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Multimedia Chemistry I & II (1996-9-11) [English].img
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chapt12.5c
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à 12.5cèElectrolytic Cells
äèPlease fïd ê unknown quantity ï ê followïg electrolytic cells.
âèHow many grams ç gold would be plated from a solution contaï-
ïg AuCl╣ú by a current ç 2.00 A for 4.00 hours?èThe reaction is
AuCl╣ú + 3eú ─¥ Au(s) + 3Clú.èOn which electrode (anode or cathode) does
ê gold plate?èThe AuCl╣ú is reduced ë Au at ê cathode.èCurrent å
time give ê moles ç electrons, from which we obtaï ê grams ç gold.
?mol eú = (2.00 A)(4.00 hr)(3600 s/hr)(1 mol eú/96485 C) = 0.298 mol eú.
? g Au = (0.298 mol eú)(1 mol Au/3 mol eú)(197.0 g Au/mol) = 58.7 g Au.
éSèIn ê previous sections, we ïvestigated voltaic cells ï which
a spontaneous reaction supplies an electric current.èReturnïg ë ê
Daniell cell, Zn(s) + Cuìó ─¥ Znìó + Cu(s), E°(cell) = 1.103 V, zïc is
ê negative anode å copper is ê positive cathode.èIf we apply a
voltage difference greater than 1.103 V at ståard conditions ë ê
electrodes, we will cause ê reverse reaction (right ë left) ë occur.
(This is what happens after a car engïe starts å ê voltage regulaër
permits ê battery ë be recharged.)èAn electrolysis is ê application
ç an external current ë cause a chemical reaction ï a cell.
èè With a voltage difference greater than 1.103 V, Zn ê negative ter-
mïal, å Cu ê positive termïal, ê cell reaction is
Znìó + Cu(s) ─¥ Zn(s) + Cuìó.èDurïg ê electrolysis ê zïc electrode
is ê cathode, å ê copper electrode is ê anode.èIn an electro-
lytic cell, ê polarity å identity ç ê electrodes change from those
ï ê voltaic cell.èWhat does not change is that oxidation still occurs
at ê anode å that reduction still occurs at ê cathode.
ElectrodeèèèèèèèVoltaic CellèèèèèèElectrolytic cell
──────────èèèè ─────────────────────èè ─────────────────────
ANODE:èèèèèè Zn, negative termïalèè Cu, positive termïal
Anode Reaction:èèZn(s) ─¥ Znìó +2eúèèèèCu(s) ─¥ Cuìó +2eú
CATHODEèèèèèèCu, positive termïalèè Zn, negative termïal
Cathode reaction:èCuìó +2eú ─¥ Cu(s)èèèèZnìó +2eú ─¥ Zn(s)
In an electrolysis we cause a chemical change by passïg an electric cur-
rent through ê cell.èThe extent ç ê chemical change depends on ê
number ç electrons that flow through ê cell.èIf we plated Ni(s) onë
a cathode from a solution contaïïg Niìó, ê reaction at ê cathode
would beèNiìó + 2eú ─¥ Ni(s).èThe reduction reaction shows that two
moles ç electrons are required ë plate one mole ç Ni.èTo determïe
ê mass ç plated nickel, we must determïe ê number ç moles ç elec-
trons transferred.
èè The charge on an electron is 1.60217733x10úîö Coulomb(C).èThe ëtal
charge ç one mole ç electrons is Avogadro's number times ê charge on
ê electron.èThe amount ç charge is called a Faraday(F).
èèè1 Faraday = (6.0221367x10ìÄ eú/mol)(1.60218x10úîö C/eú)
èèèèèè1 F = 96485.3 C/mol eú.
A current ç 1 Ampere is an electron flow ç one Coulomb per second.èIf
we know how long an electric current was flowïg, we can use Faraday's
constant ë fïd ê number ç moles ç electrons that flowed.èThe
general equation is:
èè Moles çèè (Amperes)(time ï seconds)
èè electronsè= ──────────────────────────
èèèèèèèèèèè96485.3 C/mol eú
How many grams ç Ni would be plated ï 8.00 hours at a constant current
ç 5.00 A from a solution ç Niìó?èèNiìó + 2eú ─¥ Ni(s).
First we fïd ê number ç moles ç electrons.
