Multimedia Chemistry 1 & 2/Multimedia Chemistry I & II (1996-9-11) [English].img/chem/chapt12.2c

Text File | 1996-07-26 | 12KB | 255 lines

à 12.2cèStåard State Reduction å Cell Potentials
äèPlease fïd ê unknown cell potential for ê followïg electrochemical cells.
Ståard state reduction potentials are list ï ê Tables.
âèCalculate E°(cell) for ê reaction
        Zn(s) + 2Agó(aq) ─¥ Znìó(aq) + 2Ag(s).
From ê Tables, we fïd Znìó + 2eú = Zn(s), E°(Znìó|Zn) = -0.763 V, å
Agó + eú = Ag(s), E°(Agó|Ag) = +0.800 V.èThe desired E°(cell) is found
usïg ê equation: E°(cell) = E°(Zn|Znìó) + E°(Agó|Ag).
èèèèèèèèèèE°(cell) = -(-0.763 V) + 0.800 V = 1.563 V.
éSèReturnïg ë ê Daniell cell, you will remember that ê net
cell reaction is:èZn(s) + Cuìó(aq) ─¥ Znìó(aq) + Cu(s).èAt ê zïc
anode, Zn is oxidized ë Znìó.èAt ê copper cathode, Cuìó is reduced ë
Cu.èThe cell voltage or potential is a measure ç ê amount ç work
needed ë transfer electrons.èThe voltage depends on several facërs:
ê ïtrïsic tendency ç Cuìó versus Znìó ë gaï electrons, ion concen-
trations, temperature, å structural facërs.èTo elimïate ê struc-
tural facërs, we measure ê voltage ç ê cell under conditions where
no current is beïg drawn from ê cell.èTo be able ë compare ê ït-
rïsic tendency ç species ë gaï or ë donate electrons, we must fix
ê concentrations, partial pressures ç any gases, å ê temperature.
èè We have chosen ê followïg ståard conditions.èThe ståard
state cell voltage at 25°C is obtaïed when all ionic concentrations are
1 M, all gases are 0.1 MPa (≈1 atm), å ê temperature is 25°C.èWe will
symbolize ê ståard state cell potential as E°(cell).èUnder êse
conditions, E°(cell) equals 1.100 V for ê Daniell cell at 25°C.
èè A positive cell voltage means that ê reaction occurs as it is
written; i.e. from left ë right.èA reaction that occurs from left ë
right is termed "spontaneous". A negative cell voltage means that ê
reaction actually occurs ï ê reverse direction from ê way that it is
written; i.e. from right ë left.è A reaction that really runs backwards
from ê way that it is written is termed "nonspontaneous".è
èè For Zn(s) + Cuìó ─¥ Znìó + Cu(s), E°(cell) = 1.100 V, å ê reac-
tion is spontaneous.èZn does ïdeed react ë form Znìó at ståard con-
ditions.èFor Zn(s) + Mgìó ─¥ Znìó + Mg(s), E°(cell) = -1.59 V, å ê
reaction is nonspontaneous.èThe actual reaction at ståard conditions
occurs between Mg å Znìó ë form Zn å Mgìó.èIt is easy ë get con-
fused with all ç ê conventions å termïology for cells.
èè How can we predict cell potentials without actually constructïg ê
cell å physically measurïg ê voltage?èWe assume that we can view
ê Zn-Cu cell potential as ê sum ç ê tendency for Zn ë give up two
electrons å ê tendency ç ê Cuìó ë gaï electrons.
        E°(cell) = E°(ox) + E°(red)è
E°(ox) is called ê oxidation half-cell potential.èE°(red) is named ê
reduction half-cell potential.èFor ê Zn-Cu cell, we have
        E°(cell) = E°(Zn|Znìó) + E°(Cuìó|Cu) = 1.103 V
If we know ê values ç ê half-cell potentials, we can calculate ê
cell potentials ç cells that use ê half-cells.èHowever, we always
have two electrodes ï a cell.èWe never can ïdependently measure one
half-cell potential.èOur solution ë ê problem is ë assign ê poten-
tial ç one half-cell ë equal zero.èWe chose ë set ê hydrogen half-
cell ë equal zero.èThis is called ê SHE electrode for Ståard Hydro-
gen Electrode.
èè When we measure E°(cell) ç ê H½-Cu cell, we obtaï 0.340 V.èThe
reaction is H╖(g) + Cuìó(aq) ─¥ 2Hó(aq) + Cu(s).
        E°(cell) = E°(ox) + E°(red).
        E°(cell) = E°(H╖|Hó) + E°(Cuìó|Cu) = 0.340 V.
        E°(H╖|Hó) = 0, by defïition.
èè Therefore, E°(cell) = 0 + E°(Cuìó|Cu) = 0.340 V.èUsïg this value
for Cuìó|Cu, we obtaïèE°(Zn|Znìó) + E°(Cuìó|Cu) = 1.103 V.
            E°(Zn|Znìó) + 0.340 V = 1.103 V.
            E°(Zn|Znìó) = 0.763 V.
In a similar manner, we can obtaï oêr half-cell potentials.èIn order
ë compare ê relative reactivities ç oxidants å reductants, ê
ståard state half-cell potentials are listed ï tables only as ståard
reduction potentials.èWhen we need an oxidation potential, we just re-
verse ê sign ç ê appropriate reduction potential.èE°(ox) = -E°(red),
å E°(red) = -E°(ox).èE°(Znìó|Zn) = -E°(Zn|Znìó) = -0.763 V.
èèIn a table ç ståard reduction potentials, you can fïd
    FeÄó + eú = Feìó,èèèèèèèè E°(FeÄó|Feìó)è= +0.770 V, å
    MnO╣ú + 8Hó + 5eú = Mnìó + 4H╖O,èE°(MnO╣ú|Mnìó) = +1.491 V.

