1.Redox Equilibria

7/21/2019 1.Redox Equilibria

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Chapter 8: Redox Equilibria and Electrode Potentials

8.1. Introduction

A redox reaction involves both oxidation and reduction

Redox reactions can be divided up into two half equations, one showing the reducing

agent and the other the oxidising agent.

These half equations show how many electrons the reducing agent loses and how

many the oxidising agent loses. The change in oxidation number of the species is

directly related to the number of electrons transferred. Water and hydrogen ions alsooften appear in half equations to balance any hydrogen or oxygen atoms.

e.g. the oxidation of hydrogen peroxide by dichromate(!" ions,

Cr2O72- 8! "!2O2 2Cr" 7!2O "O2

(#$"(%&" (#'" (#'"(%'" (#" (#'"(%&" ()"

*ach +r in +r &-&% gains electrons to form +r # % it is the oxidiing agent and

becomes reduced

Cr2O72- 1#! $e- 2Cr" 7!2O

*ach in /&& loses ' electron to form & % it is the reducing agent and

becomes oxidied

!2O2 O2 2! 2e-

!n this chapter, the system of electrode potentials will be introduced. At a basic level,

electrode potentials are measurements which show the relative strengths of oxidiing

agents and reducing agents.

8.2. Electrode potentials and the electroche%ical series

At 0+1* level, test%tube displacement reactions are used to demonstrate that somemetals are more reactive than others. 2y comparing the results of such experiments, a

reactivity series can be drawn up.

% $3 %

Oxidation is the loss of electrons

Reduction is the gain of electrons (Remember OILRIG)

A reducing agent is a species that loses electrons easily, becoming

oxidised in the process.

An oxidising agent is a species hich gains electrons easily, becoming

reduced in the process

copper(!!"

sulphate

solution

inc

metalcopper

metal

ppte.

inc(!!"

sulphate

solution

1o inc must be

%ore reacti&e

than copper



The displacement reaction between inc and copper(!!" sulphate is actually a redox

reaction, as shown by the changes in oxidation number4

+u15(aq" # 6n(s" → +u(s" # 6n15(aq"

(#&" ()" ()" (#&"

!n simple terms, the +u&# ions have gained two electrons and been reduced whilst the

inc atoms have lost two electrons and been oxidised. The inc is therefore a more

powerful reducing agent than copper.

At A%level, the reactivity series is called the electroche%ical series as it is listed more

accurately using 'tandard Electrode Potentials (Eo). The electrode potential of a

substance is measured in &olts and shows its ability to lose or gain electrons. 2y

convention, electrode potentials are always written next to a half equation which has

electrons on the le*t hand-side, so that4

+ s%all section o* the electroche%ical series

, e- , Eo -2.2 /

0a e- 0a Eo -2.71 /

n2 2e- n Eo -.7$ /

3e2 2e- 3e Eo -.## /

2! 2e- !2 Eo . /

Cu2 2e- Cu Eo ."# /

I2 2e- 2I- Eo .4# /

5r2 2e- 25r- Eo 1. /

Cr2O72- 1#! $e- 2Cr" 7!2O E

o 1."" /

6nO#- 8! 4e- 6n2 #!2O E

o 1.41 /

Cl2 2e- 2Cl- Eo 1."$ /

32 2e- 23- Eo 2.87 /

The %ost poer*ul reducin aent in the list is potassiu% metal % the species which

loses electrons most effectively. As the very first equation has the most negative

electrode potential, it goes bac7wards most readily and potassium loses electrons.

The %ost poer*ul oxidi9in aent is *luorine. The final equation has the most

positive electrode potential, goes forwards most readily and fluorine gains electrons.

The hydrogen equation has been given an electrode potential value of ).)) volts by

chemists so that it acts as the standard.

% $8 %

A positi!e " o shos the forard reaction is thermodynamically fa!orable,

a negati!e " o shos the re!erse reaction is thermodynamically fa!oured



8.". sin electrode potentials to predict che%ical reactions

9ust as half equations can be combined to give full redox equations, electrode

potentials can be added to show whether redox reactions are thermodynamically

favoured or not. *ven though half equations are sometimes multiplied through by

whole numbers to ensure electrons cancel, electrode potentials are not : they are

simply added together to give Eoreaction values. Again,

e..1 for the reaction between manganate(!!" ions and iodide ions,

('" ;n5% # 3/# # <e% ;n&# # 5/& *

o = #'.<'

(&" &!% !& # &e% *o = %).<5

*quation ('" is multiplied by & and equation (&" by < before combination gives4

26nO#- 1$! 1I- 26n2 8!2O 4I2

2ut Eo

reaction is simply (1.41) (-.4#) .8 /

As Eo

reaction is positi&e, the forward reaction is *easible.

e..2 for the reaction between manganate(!!" ions and fluoride ions,

(" ;n5% # 3/# # <e% ;n&# # 5/& *

o = #'.<'

(5" &>% >& # &e% *o = %&.3-

*quation (" is multiplied by & and equation (5" by < before combination gives4

26nO#- 1$! 13- 26n2 8!2O 432

and Eo

reaction (1.41) (-2.87) -1."$ /

As Eo

reaction is neati&e, the forward reaction is not *easible (but the reverse reaction is?"

