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Electrochemistry
Nernst EquationIon selective electrodes
Redox reactions
• oxidation - loss of electrons• Mn+ ⇒ Mn+1 + e-
• M is oxidized - reducing agent• reduction - gain of electrons• Nn+ + e- ⇒ Nn-1
• N is reduced - oxidizing agent
Half Reactions
• Fe3+ + e- ⇒ Fe2+
• Zn2+ + 2e ⇒ Zn• if mix these which will donate, which will
accept electrons?• can’t measure equilibrium
concentrations - only represent thisreaction
-ve, ANODE +ve CATHODE
Galvanic Cell
• Cu2+ + Zn ⇒ Cu + Zn2+
Half reactions are:• Cu2+ + 2e- ⇒ Cu• Zn - 2e- ⇒ Zn2+
• transfer of electrons - electric current -can be measured
• half-cell with greater driving forcemakes other cell accept electrons
Spontaneous reaction
• copper more easily reduced• electrons flow spontaneously from zinc half
cell to copper half cell• Zn - 2e- ⇒ Zn2+ (zinc dissolves)• Cu2+ +2e- ⇒ Cu (copper bar gains weight)• Zn electrode -ve (loses electrons) ANODE• Cu electrode +ve (gains electrons) CATHODE
Salt Bridge• Salt bridge maintains electrical
neutrality by transport of ions• Cu deposits leaves excess SO4
2- -neutralized by K+ from KCl bridge
• Zn dissolves to give excess Zn2+ insolution, neutralized by Cl- from saltbridge
• Also SO42- and Zn2+ could migrate into
bridge - does not matter which
Driving Force• Driving force of half-cell can’t be measured -
only by comparison to other half cells• All potentials quoted against hydrogen half
cell - assigned zero potential• 2H+ + 2e- ⇒ H2
• if half cell causes H2 cell to accept electrons -assigned -ve potential (Zn -0.763)
• if H2 call causes half cell to accept electrons -assigned +ve potential (Cu +0.86)
Oxidizing/Reducing agents
• Strong oxidizing agents e.g. permanganate+ve potential
• Strong reducing agents, e.g. zinc -ve potential• Potentials shown in next slide are for gases at
1 atm pressure, and 1M for solutions relativeto hydrogen electrode
Which way is spontaneous?• Fe3+ + e- ⇒ Fe2+ 0.771V• Zn2+ + 2e- ⇒ Zn -0.763V
1. Spontaneous reaction has +ve potential2. Subtract one reaction potential from
other to make difference +vei.e 0.771 - (-0.763) = +1.534 V3. potential of Zn has to be subtracted to
make final number positive - Zn goes inreverse: Zn ⇒ Zn2+ + 2e-
Overall Reaction
2Fe3+ + 2 e- ⇒ 2 Fe2+ E01 = 0.771
Zn ⇒ Zn2+ + 2e- E02 = -0.763
2 Fe3+ + Zn ⇒ Zn2+ + 2 Fe2+
E1 - E2 = + 1.534 V
The Nernst Equation• if species not in standard state, E0
changes depending on concentrations
E = E0 !