èè? moles eú = (5.00 A)(8.00 hr)(3600 sec/hr)/(96485 C/mol eú)
èè? moles eú = 1.492 mol eú.
Two moles ç electrons are needed ë plate one mole ç Ni.èThe mass ç
one mole ç Ni is its aëmic mass (58.69 g/mol).èWe have all ç ê
ïformation that we need.
èè? g Ni = 1.492 mol eú)(1 mole Ni/2 mole eú)(58.69 g Ni/mol Ni)
èè? g Ni = 43.8 g Ni.
43.8 grams ç Ni would plate on ê electrode.
We can reverse ê process å ask how long will it take ë plate 0.25 g
ç Ni at a current ç 5.00 A?èThe reduction reaction lïks ê mass å
ê current.èThe reaction, Niìó + 2eú ─¥ Ni(s), shows two moles ç elec-
trons are needed per mole ç Ni(s).èThis is anoêr conversion problem.
g Ni ¥ mol Ni ¥ mol eú ¥ time.èWe will split ê calculation ïë two
steps.
è? mol eú = (0.25 g Ni)(1 mol Ni/58.69 g Ni)( 2 mol eú/1 mol Ni)
è? mol eú = 8.519x10úÄ mol
èèèèèèèèèèèèèèèè 96485 Cèè1 secèè1 mï
è? timeè = 8.519x10úÄ mol eú x ──────── x ────── x ────── = 2.7 mï.
èèèèèèèèèèèèèèèè 1 mol eúè 5.00 Cè 60 sec
Assumïg 100% cell efficiency, 2.7 mïutes would be required ë plate
0.25 g Ni.
1èElectrolysis ç molten sodium chloride ï a Down's Cell yields
sodium å chlorïe.è2NaóClú(l) ─¥ 2Na(l) + Cl╖(g).èWhat reaction
occurs at ê anode ç ê cell?
A) Naó(l) ─¥ Na(l) + eú. B) Naó(l) + eú ─¥ Na(l).
C) 2Clú(l) + 2eú ─¥ Cl╖(g). D) 2Clú(l) ─¥ Cl╖(g) +2eú.
üèOxidation occurs at ê anode ç a cell regardless ç ê type,
voltaic or electrolytic.èOxidation is ê loss ç electrons.èAlthough
choice (A) ïdicates a loss ç electrons, ê equation is not balanced.
The correct oxidation reaction is 2Clú(l) ─¥ Cl╖(g) +2eú.èChlorïe is
formed at ê anode ç ê Down's cell.
Ç D
2èIf you wanted ë plate silver onë an object (which could be
made electrically conductïg), you would make ê object ê __________
ï an electrolytic cell.
èèA) anode B) cathode
üèTo plate silver from a solution, we need ë reduce ê silver ion
ë silver.èAgó + eú ─¥ Ag(s).èReduction occurs at ê cathode ç a cell.
The object ë be plated should be made ê cathode.
Ç B
3èThe electrolysis reaction for aqueous NaCl solution can be
written as 2NaCl(aq) + 2H╖O ─¥ 2Naó(aq) + 2OHú(aq) + H╖(g) + Cl╖(g).
What reaction occurs at ê cathode?
A) Naó(aq) + eú ─¥ Na(s).
B) 2Clú(aq) ─¥ Cl╖(g) + 2eú.
C) 2H╖O + 2eú ─¥ 2OHú(aq) + H╖(g).
D) 2Hó(aq) + 2eú ─¥ H╖(g).
üèReduction occurs at ê cathode.èNo change occurs ë ê Naó.
Chloride is oxidized ë chlorïe gas.èWater is reduced ë hydrogen gas.
The oxidation state ç hydrogen changes from +1 ï water ë 0 ï H╖.
The reduction reaction is 2H╖O + 2eú ─¥ 2OHú(aq) + H╖(g).
Ç C
4èHow many grams ç Al would be produced by a constant current
ç 120. A for 8.00 hours?èThe reaction is AlÄó + 3eú ─¥ Al.
èèA)è5.37 g Al B) 16.1 g Al
èèC) 322 g Al D) 966 g Al
üèUsïg ê current å ê time, you can fïd ê number ç moles
ç electron carryïg out ê reduction.èThe half-reaction connects ê
moles ç electrons ë ê mass ç Al.è120 A = 120 C/sec.
