M) || Cu2+ (aq, 1 M) | Cu (s) Eo = 1.10 V

Mg (s) | Mg2+ (aq, 1 M) || Cu2+ (aq, 1 M) | Cu (s) Eo = 2.90 V

The absolute potential of the electrode cannot be determined. Only relative potential can be measured. The standard reduction potential is measured against a SHE, for which Eo = 0.0000 V

Zn (s) | Zn2+ (aq, 1 M) || H+ (aq, 1 M) | H2 (1 atm)(g) Eo = 0.76 V

Zn = Zn2+ + 2 e– (oxidation, displace H+ possible) Eo = 0.76 V

Zn2+ + 2 e– Zn (reduction) Eo = – 0.76 V


Gibb’s Free Energy in a CellHow much energy is available for the cell

Zn | Zn2+ || Ag+ | Agoperating at standard condition when one mole of Zn is consumed ?

Solution Eo

Zn = Zn2+ + 2 e 0.762 V 2 Ag+ + 2e = 2 Ag 0.799 V (from table) Zn + 2 Ag+ = Zn2+ + 2 Ag Eo = 1.561 V

Go = – n F E = – 2 * 96485 C * 1.561 V = 301226 J (1J = 1 C V) = 301.2 kJ

How much silver is consumed?How much energy is available if 6.5 g of Zn is consumed?


Table of standard reduction potential Reaction E o (V)

Li+ + e– = Li (s) – 3.04Na+ + e– = Na (s) – 2.71

Mg2+ + 2 e– = Mg (s) – 2.38Zn2+ + 2 e– = Zn (s) – 0.76

(reference) 2 H+ + 2 e– = H2 (g) 0.000 (reference) H2 (g) = 2 H+ + 2 e– 0.000

Cu2+ + 2 e– = Cu (s) 0.34Cu+ + e– = Cu (s) 0.52

I2 (s) + 2e– = 2 I– (aq) 0.54Br2 (l) + 2 e– = 2 Br– (aq) 1.07Cl2 (g) + 2 e– = 2 Cl– (aq) 1.36

F2 (g) + 2 e– = 2 F– (aq) 2.87

Standard cell potentials

Cell E o

Li | Li+ || Cu2+ | Cu ____

Mg | Mg2+ || I2 | I– | Pt ____

Zn | Zn2+ || Br2 | Br– | Pt ____

Cu | Cu2+ || Zn2+ | Zn ____

Which is not spontaneous?

See 21-2


Table of standard reduction potential Reaction E o (V)

F2 (g) + 2 e– = 2 F– (aq) 2.87Cl2 (g) + 2 e– = 2 Cl– (aq) 1.36Br2 (l) + 2 e– = 2 Br– (aq) 1.07

I2 (s) + 2e– = 2 I– (aq) 0.54Cu+ + e– = Cu (s) 0.52

Cu2+ + 2 e– = Cu (s) 0.34(reference) H2 (g) = 2 H+ + 2 e– 0.000

(reference) 2 H+ + 2 e– = H2 (g) 0.000Zn2+ + 2 e– = Zn (s) – 0.76

Mg2+ + 2 e– = Mg (s) – 2.38Na+ + e– = Na (s) – 2.71

Li+ + e– = Li (s) – 3.04

The listing order in the table may be different in different text books. However, the principles and methods of application remain the same.

This is the order given on the Exam Data Sheet, that is different from the text.


Strength of oxidation

The ability of a chemical to oxidize is its ability to take electrons from other species,

Oxidizing agent + n e Reduced species

Strength of oxidation of an oxidizing agent is measured by its reduction potential.

Similarly, strength of reduction of a reducing agent is measured by its oxidation potential.

Oxidized species Reducing agent + n e

Be able to order the species according to oxidizing strength. _____


Reaction direction and emf

What is the emf for the reaction, Zn2+ (aq) + 2Fe2+ (aq) = Zn (s) + 2Fe3+ (aq)?

