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à 12.4cèThe Nernst Equation
äèPlease fïd ê nonståard state cell voltage ï ê followïg electrochemical systems.
âèWhat is ê cell potential ç ê Daniell cell when ê concen-
trations are: [Znìó] = 1.50 M å [Cuìó] = 0.50 M.èThe net reaction is
Zn(s) + Cuìó ─¥ Znìó + Cu(s), E°(cell) = 1.103 V.èThe Nernst equation
for this system isèE(cell) = E°(cell) - (0.05916/n)∙log{[Znìó]/[Cuìó]}.
Two electrons are exchanged per Zn or Cu, so n=2.èWe want E(cell).
èèE(cell) = 1.103 - (0.05916/2)∙log{1.50/0.50} = 1.089 V.
éSèIn 1889, H. W. Nernst developed an equation that describes how
concentrations, pressures, å temperature affect ê voltage ç a cell.
This equation is now known as ê Nernst Equation.èIn general form, ê
equation is
èèèR·T
E(cell) =èE°(cell) - ─── ln(Q)
èèèn·F
E(cell)èis ê nonståard state cell potential.
E°(cell) is ê ståard state cell potential.èBy now you are an
èèèèèèèè expert at calculatïg êse values.
R is ê gas constant.
T is ê absolute temperature
n is ê number ç moles ç electrons exchanged ï ê reaction.
F is ê Faraday constant, which is ê charge on one mole ç
èelectrons.
Q is ê reaction quotient, which we met ï Section 10.3.èIt is
èê concentration part ç ê equilibrium constant expression.
We will use this equation at 25°C (298 K).èSubstitutïg ï ê values ç
ê constants, R å F, å convertïg from natural logarithms ë ê
base 10 logarithm, we obtaï ê form ç Nernst equation that we will use
èèè0.05916
E(cell) =èE°(cell) - ─────── ln(Q)èat 25°C.
èèèèn
Let's apply this equation ë ê oxidation ç Feìó by Cl½.èThe
reaction isè2Feìó + Cl½è─¥ 2FeÄó + 2Clú,èE°(cell) = +0.588 V.
What is ê cell voltage under ê followïg conditions:
P(Cl½) = 0.95 atm, [Feìó] = 0.025 M, [FeÄó] = 0.050 M, [Clú] = 0.020 M?
We begï by determïïg ê value ç n.èEach Fe changes oxidation state
from +2 ë +3, which is a one electron change.èThere two Fe's ï ê
balanced chemical equation, so n = 2.èThe Q term equals ê products
divided by ê reactants, excludïg pure solids å liquids.èThe Nernst
equation is
èèèèèèèèèèèèèèèè 0.05916èè í [FeÄó]ì·[Clú]ì ┐
èèèèèèè E(cell) = 0.588 ─ ─────── log |────────────────|
èèèèèèèèèèèèèèèèèè2èèèè└ [Feìó]ì·P(Cl½) ┘
Substitutïg ê concentrations å partial pressure ç chlorïe ïë
ê equation, we get
èèèèèèèèèèèèèèèè 0.05916èè í [0.050]ì·[0.020]ì ┐
èèèèèèè E(cell) = 0.588 ─ ─────── log |───────────────────|
èèèèèèèèèèèèèèèèèè2èèèè└è[0.025]ì·(0.95)è┘
èèèèèèè E(cell) = 0.588 ─ (0.02958) log(1.684x10úÄ)
èèèèèèè E(cell) = 0.588 - (-0.082) = 0.670 V.
The cell voltage ïcreased above ê ståard state cell voltage.èThe
higher voltage means that ê drivïg force for ê reaction is greater
than it is at ståard state conditions.èWe can use LeChatelier's
prïciple ë understå qualitatively what will happen ë ê cell
voltage when conditions are altered.èThe Nernst equation gives us a
quantitative underståïg ç what happens ë ê cell potential.
The ståard reduction potential for ê reduction ç nitrate ion ë NO
isèè NO╕ú + 4Hó + 3eú ─¥ NO(g) + 2H½O, E°(red) = +0.956 V.