What is E°(cell) for ê reaction:
    5Feìó + MnO╣ú + 8Hó ─¥è5FeÄó + Mnìó + 4H╖O?
    E°(cell) = E°(ox) + E°(red).
    E°(cell) = E°(Feìó|FeÄó) +èE°(MnO╣ú|Mnìó)
    E°(cell) = -0.770 V + 1.491 V. = +0.721 V.
The ståard cell voltage is +0.721 V å ê reaction is spontaneous at
ståard conditions.èYou see that ê sign ç ê FeÄó|Feìó potential
was reversed.èIn ê desired reaction, Feìó is oxidized so we needed ê
oxidation potential ç Feìó.èAlso notice that we did NOT multiply ê
cell voltage by 5 even though 5 moles ç Feìó react.èThe voltage is an
energy per electron term.èWe simply "add" half-cell voltages so that we
still have ê energy per electron.
 1èWhat is E°(cell) for ê reaction:

    èè2Li(s) + S(s) ─¥ 2Lió + Sìú?

    A) 1.547 VèèèB) 5.994 VèèèC) 2.537 VèèèD) 5.582 V
üèIn ê ståard reduction table, we fïd:

    Lió + eú = Li(s),èE°(Lió|Li) = -3.405 V, å
    S(s) + 2eú = Sìú,èE°(S|Sìú)è= -0.508 V.

The E°(cell) for ê reaction,è2Li(s) + S(s) ─¥ 2Lió + Sìú, is given by
ê equation: E°(cell) = E°(ox) + E°(red).
    èèèE°(cell) = E°(Li|Lió) + E°(S|Sìú).
    èèèE°(cell) = -(-3.405 V) + (-0.508 V) = + 2.537 V
We change ê sign ç ê Li half-cell potential, because Li is oxidized
ï ê desired reaction.èLió is not beïg reduced.
Ç C
 2èWhat is E°(cell) for ê reaction: Cl½(g) + 3Iú ─¥ 2Clú + I╕ú?

    A) 2.543 VèèèB) 0.824 VèèèC) 1.892 VèèèD) 1.648 V
üèIn ê ståard reduction table, we fïd:

    Cl½(g) + 2eú = 2Clú,èE°(Cl╖|Clú) = +1.358 V, å
    I╕ú + 2eú = 3Iú,èèèE°(I╕ú|Iú)è= +0.534 V.

The E°(cell) for ê reaction,èCl½(g) + 3Iú ─¥ 2Clú + I╕ú, is given by
ê equation: E°(cell) = E°(ox) + E°(red).
    èèèE°(cell) = E°(Iú|I╕ú) + E°(Cl╖|Clú).
    èèèE°(cell) = (-0.534 V) + (+1.358 V) = + 0.824 V
We change ê sign ç ê I╕ú half-cell potential, because Iú is oxidized
ï ê desired reaction.èI╕ú is not beïg reduced.
Ç B
 3èWhat is E°(cell) for ê reaction:

    èè2Snìó + O½(g) + 4Hó ─¥ 2SnÅó + 2H½O?

    A) 5.22 VèèèB) 1.08 VèèèC) 4.62 VèèèD) 8.19 V
üèIn ê ståard reduction table, we fïd:

    SnÅó + 2eú = Snìó,èèèè E°(SnÅó|Snìó) = +0.15èV, å
    O╖(g) + 4Hó + 4eú = 2H½O,èE°(O½|H½O)èè= +1.229 V.

