Electrochemistry
Nernst Equation
Ion selective electrodes
Redox reactions
•
•
•
•
•
•
oxidation - loss of electrons
Mn+ ⇒ Mn+1 + eM 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 this
reaction
+ve CATHODE
-ve, ANODE
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 force
makes 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 SO42- neutralized by K+ from KCl bridge
• Zn dissolves to give excess Zn2+ in
solution, neutralized by Cl- from salt
bridge
• Also SO4 2- 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 relative
to hydrogen electrode
Which way is spontaneous?
• Fe3+ + e- ⇒ Fe2+
•
Zn2+
+
2e-
⇒ Zn
0.771V
-0.763V
1. Spontaneous reaction has +ve potential
2. Subtract one reaction potential from
other to make difference +ve
i.e 0.771 - (-0.763) = +1.534 V
3. potential of Zn has to be subtracted to
make final number positive - Zn goes in
reverse: 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
(half reactions)
E = E0 !
b
RT [red]
ln
nF [ox]a
(complete reactions)
" [products] %
$ or
'
# [reactants] &
where a ox + n e - ( b red
a, b coefficients in balanced equation
at 25 0 C :
E = E0 !
0.059 [red]b
ln
n
[ox]a
Example using cell notation
Cu Cu + (10!5 M) Sn 4 + (10!1 M)Sn 2+ (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
E cell = E right ! E left
Cu Cu + (10!5 M) Sn 4 + (10!1 M)Sn 2+ (10!4 M) Pt
write each half reaction with its potential
Cu+ + e- ⇒ Cu
0.521 Volts
Sn4+ + 2 e- ⇒ Sn2+
0.154 Volts
balanced equation is:
Sn4+ + 2Cu ⇒ Sn2+ + 2 Cu+
Nernst eqn to calculate each potential
E L = 0.521!
= 0.521!
0.059 [Cu]
ln
2
[Cu + ]2
could use 1’s here gives the same answer
0.059
[1]
ln !5 2 = 0.226 volts
2
[10 ]
Potential less than E0
Cu Cu + (10!5 M) Sn 4 + (10!1 M)Sn 2+ (10!4 M) Pt
anode
0.059 [Sn 2+ ]
ln
oxidation E R = 0.154 !
4+ 2
2
cathode
reduction
[Sn ]
0.059 [10!4 ]
= 0.154 !
ln !1 = 0.243 volts Potential > E0
2
[10 ]
E R ! E L = E cathode ! E anode = 0.243 ! 0.226 = 0.017 volts
opposite to expected using E0
Subtract to give
+ve number
Cu+ + e- ⇔ Cu
0.521 volts
4+
2+
Sn + 2e ⇔ Sn
0.154 volts
for +ve potential 0.154 must be made -ve
tin must go in reverse, Sn2+ ⇒ Sn4+ + 2 eor: 2Cu+ + Sn2+ ⇒ 2Cu + Sn4+
Cu Cu + (10!5 M) Sn 4 + (10!1 M)Sn 2+ (10!4 M) Pt
Nernst equation changes predicted
direction of the reaction
ER is the actual tin reaction = 0.243 volts
EL is the actual Cu reaction = 0.226 volts
with concentrations in
cell notation, reaction reversed
cf predicted by subtraction of E0’s
Subtract to give
+ve number (0.226
is made -ve)
Cu reaction goes in reverse
Sn4+ + 2Cu ⇒ Sn2+ + 2Cu+
Measurement of Potential
• assumed electrons actually flow during
measurement
• undesirable to have current flow
• reduction or oxidation - changes
concentration
• potentiometer principle is used
potentiometer
-
+
battery
variable resistor
-
+
galvanometer
measures current
electrodes
cell with electrodes
fraction of a standard
voltage from battery
varied until no current
flows
voltage required to stop
flow matches potential
being measured
pH meter is a
potentiometer measures voltage w/o
current flow
electrodes
• half cells that do not involve pure metal in reaction:
conducting electrode is usually inert Pt to conduct
electrons
• potentiometric measurements:
– choose a suitable electrode whose potential depends on specie
being measured
e.g. a) Ag Ag + Cu Cu ++ Zn Zn + + M M n+
b) Pt redox couple
""
"
# 2Cr 3+ + 7H 2O E 0 = 1.33 Volts
such as Cr2O72 ! + 14H + + 6e! $
"
electrodes (cont.)