Usïg LeChatelier's prïciple, we understå that if we ïcrease ê [Hó]
ê system will shift ë ê right.èThe drivïg force for ê forward
reaction is greater.èThe nitrate ion will be a stronger oxidizïg agent
at higher acid concentrations.èThe reduction potential will ïcrease,
å we could use ê Nernst equation ë calculate ê new E(red) at ê
higher [Hó].èWhat is E(red) when P(NO) = 1 atm, [NO╕ú] = 1 M, å
[Hó] = 6.0 M.è
èèèèèèèèèèè 0.05916èè íèèP(NO)èè┐
èè E(cell) = 0.956 ─ ─────── log |─────────────|
èèèèèèèèèèèèè3èèèè└ [NO╕ú][Hó]Å ┘
èè E(cell) = 0.956 ─ (0.01972) log(1/(6.0)Å) = 1.017 V.
The voltage is higher at ê higher acid strength, å NO╕ú becomes a
stronger oxidizïg agent.
1èGiven ê cell reaction for ê lead sërage battery:
èè PbO╖(s) + Pb(s) + 2Hó(aq) + 2HSO╣ú(aq) ──¥ 2PbSO╣(s) + 2H╖O(l).
èèèèè What number should be used for "n" ï ê Nernst equation?
èèA) 1èèèB) 2èèèC) 4èèèD) 8
üè"n" is ê number ç moles ç electrons exchanged ï ê reaction.
Pb changes oxidation state from +4 ï PbO½ å 0 ï Pb ë +2 ï PbSO╣.
The change ï oxidation state is -2è(PbO╖ ¥ PbSO╣) å +2 (Pb ¥ PbSO╣).
The value ç n ï ê Nernst equation is 2.
Ç B
2èGiven,è2Al(s) + 3Cuìó ─¥ 2AlÄó + Cu(s), Eò(cell) = 2.126 V.
èèèèè What number should be used for "n" ï ê Nernst equation?
èèèèè A) 2èèèB) 3èèèC) 5èèèD) 6
üè"n" is ê number ç moles ç electrons exchanged ï ê reaction.
Al changes oxidation state from 0 ï Al ë +3 ï AlÄó.èAlumïum under-
goes a three electron change, å êre are two moles ç Al ï ê bal-
anced reaction.èThe ëtal number ç moles ç electrons exchanged ï ê
reaction is 2*3 = 6.
Ç D
3èWhat is ê cell potential at 25°C ç a cell ï ê lead
sërage battery when [Hó] = [HSO╣ú] = 0.526 M?èThe cell reaction is
PbO╖(s) + Pb(s) + 2Hó(aq) + 2HSO╣ú(aq) ──¥ 2PbSO╣(s) + 2H╖O(l), E°=2.041.
èè A) 2.107 Vèèè B) 2.008 Vèèè C) 2.041 Vèèè D) 2.074 V
üèA nonståard state cell voltage is calculated usïg ê Nernst
equation:èE(cell) = E°(cell) - (0.05916/n)log(Q).èIn ê PbO╖ - Pb(s)
cell, n = 2.èOnly ion concentrations å gases appear ï ê Q term.
The Nernst equation isèèèèèèèèè 0.05916èè┌èèè1èèèè┐
èèèèèèèèèèè E(cell) = 2.041 ─ ─────── log|───────────────|
èèèèèèèèèèèèèèèèèèèèèè2èèè └ [Hó]ì[HSO╣ú]ì ┘
èèèèèèèèèèèèèèèèèèèè 0.05916
èèèèèèèèèèè E(cell) = 2.041 ─ ─────── (-2)log([Hó][HSO╣ú])
èèèèèèèèèèèèèèèèèèèèèè2
èèèèèèèèèèè E(cell) = 2.041 + 0.05916*log((0.526)(0.526))
èèèèèèèèèèè E(cell) = 2.041 - 0.033 = 2.008 V
Ç B
4èGiven: Cu(s) + 2Agó ─¥ Cuìó + 2Ag(s), E°(cell) = +0.369 V.
What is ê cell voltage when [Agó] = 1.00x10úæ M å [Cuìó] = 0.50 M?