è 1 F = 96485 C/mole eú.
èèèèèèè120 Cèèèèèè 3600 secè 1 mol eú
è ? mol eú = ───── x 8.00 hr x ──────── x ──────── = 35.8 mol
èèèèèèè1 secèèèèèèè1 hrèèè96485 C
èèèèèèèèèèèèè1 mol Alè 26.98 g Al
è ? g Al = 38.5 mol eú x ──────── x ────────── = 322 g Al
èèèèèèèèèèèèè3 mol eúèè1 mol Al
The unit conversions are: amperesxtime ¥ C ¥ mol eú ¥ mol Al ¥ g Al.
Ç C
5èHow many grams ç Pt would be plated ï 10.0 mïutes at 2.00 A
from a solution contaïïg PtCl╣ú ?èPtCl╣ú + 2eú ─¥ Pt(s) + 4Clú.
èèA) 2.42 g Pt B) 0.824 g Pt
èèC) 72.8 g Pt D) 1.21 g Pt
üèUsïg ê current å ê time, you can fïd ê number ç moles
ç electron carryïg out ê reduction.èThe half-reaction connects ê
moles ç electrons ë ê mass ç Pt.è2.00 A = 2.00 C/sec.
è 1 F = 96485 C/mole eú.
èèèèèèè2.00 Cèèèèèèè60 secè 1 mol eú
è ? mol eú = ────── x 10.0 mï x ────── x ──────── = 0.01244 mol eú
èèèèèèè1 secèèèèèèè 1 mïèè96485 C
èèèèèèèèèèèèèè 1 mol Ptè 195.1 g Pt
è ? g Pt = 0.01244 mol eú x ──────── x ────────── = 1.21 g Pt
èèèèèèèèèèèèèè 2 mol eúèè1 mol Pt
The unit conversions are: amperesxtime ¥ C ¥ mol eú ¥ mol Pt ¥ g Pt.
Ç D
6èHow many grams ç Cl½ would be produced by a current ç
75.0 A for 24.0 hours?èè2Clú + 2eú ─¥ Cl╖(g).
èèA) 2.38x10Ä g Cl╖ B) 1.90x10Ä g Cl╖
èèC) 1.19x10Ä g Cl╖ D) 1.06x10Ä g Cl╖
üèUsïg ê current å ê time, you can fïd ê number ç moles
ç electron carryïg out ê reduction.èThe half-reaction lïks ê
moles ç electrons ë ê mass ç Cl╖.è75.0 A = 75.0 C/sec.
è 1 F = 96485 C/mole eú.
èèèèèèè75.0 Cèèèèèè 3600 secè 1 mol eú
è ? mol eú = ────── x 24.0 hr x ──────── x ──────── = 67.2 mol eú
èèèèèèè1 secèèèèèèèè1 hrèèè96485 C
èèèèèèèèèèèèè 1 mol Cl╖è 70.90 g Cl╖
è ? g Cl╖ = 67.2 mol eú x ───────── x ─────────── = 2.38x10Ä g Cl╖
èèèèèèèèèèèèè 2 mol eúèè 1 mol Cl╖
The unit conversions are: amperesxtime ¥ C ¥ mol eú ¥ mol Cl╖ ¥ g Cl╖.
Ç A
7èHow many grams ç Ca would be produced from ê electrolysis
ç molten calcium chloride at 20.0 A for 4.00 hours?
èèA) 0.997 g Ca B) 1.20x10ì g Ca
èèC) 59.8 g Ca D) 13.4 g Ca
üèUsïg ê current å ê time, you can fïd ê number ç moles
ç electron carryïg out ê reduction.èThe half-reaction lïks ê
moles ç electrons ë ê mass ç Ca.èCalcium exists as a 2+ ion.èThe
reduction is Caìó + 2eú ─¥ Ca(l).è20.0 A = 20.0 C/sec.
è 1 F = 96485 C/mole eú.
èèèèèèè20.0 Cèèèèèè 3600 secè 1 mol eú
è ? mol eú = ────── x 4.00 hr x ──────── x ──────── = 2.98 mol eú
èèèèèèè1 secèèèèèèèè1 hrèèè96485 C
èèèèèèèèèèèèè1 mol Caè 40.08 g Ca
è ? g Ca = 2.98 mol eú x ──────── x ─────────── = 59.8 g Ca
èèèèèèèèèèèèè2 mol eúèè1 mol Ca
The unit conversions are: amperesxtime ¥ C ¥ mol eú ¥ mol Ca ¥èg Ca.