• Potential must be measured relative to a
reference electrode
– Ecell = Ecathode - E anode (Eright - Eleft)
• hydrogen is standard ref electrode - but
difficult to use
• need another reference electrode
– needs to have constant potential not affected by ions
in solution
Saturated calomel electrode
(SCE)
Hg Hg2Cl2 KClsat
E = 0.242V vs SHE
Hg 2Cl2 + 2e! " 2Hg + 2Cl!
mercurous chloride
0.059
E =E !
lg[Cl! ]
1
0
• potential constant
with small current
flow
• why?
If the electrode accepts electrons: Hg + ! Hg
mercurous
mercury
solid Hg 2Cl2 dissolves to resaturate the solution
If the electrode produces electrons: Hg !
mercury
Hg +
mercurous
but solution is saturated with Hg 2Cl2 so merurous ion
precipitates as Hg 2Cl2
because [Cl- ] is high - small changes in [Cl- ] do not
affect potential significantly
Cl- depresses solubility of Hg 2Cl2 by common ion
effect to maintain constant ionic strength and
constant potential
practical device
– electrode constructed
to dip directly into
analytical solution
– salt bride replaced
by fiber - acts like
salt bridge
– small [Cl- ] leaks into
solution, but not
usually 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
MnO4! + 5Fe2 + + 8H + ! Mn 2 + + 5Fe3+ + 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+ + SnCl42 ! + 2Cl! ! 2Fe2 + + SnCl62 !
then must destroy tin II with mercury II
calomel
SnCl2-4 + 2HgCl42 ! ! SnCl62 ! + Hg2Cl2 (s) + 4Cl !
enough tin II added to complete reduction of iron III
but if too much excess Sn II, Hg metal will form, not calomel
SnCl2-4 + HgCl42 ! ! SnCl62 ! + Hg(l) + 2Cl !
will react with MnO-4 and interfere with permanganate titration
Jones reductor
Amalgam of Zn and Hg in a column (zinc shot)
Zn + Hg 2+ ! Zn 2 + + Hg
pass iron Fe 3+ through column to reduce it to Fe 2+
1M H 2 SO4 as the solvent
Zn is a powerful reducing agent
will reduce almost anything
Zn 2+ + 2e! ! Zn(s) E 0 = !0.764V
Harris, 6edn p358, fig. 16-7
Redox titration calculations
MnO4! + 5Fe2 + + 8H + ! Mn 2 + + 5Fe 3+ + 4H 2O
After adding aliquot of MnO-4 - reaction comes to eq m
potentials of both half reactions are equal
Calculate potential of reaction with half reaction
for iron ...... [C] of both species known
(each mmole of MnO-4 will oxidize 5 mmole Fe 2+ )
Fe3+ + e! ! Fe2 +
E = 0.771 ! 0.059 lg
[Fe 2+ ]
[Fe3+ ]
add drop titrant - know amount Fe 2+ converted to Fe 3+
" Fe 2+ %
(1 mmole MnO-4 ! 5 mmole Fe 2+ ) known ratio $ 3+ '
# Fe &
calculate E from Nernst E = E 0 (
0.059 Fe2 +
lg 3+
1
Fe
at equivalence point
MnO4( + 5Fe2 + + 8H + !
1
x + x
5
Mn 2 +
+
1 %
"1
$# C ( x '&
5
5
3+
5Fe 3+ + 4H 2O
(C ( x )
2+
C is [Fe ] - know this because all Fe converted to Fe 3+
x is negligible compared to C, in terms of [] but not in potential
eq m will affect potential - can solve for x
by equating two Nernst equations - obtains equilibrium constant
must be equal and opposite at equilibrium
0.059 [Fe2 + ]5
0.059
[Mn 2 + ]
lg
= 1.51 !
lg
3+ 5
5
[Fe ]
5
[MnO4! ][H + ]8
E = 0.771 !
i.e.(1.51 ! 0.771) =
0.059
[Mn 2 + ]
0.059 [Fe2 + ]5
lg
!
lg
!
