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Electrolysis calculations

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    Electrolysis calculations Electrolysis calculations Presentation Transcript

    • ELECTROLYSISCALCULATIONS
    • A LOOK AT HALF -EQUATIONS  A half equation really just tells you the number of electrons removed or added to an ion: The following half-equation tells us that 1 mole of electrons is lost from sodium metal to form1 mole of sodium ions: Na – 1e- → Na+
    • 1 mole of Magnesium needs THIS MEANS THAT to lose 2 moles of electrons to form 1 mole of magnesium ionMg – 2e- → Mg2+ 1 mole of aluminium needs to lose 3 moles of electrons to form 1 mole ofAl – 3e- → Al3+ aluminium ionCl + 1e- →Cl- 1 mole of chlorine atoms need to gain 1 mole of electrons to form 1 mole of chloride ion
    • BU T S U P P O S E YO U D O N ’ T H AV E T H E SUFFICIENT AMOUNT OF MOLES? In essence,electrons tonumbermetal is 1:1,Let’s assume that for the moles of sodium of  Since the ratio between moles ofsodium metal, only moles of electrons of electrons shouldhalf the number of  then 0.5 moles is directly proportional to the number + supply 0.5 moles of Na ionsmoles of electrons of ions in electrolysis andare supplied. vice-versa
    • H OW D O YO U K N OW T H E A M O U N T O F E L E C T RO N S I N T H E C E L L ? It has been determined that 1 mole of electrons has a charge of 96,500 ColoumbsCharge of 1 mole of electrons = 96,500 C = 1F (C). This number of Coulombic charge is equal to 1 Faraday (F)
    • WHAT IS CURRENT? The word, current, refers to the rate ofTherefore, flow of electricity amps = Coulombs/second and One ampere (amp) is equal to a rate of flow of charge of 1 Coulomb seconds Coulombs = amps * every second
    • WHAT IS CURRENT? This can be denoted by the following equation: Q=IxTWhereQ = quantity of charge in coloumbsI = current in AmpsT = time in seconds
    • RELATIONSHIPS BETWEEN TimeNumber of Amount of current Mass of moles of electric passed product at electrons current through the electrodes system
    • WORKED EXAMPLES
    • EXAMPLE 1Calculate the mass of copper produced in 1.5 hours by the electrolysis ofmolten CuCl2 if the electrical current is 12.3 Amps
    • STEP 1: DETERMINE CHARGE Since the charge in Couloumbs = amps * seconds orQ = 12.3 amps * 1.5 hours * 60 min/hour * 60 sec/min Q=IxTQ = 6.64 x 104 C
    • STEP 2:DETERMINE NUMBER O F E L E C T RO N S Therefore, 6.64 x 104 Coulombs total pass into the reduction cell (cathode) during the reaction.Since 96,500 C is the charge for 1 mole of e-,96,500 C = 1 mole of e-1C = (1/ 96,500) mole of e-6.64 x 104 C = 6.64 x 104 C * (1 mole e- / 96,500C) = 0.688 moles e-
    • STEP 3:DETERMINE NUMBER OF M O L E S O F I O N / M E TA L P RO D U C E D How many moles of Cu can be reduced with 0.688 moles of electrons?The half-equation for this reaction is Cu2+ + 2e- → CuThis means that it takes2 moles of e- to produce 1 mole of Cu or1 mole of e- to produce 1/2 mole of CuTherefore0.688 moles e- = 0.688 * (1/2) = 0.344 moles Cu
    • STEP 4:DETERMINE MASS OF I O N / M E TA L P RO D U C E D Finally, how many grams of Copper are there per mole?The relative atomic mass of copper, Ar = 63.5 g/moleThis means that1 mole of Cu has a mass of 63.5 gTherefore0.344 moles of Cu = 0.344 * 63.5 = 21.8 g
    • EXAMPLE 2In the industrial production of chlorine gas, acurrent of 50,000 A was passed through a saltsolution for 1 hour. Calculate the volume of gaswhich would be produced.Take the molar volume to be 24 dm3 / mol.
    • STEP 1: DETERMINE CHARGEI = 50000 At =Sinces 3600Q the charge in Couloumbs = amps * seconds or =I×t = 50000 × 3600 = 180000000 Q=IxT = 1.8 × 108 C
    • STEP 2:DETERMINE NUMBER O F E L E C T RO N S Therefore, 1.8 × 108 Coulombs total pass into the cell (anode) during the reaction.Since 96,500 C is the charge for 1 mole of e-,96,500 C = 1 mole of e-1C = (1/ 96,500) mole of e-1.8 × 108 C = 1.8 × 108 C * (1 mole e- / 96,500C) = 1,865.28 moles e-
    • STEP 3:DETERMINE NUMBER OF M O L E S O F I O N / M E TA L P RO D U C E D How many moles of Cl2 can be produced from 1,865.28 moles of electrons?The half-equation for this reaction is 2Cl-(aq)→Cl2(g) + 2e-This means that it takes2 moles of e- for every 1 mole of Cl2(g) or1 mole of e- for every 1/2 mole of Cl2(g)Therefore1,865.28 moles e- = 1,865.28 * (1/2) = 932.64 moles Cl2(g)
    • S T E P 4 : D E T E R M I N E M A S S / VO L U M E O F I O N / M E TA L / G A S P RO D U C E D Now, assuming room temperature and pressure (RTP), (20◦C and 1 atmosphere presssure)Molar volume of chlorine gas , Cl2(g)= 24 dm3This means that1 mole of Cl2(g) occupies a volume of 24 dm3Therefore932.64 moles Cl2(g) = 932.64 x 24 = 22, 383.36 dm3