LAB GENERAL CHEMIST
LAB GENERAL CHEMIST CHEM 178L
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Chapter 20 Electrochemistry Electrochemistry is the study of the relationships between current and chemical reactions In an electrochemical reactions electrons are transferred from one species to another j LEO the Lion goes GER DO you recall thls Losing Electrons is Oxidation Gaining Electrons is Reduction Redox reaction Both Oxidation and Reduction occurs in a single electrochemical reaction 201 Oxidation Numbers Oxidation State In a chemical reactions in order to keep track of what loses electrons and what gains the electrons we assign oxidation state or oxidation numbers for each element in the chemical compound For example If you put a piece of zinc metal in an acid solution you will see brisk effervescence of H2 gas see figure The redox reaction that take place in this experiment can be v 0 9 written as follows i o m e Zn5 2 I lClaq ZnClglmy 1 135 Zns 2Haq gt Zn2aQ H2g Let us see how to assign oxidation state 2 44 Assigning Oxidation Numbers m 7 l r b Hun V u rgr m Q sxvii quotr 939 if z toga 3 a quotr lFilaeazee metUrele gaggllgzm Am ilQl3 Follow these rules 1 Elements in their elemental form have an oxidation number of O For example gt H atom in H2 molecule has an oxidation number 0 gt P atom in P4 molecule has an oxidation number 0 2 The oxidation number of a monatomic ion is the same as its charge gt K gt oxidation number 1 gt 8239 gt oxidation number 2 gt Ali gt oxidation number 3 44 Assigning Oxidation Numbers Contd 3 Nonmetals tend to have negative oxidation numbers although some are positive in certain compounds or ions gt V Oxygen has an oxidation number of 2 except in the peroxide ion 02239 ion in which the oxidation number is 1 Hydrogen is 1 when bonded to a metal NaH 1 when bonded to a nonmetal HCI Fluorine always has an oxidation number of 1 The other halogens have an oxidation number of 1 in most binary compounds HCI When combined with oxygen as in oxyanions HOCI they have positive oxidation states 44 Assigning Oxidation Numbers Contd 4 The sum of the oxidation numbers in a neutral compound is O For example NaCl Na 1 and Cl391 5 The sum of the oxidation numbers in a polyatomic ion is the charge on the ion For example H3O H 1 and O 2 What is the oxidation state of the Red element in each of the following a P205 b NaH c Cr207239 d SnBr4 e BaO2 201 Oxidation State Contd Now we will go back to slide2 and write the oxidation state of each species in the reaction as shown below Zns 2 Haq gt Zn2aq H2g db A species is oxidized when it loses electrons gt Here zinc loses two electrons to go from neutral zinc metal to the Zn2 ion So Zn is getting oxidized A species is reduced when it gains electrons gt Here each of the H gains an electron and they combine to form H2 80 H is getting reduced 201 Oxidation State Contd 2115 2 Haq gt Zn2aq H2g do Species which is getting reduced is called the oxidizing agent gt H oxidizes Zn by taking electrons from it gt Thus H is an oxidizing agent Oxidant Species which is getting oxidized is called the reducing agent gt Zn reduces H by giving it electrons gt Thus Zn is a reducing agent Reductant Look at the sample exercise 201 in page 849 to understand this concept Practice Exercise Identify the oxidizing and reducing agents in the following redox reactions a 2H20I As MnO4 aq gtAIOH4 aq Mn02s b 2Hg2aCI N2H4aCI gt 2H910 N2g 4HaCI 202 Balancing OxidationReduction Equation Whenever we balance a chemical reaction we follow law of conservation of mass number or amount of each element should be balanced on both sides of the equation In a redox reaction in addition to balancing the elements we need to balance the number of electrons lost and gained We do this task by considering the redox reaction into two halfreactions 202 Balancing Redox Equation Contd For example in the following reaction we can identify two halfreactions one corresponding to an oxidation and another corresponding to a reduction Remember Oxidation and Reduction always occur simultaneously in a redox reaction Sn2aq 2Fe3aq gt Sn4aq 2Fe2aq Oxidation Sn2aq Sn4aq 2e Reduction 2Fe3aq 2e 2Fe2aq After writing the two halfreactions follow these steps described in the next page Text book page 847 10 202 Balancing Redox Equation