èè A) 1.061 VèèèB) 0.715 VèèèC) -0.323 VèèèD) 0.023 V
üèThe Nernst equation for this cell is
èèèèèèèèèèèèè0.05916èè┌ [Cuìó] ┐
èè E(cell) = E°(cell) ─ ─────── log|────────│
èèèèèèèèèèèèèè 2èèè └ [Agó]ì ┘
n = 2, Cu undergoes a two electron change.èSubstitutïg ê numerical
values ïë ê equation, you obtaï
èè E(cell) = 0.369 - (0.02958)*log{(0.50)/(1.00x10úæ)ì}
èè E(cell) = 0.369 - (0.02958)*(11.70) = 0.023 V.
We can use LeChatelier's prïciple ë predict that ê cell voltage will
be less than E°.èE° is ê voltage when ê ionic concentrations are
1 M.èThe [Agó] is very much less than 1 M, so ê drivïg force for ê
reaction is much smaller.èThat is what ê smaller cell voltage shows.
Ç D
5.èGiven ê reaction Zn(s) + Cuìó ─¥ Znìó + Cu(s),
E°(cell) = 1.103 V,èwhat will ïcrease E(cell) above E°(cell) under
ideal conditions?
A) Increase ê size ç ê Zn å Cu electrodes.
B) Increase only [Znìó] above 1 M.
C) Increase only [Cuìó] above 1 M.
D) All ç ê above.
üèIn order ë ïcrease ê cell potential above ê ståard state
cell potential, we must ïcrease ê drivïg force for ê reaction.èWe
can eiêr reduce ê product concentrations or ïcrease ê reactant
concentrations.èUnder ideal conditions ê size ç ê electrodes has
no effect on ê potential.èWe can ïcrease ê cell potential by
ïcreasïg ê copper(II) concentration above 1 M.èIncreasïg [Znìó]
above 1 M will reduce ê cell voltage.
Ç C
6èThe overall reaction for ê mercury battery can be written as
Zn(s) + HgO(s) + H½O(l) ─¥ Zn(OH)╖(s) + Hg(l), E°(cell) = 1.36 V.
What happens ë ê cell voltage as ê battery is used?
A) The voltage decreases because [Zn] å [HgO] decrease.
B) The voltage remaïs practically constant.
C) The voltage ïcreases because [Zn(OH)╖] å [Hg] ïcrease.
D) More ïformation is needed ë determïe ê voltage change.
üèBy applyïg ê Nernst equation ë ê net cell reaction, we can
see that ê cell voltage should remaï practically constant throughout
ê life ç ê battery.èThe Q term ï ê Nernst equation equals one
because ê reactants å products are solids å liquids.èThere are no
ion concentrations or gas pressures ïvolved.èThe Nernst equation is
èè E(cell) = E°(cell) - (0.05916/n)*log(Q).
èè E(cell) = 1.36 - (0.05916/2)*log(1), but log(1) = 0.
èè E(cell) = 1.36.èThe cell voltage should remaï constant at 1.36 V.
Ç B
7èWhat is ê effect ç [Hó] on ê oxidizïg strength ç MnO╣ú?
èèèèèèèMnO╣ú + 8Hó + 5eú ─¥ Mnìó + 4H╖O, E°(cell) = 1.491 V.
èè A) MnO╣ú becomes a stronger oxidizïg agent as [Hó] ïcreases.
èè B) MnO╣ú becomes a weaker oxidizïg agent as [Hó] ïcreases.
èè C) The oxidizïg strength ç MnO╣ú is unaffected by [Hó] changes.
èè D) More ïformation is needed ë answer ê question.
üèWe can analyze this problem usïg LeChatelier's prïciple.èAs
ê concentration ç Hó ïcreases, a greater stress is placed on ê left
hå side ç ê half-reaction.èThe drivïg force for ê forward reac-
tion (left ë right) ïcreases ï order ë relieve ê stress ç ê
higher Hó concentration.èThe reduction potential will ïcrease.èMnO╣ú
becomes a stronger oxidizïg agent at higher Hó concentrations.
Ç A
8èWhat is ê effect ç [Hó] on ê oxidizïg strength ç BrO╕ú?
èèèèèèèBrO╕ú + 6Hó + 6eú ─¥ Brú + 3H╖O, E°(cell) = 1.44 V.
èè A) The oxidizïg strength ç BrO╕ú is unaffected by [Hó] changes.
èè B) BrO╕ú becomes a weaker oxidizïg agent as [Hó] ïcreases.
èè C) BrO╕ú becomes a stronger oxidizïg agent as [Hó] ïcreases.
