Gen-chem-2-2qtr
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Charles Darwin University**We aren't endorsed by this school
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CHEM 2
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Chemistry
Date
Dec 15, 2024
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17
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notes ni jettSPONTANEOUSPROCESSESfavortheformationofproductsatthe conditions underwhich the reaction is occurring.●Theychangeinonedirectionunder agiven set of conditions,without externalinfluence.NONSPONTANEOUS PROCESSESdo not favortheformationofproducts at the given set ofconditions.●Theyrequireanexternalinputofenergyto drive the change.ENTROPY●It can be thought of as a measure of thedisorderof a system.●In general, greater disorder means greaterentropy.●Rudolph Clasiuscoined the term Entropyin1865.●Lower entropydescribes theorganizedorderof a system.●Higherentropydescribestherandomness or chaosof a system.SECOND LAW OF THERMODYNAMICS●Theentropyoftheuniverse(closedsystem) isalways increasing.●Thechangein entropy will begreaterthan zero.●Entropy (S) is measured injoules perKelvin(j/K).●Entropy is not a measure of energy itselfbut ofhow energy is distributed withinthe system.●Enthalpy (H) describes the internal energyof a system.●An increase in entropy has avalue△?greaterthan zero, while a decrease inentropy has avalue less than zero.△?●Phasetransitions that produce amoredisordered arrangementof particles leadtoanincreaseinentropy:melting,vaporization, andsublimation.●Anincrease in temperatureleads to anincrease in entropybecause the additionof thermal energy brings more disorder tothe particles of the system.●Any chemical reaction that producesmoregas moleculesresults in anincrease inentropy.●Ifaprocessgeneratesentropy,itwillhappenspontaneouslyandwillbeirreversibleunlessyouinputmoreenergy.●A spontaneous process generates entropyand involves heat moving from a hotterbody to a colder body.UNIVERSE’S HEAT DEATHis a phenomenonwhere heat flowing from a hotter body to a colderbody will eventually lead to a point where all oftheatomsintheuniversearethermallyequilibrium.GIBBSFREEENERGYtellsusaboutthespontaneityof a process. Discussed by JosiahWillard Gibbs.Formula:△? = △? − ?△?Where:= Gibbs free energy△?= change in enthalpy△?= temperature (in Kelvin)?= change in entropy△?●Ifisnegative,theprocessis△?spontaneousandisconsideredexergonic.●Ifpositive,theprocessisnonspontaneousandisconsideredendergonic.●Ifzero, the process is atequilibrium.●Theofa process can be used to△?determine thespontaneityof the process.review well!
notes ni jettEXAMPLESGiven the following data, calculatefor this△?chemical reaction.1.??3??2?? + ?2→ ??3???? + ?2?Where:-495.2 kJ△? =-0.136 kJ/K△? =298 K? =△? = △? − ?△?(-495.2 kJ) – (298 K)(-0.136 kJ/K)△? =-454.672 kJ△? =-454.67 kJ△? ≈The process isspontaneous.(exergonic process)2.??4+ 2?2→ ??2+ ?2?Where:90.3 kJ△? =-0.242 kJ/K△? =298 K? =△? = △? − ?△?(90.3 kJ) – (298 K)(-0.242 kJ/K)△? =△?=162. 416 ??△?≈162. 42 ??The process isnonspontaneous.(endergonic process)●Chemical reactions do not always happenin one direction.●Most reactions happen in both directions.CHEMICAL EQUILIBRIUM●This is the state in which the rate of theforward reaction isequalto the rate of thereverse reaction.Theconditionatequilibriuminasystemischaracterized by anEQUILIBRIUM CONSTANT.●The symbol for equilibrium constant is.?●Thevalueofvarieswiththe?concentrationsof the substance presentattheequilibriumandwiththetemperature.●productoftheconcentrationsof? =products,divided bythe product of theconcentrations of reactants●, whereandare the? =[?𝑟?????]?[𝑟???????]???coefficients.EXAMPLES1.2?2(𝑔)+ ?2(𝑔)↔ 2?2?(?)? =[?𝑟?????]?[𝑟???????]?? =[?2?]2[?2]2[?2]2.2??(𝑔)+ 2?2(𝑔)↔ ?2(𝑔)+ 2?2?(?)? =[?2][?2?]2[??]2[?2]2review well!
