Electrochemistry

ElectrochemistryThe study of the interchangebetween chemical and electrical energy

Oxidation-Reduction ReactionsRedox reactions involve the transfer of electrons (e-)Reduction: gain e-Oxidation: lose e- LEO the lion says, GER OIL RIGUse oxidation states to keep track of the e-

Leo says GerLose electron oxidation Zn 2e- + Zn2+

Gain electron reduction2e- + Cu2+ Cu

My name is Leo. Grr-rrrrAssigning Oxidation StatesSpecific rules for assigning Ox #s Usually the same charge assigned by the PTH is almost always +1 O is almost always -2 F is always -1 in compoundsFor elements (H2, O2, F2, Ca, K, etc ) the oxidation state always = 0

Some exceptions do exist!

Assigning Oxidation NumbersOverall charge = sum of the oxidation states of all atoms in itNeutral Compounds (e.g. H2O, CO2, CH4)H2O : The overall charge is 2(1) + -2 = 0CO2: What is the oxidation state of C?Since C + 2 (O) = 0C + 2(-2) = 0, thusCH4: Is C still +4?H is always +1 To remain neutral 4(1) + C = 0 C must = - 4

H = +1and O = -2C = +4Assigning Oxidation NumbersCharged compounds (e.g. NO3-, CO32-)NO3- or (NO3)- : What is the oxidation # of N?O is -2, and the overall charge is -1So N + 3(O) = -1 or N + 3(-2) = -1N = + 5(CO3)2-: What is the oxidation # of C?O is -2, and the overall charge is -2So C + 3(O) = -2 or C + 3(-2) = -2C = +4The oxidation # of ions = charge of ionsMn3+ has an oxidation # of +3S2- has an oxidation # of -2

Assigning Oxidation # Practice Assign oxidation numbers to each atomCl2Fe2+ClO3-ClO4-IO2-CrO42-Fe3(PO4)2CoSO4H2CO3

Cl: 0 (element)Fe: 2+ (ion)O: 2-, 3(2-) + Cl = 1-Cl: 5+O: 2-, 4(2-) + Cl = 1-Cl: 7+O: 2-, 2(2-) + I = 1-I: 3+O: 2-, 4(2-) + Cr = 2-Cr: 6+Fe: 2+ (ion) PO4:3- (ion).O:2-, 4(2-) + P = 3-, P: 5+Co: 2+ (ion) SO4:2- (ion).O:2-, 4(2-) + S = 2-, S: 6+H: 1+ (ion) CO3:2- (ion).O:2-, 3(2-) + C = 2-, C: 4+

Assigning Oxidation Numbers ReviewTry theseMnO4-, Cr2O72-, C2O42-(MnO4)-O = -2, so [4(-2) + Mn = -1]Mn = +7(Cr2O7)2-O = -2, so [7(-2) + 2Cr = -2]2Cr = 12, therefore (C2O4)2-O = -2, so [2C + 4(-2) = -2]2C = 6, thereforeCr = +6C = +3

Oxidation-Reduction ReactionsTwo separate reactions occurring simultaneouslyOxidation: oxidation # of an atom increasese.g. Fe(s) Fe3+(aq)Reduction: oxidation # of an atom is reducede.g. O2(g) O2-(aq)When occurring togetherFe(s) + O2(g) Fe3+(aq) + O2-(aq)This is the redox reaction responsible for rust!But, how do we balance this?(ox # goes from 0 +3)(oxidation # goes from 0 -2)

Balancing by Half-Reactions*in acidic solution

Assign oxidation states for each element.Write separate half-reactions for the reduction/oxidation reactions.Balance all the atoms EXCEPT O and H.Balance the oxygen with water (H2O).Balance the hydrogen with hydrogen ions (H+).Balance the charge with electrons.Multiply each half-reaction by an appropriate number to make the electrons equal.Combine both reactions into one and cancel the e -

Balancing by Half-Reactions*in acidic solutionCH3OH (aq) + Cr2O72-(aq) CH2O(aq) + Cr3+(aq)1.Assign oxidation states.C-2H4+O2- + (Cr26+O72-)2- C0H2+O2- + Cr3+2. Write separate half-reactions for the reduction and oxidation reactions. (only keep charges that are changing)Ox: C-2H4O C0H2O (C is going from -2 to 0)Red: (Cr26+O7)2- Cr3+ (Cr is being reduced from +6 to +3)Ox: C2-H4O C0H2O3. For each half reaction, balance all the atoms EXCEPT O and H. 4. Balance the oxygen by adding water (H2O).