RT
nFln
[red]b
[ox]a or
[products]
[reactants]
"
#
$
%
&
'
where a ox + n e- ( b red
a, b coefficients in balanced equation
at 250C : E = E
0 !0.059
nln
[red]b
[ox]a
(half reactions) (complete reactions)
Example using cell notation
CuCu+ (10!5
M) Sn4 + (10!1M)Sn2+ (10!4
M)Pt
CONVENTION :
reaction proceeds from Left to Right
Anode on LHS Cathode on RHS
Oxidation Reduction
" loss of electrons " gain of electrons
Ecell = Eright ! Eleft
write each half reaction with its potentialCu+ + e- ⇒ Cu 0.521 VoltsSn4+ + 2 e- ⇒ Sn2+ 0.154 Voltsbalanced equation is:Sn4+ + 2Cu ⇒ Sn2+ + 2 Cu+
Nernst eqn to calculate each potential
CuCu+(10
!5M) Sn
4+(10
!1M)Sn
2+(10
!4M)Pt
EL
= 0.521!0.059
2ln
[Cu]
[Cu+]
2
= 0.521!0.059
2ln
[1]
[10!5
]2
= 0.226 volts Potential less than E0
could use 1’s here - gives the same answer
CuCu+(10
!5M) Sn
4+(10
!1M)Sn
2+(10
!4M)Pt
ER
= 0.154 !0.059
2ln
[Sn2+
]
[Sn4 +
]2
= 0.154 !0.059
2ln
[10!4
]
[10!1
]= 0.243 volts
ER! E
L= E
cathode! E
anode= 0.243! 0.226 = 0.017 volts
Potential > E0
anodeoxidation
cathodereduction
opposite to expected using E0
Cu+ + e- ⇔ Cu 0.521 voltsSn4+ + 2e- ⇔ Sn2+ 0.154 voltsfor +ve potential 0.154 must be made -vetin must go in reverse, Sn2+ ⇒ Sn4+ + 2 e-
or: 2Cu+ + Sn2+ ⇒ 2Cu + Sn4+
Subtract to give+ve number
Nernst equation changes predicteddirection of the reaction
ER is the actual tin reaction = 0.243 voltsEL is the actual Cu reaction = 0.226 volts
with concentrations incell notation, reaction reversedcf predicted by subtraction of E0’s
Cu reaction goes in reverseSn4+ + 2Cu ⇒ Sn2+ + 2Cu+
CuCu+(10
!5M) Sn
4+(10
!1M)Sn
2+(10
!4M)Pt
Subtract to give+ve number (0.226is made -ve)
Measurement of Potential
• assumed electrons actually flow duringmeasurement
• undesirable to have current flow• reduction or oxidation - changes
concentration• potentiometer principle is used
potentiometer
- +
- +
galvanometermeasures current
cell with electrodes
variable resistor
battery
electrodes
fraction of a standardvoltage from batteryvaried until no currentflows
voltage required to stopflow matches potentialbeing measured
pH meter is apotentiometer -measures voltage w/ocurrent flow
electrodes• half cells that do not involve pure metal in reaction:
conducting electrode is usually inert Pt to conductelectrons
• potentiometric measurements:– choose a suitable electrode whose potential depends on specie
being measured
e.g. a) Ag Ag+ Cu Cu++ Zn Zn++ M Mn+
b) Pt redox couple
such as Cr2O7
2!+14H +
+ 6e! "#"$"" 2Cr3+
+ 7H2O E0= 1.33 Volts
electrodes (cont.)• Potential must be measured relative to a
reference electrode– Ecell = Ecathode - Eanode (Eright - Eleft)
• hydrogen is standard ref electrode - butdifficult to use
• need another reference electrode– needs to have constant potential not affected by ions
in solution
Saturated calomel electrode(SCE)
HgHg2Cl2 KClsat E = 0.242V vs SHE
Hg2Cl2 + 2e! " 2Hg+ 2Cl!
mercurous chloride
E = E0!
0.059
1lg[Cl!]
• potential constantwith small currentflow
• why?
If the electrode accepts electrons: Hg+
mercurous
! Hgmercury
solid Hg2Cl2 dissolves to resaturate the solution
If the electrode produces electrons: Hgmercury
! Hg+
mercurous
but solution is saturated with Hg2Cl2 so merurous ion
precipitates as Hg2Cl2
because [Cl- ] is high - small changes in [Cl- ] do not
affect potential significantly
Cl- depresses solubility of Hg2Cl2 by common ion
effect to maintain constant ionic strength and
constant potential
practical device
– electrode constructedto dip directly intoanalytical solution
– salt bride replacedby fiber - acts likesalt bridge
– small [Cl-] leaks intosolution, but notusually important
silver/silver chloride ref electrode
Ag AgCl Cl! + 0.97 volts at 25 C, w.r.t. SHE
silver chloride immersed in saturated KCl saturated with AgCl
As long as Cl- doesn't take part in reaction can be used as
a reference electrode.