+ 8
5
[MnO4 ][H ]
5
[Fe 3+ ]5
0.059
[Mn 2 + ][Fe 3+ ]5
log
5
[MnO4! ][H + ]8 [Fe2 + ]5
0.059
0.739 =
lg K eqm
lg K eqm = 62.6,
K eqm = 5 " 10 62
5
1 &
#1
substitute x, (C-x) and % C ! x ( into eq m constant exp n to calc x
$5
5 '
use either half reaction to calculate potential using Nernst
0.739 =
after equivalence point
have Mn
2+
E = E0 !
formed and excess MnO-4
0.059 [Mn 2 + ]
lg
1
[MnO4! ]
calculate
from
volume
We want a difference in potential of 0.2 V in E 02 and E10
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
Rules for redox titrations
• equilibrium constant must be high so that x
is small - reaction well to right (difference
in E0 of about 0.2 V should do it)
• measure potential for observation of the
end point, or use an indicator such as MnO4
Reagents for redox titrations
• Oxidizing agents
Potassium Permanganate - KMnO 4
MnO4! + 5Fe2 + + 8H + ! Mn 2 + + 5Fe 3+ + 4H 2O,
purple solution - self indicating
Potassium Dichromate - primary standard
Cr2O72 ! + 14H + + 6e! ! 2Cr 3+ + 7H 2O
Ceric ion
Ce 4+ + e! ! Ce3+
E 0 = 1.51V
E 0 = 1.33 Volts
E 0 = 0.771 V
• reducing agents
Fe2 + stable for short periods
Fe 3+ + e! ! Fe2 + E 0 = 0.771 V
Thiosulpate S2O32 ! not oxidized by air (rare)
S2O62 ! + 2e! ! 2S2O32 !
Note: you should study the iron/cerium system
Fe 2+ + Ce4 + ! Fe 3+ + Ce4 +
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 SCE
placed 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:
2.303RT
Ecell = k +
lg aH +
unknown
F
• 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 a H+
2.303RT
E !k
pH , i.e. pH = cell
2.303RT
F
F
k is determined using a buffer of known pH
2.303RT
i.e. k = Ecellstd !
lg pH std
F
Ecellunknown ! Ecellstd
then pH unknown = pH std +
2.303RT
F
Ecell = k !
acid/alkaline error
membrane glass made of Na 2O and SiO 2
glass surface -SiO - Na + + H + ! SiO ! H + + Na +
K for this equilibrium is large - gives silicic acid
• 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 solely
to H+ (xpt very high [H+] acid error)
• basic (>pH 9) H+ activity small - glass electrode begins to
respond to other monovalent cations e.g. Na - alkaline error
Alkaline error
of glass
electrode
Acid error of glass electrode
• activity, a, different from [H+]
– electrode behaves like there are less protons
available then actually added
Other ISE’s
e.g. LiO-Si
instead of NaO-Si
• found that different membrane
compositions can enhance alkaline error
• electrode can be made to be more
specific for Na, K, Li, etc.
• construction similar to H+ responding
glass electrode
• internal solution usually chloride salt of
cation of interest
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 ratio
potential of ISE for an ion on its own:
2.303RT
Eelectrode = EM0 ' !
lg aM n+
nF
where EM0 ' depends on internal ref. electrode, filling
solution, and construction of membrane
EM0 ' 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 if
determining Na
0'
ENaK = ENa
!
2.303RT
lg(aNa+ + kNaK aK + )
F
– ENaK is measured potential
– kNaK is the selectivity ratio for potassium
over sodium
– selectivity ratio is the fraction of the sodium
potential that is due to potassium
0'
– KNaK and E Na determined by use of known
solutions of Na and K, and by solving
0'
simultaneous equations for KNaK and E Na
• know how to calculate selectivity ratios and
potential of ISE’s