notes ni jettThis tells us whether the reactants or productsare favored in the equilibrium.●If, that means that the chemical? > 1reaction is producingmore products.●If, that means that the chemical? < 1reaction is producingmore reactants.EXAMPLES1.?2(𝑔)+ ?2(𝑔)↔ ??3(𝑔)At equilibrium,,,?2=4. 26 ??2=2. 09 ?.Whatistheequilibrium? = 0. 060concentration of???3?2(𝑔)+ 3?2(𝑔)↔ 2??3(𝑔)? =[??3]2[?2][?2]30. 060 =[??3]2[4.26 ?][2.09 ?]3[??3]2=[0. 060][4. 26 ?][9. 129329 ?][??3]2=2. 333456492 ???3=1. 527565544 ???3≈1. 53 ?2.2??2(𝑔)↔ ?2?4(𝑔)If 0.95 M ofis added to a sealed 1.0?2?4L container and kept at 100°C, calculatethe equilibrium concentrations ofand??2. (take note that)?2?4? = 4. 7I.C.E. TABLE METHOD2??2?2?4Initial00. 95Change+ 2?− ?Equilibrium2?0. 95 − ?? =[?2?4][??2]24. 7 =[0.95−?][2?]20. 95 − ? = 4. 7(2?)20. 95 − ? = 18. 8?20 = 18. 8?2+ ? − 0. 95??2+ ??2+ ? = 018.8,1,-0.95? =? =? =Quadratic Formula? =−?±?2−4??2?? =−1±12−4(18.8)(−0.95)2(18.8)? = 0. 20Any negativefactors are not counted.???2= 2???2= 2(0. 20)??2=0. 40 ??2?4= 0. 95 − ??2?4= 0. 95 − 0. 20?2?4=0. 75 ?3.?2?4(𝑔)↔ 2??2(𝑔)The equilibrium constant of the chemicalreaction at 373 K is 0.36. If the initialconcentration ofis 0.10 M, what is?2?4the equilibrium concentration of???2?2?42??2Initial0. 100Change− ?+ 2?Equilibrium0. 10 − ?2?? =[??2]2[?2?4]0. 36 =[2?]2[0.10−?][2?]2= [0. 36][0. 10 − ?]4?2= 0. 036 − 0. 36?4?2+ 0. 36? − 0. 036 = 0review well!
notes ni jett? =−0.36±0.362−4(4)(−0.036)2(4)? = 0. 06??2= 2???2= 2(0. 06)??2=0. 12 ?4.Nitrogen reacts with chlorine to producenitrogentrichloride.Atequilibrium,theconcentrations of each gas were found tobe,,and[?2]=0. 15 ?[??2]=0. 25 ?. Calculate the value of[???3]=0. 50 ?the equilibrium constant.?2+ ??2↔ ???3?2+ 3??2↔ 2???3? =[???3]2[?2][??2]3? =[0.50 ?]2[0.15 ?][0.25 ?]3? =0.25 ?0.00234375 ?? ≈ 106. 675.20 mol ofis placed inside an empty????4-L container. At equilibrium, 8 mol of??2was found to be in the container. Calculatethe value of K for this reaction.2????(𝑔)↔ 2??(𝑔)+ ??2(𝑔)20 ???????÷4 ?=5 ??????=8 ?????2÷4 ?=2 ???22????2????2500− 2(2)+ 2(2)+ 2142? =[??]2[??2][????]2? =[4 ?]2[2 ?][1 ?]2? =321? = 326.Carbon monoxide reacts with oxygen gasto produce carbon dioxide. At equilibrium,theconcentrationsofandare?2??2andrespectively.If0. 10 ?0. 75 ?, what is the concentration of? = 4 × 103at equilibrium????? + ?2↔ 2??22?? + ?2↔ 2??2? =[??2]2[??]2[?2]Note thatand?? = ?4 × 103= 40004000 =[0.75 ?]2[?]2[0.10 ?][4000][?2][0. 10 ?]=[0. 75 ?]2400?2= 0. 5625?2= 0. 00140625𝑥 = 0. 03757.A 0.200 mol of iodine and 0.20 mol ofchlorine are placed in a 5.00-L flask at25°C. Calculate the concentrations of allreactants and products at equilibrium if thefollowing reaction takes place.?2+ ??2↔ 2???? = 81. 90. 200 = 0. 20. 20 = 0. 20. 2 ???÷5 ?=0. 04 ??2??22???0. 040. 040− ?− ?+ 2?0. 04 − ?0. 04 − ?2?review well!