Contd Suppose the redox reaction is happening in an acidic aqueous solution 1 Identify two halfreactions 2 Balance each halfreaction a Balance elements other than H and O b Balance 0 by adding H20 0 Balance H by adding H d Balance charge by adding electrons 3 Multiply the halfreactions by integers so that the electrons gained and lost are the same 11 202 Balancing Redox Equation Contd 4 Add the halfreactions and cancel out species that appear on both sides 5 Check a Number of elements should be equal on both sides b Number of positive or negative charges should be equal on both sides Balance the following redox reaction taking place in an acidic aqueous solution MnO439aq C2042 aq gt Mn2aq C02aq 12 Step 1 First assign oxidation num logers MnO4 02042 gt Mn2 002 Since the manganese goes from 7 to 2 it is reduced Since the carbon goes from 3 to 4 it is oxidized Step 2 Write the two halfreactions separately Oxidation C2042 aq gt CO2 aq Reduction MnO4 gt Mn2aq Step 3 Balance each halfreaction by following rules 13 Step 3 Balance each halfreaction by following rules Halfreaction 1 Halfreaction 2 020 gt CO2 MnO4 gt Mn2 To balance the carbon we add a coefficient of 2 02042 gt 2 002 The manganese is balanced to balance the oxygen we must add 4 waters to the right side MnO4 gt Mn2 4 H20 The oxygen IS now balanced as well To balance the To balance the hydrogen we charge we must add 2 add 8 H to the left side electrons to the right side 8 H MnO439 gt Mn2 4 H20 02042 gt 2 CO2 2 e To balance the charge we add 5 e to the left side 5 e39 8 H MnO439 gt Mn2 4 H20 Step 4 Add both balanced halfreactions Before adding the two halfreactions make sure that there is equal number of electrons on both sides There should NOT be any left over electrons on any side of equation at the end 1 2 02042 gt 2 002 2 e39 5 e 8 H MnO4 gt Mn2 4 H20 To obtain the same number of electrons on each side multiply eqn 1 by 5 and eqn 2 by 2 5czo42 gt 10 co2 3 106 16 H 2 MnO4 gt 2 Mn2 8 H20 4 16 H 2 MnO4 5czo42 gt 2 Mn2 10 002 8 H20 15 Practice Exercise Complete and balance the following redox reaction CF2072 aq aq gt Cr3aq 03 aq acidic solution 16 Balancing redox reaction in basic medium To do this treat the redox reaction as if it is in acid medium and balance the redox reaction as you normally do by balancing atoms oxygens and hydrogens and charges Finally count the number of H ions in the balanced redox equation then add equal number of OH39 ions on both sides of the equation Thus H and OH39 would make H20 on one side neutralizes and leaving OH39 ions on the other side of the equation At the end make sure that you balance the number of water molecules Look at the following example The following redox reaction is balanced for an acidic solution Cue 4Haq 2NosYacl 9 C 2aq 2N02g 2Hzo Balance this reaction for a basic solution 17 Balancing redox reaction in basic medium Contd Cus 4Haq 2NO3aq 9 Cu2aq 2N02g 2H20 Since this equation has 4 H ions we will add 4 OH ion on both sides so the eqn becomes 2NO3aq 9 Cu2aq 2N02g 2H20 40Hhm This would result in 4H20 on the left so the eqn becomes Now balance the water molecules you will get the balanced redox reaction for a basic solution 800 18 Practice Exercise Complete and balance the following redox reaction that occur in a basic solution CVOH3s C39O39aq 9 CrO42M 29 19 203 Voltaic Cells Consider a spontaneous oxidationreduction redox reactions For example Zn strip immersed in CuSO4 solution Electrons are transferred and energy is released l7 r V x V an mq Cus But no useful electric current produced 20 203 Voltaic Cells Contd If we physically separate the oxidation and reduction process the electrons flows from one side to another through a wire producing electric current Of course the circuit should be closed by using a salt bridge or some porous glass disc connection We call such a setup a voltaic cell or Galvanic cell Thus in a voltaic cell the chemical reaction produces the electric current 21 Things to remember in Voltaic Cells A typical cell looks like this The two solid plates S t h that are connected me We by external circuit are called electrodes Voltmeter l 2m K 31J Electrode at which ms 2112 aq 2e cuzwaq 2e gt we the reduction occurs is called the cathode Electrode at which oxidation occurs is cca ihode called the anode