èè D) More ïformation is needed ë answer ê question.
üèWe can apply LeChatelier's prïciple ë this problem.èAs ê
concentration ç Hó ïcreases, a greater stress is placed on ê left-
hå side ç ê half-reaction.èThe drivïg force for ê forward reac-
tion (left ë right) ïcreases ï order ë relieve ê stress ç ê
higher [Hó].èThe reduction potential will ïcrease.èBrO╕ú becomes a
stronger oxidizïg agent at higher Hó concentrations.
Ç C
äèPlease fïd ê unknown concentration ï ê followïg electrochemical cells.
âèGiven ê cell Zn(s) + Feìó ─¥ Znìó + Fe(s), E°(cell) = 0.354 V.
What is ê concentration ç Feìó ï a cell that has a cell voltage ç
0.366 V when [Znìó] = 0.050 M?èThe Nernst equation relates ê cell voltages å
concentrations.èE(cell) = E°(cell) - (0.05916/n)*log(Q).èIn this reac-
tion, n = 2.è0.366 = 0.354 - (0.05916/2)*log([Znìó]/[Feìó]).
(0.366-0.354)*2/(-0.05916) = -0.4057 = log(0.050/[Feìó]).
+0.4057 = log([Feìó]/0.050).è[Feìó] = .050*10òñÅòÉÆ = 0.13 M.
éSèSo far we have used ê Nernst equation ë fïd nonståard
state cell potentials at nonståard state conditions.èI hope that it is
obvious that we can turn this around ë use nonståard state cell pot-
entials ë fïd nonståard state concentrations or pressures.
èè Let's consider ê a copper-silver cell.èCopper will reduce Agó ë
Ag.èThe reaction is Cu(s) + 2Agó ─¥ Cuìó + 2Ag(s), E°(cell) = 0.4594 V.
The Nernst equation for ê cell is:
èèèèèèèèèèèèèèè 0.05916èè í [Cuìó] ┐
èèèèèèE(cell) = 0.4594 ─ ─────── log |────────|
èèèèèèèèèèèèèèèèè2èèèè└ [Agó]ì ┘
The value ç n is 2 because each copper loses 2 electrons.èEquivalently,
each silver ion gaïs one electron, å êre are two silver ions ï ê
reaction.èThe Cu(s) å Ag(s) do not appear ï ê Q term because êy
are solids.
èè What is ê concentration ç Agó ï a cell ï which ê [Cuìó] is
1.00 M å for which ê cell potential is 0.2511 V?èSubstitutïg ê
known values ïë ê equation, we obtaï
èèèèèèèèèèèèèèè0.05916èè í (1.00) ┐
èèèèèè0.2511 = 0.4594 ─ ─────── log |────────|
èèèèèèèèèèèèèèèè 2èèèè└ [Agó]ì ┘
We see that ê only unknown ï ê equation is ê concentration ç ê
silver ion.èLet's manipulate ê log term.
èè í (1.00) ┐
log |────────|è=èlog([Agó]úì)è= -2·log([Agó]).èPluggïg ê result
èè └ [Agó]ì ┘
ïë our equation yields:
èèèèèèèèèèèèèèè 0.05916
èèèèèè0.2511 = 0.4594 ─ ─────── (-2)log([Agó])
èèèèèèèèèèèèèèèèè2
èèèèèè0.2511 = 0.4594 ─ 0.05916·log([Agó])
èèèèèèèèèèèèèèèèèèèèèèèèè0.2511 - 0.4594
Rearrangïg ë get log([Agó] gives:èlog([Agó]) = ────────────────.
èèèèèèèèèèèèèèèèèèèèèèèèèè0.05916
èèèèèèèèèèèèèèèèèè log([Agó]) = -3.521
The silver ion concentration is found by takïg ê antilog ç -3.521.
[Agó] = ╢╡-3.521è=è3.01x10úÅ M.
By measurïg ê cell voltage, we have been able ë obtaï ê concen-
tration ç ê silver ion.èDo you know a routïe application ç this
concept?èThe pH ç a solution is routïely obtaïed by measurïg ê
potential ç a "glass electrode" that is placed ï ê solution.