notes ni jett? =[???]2[?2][??2]81. 9 =[2?]2[0.04−?]281. 9[0. 04 − ?]2= [2?]281. 9[0. 0016 − 0. 08? + ?2] = [2?]20. 13104 − 6. 552? + 81. 9?2= 4?277. 9?2− 6. 552? + 0. 13104 = 0? =−(−6.552)±(−6.552)2−4(77.9)(0.13104)2(77.9)?1≈ 0. 05?2≈ 0. 03The condition of the fraction is.? < 0. 04? = 0. 032760090? ≈ 0. 033?2= 0. 04 − ??2= 0. 04 − (~0. 033)?2=0. 007239910 ??2≈0. 007 ??2= ??2??2≈0. 007 ?2??? = 2?2??? = 2(~0. 033)2???=0. 065520180 ?2???≈0. 07 ?8.At, thevalue is 0.91. Initially,325 ??isplacedina1.00L0. 34 ??? ?2?4container and equilibrium is established.Determine the equilibrium concentration ofeach gas in the container.?2?4↔ 2??20. 34 ????2?4÷1 ?=0. 34 ??2?4?2?42??2Initial0. 340Change− ?+ 2?Equilibrium0. 34 − ?2?? =[??2]2[?2?4]0. 91 =[2?]2[0.34−?][0. 91][0. 34 − ?] = [2?]20. 3094 − 0. 91? = 2?20 = 2?2+ 0. 91? − 0. 3094? =−0.91±0.912−4(2)(−0.3094)2(2)? = 0. 226874570? ≈ 0. 23?2?4= 0. 34 − ??2?4= 0. 34 − (0. 23)?2?4=0. 113125430 ??2?4≈0. 11 ?2??2= 2?2??2= 2(0. 23)2??2=0. 453749140 ?2??2≈0. 45 ?review well!
notes ni jett●In 1884,Henri Le Chatelierstudied theeffect of the changes in the conditions ofthe systems in equilibrium.●LE CHATELIER’S PRINCIPLEstates that“asysteminequilibriumreactstoastress(condition changes) in a way thatcounteractsthe stress andestablishes anew state of equilibrium.”1. CHANGE IN CONCENTRATION●If the concentration of a substance hasincreased,theequilibriumwillfavor areaction that will decrease(or consume)theexcessconcentrationofthatsubstance.●Solids and liquids do not appearin theequilibrium constant expression becausetheirconcentrationdoesnot changesignificantly.Thus,thepositionofequilibrium remains unchanged.●Example:-2???(𝑔)↔ ?2(𝑔)+ ??2(𝑔)2. CHANGE IN PRESSURE●Changeinpressureaffectsonlytheequilibriumsysteminvolvinggaseoussubstances.●Anincreaseinthepressureofthesystem favors the side of the reaction thatproducesgreater to lesser numberofmolesofgaseoussubstances,anddecrease in the pressure favorslesser togreatermoles.●Examples:-?2(𝑔)+ 3?2(𝑔)↔ 2??3(𝑔)-?2(𝑔)+ ?2(𝑔)↔ 2??(𝑔)3. CHANGE IN TEMPERATURE●A change in energy can beendothermicorexothermic.●The heat formation of anendothermicreaction is positive.△?=(+)●Theheatformationofanexothermicreaction is negative.△?=(−)●If the equilibrium of a system is disturbedbyanincreaseintemperature,thesystem willfavor the forward reaction.●If the equilibrium of a system is disturbedbyadecreaseintemperature,thesystem willfavor the reverse reaction.●Examples:-2??3(𝑔)↔ ?2(𝑔)+ 3?2(𝑔)△? = (+)-??(𝑔)+ ?2(𝑔)↔ ?(𝑔)+ ?2?(𝑔)△? = (−)4. CATALYST●Itgenerallyincreasestherateofachemical reaction.●If it is added to a system in equilibrium, itwill accelerate both the forward and thebackward reactions.●The presence of a catalystwill not affectthe equilibrium system.EXAMPLEConsider the following equilibrium:?2?2(𝑔)↔ ?2(𝑔)+ ?2(𝑔)△?=150 ??Which way will the equilibrium shift if:a.Moreis added.?2Reverse reaction(left)(use up)?2b.is removed.?2Forward reaction(right)(restore)?2c.Temperature has increased.Forward reaction(right)(endothermic, heat is a reactant)d.The volume is reduced.Reverse reaction(left)(reduce moles of gas to relieve excesspressure)review well!
notes ni jettARRHENIUS THEORY OF ACIDS AND BASESdefinedacidsassubstancesthatreleaseprotons in a solutionandbasesas substancesthatrelease hydroxide ions in a solution.BRØNSTED-LOWRYTHEORYinvolvesthetransfer of a protonas a function of pH.●BRØNSTED-LOWRYACIDSdonateprotons,whileBRØNSTED-LOWRYBASESacceptprotons.●Substancesthatcan act as either anacidorabaseareconsideredAMPHOTERIC. Example: Water?2?●HYDRONIUMIONS(akaoxonium?3?ions) are formed whenwater moleculesaccept hydrogenions.?●Whenever ahydrogen atom is addedtoa molecule, it becomes aCONJUGATEACID.●Wheneverahydrogenatomis takenawayfromamolecule,itbecomesaCONJUGATE BASE.EXAMPLESSubstancesConjugateacidConjugatebase???3?2??3??3??3??4??2???4?2??4??4●When a Brønsted-Lowry acidreleasesaproton,theresultingsubstanceisitsconjugate base.●When a Brønsted-Lowry baseacceptsaproton,theresultingsubstanceisitsconjugate acid.EXAMPLESConjugate Pair 1Conjugate Pair 2BL acidC. baseBL baseC. acid1.?? + ?2? ↔ ? + ?3?????2??3?2.??3+ ?2? ↔ ???3+ ???2?????3???33.?2??4+ ??3↔ ???4+ ??4?2𝑆?4?𝑆?4??3??44.??3+ ?2? ↔ ??4+ ???2?????3??45.??3?? + ?2? ↔ ?3? + ??3???3????3??2??3?review well!