Solution: Know what data to look for

Zn2+ + 2 e Zn Eo = – 0.76 VFe3+ + e Fe2+ Eo = + 0.77 V +

2 Fe2+ 2 Fe3+ + 2 e Eo = – 0.77 VZn2+ (aq) + 2 Fe2+ (aq) Zn (s) + 2Fe3+ (aq) Eo = – 1.53 V

non-spontaneous Pt | Zn2+ | Zn || Fe2+ | Fe3+ | Pt impractical

The reverse reaction is spontaneous, Zn (s) + 2Fe3+ (aq) Zn2+ (aq) + 2Fe2+ (aq) Eo = + 1.53 V

Zn | Zn2+ || Fe3+ | Fe2+ | Pt


Free energy and emfWhat is the free-energy change for the cell,

Zn | Zn2+ (aq) 1 M || Ag+ (aq) 1 M | Ag?

Solution: Reduction potential required,Zn Zn2+ + 2e Eo = 0.76

Ag+ + e Ag Eo = 0.80 2 Ag+ + 2e 2Ag Eo = 0.80

Cell reaction 2 Ag+ + Zn 2Ag + Zn2+ Eo = 1.56 V

Go = – n F Eo = – 2 * 96485 * 1.56 = – 3.01e5 J or – 302 kJ

Negative indicate energy is released.

The free energy for the cell is –301 kJ per mole of Zn, what is the emf?

Condition for spontaneous reaction is – G or + E.


General cell emfG o is the standard energy change. G is for non-standard conditions.

G = – n F E

Similarly, E o is the standard emf whereas E is general emf.

G = G o + R T ln Q E = Eo – R T / n F ln Q reaction quotient

When a system is at equilibrium (Q = K), G = 0. Therefore,

G = G o + R T ln K = 0 E = Eo – R T / n F ln K = 0 equilibrium constant

G o = – R T ln K E o = R T / n F ln K

or G o = – ln(10) R T log K E o = 2.303 R T / n F log K

At 298 K 0.0592 E o = ———— log K

n

Text uses Ecell instead of E


The Nernst equation

R T [C]c [D]d

E = E o – ——— ln ———— n F [A]a [B]b

For a general reaction,a A + b B = c C + d D

This Nernst equation correlates cell emf with [ ] or reactivities of reactants and products as well as T

Units for [ ]: mol L-1 for aqueous solution, atm (for gas), and constant for solid and liquid.

See 21-4


Evaluating EAt 300 K, evaluate the cell emf for

Zn | Zn2+ (0.100 M) || H+ (0.200 M) | H2 (1.111 atm) | Pt

Solution:Look up: Zn Zn2+ (aq) + 2e, E o = 0.76 V

2 H+ (aq) + 2 e H2 (g), E o = 0.00 V (R = 8.314 J mol-1 K-1, F = 96485 C mol-1) E o = 0.76 VThe reaction is

Zn(s) + 2 H+ (aq) Zn2+ (aq) + H2(g)

8.314 J mol-1 K-1 * 300 K (0.100) (1.111) E = 0.76 – —————————— ln ——————

2 * 96485 C mol-1 (0.200)2

= 0.76 – 0.0129 * (1.02) = 0.76 – 0.013 = 0.75 V

R T [C]c [D]d

E = E o – ——— ln ———— n F [A]a [B]b

See example 19.12


Concentration cellProblem: At 298 K, evaluate the emf of the cell

Cu | Cu2+ (0.10 M) | | Cu2+ (1.0 M) | Cu Cu(s) Cu2+(0.1 M) + 2e; Cu2+ (1.0 M) + 2e Cu(s)

Solution:The standard emf (Eo = 0.00)for Cu | Cu2+ || Cu2+ | Cu The reaction is actually Cu (s) + Cu2+ (1.0 M) = Cu2+ (0.1 M) + Cu (s)

R T [Cu2+] R T 0.10 E = 0.00 – ——- ln ——— = – –––– ln –––––

2 F [Cu2+] 2 F 1.00

8.3145 * 298 1.0 = + –––––––––– ln –––– = 0.0295 V

2*96485 0.1The voltage is purely due to concentration difference. Solutions in the two compartment try to become equal.