Movement of cations r 4 Movement of anions For remembering Anode and Oxidation start with vowels Things to remember in Voltaic Cells Contd Anode in voltaic cell is labeled with a negative sign For remembering an in anode stands for a negative electrode Cathode is labeled with a positive sign In any voltaic cell electrons flow from the anode to the cathode through an external circuit We can imagine as if a positive cathode is attracting electrons from a negative anode 23 Things to remember in Voltaic Cells Contd Once the oxidation occurs at the anode electrons will be released and it will flow from the anode to the cathode This would create a charge imbalance in the cell and the cell will not operate That is why we need a salt bridge If we use a salt bridge usually a Ushaped tube that contains a salt solution or a porous glass disc then gt Cations in the salt bridge move toward the cathode gt Anions in the salt bridge move toward the anode 24 Things to remember in Voltaic Cells Contd For example for the working of following voltaic cell Zn CuNO32 Cu ZnNO32 one would use NaNO3 salt bridge gt The Zn electrode loses mass B cos Zn undergoes oxidation to form Zn2 gt Zn goes Into the Switch 39 e F Ag ZnNO32 solution and 5 mm move towards cathode through the salt bridge Cu or cath degt NO3 ion from salt bridge 3 7 flows into the ZnN032 2ns gt2n2ltaq2e cu2ltaqgt gt cum solution towards anode to Mmmmfca m balance the charge NO3 Na i f 5 Q i i 25 Movement of anions Things to remember in Voltaic Cells Contd gt The Cu electrode gains mass B cos Cu2 from solution undergoes reduction to form Cus and deposits at the Cu cathode gt Since Cu2 comes out of solution Na from salt bridge flows into the CuNOs2 solution to balance the charge The net effect is Electrons flow from the Zn electrode to the Cu electrode But Why do electrons flow spontaneously from a Zn atom to Cu2 ion Or what is the driving force behind this electron transfer process 6 204 Electromotive Force emf As water flows spontaneously one way in a waterfall because of the potential energy difference electrons flow spontaneously one way in a redox reaction from higher to lower potential energy Electrons at the anode has higher potential energy th The potential difference 1 potential energy between the anode and Anode gm cathode in a cell is called the electromotive force emf 3 i5 is It Is also called the cell calm potential or cell voltage Low my deSIgnated as Ecequot 27 Electromotive Force Contd emf or cell potential Cell potential is measured in volts V L 1v 1C Which means It requires one joule of energy to move 1 coulomb of charge across a potential difference of 1 volt Remember that electron is a negatively charged particle and charge has units of coulomb C Thus cell potential is the tendency of electrons to move from one electrode to another in a voltaic cell 28 Electromotive Force Contd emf or cell potential The emf of a voltaic cell depends on 1 Specific reactions at the electrodes anode amp cathode 2 The concentration of reactants and products 3 The temperature Under standard conditions 25 C 1M for solutions 1 atm for gases Ecequot Ecequot E29 is the standard cell potential 29 HalfCell Potential Standard reduction potential In principle if we know the potential for each halfcells we should be able to calculate the cell39 The potential for a halfcell can be measured with reference to a standard hydrogen electrode SHE for which the lied is taken as zero 2 H aq 1M 2equot H2 g 1atm Ered 000 v By convention the halfcell potentials are given as reduction potentials Ered at standard conditions Read page 856 in BLB to understand the working of SHE and to measure halfcell potential 30 Standard Reduction Potentials Contd Reduction potentials for many electrodes have been measured and tabulated Potential V Reduction HalfReaction 287 F2g 2 equot gt 2 F m 0 151 MnO4aq 8 Haq 5 e gt Mn2aq 4 H200 136 023 2 e gt 2Cl39nq 9 1 133 212072 0111 14 Haq 6 e 2Cr3aq 711200 9 123 02g 4 Haq 4e gt 2 H200 66 90 106 Br2l 2e 2 Braq 9 Q9 096 No3 aq 4 Haq 3 9 gt NOg 2 H200 080 Agaq e7 gt Ags 077 Fe3aq e39 gt Fe2aq 068 02g 2 Haq 2e gt H202aq 6 059 MnO4aq 2 H200 3 e gt Mn02s 4 OH39aq 1 054 125 2 e39 gt 21 uq 60 1 040 02g 2 H200 4 e gt 40H uq 99 9 0 34 r112er 2 p gt Cum 6 p 0 I 0 de ned 2 11014 2 e gt 133 I 