notes ni jett●LEVELaffects the equilibrium state??of a reversible reaction.●Søren Peder Lauritz Sørensencoinedtheterm,wherestandsfor???Hydrogen.●isameasureofthestrengthof??Hydrogen ionsin an acid or base[?+]character of a substance.●isameasure of thestrength of???Hydroxide ionsin an acid or base[??−]character of a substance.●When the concentration ofHydrogen[?+]ions increase, thelevel gets lower,𝑝?and vice versa.●WhentheconcentrationofHydroxideions increase, thelevel gets[??−]𝑝?higher, and vice versa.●Thesumofandisalways equal?????to 14.●Litmus paper turns into redwhen thesolution isacidandturns into bluewhenthe solution isbase.●MONOPROTIC ACIDis a substance thatdonates anioninto a solution.[?+]●POLYPROTIC ACIDis a substance thatdonatesmultipleionsintoa[?+]solution.FORMULAS●?? = − ??𝑔[?+/?3?+]●??? = − ??𝑔[??−]●[?+/?3?+] = 10−??●[??−] = 10−???●?? + ??? = 14●(?+/?3?+)(??−) = 1 × 10−14●Unit forismolar[?+/?3?+/??−]?●andhave no unit?????EXAMPLES1.What is theof a solution if theis???+?1 × 10−3??? = − ??𝑔[?+]?? = − ??𝑔[1 × 10−3?]𝑝? = 32.What is thelevel of a solution if the??is?[?+]1. 4 × 10−5??? = − ??𝑔[1. 4 × 10−5?]𝑝? ≈ 4. 853.Theof the solution is. Calculate???4. 5theof the solution.???? + ??? = 14?? = 14 − ????? = 14 − 4. 5𝑝? = 9. 54.Iftheis,whatisthe???3. 8[??−]concentration?[??−] = 10−???[??−] = 10−3.8[??−] ≈ 1. 58 × 10−4?5.What is theof asolution with a???3??4molarity of?1. 5 × 10−4?has threeatoms. Therefore, the?3??4?molarity must be multiplied by 3.?? = − ??𝑔[(3)(1. 5 × 10−4?)]𝑝? ≈ 3. 356.What is theof the solution if the???is?[?3?+]4. 2 × 10−3??? = − ??𝑔[4. 2 × 10−3?]?? ≈ 2. 38??? = 14 − ????? = 14 − (2. 38)𝑝?? ≈ 11. 627.Ifis, calculate the[?3?+]7. 1 × 10−2?concentration.[??−]?? = − ??𝑔[7. 1 × 10−2?]?? ≈ 1. 15??? = 14 − (1. 15)??? ≈ 12. 85[??−] = 10−(12.85)[??−] ≈ 1. 41 × 10−13review well!
notes ni jett●It is a solution that canresistchange𝑝?upon theaddition of an acidic or basiccomponent.●It is able toneutralize small amounts ofadded acid or base, thus maintaining theof the solution relatively stable.𝑝?●Madeupofaweakacidanditsconjugate baseACIDIC BUFFER.●Theweak acid consumes theions??−andconjugate base consumesions?+added to the solution.●Madeupofaweakbaseanditsconjugate acid (basic buffer).●Theweak base consumes theions?+andconjugate acid consumesions??−added to the solution.review well!
notes ni jettHENDERSON-HASSELBACH FORMULA●Thisequationprovidesarelationshipbetweentheofacids (in aqueous𝑝?solutions)andtheiraciddissociationconstant.𝑝??●ItwasfirstderivedbytheAmericanchemistLawrence Joseph Henderson,then re-expressed in logarithmic terms byDanish chemistKarl Albert Hasselbalch.FORMULAS●?? = ???+ ??𝑔([?−] ÷ [??])●???= − ??𝑔(??)●???+ ???= 14●1 × 10−14= ????●If moles of a strong base likeare??−added to a buffer solution, it will consumethe acidand produce the base.[??][?−]As such, the calculation should have themoles added to the base and subtractedfrom the acid, given their initial molarities.●However, if moles of a strong acid like?+areaddedtoabuffersolution,theconsumption will be the opposite. As suchthecalculationshouldhavethemolesadded to the acid instead and subtractedfrom the base, given their initial molarities.Weak acids???????6. 8 × 10−43. 17???2?1. 8 × 10−43. 74??2?3?21. 8 × 10−54. 74???4. 9 × 10−109. 31EXAMPLES1.What is theof a solution consisting of??and?0. 75 ? ??2?3?20. 50 ? ?2?3?2[?−] = ?2?3?2[??] = ??2?3?2??= 1. 8 × 10−5???= − ??𝑔(??)???= − ??𝑔(1. 8 × 10−5)???≈ 4. 74?? = ???+ ??𝑔([?−] ÷ [??])??=(4. 74)+??𝑔(0. 50 ?÷0. 75 ?)𝑝? ≈ 4. 572.What is theof a solution consisting of??ofandof?0. 15 ?????1. 5 ????[?−] = ???[??] = ??????= 1. 5 × 10−4???= − ??𝑔(1. 5 × 10−4)???≈ 3. 82??=(3. 82)+??𝑔(1. 5 ?÷0. 15 ?)𝑝? ≈ 4. 82review well!