When 2 [ ]’s are equal, E = 0

See p. 841


Equilibrium Constant K and Eocell

Calculate the solubility product of AgCl from data of standard cell

Solution: Look up desirable data Ag+Cl– (s) + e Ag0(s) + Cl– E° = 0.2223 V

Ag+ (aq) + e Ag (s) E° = 0.799 V Ag (s) Ag+ (aq) + e E° = – 0.799 V

Get the desirable eq’n AgCl (s) Ag+ (aq) + Cl– (aq) E° = – 0.577 V

Ksp = [Ag+][Cl– ] log Ksp = – 0.577 / 0.0592 = – 9.75 Ksp = 10– 9.75 = 1.8e–10

At 298 K 0.0592 E o = ———— log K n See p. 837

Show that for this cellAg | Ag+, 1 M || Cl–, 1 M | AgCl | Ag

Eo = – 0.577 Vbut for this cell

Ag | AgCl |Cl–, 1 M || Ag+, 1 M | AgEo = +0.577 V


Evaluate free-energy change Evaluate G o for the reaction

Zn (s) + 2 Ag+ (aq) = Zn2+(aq) + 2 Ag (s)Solution:Required to look up: Eo V

Ag+ + e = Ag 0.80Zn2+ + 2 e = Zn – 0.76

2 Ag+ + 2 e 2Ag 0.80 +Zn Zn2+ + 2 e + 0.76

Zn (s) + 2 Ag+ (aq) Zn2+(aq) + 2 Ag (s) 1.56 (= E o)

G o = – n F E o

= – 2 * 96485 C * 1.56 V = – 301000 J = – 301 kJSee slides 17 and 22

Write the cell for this rxn


Summary of thermodynamicsChemical energy

Ho, So

Go = Ho – T SoElectric energy

Go = – n F Eo

Reaction quotient &equilibrium constantG = G o + R T ln Q

G o = – R T ln K

Reaction quotient &equilibrium constant

E = E o – R T/ n F ln Q

E o = R T/ n F ln K

Stoichiometry

n


insight from cold denaturation and a two-state water structure

By Tsai CJ, Maizel JV Jr, Nussinov R.

…The exposure of non-polar surface reduces the entropy and enthalpy of the system, at low and at high temperatures. At low temperatures the favorable reduction in enthalpy overcomes the unfavorable reduction in entropy, leading to cold denaturation. At high temperatures, folding/unfolding is a two-step process: in the first, the entropy gain leads to hydrophobic collapse, in the second, the reduction in enthalpy due to protein-protein interactions leads to the native state. The different entropy and enthalpy contributions to the Gibbs energy change at each step at high, and at low, temperatures can be conveniently explained by a two-state model of the water structure….

Biochem Mol Biol. 2002;37(2):55-69


pH and emfConsider the cell,

Zn | Zn2+ (1.00 M) || H+ (x M) | H2 (1.00 atm) | Pt

From table data, Zn2+ + 2e = Zn Eo = – 0.762 H+ + 2 e = H2 Eo = 0.00Zn = Zn2+ + 2e Eo = 0.76Zn + 2H+ = Zn2+ + H2 Eo = 0.76

R T [Zn2+] PH2

E = Eo – —— ln ————2 F [H+]2

= Eo + 0.0592 log [H+]

= 0.76 – 0.0592 pH

At 298 K (pH meters)

0.76 – E

pH = —————0.0592


pH electrodes

pH Range: 0-14Temp. Range: 0-100 CInternal Ref: ROSS

Junction: CeramicDimensions: 120 mm x 12 mm

Slope: 92 - 102%Temp. Accuracy: 0.5 CCatalog Number: 8202BN

(BNC Connector, 1 meter cable)


Ion selective electrode

1. Concentration; 2. Temperature; 3. Electrode surface conditions; 4. Number of charges of ions (8); 5. Stirring (6); 6. Suspension (7); 7. Zwitterionic nature, net charge density; 8. Anything changing ionic adsorption; 9. Isoelectric nature of surface material; 10. The Nernst equation deals only with concentration and temperature

More research has gone into pH measurements. Nernst started it.


