Electrochemistry Nernst Equation Ion selective electrodes Redox reactions • oxidation - loss of electrons • M n+ ⇒ M n+1 + e - • M is oxidized - reducing agent • reduction - gain of electrons • N n+ + e - ⇒ N n-1 • N is reduced - oxidizing agent Half Reactions • Fe 3+ + e - ⇒ Fe 2+ • Zn 2+ + 2e ⇒ Zn • if mix these which will donate, which will accept electrons? • can’t measure equilibrium concentrations - only represent this reaction
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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
– 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