notes ni jett3.What is theof a solution of??0. 2 ? ??andif its?0. 1 ? ???= 6. 3 × 10−4[?−] = ?[??] = ?????= − ??𝑔(6. 3 × 10−4)???≈ 3. 20??=(3. 20)+??𝑔(0. 1 ?÷0. 2 ?)𝑝? ≈ 2. 904.What is theof a buffer solution that is??andwhose0. 24 ? ??30. 20 ? ??4??? Ifof??= 5. 6 × 10−100. 005 ???????is added toof the buffer solution,0. 50 ?what is the resulting???[?−] = ??3[??] = ??4?????= − ??𝑔(5. 6 × 10−10)???≈ 9. 25??=(9. 25)+??𝑔(0. 24 ?÷0. 20 ?)𝑝?𝑖𝑛𝑖?𝑖??≈ 9. 33??3=(0. 24 ?×0. 50 ?)+0. 005 ?????3=0. 12 ???+0. 005 ?????3=0. 125 ???÷0. 50 ???3=0. 25 ???4??=(0. 20 ?×0. 50 ?)−0. 005 ?????4??=0. 1 ???−0. 005 ?????4??=0. 095 ???÷0. 50 ???4??=0. 19 ???=(9. 25)+??𝑔(0. 25 ?÷0. 19 ?)𝑝??𝑖𝑛??≈ 9. 375.Refer to the first part of Question 4. Ifofis added toof the0. 03 ??????0. 50 ?buffer solution, what is the resulting?????3=0. 12 ???−0. 03 ?????3=0. 09 ???÷0. 50 ???3=0. 18 ???4??=0. 1 ???+0. 03 ?????4??=0. 13 ???÷0. 50 ???4??=0. 26 ???=(9. 25)+??𝑔(0. 13 ?÷0. 26 ?)𝑝??𝑖𝑛??≈ 9. 096.Calculate the finalof a buffer solution??containingsodiumacetateand0. 1 ?acetic acidfollowing0. 2 ?[???= 4. 76]the addition of.0. 05 ? ????[?−] = ??3?????[??] = ??3????[?−]=0. 1 ?+0. 05 ?[?−]=0. 15 ?[??]=0. 2 ?−0. 05 ?[??]=0. 15 ???=4. 76+??𝑔(0. 15 ?÷0. 15 ?)𝑝? = 4. 76If both the acid and the base have thesame molarity afteris added to the??−buffer solution,.𝑝? = 𝑝??7.What is theof a solution containing??ofandof?0. 15 ???4??1. 5 ???3??= 1. 8 × 10−5[?−] = ??3[??] = ??4?????= − ??𝑔(??)???= − ??𝑔(1. 8 × 10−5)???≈ 4. 74???+ ???= 14???= 14 − ??????= 14 − (4. 74)???≈ 9. 26??=(9. 26)+??𝑔(1. 5 ?÷0. 15 ?)𝑝? ≈ 10. 268.Calculate theof a buffer that is??0. 12 ?lactic acid andsodium lactate.0. 11 ???= 1. 4 × 10−4[?−] = ???3?5?3[??] = ?3?6?3???= − ??𝑔(1. 4 × 10−4)???≈ 3. 85??=(3. 85)+??𝑔(0. 11 ?÷0. 12 ?)𝑝? ≈ 3. 82review well!