Balance the hydrogen by adding hydrogen ions (H+)Balance the charge by adding electrons.use the oxidation state as a guideBalancing the Oxidation

Carbon is already balanced!+ 2H+ + 2e-Oxygen is already balanced!On to the reduction (Cr26+O7)2- Cr3+

Balance all elements except H and O Balance O by adding H2O, if necessaryBalance H by adding H+,if necessaryBalance charge by adding e-Remember, you only care about the charges that are changing

2+ 7H2O14H+ +6e- +

Adding Half-Reactions*in acidic solutionNow add the two reactions togetherOx: CH4O CH2O + 2H+ + 2e-Red: 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O7. Multiply each half-reaction by an appropriate number to make the electrons equal. CH4O CH2O + 2H+ + 2e-6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O 3CH4O 3CH2O + 6H+ + 6e-

3 () 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O 3CH4O 3CH2O + 6H+ + 6e- 3CH4O + + Cr2O72- 3CH2O + 2Cr3+ + 7H2Oand the reaction is now balanced! 8H+Adding Half-Reactions*in acidic solution8. Combine both reactions into one and cancel.

Practice Balancing Redox ReactionsUnbalanced reaction (in acid):MnO4 + Fe2+ Mn2+ + Fe3+Balanced Reduction half-reaction:8H+ + MnO4 + 5e Mn2+ + 4H2OBalanced Oxidation half-reaction: Fe2+ Fe3+ + eBalanced overall reaction:8H+ + MnO4 + 5Fe2+ Mn2+ + 5Fe3+ + 4H2O5( )Balancing by Half-Reactions*in basic solutionAssign oxidation states.Write separate half-reactions for the reduction/oxidation reactions.Balance all the atoms EXCEPT O and H.Balance the oxygen by adding water (H2O).Balance the hydrogen by adding H+.Balance the charge by adding electrons.Multiply each half-reaction by an appropriate number to make the electrons equal.Combine both reactions into one and cancel.Add OH- to both sides to cancel out H+ and create H2O. Simplify further, if necessary.

Balancing by Half-Reactions(in basic solution)Lets balance a previous example in basic solution

Remember, it is all the same steps up to this point3CH4O + 8H+ + Cr2O72- 3CH2O + 2Cr3+ + 7H2O

3CH4O + + Cr2O72- 3CH2O + 2Cr3+ + 7H2O + 8OH-

3CH4O + H2O + Cr2O72- 3CH2O + 2Cr3+ + 8OH-+ 8OH-+ 8OH-8H2OPractice Balancing Basic Redox RxnsUnbalanced reaction:ClO + Zn Cl- + Zn2+Balanced Reduction half-reaction:2e- + 2H+ + ClO- Cl- + H2O Balanced Oxidation half-reaction: Zn Zn2+ + 2e- Balanced overall reaction (acidic):2H+ + ClO + Zn Zn2+ + Cl- + H2OBalanced overall reaction (basic):H2O + ClO + Zn Zn2+ + Cl- + 2OH-

Redox VocabularyClO + Zn Cl- + Zn2+Oxidized species (atom, ion, molecule, or compound)whichever species is being oxidized ( oxidation #)Reduced specieswhichever species is being reduced ( oxidation #)Oxidizing agentwhichever species CAUSES oxidation to occurin other words, the oxidizing agent IS the reduced speciesReducing agentwhichever species CAUSES reduction to occurin other words, the reducing agent IS the oxidized species

ClO + Zn Cl- + Zn2+Oxidized speciesReduced speciesReducing AgentOxidizing AgentClO-ZnNotice that these terms only ever apply to reactants

Sample electrochemical processes:1. Corrosione.g. 4 Fe(s) + 3 O2(g) 2 Fe2O3(s)2. Biological processese.g. C6H12O6 + 6 O2 6 CO2 + 6 H2O3. Batteries (Galvanic or Voltaic cells)Electrochemical cells that produce a current (flow of electrons) as a result of a redox reaction4. Electrolytic cellsElectrical energy is used to produce chemical changeUsed to prepare or purify metals (such as sodium, aluminum, copper)ElectrochemistryThe study of the interchangeof chemical and electrical energy

Chemical Change Electron FlowCopper: Cu(s), Cu2+(aq)Cu(s) Cu2+(aq) + 2e- Grxn = Gf(Cu2+) = 65.6 kJSilver: Ag(s), Ag+(aq)Ag(s) Ag+(aq) + e-Grxn = Gf(Ag+) = 77.2 kJCu(s) Cu2+(aq) + 2e-G = +65.6 kJAg+(aq) + e- Ag(s)G = -77.2 kJCu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s)G = -88.8 kJSpontaneouswmax = -88.8 kJCu2+ in solution2() 2( )Ag(s)Harnessing the EnergySeparate the half-reactionsCreates a galvanic or voltaic cellCuAg1 M CuSO4Cu(s) Cu2+(aq) + 2e-1 M AgNO3Ag+(aq) + e- Ag(s)