Titration MnO4- with Fe2+
• determn of Fe in soln - titrate w/ std permanganate
MnO
4
!+ 5Fe
2++ 8H
+! Mn
2++ 5Fe
3++ 4H
2O
– Fe must be in Fe2+ state - reduce w/ stannous (see next)– add known increments KMnO4 , measure potential of Pt
electrode vs SCE as titration proceeds.– plot of potential vs mLs of titrant– potential determined by Nernst eqn at different concns
reduction Fe3+ to Fe2+
2Fe3++ SnCl4
2!+ 2Cl!! 2Fe2+
+ SnCl62!
then must destroy tin II with mercury II
SnCl4
2-+ 2HgCl4
2!! SnCl6
2!+ Hg2Cl2 (s) + 4Cl!
enough tin II added to complete reduction of iron III
but if too much excess Sn II, Hgmetal will form, not calomel
SnCl4
2-+ HgCl4
2!! SnCl6
2!+ Hg(l) + 2Cl!
will react with MnO4
- and interfere with permanganate titration
calomel
Jones reductor
Amalgam of Zn and Hg in a column (zinc shot)
Zn + Hg2+! Zn
2++ Hg
pass iron Fe3+ through column to reduce it to Fe2+
1M H2SO4 as the solvent
Zn is a powerful reducing agent
will reduce almost anything
Zn2++ 2e! ! Zn(s) E0
= !0.764V
Harris, 6edn p358, fig. 16-7
Redox titration calculations MnO
4
!+ 5Fe
2++ 8H
+! Mn
2++ 5Fe
3++ 4H
2O
After adding aliquot of MnO4
- - reaction comes to eqm
potentials of both half reactions are equal
Calculate potential of reaction with half reaction
for iron ...... [C] of both species known
(each mmole of MnO4
- will oxidize 5 mmole Fe2+ )
Fe3++ e!! Fe
2+
E = 0.771! 0.059 lg[Fe2+ ]
[Fe3+ ]
add drop titrant - know amount Fe2+ converted to Fe3+
(1 mmole MnO4
- ! 5 mmole Fe2+ ) known ratio Fe2+
Fe3+
"#$
%&'
calculate E from Nernst E = E0 (
0.059
1lgFe
2+
Fe3+
at equivalence point
MnO4
(+ 5Fe2+
+ 8H +! Mn2+ + 5Fe3+
+ 4H2O
1
5x + x
1
5C (
1
5x
"#$
%&'
C ( x( )
C is [Fe3+ ] - know this because all Fe2+ converted to Fe3+
x is negligible compared to C, in terms of [] but not in potential
eqm will affect potential - can solve for x
by equating two Nernst equations - obtains equilibrium constant
must be equal and opposite at equilibrium
E = 0.771!0.059
5lg
[Fe2+ ]5
[Fe3+ ]5= 1.51!
0.059
5lg
[Mn2+ ]
[MnO4
! ][H + ]8
i.e.(1.51! 0.771) =0.059
5lg
[Mn2+ ]
[MnO4
! ][H + ]8!
0.059
5lg
[Fe2+ ]5
[Fe3+ ]5
0.739 =0.059
5log
[Mn2+ ][Fe3+ ]5
[MnO4
! ][H + ]8[Fe2+ ]5
0.739 =0.059
5lgKeqm lgKeqm = 62.6, Keqm = 5 "1062
substitute x, (C-x) and 1
5C !
1
5x
#$%
&'(
into eqm constant expn to calc x
use either half reaction to calculate potential using Nernst
after equivalence pointhave Mn2+ formed and excess MnO4
-
E = E0!
0.059
1lg
[Mn2+ ]
[MnO4
! ]
We want a difference in potential of 0.2 V in E2
0 and E1
0
for a sharp endpoint break
Note: in advanced calculations, activity must be taken into
account rather than just [] a = f [C]
f is the activity coefficient, and depends on charge
on the ion, which affects ionic strength of solution
calculatefrom
volume
Rules for redox titrations
• equilibrium constant must be high so that xis small - reaction well to right (differencein E0 of about 0.2 V should do it)
• measure potential for observation of theend point, or use an indicator such as MnO4
Reagents for redox titrations• Oxidizing agents
Potassium Permanganate - KMnO4
MnO4
!+ 5Fe2+
+ 8H +! Mn
2++ 5Fe3+
+ 4H2O, E0= 1.51V
purple solution - self indicating
Potassium Dichromate - primary standard
Cr2O7
2!+14H +
+ 6e! ! 2Cr3++ 7H2O E0
= 1.33 Volts
Ceric ion
Ce4++ e
!! Ce
3+ E0= 0.771 V
• reducing agents
Fe2+ stable for short periods
Fe3++ e
!! Fe
2+ E0= 0.771 V
Thiosulpate S2O3
2! not oxidized by air (rare)
S2O6
2!+ 2e! ! 2S2O3
2!