Ç C
8èHow many mïutes would be required ë plate 2.50 g ç Ag at a
constant current ç 5.00 A?èThe reaction is Ag(CN)╖ú + eú ─¥ Ag + 2CNú.
èèA) 6.95 mï B) 7.45 mï
èèC) 29.1 mï D) 37.8 mï
üèThe reduction reaction, Ag(CN)╖ú + eú ─¥ Ag + 2CNú, shows that
platïg one mole ç Ag requires one mole ç electrons.èWe want ë
achieve ê conversion: g Ag ─¥ mol eú ─¥ Coulombs ¥ time.èLet's split
it ïë two steps.èAmperes = Coulomb/sec.è1 F = 95485 C/mol eú.
èèèèèèèèèèèè 1 mol Agèè 1 mol eú
è? mol eú = 2.50 g Ag x ────────── x ──────── = 0.02317 mol eú
èèèèèèèèèèèè 107.9 g Agè 1 mol Ag
èèèèèèèèèèèèè 96485 Cèè1 secèè1 mï
è? mï = 0.02317 mol eú x ──────── x ────── x ────── = 7.45 mï
èèèèèèèèèèèèè 1 mol eúè 5.00 Cè 60 sec
Ç B
9èHow many mïutes would be required ë plate 0.1700 g ç Fe
from a solution contaïïg FeÄó at a constant current ç 2.00 A?
èèA) 2.45 mï B) 7.34 mï
èèC) 410 mï D) 29.3 mï
üèThe reduction reaction, FeÄó + 3eú ─¥ Fe(s), shows that platïg
one mole ç Fe requires three moles ç electrons.èWe want ë achieve ê
conversion: g Fe ─¥ mol eú ─¥ Coulombs ¥ time.èLet's split
it ïë two steps.èAmperes = Coulomb/sec.è1 F = 95485 C/mol eú.
èèèèèèèèèèèèè 1 mol Feèè 3 mol eú
è? mol eú = 0.1700 g Fe x ────────── x ──────── = 9.132x10úÄ mol eú
èèèèèèèèèèèèè 55.85 g Feè 1 mol Fe
èèèèèèèèèèèèèèè96485 Cèè1 secèè1 mï
è? mï = 9.132x10úÄ mol eú x ──────── x ────── x ────── = 7.34 mï
èèèèèèèèèèèèèèè1 mol eúè 2.00 Cè 60 sec
Ç B
10èThe electrolysis ç cold 40% H╖SO╣ produces peroxydisulfuric
acid, H╖S╖O╜, which is hydrolyzed ë manufacture hydrogen peroxide.èHow
many hours would be required ë produce 5000. g ç H╖S╖O╜ at a constant
current ç 30.0 A assumïg 100% current efficiency?èThe reaction is
2HSO╣ú ─¥ H╖S╖O╜ + 2eú.
è A) 23.0 hrèèè B) 57.6 hrèèè C) 46.0 hrèèè D) 17.2 hrèè
üèThe half-cell reaction, 2HSO╣ú ─¥ H╖S╖O╜ + 2eú, shows makïg one
mole ç H╖S╖O╜ê requires ê removal ç two moles ç electrons.èThe
conversions are: g H╖S╖O╜ ─¥ mol eú ─¥ Coulombs ¥ time.èLet's split
it ïë two steps.èAmperes = Coulomb/sec.è1 F = 95485 C/mol eú.
èèèèèèèèèèèèèè1 mol H╖S╖O╜èèè 2 mol eú
? mol eú = 5000 g H╖S╖O╜ x ─────────────── x ──────────── = 51.50 mol eú
èèèèèèèèèèèèè 194.16 g H╖S╖O╜è 1 mol H╖S╖O╜
èèèèèèèèèèè96485 Cèè1 secèèè1 hr
? hr = 51.50 mol eú x ──────── x ────── x ──────── = 46.0 hr.
èèèèèèèèèèè1 mol eúè 30.0 Cè 3600 sec
Ç C