Luigi Galvani

AlessandroVoltaCu2+SO42-Ag+NO3-KNO3(aq)K+NO3-e-salt bridgeOxidationReductionAnodeCathodecathode and reduction begin with consonantsanode and oxidation begin with vowels(produces electrons)+(attracts electrons)

Red Cat

Standard Reduction Potentials The cell potential, cell ,can be determined from the standard reduction potentials (red) for the half-reactions: Reduction potential = tendency for reduction to happenPositive red spontaneous reduction reactionNegative red non-spontaneous reduction (use reverse reaction)Standard (o) = standard conditions (1 M solutions)Standard Reduction PotentialsHalf-Reaction (V)F2 + 2e- 2F- 2.87Au3+ + 3e- Au 1.50Ag+ + e- Ag 0.80Cu2+ + 2e- Cu 0.3372H+ + 2e- H2 0.00Ni2+ + 2e- Ni-0.28Zn2+ + 2e- Zn-0.763Al3+ + 3e- Al-1.66Li+ + e- Li-3.05 = 0 Standard Hydrogen Electrode > 0Spontaneous reduction(Oxidizing Agents!) < 0Non-Spontaneous reduction(Reducing Agents!)Spontaneous oxidation (reverse rxn)Ni Ni2+ + 2e- +0.28Zn Zn2+ + 2e- +0.763Al Al3+ + 3e- +1.66Li Li+ + e- +3.05Remember: an oxidation CANNOT happen without a reduction!Cell Potentialcell = reduction + oxidationAg+(aq) + e- Ag(s) = 0.80 VCu2+(aq) + 2 e- Cu(s) = 0.34 VReduction reaction:2(Ag+(aq) + e- Ag(s)) = +0.80 VOxidation reaction:Cu(s) Cu2+(aq) + 2 e- = - 0.34 VCu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s)cell = +0.46 VThe cell MUST be + and thus spontaneous for Galvanic cellsBrain WarmupHalf-Reaction (V)Ag+ + e- Ag 0.80Cu2+ + 2e- Cu 0.34Zn2+ + 2e- Zn-0.76Al3+ + 3e- Al-1.66What is for each of the following reactions?Which reaction(s) are spontaneous?3 Ag+(aq) + Al(s) 3 Ag(s) + Al3+(aq)Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq)2 Al3+(aq) + 3 Zn(s) 2 Al(s) + 3 Zn2+(aq)2.46 V1.10 V-0.90 VSpontaneous?YYNZn can reduce Cu2+, but not Al3+Line Notation for Galvanic CellsCuAg1 M CuSO4Cu(s) Cu2+(aq) + 2e-1 M AgNO3Ag+(aq) + e- Ag(s)Cu2+SO42-Ag+NO3-K+NO3-e-OxidationReductionAnode()Cathode(+)Anode always on the leftCu(s)Cu2+ (1 M)Ag+ (1 M)Ag(s)Cathode always on the rightPractice TimeGiven the following information, draw a galvanic cell.Fe(s)Fe2+(1 M)Au3+(1 M)Au(s)Be sure to include the following:Anode/Cathode reactionsBalanced overall reaction with potentialComplete circuit (external wire with e- flow direction, salt bridge)Label all parts of the cell (solution, electrode, etc.)

Fe(s)Fe2+(1 M)Au3+(1 M)Au(s)FeAu1 M Fe2+Fe(s) Fe2+(aq) + 2e-1 M Au3+Au3+(aq) + 3e- Au(s)Fe2+

Au3+

anionse-OxidationReductionAnodeCathode3Fe(s) + 2Au3+(aq) 3Fe2+(aq) + 2Au(s)cationscell = +0.440V (Fe rxn) + 1.50 V (Au rxn) = 1.94 VPractice TimeGiven the following information, draw a galvanic cell.C(gr)Cr2+(1 M), Cr3+(1 M)Cl-(1M)Cl2(g)Pt(s)Be sure to include the following:Anode/Cathode reactionsBalanced overall reaction with potentialComplete circuit (external wire with e- flow direction, salt bridge)Label all parts of the cell (solution, electrode, etc.)