notes ni jett9.Howmanygramsofsodiumlactateisbeingaddedtoof???3?5?31. 0 ?lactic acid to form a buffer solution0. 15 ?withof???4. 00??= 1. 4 × 10−4[?−] = ???3?5?3[??] = ?3?6?3???= − ??𝑔(1. 4 × 10−4)???≈ 3. 85?? = ???+ ??𝑔([?−] ÷ [??])?? − ???= ??𝑔([?−] ÷ [??])[?−][??]= 10??−???[?−]0.15 ?= 104.00−(3.85)[?−]0.15 ?= 1. 40. 15 ?×1. 0 ?=0. 15 ???[?−]=(1. 4)(0. 15 ???)[?−]=0. 21 ???[?−]=22. 99 𝑔+3(12. 01 𝑔)+5(1. 01 𝑔)+3(16 𝑔)[?−]=22. 99 𝑔+36. 03 𝑔+5. 05 𝑔+48 𝑔[?−]=112. 07 𝑔/???×0. 21 ???[?−]=23. 5347 ?≈23. 53 ?10. Calculate theof a solution formed by??mixingofwith250. 0 ??0. 15 ? ??4??of. Thefor100. 0 ??0. 20 ? ??3????3is.1. 8 × 10−5[?−] = ??3[??] = ??4??100. 0 ??=0. 1 ?250. 0 ??=0. 25 ?0. 25 ?+0. 1 ?=0. 35 ?[?−]=0. 20 ?×0. 1 ?[?−]=0. 02 ???÷0. 35 ?[?−]≈0. 06 ?[??]=0. 15 ?×0. 25 ?[??]=0. 0375 ???÷0. 35 ?[??]≈0. 11 ????= − ??𝑔(1. 8 × 10−5)???≈ 4. 74???= 14 − (4. 74)???≈ 9. 26??=(9. 26)+??𝑔[(0. 06 ?)÷(0. 11 ?)]𝑝? ≈ 8. 98review well!
notes ni jettOXIDATION REACTIONis defined as a chemicalchange in whichelectrons are lostby an atom ora group of atoms.●In this reaction, theoxidation numberofan elementincreases.REDUCTIONREACTIONisdefinedasachemical change in whichelectrons are gainedby an atom or a group of atoms.●In this reaction, theoxidation numberofan elementdecreases.Therearecertainchemicalreactionsthatinvolvechanges in the oxidation numbersofthe elements.●ThisispopularlyknownasREDUCTION-OXIDATIONREACTIONSor simplyREDOX REACTIONS.●Oxidation and reduction reactionsalwaystake place simultaneously.RULES IN WRITING OXIDATION NUMBERS1.The oxidation number of an atom iszeroinaneutralsubstancethatcontainsatoms ofonly one element.2.If it is amonatomic ion, the oxidationnumber isthe same with its ion charge.3.Oxygeninamoleculeisalmostandalways equal to.− 24.The oxidation number ofHydrogencaneitherbe(whencombinedto+ 1nonmetal) and(when combined to− 1metal).5.Forneutralmolecules,oxidationnumbers mustadd up to zero.6.Forpolyatomicions,theoxidationnumber mustadd up to the charge ofthe ion.Examples1.???Because, then.?? = − 1? = + 12.?𝑔?Because, then.? = − 2?? = + 23.??3isanonmetal,therefore,?? = + 1whichmeans.Assuch,?3= + 3.? = − 34.???22−has a charge of. Becausehas??+ 2?2anadditionalchargeof,the− 2molecule isneutral.5.??2−Treatas.??? + 2(?) = − 1? + 2(− 2) = − 1? − 4 = − 1? = + 3? = + 36.?𝑟?3−Treatas.?𝑟??𝑟 + 3(?) = − 1? + 3(− 2) = − 1? − 6 = − 1? = + 5?𝑟 = + 57.4?? + 3?2→ 2??2?3oxidized,reduced???2(Reducingagentsoxidize,whileoxidizing agents reduce)8.2?𝑔 + ?2→ 2?𝑔?oxidized,reduced???29.?? + 2??? → ?2+ ????2oxidized,reduced?𝑛???10.2???𝑟 + ??2→ 2???? + ?𝑟2oxidized,reduced???𝑟??211.?? + ????2→ ????3+ ??oxidized,reduced??????212.2?? + 2???3→ ??2?3+ ?? + ?2?oxidized,reduced𝑆????3review well!