Note: you should study the iron/cerium system
Fe2++ Ce
4+! Fe
3++ Ce
4+
Ion Selective Electrodes
• The glass electrode - for pH measurement -specific for H+ ions.– potential difference develops across thin glass
membrane w/ solns of diff. pH on either side– potential measured by placing ref. electrodes on
each side of membrane– on ref. electrode is incorporated in the glass
electrode (Ag/AgCl) and the other is usually an SCEplaced in soln whose pH is to be measured.
Ag AgCl HCl (H+ internal) glass membrane H+ (unknown) SCE
The potential of this cell is given by:
Ecell
= k +2.303RT
Flga
Hunknown
+
• k is a constant - contains:– potentials of the two reference electrodes– potential due to H+ inside the glass membrane– asymmetry potential - due to non-perfect
behavior of glass membrane• potential not same when pH is same on both sides of
membrane• changes if physical condition of membrane changes
Since pH = - lg aH+
Ecell = k !2.303RT
FpH , i.e. pH =
Ecell ! k
2.303RT
F
k is determined using a buffer of known pH
i.e. k = Ecellstd!
2.303RT
Flg pHstd
then pHunknown = pHstd +Ecellunknown
! Ecellstd
2.303RT
F
acid/alkaline error
• potential due to ion exchange between H+ in soln & (Na+)ions in hydrated glass layer at solution/membrane boundary
• acid solns, H+ concn high - glass electrode responds solelyto H+ (xpt very high [H+] acid error)
• basic (>pH 9) H+ activity small - glass electrode begins torespond to other monovalent cations e.g. Na - alkaline error
membrane glass made of Na2O and SiO2
glass surface -SiO-Na
++ H
+! SiO
!H
++ Na
+
K for this equilibrium is large - gives silicic acid
Alkaline errorof glass
electrode
Acid error of glass electrode• activity, a, different from [H+]
– electrode behaves like there are less protonsavailable then actually added
Other ISE’s
• found that different membranecompositions can enhance alkaline error
• electrode can be made to be morespecific for Na, K, Li, etc.
• construction similar to H+ respondingglass electrode
• internal solution usually chloride salt ofcation of interest
e.g. LiO-Si
instead of NaO-Si
Solid State Membrane electrodes
• solid state fluoride electrode– membrane single LaF3 crystal + small qty of Eu II– creates disorders in crystal, lattice defects– defects correct size for F- ion– F- in lattice - mobile– lattice acts as semi-permeable membrane for F- alone– construction similar to glass electrode– Ag/AgCl internal ref. electrode
Fluoride Electrode
selectivity ratiopotential of ISE for an ion on its own:
Eelectrode
= EM
0 '!
2.303RT
nFlga
Mn+
where EM
0 ' depends on internal ref. electrode, filling
solution, and construction of membrane
EM
0 ' is a constant
determined by measuring solution of
known concentration
• if soln contains mixture of cations– may respond to other cations– eg. mixture of Na and K
• Nernst eqn has additive term for K ifdetermining Na
ENaK
= ENa
0 '!2.303RT
Flg(a
Na+ + kNaKaK + )
– ENaK is measured potential– kNaK is the selectivity ratio for potassium
over sodium
– selectivity ratio is the fraction of the sodiumpotential that is due to potassium
– KNaK and determined by use of knownsolutions of Na and K, and by solvingsimultaneous equations for KNaK and
• know how to calculate selectivity ratios andpotential of ISE’s
ENa
0 '
ENa
0 '