C(gr)Cr2+(1 M), Cr3+(1 M)Cl-(1M)Cl2(g)Pt(s)CPt1 M Cr2+/Cr3+Cr2+ Cr3+ + e-1 M Cl-, 1 atm Cl2(g)Cl2(g) + 2e- 2Cl -Cr2+ Cr3+Cl- Cl2

anionse-OxidationReductionAnodeCathodecations

The First Battery

Alessandro Volta (1745 1827)Generated electricity by putting a layer of cardboard soaked in brine between discs of copper and zinc a voltaic cell.When he made a pile of these cells, heincreased the amount of electricity generated. This was the first battery a collection of cells.Newmark, CHEMISTRY, 1993, page 46Zinc discCopper discCar Battery (Lead storage battery)Anode: Pb + HSO4- PbSO4 + H+ + 2e-Cathode: PbO2 + HSO4- + 3H+ + 2e- PbSO4 + 2H2OCell: Pb + PbO2 + 2H+ + 2 HSO4- 2 PbSO4 + 2H2O2 volts per cell, 6 cells to a battery 12 volt batteryAlkaline BatteryAnode: Zn Zn2+ + 2e-Cathode: 2 MnO2 + H2O + 2e- Mn2O3 + 2OH-cell = 1.5 voltsLemon BatteryAnode: Zn Zn2+ + 2e-Cathode: Cu2+ + 2e- CuCell reaction: Zn + Cu2+ Zn2+ + Cucell = 1.1 volts (if all goes well)Want more volts? Link cells in series...

Batteries

Onion Battery?According to YouTube, you can charge your iPod with just an onion and GatoradeAs Michael Scott (from the Office) points out, anyone can put anything on the Internet, so you know the information is good

One Cell of a Lead Battery

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 595Mercury Battery

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 596Common Alkaline Battery

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 596Cathodic Protection of an Underground Pipe

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 598Free Energy and Cell PotentialG = wmax = nFn= moles of e- transferredF= Faradays constant = 96,485 C/mol e-= standard cell potential (V or J/C)

Michael FaradayCu(s)Cu2+ (1 M)Ag+ (1 M)Ag(s)cell = +0.46 VG = -nFcellG = -(2 mol e-)(96485 C/mol e-)(0.46 V)G = -88,800 J or -88.8 kJ3 Ag+(aq) + Al(s) 3 Ag(s) + Al3+(aq)Polishing Silver with Aluminum Foil = 2.46 V3 Ag2S + 2 Al + 3 H2O 6 Ag + Al2O3 + 3 H2SStinky!!(Add NaHCO3 to neutralize H2S)

Reaction QuotientThe reaction quotient (Q) sets up a ratio of products and reactantsFor a reaction, A + 2B 3C + 4D[C]3[D]4 [A]1[B]2

Only include concentrations (aq) OR pressures (g)Solids (s) and liquids (l) are not included

Q = Reaction Quotient practiceWrite the Q expression for the following reactionCH4(g) + O2(g) CO2(g) + H2O(g)Reaction must be balanced firstCH4(g) + 2O2(g) CO2(g) + 2H2O(g)

(CO2)(H2O)2 (CH4)(O2)2

Q =

Reaction Quotient practiceWrite the Q expression for the following reaction Cu(s) + 2Ag+(aq) Cu2+(aq) + 2Ag(s)

(Cu2+)(Ag)2(Cu)(Ag+)2Is this correct?NO: Solids arent included in the equation!(Cu2+)(Ag+)2Q =Q =

Under Non-standard conditions:The Nernst EquationIf we dont have 1 M concentration or 1 atm pressure,we must take a different approachG = G + RT lnQG = -nFG = -nF-nF = -nF + RT lnQ

Walther Nernstcell = cell -

R= 8.3145 J/mol KT= temperature (in K)n= moles of e- transferredF= Faradays constant 96,485 C/mol e-

Practice with the Nernst EquationWhat will be the cell potential of a Cu/Ag cell using 0.10 M Cu2+ and 1.0 M Ag+ solutions at 25C?Cu(s) Cu2+(aq) + 2e-Ag+(aq) + e- Ag(s)CuAgCu2+SO42-Ag+NO3-Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s)cell = 0.46 V (-0.03 V)Cu(s)Cu2+ (0.10 M)Ag+ (1.0 M)Ag(s)

cellcell = 0.49 V

2( )cell = cell -

3( ) 5( )More PracticeWhat will be the cell potential of a Al/MnO4-,Mn2+ cell using 0.5 M Al3+, 1.5 M Mn2+, 1.0 M MnO4- and 2.0 M H+ solutions at 15C?Al(s) Al3+(aq) + 3e-MnO4- + 8H+ + 5e- Mn2+ + 4 H2OAlPtAl3+H+MnO4-, Mn2+5Al(s) + 3MnO4- + 24H+ 3Mn2++ 5 Al3+ + 12 H2Ocell = 3.17 V (-0.03 V)Al(s)Al3+ (0.5 M)Mn2+ (1.5 M), MnO4- (1.0 M), H+ (2.0 M)Pt(s)

cellcell = 3.20 V

cell = cell -

= 6.285 x 10-9Remove NH3.. NH3 H2..Add more N2..Le Chateliers principleN2(g) + 3 H2(g) 2 NH3(g)DisturbanceEquilibrium Shiftno shiftWhen a system at equilibrium is disturbed, it shifts to a new equilibrium that balances the disturbanceAdd a solid/liquidRemove either reactant.