notes ni jettREDOX REACTIONSare reactions in whichonereactantisoxidizedandonereactantisreduced simultaneously.●Tomaintaina balanced reaction, theredox reaction willentailbotha reductioncomponentandanoxidationcomponent.●Theseareoftenseparatedintoindependenttwohypotheticalhalf-reactions. This requiresidentifyingwhichelementisoxidizedandwhichelement isreduced.InACIDICSOLUTIONS,eachreactionisbalanced byadjusting coefficientsandadding,, andin this order:?2??+?−1.Balance elementsin the equationotherthanand.??2.Balance the oxygen atomsbyaddingthe appropriate number of water?2?moleculesto theopposite sideof theequation.3.Balance the hydrogen atoms(includingthoseaddedinstep2to balance theoxygen atom)by addingionsto the?+opposite sideof the equation.4.Add up the charges on each side.Makethem equal byadding enough electronsto themore positive side.As a rule of?−thumb,bothandarealmost?−?+always on the same side.5.Theoneachsidemustbemade?−equal; if they arenot equal, they must bemultiplied by appropriate integers(thelowest common multiple) to be madethe same.6.The half-equations areadded together,canceling out the electronsto form onebalancedequation.Commontermsshould also be canceled out.Examples1.?𝑔 + ??3−→ ?𝑔++ ??(oxidation HR)?𝑔 → ?𝑔+is balanced. Noor. Only add.?𝑔???−?𝑔 → ?𝑔++ ?−Charges:0 → 0(reduction HR)??3−→ ??is balanced. Addto the product.?2?2?..2?2= 4?+??3−= − 1. Therefore, add.??3−+ 4?+= + 33?−??3−+ 4?++ 3?−→ ?? + 2?2?Charges:0 → 0BecauseofinthereductionHR,3?−multiply the oxidation HR by 3 to cancelout the electrons.3?𝑔 → 3?𝑔++ 3?−??3−+ 4?++ 3?−→ ?? + ?2?Add the two half-reactions.3?? + ??3−+ 4?+→ 3??++ ?? + ?2?Charges:+ 3 → + 32.?? + ???3−→ ??2++ ??(oxidation HR)?? → ??2+is balanced. Noor. Add.????2?−?? → ??2++ 2?−Charges:0 → 0(reduction HR)???3−→ ??is balanced.to the product.????? 3?2?..3?2= 6?+???3−= − 1. Therefore, add.???3−+ 6?+= + 55?−???3−+ 6?++ 5?−→ ?? + 3?2?Charges:0 → 0. Multiply oxidation HR by???(2, 5)= 10, reduction HR by.525?? → 5??2++ 10?−2???3−+ 12?++ 10?−→ 2?? + 6?2?Add the two half-reactions.5?𝑛 + 2???3−+ 12?+→ 5?𝑛2++ 2?? + 6?2?Charges:+ 10 → + 10review well!
notes ni jett3.?? + ???4−→ ??3++ ??2+(oxidation HR)?? → ??3+?? → ??3++ 3?−Charges:0 → 0(reduction HR)???4−→ ??2+???4−+ 8?+→ ??2++ 4?2????4−+ 8?+− ??2+= + 5???4−+ 8?++ 5?−→ ??2++ 4?2?Charges:+ 2 → + 2???(3, 5)=15; ???×5; ???× 35?? → 5??3++ 15?−3???4−+ 24?++ 15?−→ 3??2++ 12?2?——————————————————5?? + 3?𝑛?4−+ 24?+→ 5??3++ 3?𝑛2++ 12?2?Charges:+ 21 → + 214.?𝑟2?72−+ ???2→ ?𝑟3++ ??3−(oxidation HR)???2→ ??3−???2+ ?2? → ??3−+ 3?++ 2?−Charges:0 → 0(reduction HR)?𝑟2?72−→ ?𝑟3+?𝑟2?72−→ 2?𝑟3+?𝑟2?72−+ 14?+→ 2?𝑟3++ 7?2??𝑟2?72−= − 2?𝑟2?72−+ 14?+− 2?𝑟3+= + 6?𝑟2?72−+ 14?++ 6?−→ 2?𝑟3++ 7?2?Charges:+ 6 → + 6???(2, 6)=6; ???×3; ???× 13???2+ 3?2? → 3??3−+ 9?++ 6?−?𝑟2?72−+ 14?++ 6?−→ 2?𝑟3++ 7?2?——————————————————?𝑟2?72−+ 5?++ 3???2→ 2?𝑟3++ 4?2? + 3??3−Charges:+ 3 → + 35.???4−+ ?? → ???2+ ??(??)42−(oxidation HR)?? → ??(??)42−?? + 4?2? → ??(??)42−4?2− ?4= 4?+?? + 4?2? → ??(??)42−+ 4?+??(??)42−= ???4?42−??(??)42−+ 4?+= + 2.?? + 4?2? → ??(??)42−+ 4?++ 2?−Charges:0 → 0(reduction HR)???4−→ ???2???4−+ 4?+→ ???2+ 2?2????4−+ 4?+= + 3???4−+ 4?++ 3?−→ ???2+ 2?2?Charges:0 → 0???(2, 3)=6; ???×3; ???× 23?? + 12?2? → 3??(??)42−+ 12?++ 6?−2???4−+ 8?++ 6?−→ 2???2+ 4?2?——————————————————2?𝑛?4−+ 3?𝑛 + 8?2?→ 2?𝑛?2+ 3?𝑛(??)42−+ 4?+Charges:− 2 → − 2InBASIC SOLUTIONS, the steps are similar toacidic balancing, but with one additional step.●Refer to steps 3 and 4 from earlier. Dothis after step 3 but before step 4.●Add the appropriate number ofto??−neutralizealland toconvert into?+watermolecules.●Basesdissolveintoionsina??−solution. Hence, balancing redox reactionsinbasic conditions requires.??−review well!