66 028 N1 aq 2 e gt N1s 044 Fe2aq 2 e gt Fes 076 Zn2aq 2 e39 gt Zns 083 2 H200 2 e gt H2g 20Haq 166 A13aq 3 e gt Als 7271 Naaq e gt Nas 305 Liaq e gt Lis Standard Reduction Potentials Contd 0 Use of Ereel gt We can predict the direction of reaction Halfreaction with more positive or less negative Eed is the reduction process more tendency for reduction Occurs at the cathode Half reaction with less positive or more negative Eed is the oxidation process more tendency for oxidation Occurs at the anode gt We can calculate standard cell emf Ege for any voltaic cell El Eiedmathode EZanode 32 Standard Reduction Potentials Contd The greater the difference between the two half reactions the greater the voltage of the cell ce Thus in a given set of reactions we can choose any two combinations which will give largest positive cell potential or the smallest positive cell potential This diagram tells you how to calculate E ce More positive A Cu2 2 e7 6 Cu Cathode 034 V 0 red E2811 034 076 110 V l Anode O76 Zn 6 Zn2 2 e 33 Sample Exercise Type 1 Suppose we construct a voltaic cell by placing a Ni strip into 1M NiN032 and by placing a Pt wire into a solution that contains 1M Fei and 1M Fe2 What reaction takes place at the anode What reaction takes place at the cathode What is the overall reaction What is the cell emf Ecen For such problems first of all look for E for each halfcell E reel 077 V for Fe3aq e39 gt Fe2 aq E red 028 V for Ni2aq 2e39 gt Ni 3 The more positive value indicate that this will be a favorable halfcell reaction for a reductive process Remember the reduction Will occur at the cathode and oxidation at the anode 34 Sample Exercise Type 1 Contd At cathode Fe3aq e39 gt Fe2 aq At anode Ni 3 gt Ni2aq 2e 1 2 For writing overall reaction multiply eq1 by 2 and add with eq 2 2Fe3aq 2e39 gt 2Fe2 aq Ni 3 gt Ni2aq 2e 2Fe3aq Ni 3 gt 2Fe2 Ni2aq EL 6 Eiedicathode Eiedianode 077v 028 V 105 v Know that The standard reduction potentials are intensive properties It does not depend on the stoichiometric coefficient of the cellreaction So do not multiply E0 with the stoichiometric coefficient vv Sample Exercise Type 2 Given a redox reaction we should be able to calculate the standard emf for a voltaic cell that could be constructed using that reaction For example What is the standard emf for voltaic cell using the reaction Fes 2 Fe3 aq gt 3Fe2aq Strategy Determine the halfcell reactions Identify which reaction could take place at the anode and at the cathode Calculate Ece using E cell EredCathOde EredanOde 36 Sample Exercise Type 2 Contd Halfcell reactions are as follows Fe 3 gt Fe2 aq 2e39 This is an oxidation It will occur at the anode E red o44 v Fe e39 gt F92 3C1 This is a reduction red 077 V It will occur at the cathode Now Ece Eredcathode Eredanode 077 V 044 V 121 V If you make a wrong choice of What to call the cathode and the anode you will get a negative voltage Remember the net voltage for a votaic cell should be a positive value Predicting Oxidizing and Reducing properties Yet another use of Ere We know that More positive or less negative value of Eoreel means more tendency for reduction Consider again the voltaic ce involving redox reaction of Zn and Cu Ered 034 V for Cu2aq 2e39 gt Cu S Ered O76 V for Zn2aq 2e39 gt Zn 8 Thus Cu2 has more tendency to get reduced This is the driving force for the Zn to get oxidized In other words Cu oxidizes Zn Hence we can say Cu2 is a strong oxidizing agent or Zn is a good reducing agent 38 Predicting Oxidizing and Reducing properties Contd Strongest o oxidizing agent Most posmve values of Ered The strongest oxidizers have the most positive reduction potentials The strongest reducers have the most negative reduction potentials Increasing strength of oxidizing agent Increasing strength of reducing agent Liaq e gt Strongest reducing 3 9 Most negative values of Eject agent Practice Exercise From each of the following pairs of substances use data in Appendix E to choose the one that is the stronger oxidizing agent aCI2 g or Brz I bNi2 aq or Cd2 aq C Bros39 3C1 or 3903 3C1 d H202 3C1 or 03 g 40 205 Free Energy and Redox Reactions EMF amp AG Any redox reaction that occur in a voltaic cell to produce a positive emf must be a spontaneous process In other words We can predict the spontaneity of a redox