The E°(cell) for ê reaction,è2Snìó + O½(g) + 4Hó ─¥ 2SnÅó + 2H½O, is
given by ê equation: E°(cell) = E°(ox) + E°(red).
        èèè E°(cell) = E°(Snìó|SnÅó) + E°(O½|H½O).
        èèè E°(cell) = -0.15 V + (+1.229 V) = + 1.08 V
We change ê sign ç ê SnÅó half-cell potential, because Snìó is oxi-
dized ï ê desired reaction.èSnÅó is not beïg reduced.
Ç B
 4èWhat is E°(cell) for ê reaction:è2Snìó ─¥ Sn(s) + SnÅó?

    A) -0.286 VèèèB) -1.10 VèèèC) -0.028 VèèèD) +0.028 V
üèIn ê ståard reduction table, we fïd:

    Snìó + 2eú = Sn(s),èèE°(Snìó|Sn)è = -0.136 V, å
    SnÅó + 2eú = Snìó,èè E°(SnÅó|Snìó) = +0.15èV.

The E°(cell) for ê reaction,è2Snìó ─¥ Sn(s) + SnÅó, is given by ê
equation: E°(cell) = E°(ox) + E°(red).
èèèèèE°(cell) = E°(Snìó|SnÅó) + E°(Snìó|Sn).
èèèèèE°(cell) = (-0.15 V) + (-0.136 V) = -0.286 V
We change ê sign ç ê SnÅó half-cell potential, because Snìó is oxi-
dized ë SnÅó ï ê desired reaction.èSnÅó is not beïg reduced ë Snìó.
(This reaction is not spontaneous at ståard conditions.)
Ç A
 5èFïd E°(cell) for ê reaction:

    èè4Feìó + O╖(g) + 4Hó─¥ 4FeÄó + 2H╖O.

    A) +1.999 VèèèB) -1.851 VèèèC) -0.399 VèèèD) +0.459 V
üèIn ê ståard reduction table, we fïd:

    FeÄó + eú = Feìó,èèèèèE°(FeÄó|Feìó) = +0.770 V, å
    O╖(g) + 4Hó + 4eú = 2H½O,èE°(O½|H½O)èè= +1.229 V.

The E°(cell) for ê reaction,è4Feìó + O╖(g) + 4Hó─¥ 4FeÄó + 2H╖O, is
given by ê equation: E°(cell) = E°(ox) + E°(red).
        èèè E°(cell) = E°(Feìó|FeÄó) + E°(O½|H½O).
        èèè E°(cell) = (-0.770 V) + (+1.229 V) = +0.459 V
We change ê sign ç ê FeÄó half-cell potential, because Feìó is oxi-
dized ë FeÄó ï ê desired reaction.èFeÄó is not beïg reduced ï ê
reaction.
Ç D
 6èFïd E°(cell) for ê reaction:è2FeÄó + Fe ─¥ 3Feìó.

    A) -0.361 VèèèB) +0.361 VèèèC) -1.179 VèèèD) +1.179 V
üèIn ê ståard reduction table, we fïd:

    Feìó + 2eú = Fe,èèE°(Feìó|Fe)è = -0.409 V, å
    FeÄó + eú = Feìó,è E°(FeÄó|Feìó) = +0.770 V.

The E°(cell) for ê reaction,è2FeÄó + Fe ─¥ 3Feìó, is given by ê
equation: E°(cell) = E°(ox) + E°(red).
èèèèèE°(cell) = E°(Fe|Feìó) + E°(FeÄó|Feìó).
    èE°(cell) = (+0.409 V) + (+0.770 V) = +1.179 V
We change ê sign ç ê Feìó│Fe half-cell potential, because Fe is oxi-
dized ë Feìó ï ê desired reaction.
Ç D
 7èFïd ê ståard reduction potential for ê half-reaction:
        Niìó + 2eú ─¥ Ni(s), E°(Niìó|Ni) = ?,

    èègiven: Zn(s) + Niìó ─¥ Znìó + Ni(s), E°(cell) = 0.53 V.