9èWhat is ê concentration ç Cuìó ï a cell that has a cell
voltage ç +0.614 V when [Agó] = 0.10 M?èThe cell reaction is
Cu(s) + 2Agó ─¥ Cuìó + 2Ag(s), E°(cell) = 0.459 V.
è A) 2.4x10úÉ Mèè B) 7.0x10úÄ Mèè C) 7x10úìî Mèè D) 5.8x10úô M
üèConcentrations are related ë ê cell voltage through ê
Nernst equation.èèèèèèè0.05916èè í [Cuìó] ┐
èèèèèèE(cell) = 0.459 ─ ─────── log |────────|
èèèèèèèèèèèèèèèè 2èèèè└ [Agó]ì ┘
èèèèèè+0.614 =è0.459 - 0.02958*log{[Cuìó]/(0.10)ì}
èèè 0.614-0.459
èèè ─────────── =èlog{[Cuìó]/0.010}.è-5.24 = log{[Cuìó]/0.01}.
èèèè-0.02958
Takïg ê antilog yields [Cuìó]/0.010 = ╢╡-5.24 = 5.8x10úæ.
èèèè[Cuìó]= (0.010)(5.8x10úæ) = 5.8x10úô M
Ç D
10èWhat is ê concentration ç Brú ï a cell that has a cell
voltage ç +0.906 V when [Znìó] = 0.20 M?èThe cell reaction is
èè Zn(s) + 2AgBr(s) ─¥ Znìó + 2Ag(s) + 2Brú, E°(cell) = 0.834 V.
è A) 0.25 Mèè B) 0.061 Mèè C) 0.14 Mèè D) 0.33 M
üèConcentrations are related ë ê cell voltage through ê
Nernst equation.èèèèèèè0.05916
èèèèèèE(cell) = 0.834 ─ ─────── log{[Znìó][Brú]ì}
èèèèèèèèèèèèèèèè 2
èèèèèè+0.906 =è0.834 - 0.02958*log{(0.20)[Brú]ì}
èèè 0.906-0.834
èèè ─────────── =è-2.434 = log{(0.20)[Brú]ì}.
èèèè-0.02958
Takïg ê antilog yields (0.20)[Brú]ì = ╢╡-2.434 = 3.68x10úÄ
èèèèèèèèí────────────────
èèèè[Brú] = á(3.68x10úÄ/0.20)è=è0.14 M
Ç C
11èWhat is ê value ç ê ratio [Feìó]/[FeÄó] ï a cell that
has a cell voltage ç 0.O V when [Iú] = 0.10 M å [I╕ú] = 0.10 M?èThe
cell reaction isè2FeÄó + 3Iú ─¥ 2Feìó + I╕ú, E°(cell) = 0.236 V.
è A) 9.8x10ìèèèèB) 0.63èèèèC) 9.5x10ÉèèèèD) 0.13
üèConcentrations are related ë ê cell voltage through ê
Nernst equation.èèèèèèè0.05916èè┌ [Feìó]ì[I╕ú] ┐
èèèèèèE(cell) = 0.236 ─ ─────── log|──────────────|
èèèèèèèèèèèèèèèè 2èèè └ [FeÄó]ì[Iú]Ä ┘
èèèèèèèèèèèèèèèèèèèè ┌ [Feìó]ì(0.10)è┐
èèèèèèèèè0 = 0.236 - 0.02958*log|────────────────|
èèèèèèèèèèèèèèèèèèèè └ [FeÄó]ì(0.10)Ä ┘
èèè-0.236 = -0.02958 *log{100*([Feìó]/[FeÄó])ì}
èè (-0.236/-0.02958) = 7.98 = log{100*([Feìó]/[FeÄó])ì}
Takïg ê antilog yields 100*([Feìó]/[FeÄó])ì = ╢╡7.98 = 9.5x10Æ
èèèèèèèèèèè í─────────────
èèè [Feìó]/[FeÄó] = á(9.5x10Æ/100)è=è9.8x10ì
Ç A
äèPlease fïd ê value ç ê equilibrium constant from ê ståard state cell potential.
âèWhat is ê value ç ê equilibrium constant at 25°C for ê
system:èCu(s) + 2Agó ─¥ Cuìó + 2Ag(s), E°(cell) = 0.459 V.èThe equili-
brium constant is found usïg ê Nernst equation with ê cell potential
set ë zero.èK = ╢╡(n·E°/0.05916).