reaction just by looking at Ecell value Cell potential Ece gt 0 The process is spontaneous Ecell lt 0 The process is nonspontaneous Ece 0 The process is at equilibrium 41 Free Energy and E We will see in Chapter19 that AG is the criterion for spontaneity of a reaction G is the Gibbs Free energy AG and E06 are related by the following relationship cell AG nFEce or AGquot nFEcel Under standard conditions n number of moles of electrons transferred F Faraday constant the charge on 1 mole of electrons 1 F 96485 Cmol e39 96485 JVmol e39 42 Sample Exercise Type 3 Calculate AG for the voltaic cell based on the reaction Zn s Cu2 aq gt Zn2 aq Cu s E26 110 V Is this reaction spontaneous AG 0 nFEcel 2 96485 Cmol 110 V 212 x 105 CVmol 212 x 105 Jmol 212 kJmol The positive value of E009 and a negative value ofAG 0 indicate that the reaction is spontaneous 43 Practice Exercise Consider the following reaction 3Nl2aq ZCrOH3s 1OHaq 9 3Nis 2Cl 04239aq 8H20I a What is the value of n b Use the data in Appendix E to calculate AGO c Calculate K at 298K To solve this problem look at the sample exercise in page 864 44 206 EMF at nonstandard conditions Nernst Equation Walther Nernst 1864 1941 a German chemist related Ecell with Ece as shown below Recall that Q has same expression as equilibrium 1 constant Keq 0 RT EceI Ecel n In Q Where Q reaction quotient R 8314 JKmol n number of moles of electrons transferred F Faraday constant 96485 Cmol Look at the next slide to know how this equation was derived 45 Nernst Equation Contd In Chapter 19 AG is related to AGO in the following eqn AGAGORTnQ 1 Now we know that AG nFE and AGO nFE Thus eqn 1 becomes nFE nFEO RTln Q Dividing both sides by nF we get the Nernst equation RT 2303RT Ecell ce nF In Q OR Ece Ecel3 TIOQ Q At 298 K the value of 2303RTF is 00592 506 E06 80 we can write 00592 log Q E29 and Equilibrium constant Again from Chapter 19 AGquot RTInK 1 We now know AG nFEce 2 On comparing equations 1 and 2 we get RT an nFEce o Rearranging we get quot7K 7736 or using base1O logarithms F 1 W 222 0232 At T 298 K 0900 W Sample Exercise Type 4 A car battery uses the following reaction Pbs Pb02s 2H2804aq gt 2 PbSO4s 2H20l lfthe E 204 V cell 0 RT What is Ecell at 25 c and 6M stofce EceI nF 39 Q What is Ecell at 25 C and 6M H2804 r At 25 c with 6M H2804 831 JKmol 298K 2 209 v ECG 23904 V 2mol96485Cmol 39 16 At 25 c with 6M H2804 831 JKmol248K 2 208 v E06 23904 V 2mol96485Cmol 39 16 In reality the temperature effect on the performance of the car battery comes from the fact that at low temperature the electrolytes becomes more viscous preventing ion movement and generates less voltage What we can learn from Nernst Equation 0 RT E E nF InQ cell cell This equation tells us that under nonstandard condition we can have different Ecell value depending upon the concentration gradient This concept takes us to another important cell type called concentration cells As the term mean we Will get a cell potential due to concentration gradient 49 Concentration Cells Cell potential due to concentration gradient is an important concept to understand cellular processes Forexample gtMuscular contractions and nerve impulses are the result of concentration gradient of ions inside and outside the cell membrane Want to know how our heart beats It is interesting to know that we have a voltaic cell in our body due to concentration gradient If you are interested read chemistry and life Page 868 Concentration Cells Contd Nernst equation implies that we can have cell potential with two halfcells made out of same materials as long as the solutions have different concentrations see next slide 0 For such a cell Ecel would be 0 but Q would not Therefore as long as the concentrations are different E66 will not be 0 Look at the following cell setup with two differently concentrated Ni2 solutions 51 Concentration Cells Contd 8 we V I a Ni2 100 gtlt 10T3M NiZT 100M Ni2 05 M Nilt 05 M The cell would operate until the concentration on both compartments becomes equal That means oxidation should occur Where Ni2 is low to increase the Ni2 ions Thus it will be the anode compartment Reduction takes place at the more concentrated compartment to reduce the Ni2 ions Thus it will be the cathode compartment 52 Practice Exercise A concentration cell is