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 '