K = ╢╡(2·0.459/.05916) = 3.4x10îÉ.
éSèApplyïg ê Nernst equation ë a system at equilibrium allows
us ë calculate ê value ç ê equilibrium constant for ê system.
The fïal state ç a process is ê equilibrium state.èAs a cell reaches
equilibrium, its cell voltage goes ë zero.èAt equilibrium, E(cell) = 0.
The Q term ï ê Nernst equation will contaï ê equilibrium concentra-
tions å pressures.èAt equilibrium, E(cell) = 0 å Q = K, ê equili-
brium constant.èReturnïg ë ê Nernst equation,
E(cell) = E°(cell) - (0.05916/n)*log(Q) at 25°C.
Substitutïg ê equilibrium condition ïë ê Nernst equation gives
èèè0 = E°(cell) - (0.05916/n)*log(K) at 25°C.
Rearrangïg, log(K) = n·E°(cell)/0.05916èat 25°C.èTakïg ê antilog,
we obtaïèK = ╢╡(n·E°(cell)/0.05916)èat 25°C.
èè What is ê value ç ê equilibrium constant at 25°C for ê
system: 2Feìó + Cl½è─¥ 2FeÄó + 2Clú,èE°(cell) = +0.588 V.
You know that you can calculate K usïgèK = ╢╡(n·E°(cell)/0.05916).
You know E°(cell).èYou need ë determïe ê value ç n.èIron undergoes
a one electron change å êre are two iron aëms ï ê reaction.
n = 2.
èèèèèK = ╢╡(2*0.588/0.05916) = ╢╡(19.878)
There are only three significant figures ï ê power ç ten.èThe 19
fixes ê decimal poït leavïg only one significant figure.
èèèèèK = 7.55x10îö = 8x10îöèë ê correct number ç significant
èèèèèèèèèèèèèèèèèfigures.
12èWhat is ê value ç K at 25°C for ê reaction ç ê
èè Daniell cell:Zn(s) + Cuìó ─¥ Znìó + Cu(s), E°(cell) = 1.103 V?
èè A) 1.9x10ÄÆèè B) 4.4x10îôèè C) 1.6x10ìèè D) 4.1x10îò
üèWe can rearrange ê Nernst equation ë fïd ê equilibrium
constant by solvïg for Q (equal ë K) when E(cell) = 0.èThe result is
èè K = ╢╡(n·E°(cell)/0.05916) at 25°C.
n = 2 ï ê reaction, because zïc loses two electrons.
èè K = ╢╡(2*1.103/0.05916).èK = 1.9x10ÄÆ.
Ç A
13èWhat is ê value ç K at 25°C for ê reaction ç ê lead
sërage battery:
PbO╖(s) + Pb(s) + 2Hó(aq) + 2HSO╣ú(aq) ──¥ 2PbSO╣(s) + 2H╖O(l), E°=2.041?
èè A) 3.2x10ÄÅèè B) 1.00x10æöèè C) 1.5x10ôèè D) 1.0x10îÄô
üèWe can rearrange ê Nernst equation ë fïd ê equilibrium
constant by solvïg for Q (equal ë K) when E(cell) = 0.èThe result is
èè K = ╢╡(n·E°(cell)/0.05916) at 25°C.
n = 2 ï ê reaction, because lead loses two electrons when formïg
PbSO╣.
èè K = ╢╡(2*2.041/0.05916).èK = 9.98x10æô.èWe are justified ï two
significant figures ï our answer.èK = 1.00x10æö
Ç B
14èWhat is ê value ç K at 25°C for ê reactionè
èè5Feìó + MnO╣ú + 8Hó ─¥ 5FeÄó + Mnìó + 4H╖O, E°(cell) = 0.721 V?
èè A) 2x10îìèè B) 2x10Äôèè C) 1x10îöîèè D) 9x10æò
üèWe can rearrange ê Nernst equation ë fïd ê equilibrium
constant by solvïg for Q (equal ë K) when E(cell) = 0.èThe result is
èè K = ╢╡(n·E°(cell)/0.05916) at 25°C.
n = 5 ï ê reaction, because Feìó loses one electron å êre are five
Feìó's ï ê reaction.
èè K = ╢╡(5*0.721/0.05916).èK = 9x10æò.
Ç D