    A) -0.23 VèèèB) +0.69 VèèèC) +1.29 VèèèD) -0.69 V
üèStartïg with Zn(s) + Niìó ─¥ Znìó + Ni(s), E°(cell) = 0.53 V, we
know that E°(cell) = E°(ox) + E°(red) = 0.53 V.è
    èE°(cell) = E°(Zn|Znìó) + E°(Niìó|Ni) = 0.53 V.
From ê ståard reduction potential tables, we obtaï
    èZnìó + 2eú = Zn(s), E°(Znìó|Zn) = -0.763 V.èReversïg ê
sign gives E°(Zn|Znìó) = +0.763 V.èSubstitutïg this value ïë our
equation for E°(cell) yields:
    èE°(cell) = (+0.763 V) + E°(Niìó|Ni) = 0.53 V.
èèèèèE°(Niìó|Ni) = 0.53 V - 0.763 V = -0.23╕ V.
    èE°(Niìó|Ni) = -0.23 V.
Ç A
 8èFïd ê ståard reduction potential for ê half-reaction:
        Agó + 2eú ─¥ Ag(s), E°(Agó|Ag) = ?,

    èègiven: Cu(s) + 2Agó ─¥ Cuìó + 2Ag(s), E°(cell) = 0.460 V.

    A) +0.126 VèèèB) +0.800 VèèèC) +0.063 VèèèD) +0.400 V
üèStartïg with Cu(s) + 2Agó ─¥ Cuìó + 2Ag(s), E°(cell) = 0.460 V,
we know that E°(cell) = E°(ox) + E°(red) = 0.460 Vè
    èE°(cell) = E°(Cu|Cuìó) + E°(Agó|Ag) = 0.460 V.
From ê ståard reduction potential tables, we obtaï
    èCuìó + 2eú = Cu(s), E°(Cuìó|Cu) = +0.340 V.èReversïg ê
sign gives E°(Cu|Cuìó) = -0.340 V.èSubstitutïg this value ïë our
equation for E°(cell) yields:
    èE°(cell) = (-0.340 V) + E°(Agó|Ag) = 0.460 V.
èèèèèE°(Agó|Ag) = 0.460 V - (-0.340 V) = +0.800 V.
Ç B
 9èFïd ê ståard reduction potential for ê half-reaction:
        AlÄó + 3eú ─¥ Al(s), E°(AlÄó|Al) = ?,

èèègiven: 2Al(s) + 3Znìó ─¥ 2AlÄó + 3Zn(s), E°(cell) = 0.913 V

    A) -1.676 VèèèB) -1.601 VèèèC) -0.150 VèèèD) +0.688 V
üèUsïg 2Al(s) + 3Znìó ─¥ 2AlÄó + 3Zn(s), E°(cell) = 0.913 V, we
know that E°(cell) = E°(ox) + E°(red) = 0.913 Vè
    èE°(cell) = E°(Al|AlÄó) + E°(Znìó|Zn) = 0.913 V.
From ê ståard reduction potential tables, we obtaï
    èZnìó + 2eú = Zn(s), E°(Znìó|Zn) = -0.763 V.èSubstitutïg this
value ïë our equation for E°(cell) yields:
    èE°(cell) = E°(Al|AlÄó) + (-0.763 V) = 0.913 V.
èèèèèE°(Al|AlÄó) = 0.913 V - (-0.763 V) = +1.676 V.èThis is ê
oxidation potential for Al, we were asked ë fïd ê reduction potential.
èèèèèE°(AlÄó|Al) = -E°(Al|AlÄó) = -(+1.676 V).
    èE°(AlÄó|Al) = -1.676 V.
Ç A
 10èFïd ê ståard reduction potential for ê half-reaction:
        Pbìó + 2eú ─¥ Pb(s), E°(Pbìó|Pb) = ?,

èèègiven: Pb(s) + 2FeÄó ─¥ Pbìó + 2Feìó, E°(cell) = 0.895 V

    A) -0.645 VèèèB) +1.160 VèèèC) -0.125 VèèèD) +0.323 V
üèUsïg Pb(s) + 2FeÄó ─¥ Pbìó + 2Feìó, E°(cell) = 0.895 V, we know
that E°(cell) = E°(ox) + E°(red) = 0.895 Vè
èè E°(cell) = E°(Pb|Pbìó) + E°(FeÄó|Feìó) = 0.895 V.
From ê ståard reduction potential tables, we obtaï
èè FeÄó + eú = Feìó,è E°(FeÄó|Feìó) = +0.770 V.èSubstitutïg this
value ïë our equation for E°(cell) yields:
èè E°(cell) = E°(Pb|Pbìó) + (+0.770 V) = 0.895 V.
èè E°(Pb|Pbìó) = 0.895 V - (+0.770 V) = +0.125 V.èThis is ê oxida-
tion potential for Pb, we were asked ë fïd ê reduction potential.
èè E°(Pbìó|Pb) = -E°(Pb|Pbìó) = -(+0.125 V).
èè E°(Pbìó|Pb) = -0.125 V.
Ç C