Battery technologyBy use: automobile, flash light, radio, computer, camera, watch, emergency equipment, artificial heart machine, pace makers, hearing aids, calculators, … (portable energy)

By type: alkaline, dry, wet, storage, rechargeable, etc

By chemistry: alkaline, carbon zinc, dry, cell, lithium, lithium ion, lithium polymer, NiCd, etc.

By material: anode material, cathode material, electrode, etc.

Aluminum for battery manufacture

See 21-5


Lead Storage Battery for Autos

Anode – Negative platePb + SO4

2- PbSO4 + 2e-

PbO2 + 4H+ + SO42- + 2e-

PbSO4 + 2 H2O

SeparatorCathode –

Positive plate

H2SO4

A 12-V battery consists of 6 such cells

Net reaction: Pb + PbO2 + 2 H2SO4 = 2 PbSO4 + H2O

21 Electrochemistry 36Copyright© by Houghton Mifflin Company. All rights reserved. 10

Figure 18.6: Schematic of one cell of the lead battery.Figure 18.6: Schematic of one cell of the lead battery.


A mercury battery.


Corrosion: Unwanted Voltaic Cells

Fe(s) Fe2+ (aq) + 2e–

O2 + H2O (l) + 4e– 4 OH– (aq)

2 Fe(s) + O2 + H2O 2 Fe2+ (aq) + 4 OH– (aq)

What are effective corrosion prevention methods?

CoatingUse sacrifice electrode

See 21-6


Cathodic protection of an underground pipe.


Ion displacement reactions (corrosion)

Zn + 2 Ag+ Zn2+ + 2 AgZn + 2 Cu2+ Zn2+ + Cu

Reaction E o (V)

Li+ + e– = Li (s) – 3.04Na+ + e– = Na (s) – 2.71

Mg2+ + 2 e– = Mg (s) – 2.38Zn2+ + 2 e– = Zn (s) – 0.76

(reference) 2 H+ + 2 e– = H2 (g) 0.000 (reference) H2 (g) = 2 H+ + 2 e– 0.000

Cu2+ + 2 e– = Cu (s) 0.34Cu+ + e– = Cu (s) 0.52

I2 (s) + 2e– = 2 I– (aq) 0.54Ag+ (aq) + e- = Ag (s) 0.80

Br2 (l) + 2 e– = 2 Br– (aq) 1.07Cl2 (g) + 2 e– = 2 Cl– (aq) 1.36

F2 (g) + 2 e– = 2 F– (aq) 2.87

Zn2+ + 2 Ag Zn + 2 Ag+ Zn2+ + Cu Zn + 2 Cu2+

What metal will react with certain ions?

See 21-1


Electrolysis of molten saltsBattery

2 Cl– Cl2 + 2e–

2 Na+ + 2 e– 2 Na

Net2 NaCl 2 Na + Cl2

805o C

Molten salts consists of Na+ and Cl– ions

CATHODE

ee

Eo=-2.71V;2Na+ + 2e– 2NaEo=-1.36 V; 2 Cl– Cl2 + 2e–

Charges required to produce 1 mole Cl2 and 2 moles Na = 2F

Energy required = 2 F E; (E > 4.07 V)

reductionoxidation

ANODE

See 21-7


Electrometallurgy of Sodium


Electrolysis of NaCl solution

Battery

Salt solution consists of Na+ and Cl– ions

CATHODE

ee

2Na+ + 2e– = 2Na2Na + 2H+ = H2 + 2Na+

2 Cl– = Cl2 + 2e– Cl2 + H2O = HCl + ½ O2

reductionoxidation

ANODE


Refining Copper by ElectrolsisCopper can be purified by electrolysis. Raw copper is oxidized

Cu = Cu2+ + 2e

and purer copper deposited on to the cathode from a solution containing CuSO4

Cu2+ + 2e = Cu

If a current of 2 amperes pass through the cell, how long will it take to deposit 5.00 g of copper on the cathode? (1 ampere = 1 C s–1)