Fritz Haber

49The Nobel Prize in Chemistry 1918Presentation SpeechPresentation Speech by Doctor .G. Ekstrand, President of the Royal Swedish Academy of Sciences, on June 1, 1920*Ladies and Gentlemen.**

The Royal Swedish Academy of Sciences has decided to confer the Nobel Prize in Chemistry for 1918 upon the Director of the Kaiser Wilhelm Institute at Dahlem near Berlin, Geheimrat Professor Dr. Fritz Haber, for his method of synthesizing ammonia from its elements, nitrogen and hydrogen.

In accordance with Nature's plan of economy, soil fertility under normal circumstances is maintained at an even level if the waste products from the crop are returned to the soil; if, however, substantially increased productivity is required from the soil, then additional fertilizer must be used. Since meanwhile a large proportion of the annual harvest is consumed by the yearly increasing population of towns, and since the towns' waste products are returned to land under cultivation only to a very incomplete extent, the inevitable consequence is that the soil becomes exhausted and the harvest yield diminishes. This has, in turn, led to the manufacture of artificial fertilizers which has also increased year by year in importance to such an extent that, at least in Europe, hardly a country exists which can do entirely without them.

Among these substances nitrogenous compounds occupy an important position, since usually the soil does not possess a large store of these to be released to suit the plants' needs by weathering as in the case of phosphoric acid and potash; added to which there is the fact that part of the effective nitrogen turns into inactive atmospheric nitrogen during the cyclic process. Admittedly a part of this loss is compensated by rainfall and through the activity of bacteria, but so far experience has shown that intensive cultivation cannot be maintained without artificial nitrogenous fertilizers. This applies, above all, to one of today's most important crops, sugar-beet.

For many years only two artificial nitrogenous compounds existed, namely potassium nitrate and ammonium chloride. The older methods by which these were made, however, ceased to play a part, at least in Europe and America, when Chile saltpetre (sodium nitrate) came into the picture and use was made of the by-products from dry distillation of mineral coal for this purpose.

The consumption of Chile saltpetre, calculated in terms of nitrogen, amounts to about 500,000 or more tons per annum. Under normal circumstances the vast majority of this saltpetre is used for fertilizer purposes. The burning question, therefore, has long been: how long will the saltpetre deposits in Chile last? The Chilean authorities give very widely varying estimates, and experts in Europe are of the opinion that at current production rates the deposits will be exhausted within the foreseeable future.

Be that as it may. The protracted World War has sufficiently demonstrated to every country the need of organizing, wherever possible, production of essential commodities within its own borders in sufficient quantities to meet its own needs.

Now, since saltpetre is among the most important of these substances, particularly in those countries which possess neither large mineral coal deposits nor cheap hydro-electric power, the artificial production of ammonia and nitric acid has reached an unprecedented degree of importance.

A substance on the borderline between natural and artificial products is the ammonia obtained by dry distillation of bituminous and brown coal. This ammonia comes from the nitrogen content of these minerals, amounting to approximately 1.3 % by weight, of which however the largest portion (around 85%) remains behind in the coke or is liberated as nitrogen during distillation.

During the first ten years of this century several methods were published, based on binding the nitrogen from the air, but few of these survived the trial stage. The first of these was Frank-Caro's cyanamide method. Indeed it appears that calcium cyanamide did not come fully up to expectations as a fertilizer, but since its nitrogen content can be converted to ammonia relatively easily, this has not so far proved to be an obstacle to the application of the method to an ever-increasing extent.

Using the main principles of thermodynamics every quantitative condition with regard to the combustion of atmospheric nitrogen to produce nitric oxide can be calculated. Birkeland and Eyde were, of course, the first to apply this technically with successful results.

Until 1904 nobody had been able to bring about a direct combination of nitrogen and hydrogen to form ammonia without the help of dark electrical discharge, although the experiments of Berthelot and Thomson proved that the combination occurred exothermically. With the experience we now have we can easily see that this negative result was due to the slowness of the reaction at low temperatures, and unfavourable equilibrium conditions at high temperatures. Admittedly, in 1884 Ramsay and Young had conducted some experiments on this, using iron fillings as a catalyst, but these yielded only uncertain results.