notes ni jettExamples1.?𝑔 + ??2+→ ?𝑔2? + ??(oxidation HR)?𝑔 → ?𝑔2?is not balanced.?𝑔2?𝑔 → ?𝑔2?Addto the reactant..?2??2= 2?+Addto both sides to neutralize.2??−2?𝑔 + ?2? + 2??−→ ?𝑔2? + 2?++ 2??−Combinethetwohydrogenandtwohydroxide to form two water molecules.Simplifytheequation by cancelling outcommonmolecules.Then,balancethecharges by adding two electrons.2?𝑔 + 2??−→ ?𝑔2? + ?2? + 2?−Charges:− 2 → − 2(reduction HR)??2+→ ??is balanced. Noor. Only add.????2?−??2++ 2?−→ ??Charges:0 → 0Add the two half-reactions.2?? + 2??−+ ?𝑛2+→ ??2? + ?2? + ?𝑛Charges:0 → 02.?? + ?𝑟?3−→ ??2++ ?𝑟−(oxidation HR)?? → ??2+is balanced. Noor. Only add.????2?−?? → ??2++ 2?−Charges:0 → 0(reduction HR)?𝑟?3−→ ?𝑟−is balanced. Addto the product.?𝑟3?2?. Addon both sides.3?2= 6?+6??−?𝑟?3−+ 6?++ 6??−→ ?𝑟−+ 3?2? + 6??−Combine hydrogen and hydroxide to formwater. Simplify, then add electrons.?𝑟?3−+ 3?2? + 6?−→ ?𝑟−+ 6??−Charges:− 7 → − 7. Multiply the oxidation HR???(2, 6)= 6by. Keep the reduction HR as is.33?? → 3??2++ 6?−?𝑟?3−+ 3?2? + 6?−→ ?𝑟−+ 6??−Add the two half-reactions.3?𝑛 + ?𝑟?3−+ 3?2?→ 3?𝑛2++ ?𝑟−+ 6??−Charges:− 1 → − 13.?? + ???4−→ ??3++ ??−(oxidation HR)?? → ??3+?? → ??3++ 3?−Charges:0 → 0(reduction HR)???4−→ ??−???4−+ 4?+→ ??−+ 4?2????4−+ 4?2? → ??−+ 4?2? + 4??−???4−+ 4?−→ ??−+ 4??−Charges:− 5 → − 5???(3, 4)=12; ???×4; ???× 34?? → 4??3++ 12?−3???4−+ 12?−→ 3??−+ 12??−——————————————————4?? + 3???4−→ 4??3++ 3??−+ 12??−Charges:− 3 → − 3review well!
notes ni jett4.?? + ??3−→ ??2++ ??2(oxidation HR)?? → ??2+?? → ??2++ 2?−Charges:0 → 0(reduction HR)??3−→ ??2??3−+ 2?+→ ??2+ ?2???3−+ 2?2? → ??2+ ?2? + 2??−??3−+ ?2? + ?−→ ??2+ 2??−Charges:− 2 → − 2???(2, 1)=2; ???×1; ???× 2?? → ??2++ 2?−2??3−+ 2?2? + 2?−→ 2??2+ 4??−——————————————————?𝑛 + 2??3−+ 2?2?→ ?𝑛2++ 2??2+ 4??−Charges:− 2 → − 25.???4−+ ?? → ???2+ ??(??)42−(oxidation HR)?? → ??(??)42−?? + 4?2? → ??(??)42−+ 4?+?? + 4?2? + 4??−→ ??(??)42−+ 4?2??? + 4??−→ ??(??)42−+ 2?−Charges:− 4 → − 4(reduction HR)???4−→ ???2???4−+ 4?+→ ???2+ 2?2????4−+ 4?2? → ???2+ 2?2? + 4??−???4−+ 2?2? + 3?−→ ???2+ 4??−Charges:− 4 → − 4???(2, 3)=6; ???×3; ???× 23?? + 12??−→ 3??(??)42−+ 6?−2???4−+ 4?2? + 6?−→ 2???2+ 8??−——————————————————2?𝑛?4−+ 4?2? + 3?𝑛 + 4??−→ 2?𝑛?2+ 3?𝑛(??)42−Charges:− 6 → − 6EDITOR’SNOTE:Theexammayalsoaskquestions concerning general information onbatteries (Lelanche, Button, Fuel Cell, LSB).Duetopossibledeviationsofinformationacrossfoursections,asbatterieswerediscussed by way of group reports, it is notincluded in this reviewer. Let’s just sayI ranout of juice in my battery.I know, corny…Let’s pretend that’s Chemistry and not Physics lolreview well!