constructed with two ZnsZn2aq halfcells The first halfcell has Zn2 135 M and the second halfcell has Zn2 375 x 104 M a Which halfcell is the anode of the cell b Calculate the emf of the cell 53 Sample ExerCIse Type 5 Avoltaic cell uses the following reaction at 298K Zns Ni2aq gt Zn2aq Nis a Calculate the emf of this cell under standard conditions Ecell EredCathOde Eoredan0de Zn undergoes oxidation anode and Ni undergoes reduction cathode 028 0763 0483 048 V b Calculate the emf of this cell when Ni2 300 M and Zn2 0100 M 202 o Q n 2 2 e transfer 506 506 n ln Q Ni2 8314JKmo298K In 0100 M ECG V 2mol96485Cmo 300 M 54 052 V 209 Electrolysis Voltaic cells are based on spontaneous oxidation reduction reactions That means the reverse process is nonspontaneous under normal conditions But it is possible to use electrical energy to force or push the nonspontaneous redox reactions to occur This concept takes us to another important topic called Electrolysis Chemical reactions under electrical current Such process are done in an electrolytic cell 55 Applications of Electrolysis Production of chemicals For e g Production of Na 02 Bleach Refining Of metals 3 ElectrometaIurgy For e g Production of Aluminum from Al203 Electroplating of metal For e g Coating of nickel or chromium onto a steel surface 56 Production of Chlorine and Sodium This is achieved by the electrolysis of molten NaCl Anode At the cathode Nal is produced zl quot I Na e39 gt Na Molten l At the anode e to a e L Cl2g is produced 8 tot A 91 t 2C1 gt C12g 2e 2NaJr 2 e gt 2Nal 20quot gt Clz 2639 Electrolytic cell 57 Production ofAIuminum metal gtAluminum Al is one of the most active metals produced using electrolysis gtBefore electrolytic methods Al was more costly than gold Au Al203 is dissolved in I H 3969 molten cryolite Gaff eNllt Na2AlF6 at 1000 C I l I l and Al3 is reduced to l m quot1 ii 9 molten h J 39 I 39 gl glfleilslsolved Molten quot o nth 39 Iii cryolite aluminum ar on i118 Al33e39 gtAl Cb 1d iron 58 Electroplatmg Electroplating uses electrolysis to deposit thin layer of one metal on to another metal surface For example gt Steel can be plated with Cr Ni etc gt Au and Ag can be plated on jewelry or table ware In this section the typical questions are a How much metal can be plated for a given amount of current and time b How much current is need to produce certain quantity of metal c How long Will it take to plate a surface With certain thickness of metal With a given amount of current 59 Current amp Eleotroplating Let us see the relationships between these quan es Current Amount of Charge that flows per unit time Current is usually measured in amperes 1 ampere 1 Coulomb second or in other words Amperes gtlt Time in seconds Coulombs Remember Coulomb is the unit of Charge 60 Current amp Electroplating Contd We know that 1 Faraday is the charge of 1 mole of electrons 1 F 96485 Cmol e In other words 1 mole of electron has 96485 C of charge Thus Given current and time we can find number of coulombs and then number of moles of electrons Given a balanced halfreaction we can then go from moles of electrons to moles of a chemical substance 61 Current amp Electroplating Contd To understand further consider the following scheme which gives the relationship between gt Current gt Time gt Coulomb gt Mole of electrons gt Mole of substance gt Grams of substance Sample Exercise 1 A solution of Cr3aq is electrolyzed using a current of 760 A What mass of Crs is plated out after 200 days Strategy 1 Write a balanced halfreaction and nd the number of moles of electrons needed for the reduction Cr3aq 3e gt Crs 3 mole of electrons Convert time into seconds 200 days X 2400 h 600 mm 600 sec X X 100 day 100 h 100 min 173 x 105 seconds 63 Sample Exercise 1 Contd ampere 760 03 x 173 x 105 S 131 x106 C 1mol e 96485 C x 131 gtlt1O6 C 136 mole e 1 mol Cr3 mol e x 136 mol e 453 mol Cr 52 9 Cr 1 mol Cr x 453 mol Cr 2357 9 Cr 64 Sample Exercise 2 Elemental Aluminum is produced by the electrolysis of molten AI203 What mass of AI can be produced by passing a current of 350 000 amperes for 10 h Al203 3 e gtA 600 min 600 sec 350 000 Cs X 9 100h X100min 1393X1OC 1 mol e96485 C x 13 x 109 C 13 x 104 mol e 1molAI3 mol e x 13 x 104 mol e 44 x103 moIAI 27 gAImoIAI gtlt 44x 103 moIAI 12 x 105 gAI 012 metric ton 65