In 1904 Haber and van Oordt began a methodical study of this relevant field, based on modern physico-chemical methods, after a single previous experiment had given Haber a hope of finding a technical solution to the problem. They worked at a temperature of about 1,000 C and normal pressure, using iron as a catalyst. From these experiments it emerged that from red heat onwards, and also at higher pressures, only traces of ammonia could be formed.

During this work it was also shown experimentally for the first time that a real state of equilibrium existed in the systemN2+ 3H2 D2NH3, which is in fact the real basis for the synthesis of ammonia.

In the "Zeitschrift fr Elektrochemie" of 1913 can be found the treatment of this question, by Haber and Le Rossignol which has the most important practical meaning: "ber die technische Darstellung von Ammoniak aus Elementen" (On the technical production of ammonia from the elements). This treatise provided the groundwork for the development of the method on a factory scale at the "Badische Anilin- und Sodafabrik" in Ludwigshafen, the main development occurring under the guidance of Dr. C. Bosch.

Earlier experiments had shown the pointlessness of exceeding dark red heat, i.e. about 600 C. On the other hand, the reaction formula showed that combination occurs with a contraction of from 4 to 2 volumes.

From the law of equilibrium it follows that the higher the pressure is the more the equilibrium must shift to the ammonia side. This provided the basic principles. A temperature of about 500 C had to be used at the highest possible pressure, which in practice meant at about 150-200 atmospheres. It could also be assumed that this high pressure speeded up the reaction. But work with a flow of gas in a circulation system at such high pressure and at a temperature approaching red heat posed very severe difficulties and up to then had never been tried. It was, however, completely successful. The treatise in question contains detailed drawings of the equipment used with which, using iron as a catalyst, about 250 grams of ammonia were produced per hour and per litre of contact volume; with uranium or osmium as a catalyst considerably more was produced.

The heating is done electrically. Since however the heat escaping from the equipment is largely regenerated in the entrant gases the required temperature can largely be maintained by the regenerated heat and by the heat liberated during the formation of ammonia. A very important point in Haber's observations is that the gases can be given a greater flow rate during the reaction which of course substantially increases the amount of ammonia produced per unit of time.

Haber found the best catalyst to be osmium, followed by uranium or uranium carbide. According to tests conducted mostly at the factories of the "Badische", the activity of the catalyst may be increased by oxides or certain salts of alkalis and alkaline earth metals, just as it may be decreased by catalytic poisons. Gradually more active catalysts have been discovered, and by this means it has been found possible to reduce substantially the pressure in the chamber.

In 1910 construction work was begun on the first large factory near Oppau in the neighbourhood of Frankfurt am Main, with an estimated annual output of 30,000 tons of ammonia.

The basic materials, nitrogen and hydrogen, are produced by standard methods.

Power consumption in the ammonia process is very low, amounting to no more than 0.5 kilowatt-hours per kilogram of ammonia. Per kilowattyear, therefore, no less than 10,000 kilograms of nitrogen are bound.

Since the position of the equilibrium of the reaction depends, among other things, upon the heat of formation of ammonia and its specific heat, Haber in a series of seven articles in the "Zeitschrift fr Elektrochemie" of 1914-1915, has extensively described experiments carried out to confirm these figures with the greatest possible accuracy.

As, according to Ostwald's modified method, ammonia can be converted into nitric acid and the latter into calcium nitrate, the ratio between the overall costs of producing calcium nitrate is, according to the available calculations, approximately as follows:

Norwegian Hydro: 100Haber: 103Frank-Caro: 117

in other words, they are the same for the first two methods but approximately 15% higher for the last.

Since, however, of the three existing nitrogen methods, Haber's is the only one capable of operating independently of any available source of cheap hydroelectric power it can in future be applied in all countries; since furthermore it can be utilized on any convenient scale and because it can produce ammonia very much more cheaply and nitrate equally as cheaply as any other method, as explained above, it is of universal significance for the improvement of human nutrition and so of the greatest benefit to mankind.

German Haber factories, especially the recently built Leuna Works near Merseburg, are also in full production, providing the vast majority of all nitrogenous fertilizers obtainable in Germany. Moreover, the method has already been extensively applied in the United States of America.Geheimrat Professor Haber. This country's Academy of Sciences has awarded you the 1918 Nobel Prize for Chemistry in recognition of your great services in the solution of the problem of directly combining atmospheric nitrogen with hydrogen. A solution to this problem has been repeatedly attempted before, but you were the first to provide the industrial solution and thus to create an exceedingly important means of improving the standards of agriculture and the well-being of mankind. We congratulate you on this triumph in the service of your country and the whole of humanity. Please, accept now your prize from the President of the Nobel Foundation.* The Nobel Prize in Chemistry 1918 was announced on November 13, 1919.** Owing to the death of the crown-princess Margaret, no member of the Royal family was present at the Prize ceremony, which for special reasons had been postponed to early in June, 1920.From Nobel Lectures, Chemistry 1901-1921, Elsevier Publishing Company, Amsterdam, 1966 Copyright The Nobel Foundation 1918 Printer Friendly Comments & Questions Tell a Friend