Solution

5.00 2*96485 C 1 s------ mol ----------- ------ = work out your answer New65.5 1 mol 2 C _______


Production of aluminum

AlF63– + 3 e– Al + 6 F– . . . Cathode

2 Al2OF62– + C(s) + 12 F– + 4 AlF6

3– + CO2 + 4 e– . . . Anode

2 Al2O3 + 3 C 4 Al + 3 CO2 . . . Overall cell reaction

Hall and Heroult tried to mix about 5% alumina in their molten cryolite, and obtained Al metal. This is the Hall process.

Charge required for each mole Al = 3 F

Energy required = 3 F E

Aluminum (Al), the third most abundant elements on Earth crust as bauxite or alumina Al2O3, remain unknown to man until 1827, because it is very reactive. By then, Wohler obtained some Al metal by reducing Al2O3 with potassium vapore. In 1886, two young men electrolyzed molten cryolite Na3AlF6 (melting point 1000° C), but did not get aluminum.


Electrometallurgy of Aluminum


Electrolysis of acid solutionBattery

Solutions containing H+ and SO42– ions

CATHODE

ee2H+ + 2e– = H2

H2O = ½ O2 + 2e– + 2 H+

Charges required to produce 1 mole H2 and ½ moles O2 = 2F

Energy required = 2 F E

reductionoxidation

ANODE


Electrolysis of H2SO4 solution

Pure water is not a good electric conductor. In the presence of electrolytes, water can be decomposed by electrolysis.

On the other hand, electrolysis of electrolyte solutions may reduce H+ and oxidize O2– in H2O.

In an H2SO4 solution,cathode reductions are

2 H2O (l) + 2 e– = H2 (g) + 2 OH– (same as 2H+ + 2e– = H2)

Anode oxidation:2 H2O (l) = 4 e– + O2 (g) + 4 H+ E o = – 1.23 V (observed)2 SO4

2– = [SO3O–OSO3]2– + 2 e– E o = – 2.01 V (not observed)


Electrolysis of H2SO4 solution

BatteryE o = – 1.23 V 2 H2O (l) = 4 e– + O2 (g) + 4 H+ E o = – 2.01 V 2 SO4

2– = [SO3O–OSO3]2– + 2 e–

Solution consists of H+ and SO42– ions

CATHODE

ANODE

ee2H+ + 2e = H2

reductionoxidation


Electroplating of metalsGalvanizing Zn2+ + 2 e– Zn onto metal surface

Copper purification Cu2+ + 2 e– Cu onto pure Cu electrode

Silver plating Ag+ + e– Ag onto metal surface

Since 1971, Cal-Aurum has provided electroplating services to the electronic component industry with the highest standards of quality, performance, and competitive pricing.

Miller specializes in plating metals such as magnesium, aluminum, zinc, copper, powdered metals, steel and various other substrates.

Over a half century of extensive and innovative research has made us one of the

nation's leading experts in plating on magnesium


SummaryThe 20th century belongs to electrons. They continue affecting our lives the 21st century.

Chemistry studies the drama played by electrons, and electrochemistry is the finale.

Energy directs and produces the show, but you set the magic stage for a great performance.

Leo and Ger tell electrons to get in and out of your stage, and you must skillfully provide paths to balance the flow.

Cells are the stages for the performance, you must construct, represent, figure out the potentials, and control the show.

Chemical reaction, equilibrium, (acid, base, heterogeneous, and complex formation) and electrochemistry guide us using simple rules.

Apply rules you have learned in Chem1235 to understand what is

happening around you and may your live be full of happiness.


Skills for Electrochemistry (review)

Make up a Daniel cell using Pb and Ag as the electrodes. Draw a diagram for it.

Use short notation to represent the cell for the spontaneous reaction

Write half reaction equations for both cathode and anode and explain the reactions

Write balanced redox equations

Calculate emf for a nonstandard cell and its energy

Calculate equilibrium constant K from Eo

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21elecchem3weq