The 1918 Prize in: Prev. year Next year The Nobel Prize in Chemistry 1918Presentation Speech Fritz HaberBiography Nobel Lecture Other Resources Articles - A glimpse into their world Listen to Chemistry Laureates Sign up for News from Nobelprize.org The Official Web Site of the Nobel FoundationCopyright Nobel Web AB 2007In a chicken CaO + CO2 CaCO3 (eggshells)[ CaO ] , shift[ CO2 ] , shift-- shift ; eggshells are thinner

In summer, [ CO2 ] in a chickens blood due to panting.How could we increase eggshell thickness in summer?-- give chickens carbonated water-- put CaO additives in chicken feed-- air condition the chicken houseTOO much $$$-- pump CO2 gas into the chicken housewould kill all the chickens!

I wish I had sweat glands.

50Other applications with LeChateliers PrincipleN2 + 3 H2 2 NH3 + heat Raising the temperaturefavors the endothermic reaction (the reverse reaction) in which the rise in temperature is counteracted by the absorption of heat.Increasing the pressurefavors the forward reaction in which 4 mol of gas molecules is converted to 2 mol.Dorin, Demmin, Gabel, Chemistry The Study of Matter 3rd Edition, page 53251AgCl + energy Ago + Clo

shift to a new equilibrium:Then go inside shift to a new equilibrium:

Light-Darkening EyeglassesenergyGo outsideSunlight more intense than inside light;GLASSES DARKEN(clear) (dark)energyGLASSES LIGHTEN

52Sensitive Sunglasses Oxidation-reduction reactions are the basis for many interesting and useful applications of technology. One such application is photochromic glass, which is used for the lenses in light sensitive glasses. Lenses manufactured by the Corning Glass Company can change from transmitting 85% of light to only transmitting 22% of light when exposed to bright sunlight. Photochromic glass is composed of linked tetrahedrons of silicon and oxygen atoms jumbled together in a disorderly array, with crystals of silver chloride caught in between the silica tetrahedrons. When the glass is clear, the visible light passes right through the molecules. The glass absorbs ultraviolet light, however, and this energy triggers an oxidation-reduction reaction between Ag+ and Cl-:Ag+ + Cl- --> Ag0 + Cl0

To prevent the reaction from reversing itself immediately, a few ions of Cu+ are incorporated into the silver chloride crystal. These Cu+ ions react with the newly formed chlorine atoms:

Cu+ + Cl0 --> Cu2+ + Cl-

The silver atoms move to the surface of the crystal and form small colloidal clusters of silver metal. This metallic silver absorbs visible light, making the lens appear dark (colored).

As the glass s removed from the light, the Cu2+ ions slowly move to the surface of the crystal where they interact with the silver metal:Cu2+ + Ag0 --> Cu+ + Ag+

The glass clears as the silver ions rejoin chloride ions in the crystals.Draw the following Galvanic Cell:Cu(s)Cu2+(aq) Cu2+(aq) Cu(s)Cu2+(aq) + 2e- Cu(s)Cu(s) Cu2+(aq) + 2e- OxidationReductionox = -0.337 VWrite the two half reactions and determine the cellred = +0.337 Vcell = 0??Concentration CellsA voltage is generated just by a difference in concentrationbetween the two half-cellsCu(s)Cu2+(aq) (0.5 M)Cu2+(aq) (2.5 M)Cu(s) = -

Cu2+(aq) + 2e- Cu(s)Cu(s) Cu2+(aq) + 2e- = 0

= -

= 0.02 V

= 0 -Cu(s)an + Cu2+(aq)cat Cu2+(aq)an + Cu(s)catConcentration Cells in NatureLiving cells contain low-pH vesicles surrounded byneutral-pH cytoplasmThe [H+] concentration gradient creates a voltagepH = 3pH = 7cell = cell -

4 H+ + O2 + 4e- 2 H2O = 1.23 V(10-7)4(10-3)4Q =([H+]outside)4([H+]inside)4=cell = -(0.059/4) log(10-16) = 0.24 VUsing LeChatliersCu(s)Cu2+(aq) (0.5 M)Cu2+(aq) (2.5 M)Cu(s)Cu2+(aq) + 2e- Cu(s)Cu(s) Cu2+(aq) + 2e- AnodeCathodeWrite the overall balanced equationCu(s)an + Cu2+(aq)cat Cu2+(aq)an + Cu(s)cat cell = +What effect would these changes have on the cell voltage (or cell)?Increase [Cu2+]catIncrease [Cu2+]anDecrease [Cu2+]anDecrease [Cu2+]catElectrolytic CellsGalvanic cell (battery): spontaneous reaction, o = +Electrolytic cell: non-spontaneous reaction, = -An external power source is used to force the reaction to occurUsed for :Charging (rechargeable) batteriesProducing or purifying metals (aluminum, copper)ElectroplatingCharles HallDiscovered how to produce aluminum by electrolysis

Electrolysis of WaterSpontaneous reaction: formation of water2 H2(g) + O2(g) 2 H2O(l) 0 0 +1 -2As a galvanic cell:Cathode: O2 H2O(O2(g) + 4 H+(aq) + 4 e- 2 H2O(l))Anode: H2 H2O(2 H2(g) + 4 OH-(aq) 4 H2O(l) + 4 e-) = 2.06 V (voltage that can be used)Non-spontaneous reaction: electrolysis of water2 H2O(l) 2 H2(g) + O2(g)As an electrolytic cell:Anode (oxidation):2 H2O(l) O2(g) + 4 H+(aq) + 4 e-Cathode (reduction):4 H2O(l) + 4 e- 2 H2(g) + 4 OH-(aq) = -2.06 V (the voltage added for rxn to occur)Electrolysis of Water2 H2O(l) 2 H2(g) + O2(g) = -2.06 VCathode (reduction):4 H2O + 4 e- 2 H2 + 4 OH-Produces:2 mol H2 gasBase (OH-)Anode (oxidation):2 H2O O2 + 4 H+ + 4 e-Produces:1 mol O2 gasAcid (H+)Battery> 2.06 V+Pt electrodese- e-

60 s1 min31.998 gmol O21 mol O24 mol e-mol e-96,485 C3.0 CsElectrolysis CalculationsHow many grams of O2(g) will a 3.0 amp power source produce in 5 minutes?Amperes (A) = electric current = coulombs/second (C/s)

Andr-MarieAmpreOxygen half-reaction: 2 H2O O2 + 4 H+ + 4 e-Time and CurrentCharge(F)Moles of e-Moles ofproductGrams ofproduct=5.0 min0.075 g O23.0 CsElectroplatingCathodeCu2+(aq) + 2e- Cu(s)AnodeCu(s) Cu2+(aq) + 2e-Battery+Cu Cu2+Cu2+ e- e-How long will it take a 15 amp power source to deposit 5.9 g of copper?mol Cu63.55 g Cus15 C96,485 Cmol e -2 mol e - mol Cu=5.9 g Cu1200 s or 20. min or 0.33 hr or 0.014 d or 3.8 x 10-5 yr Time and CurrentCharge(F)Moles of e-Moles ofproductGrams ofproduct

Electric Vehicle Charging StationsBloomington-Normal, IL (as of 5/8/12)

Level II charger = full charge 6-8 hrsLevel III = 80% charge 25 minFree of charge (so far)First hour parking freeCity Hall Annex (6 units)Fire HQ (1 unit)Public Works (1 unit)Marriott Deck (2 units)College Ave. Parking Deck (3 units)Heartland C.C. (2 units)

ISU (2 units)City of Bloomington (2unitsLincoln Deck)IWU (2 units)CIRA (2 units)Commerce Bank (1 unit)Constitution Trail Center (2 units)Holiday Inn Express (1 unit)Advocate BroMenn (3 units)

Mitsubishi commerical

The New Normal

Electric Vehicle Comparison: 2012Nissan Leaf$36,000 MSRP base99 mpg eq. 70 mi hwy range*Home charging = 20 hrs from dead battery220 V = 6-7 hrs Battery cells: 8 yr/100,000 mi warranty

Mitsubishi i-MiEV$30,000 MSRP base112 mpg eq.62 mi hwy range*Home charging = 20 hrs from dead battery220 V = 6-7 hrs Battery cells: 8 yr/100,000 mi warranty

Both eligible for $7500 Federal and $3000 State tax creditwww.fueleconomy.govHybrid Vehicle Comparison: 2012Chevy Volt$39,145 MSRP base94 mpg eq. First 35 mi EVThen gas engine startsGas engine = 37 mpg (prem.)*Home charging = 8 hrs from dead battery220 V = 4 hrs Battery cells: 8 yr/100,000 mi warranty

Toyota Prius Plug-in Hybrid$35,760 MSRP base95 mpg eq.First 11 mi EVThen gas engine startsGas engine = 50 mpg*Home charging = 6 hrs from dead battery220 V = 1.5 hrs Battery cells: 8 yr/100,000 mi warranty