Proble%s ith Eo predictions

A positive value for *o

reaction shows that a reaction is feasible, not that it definitely will

happen. There are & main reasons why reactions that are feasible do not actually

occur in reality4

a" *o predictions ta7e no account of the sie of the acti&ation ener;.

@inetically stable reactions may be so slow that they appear not to occur at all.

% -) %

A positi!e " o

reaction shos the forard reaction is thermodynamically feasible,

a negati!e " o

reaction shos the re!erse reaction is thermodynamically feasible.



b" *o predictions are only valid for reactions mixed under standard conditions.

Real reactions are rarely mixed at &<°+, ' atm and ' moldm%.

6ethods *or anserin exa% questions

There are two general methods for answering exam questions on *o

predictions. The

second method is much faster than the first, but it is best to be able to master both.

6ethod 1 < calculatin Eo

reaction

To see if a reaction is feasible or not, the half equations are combined and *o

reaction

calculated as before. /owever, before this can be done, one of the half%equations

must be re&ersed and the sin o* its electrode potential chaned. To decide which

half%equation to reverse, loo7 closely at the reactants in the question : these must

both appear on the left of the final overall equation.

e.g.'. Will inc reduce copper(!!" ions to copper metal

('" 6n&#

# &e%

6n *

o

= %).-$ (&" +u&# # &e% +u *

o = #).5

As inc and copper(!!" ions are the reactants, equation ('" must be flipped

Binc is on the right in equation ('"C.

1o, 6n # +u&# → 6n&# # +u

and *o

reaction = ().-$" # (#).5" = #'.') ,

positi&e= therefore reaction is *easible.

e.g.&. !s the following reaction feasible2r & # &+l% → &2r % # +l&

('" 2r & # &e% &2r % *o = #'.)8

(&" +l& # &e% &+l% *o = #'.$

As chloride ions are reactants, equation (&" has been flipped.

1o *o

reaction = ('.)8" # (%'.$" = %).&- , neati&e= therefore no reaction.

6ethod 2 < direct co%parison o* Eo &alues

A simple comparison of the *o

values of two half equations will show which will

proceed forwards (the most positive" and which will go bac7wards (the most

negative". Again, a close inspection of the reagents in the question will then show

whether or not reaction will occur.

e.g.'. Will manganate(!!" ions oxidie chloride ions to chlorine

('" ;n5% # 3/# # <e% ;n&# # 5/& *

o = #'.<'

(&" +l& # &e% &+l% *o = #'.$

% -' %



*quation ('" will go forwards as its *o is more positive, and (&" will go bac7wards.

'o 6nO#- ill oxidise Cl- as E

o6nO#->6n2 is more positi!e than E

oCl2>Cl-

e.g.&. Will inc reduce magnesium ions to magnesium metal

('" 6n&# # &e% 6n *o = %).-$

(&" ;g&# # &e% ;g *o = %&.-

*quation ('" will go forwards as its *o is more positive, and (&" will go bac7wards.

'o n ill 0O? reduce 62 as Eo

n2>n is less neati&e than Eo

62>6

There are always four possible ways to express the answer using this method (*o

AD2

more positive than *o

+DE or *o

+DE less positive than *o

AD2 etc.". +are must be ta7en to

avoid silly errors. The advantage of this method lies in the fact that it allows a large

range of reactants to be compared quic7ly. ;any A%level questions rely on this s7ill.e.g. Which metal ions can be reduced by inc metal

@ # # e% @ *o = %&.8&

Fa# # e% Fa *o = %&.-'

6n&# # &e% 6n *o = %).-$

>e&# # &e% >e *o = %).55

Fi&# # &e% Fi *o = %).&<

+u&# # &e% +u *o = #).5

Ag# # e% Ag *o = #).3)

The inc equation must go bac7wards and be more negative than the other half

equation if reaction is to occur.

'o n ill reduce 3e2= 0i2= Cu2 and + because Eo

n2>n is

%ore neati&e than Eo

3e2>3e= Eo

0i2>0i= Eo

Cu2>Cu and Eo

+>+.

8.#. @isproportionation

9ust as electrode potentials can be used to predict the feasibility of normal redox

reactions, they can also reveal the possibility of a species undergoing

disproportionation.

A species must be able to undergo oxidation and reduction and will therefore appear

on the left in one half equation and on the right in another. Reversing one of these

and adding it to another then gives a disproportionation equation : the overall sign of

*o then shows feasibility.

e.g. Will copper(!" ions disproportionate

('" +u&# # e% +u# *o = #).'<

(&" +u# # e% +u *o = #).<&

% -& %

#isproportionation is the simultaneous oxidation and

reduction of a single species



Reversing ('" and adding gives +u# # +u# → +u # +u&#

and *o

reaction = #).- ∴disproportionation is feasible.

8.4. Electroche%ical cells

An electrochemical cell is the scientific term for the basic component in a battery.+ells are basically powered by redox reactions.

+onsider again the displacement reaction between inc metal and copper(!!" sulphate.

n(s) Cu'O#(aq) n'O#(aq) Cu(s)

This is a simple redox reaction4 The inc loses electrons and is oxidised,

the copper(!!" ions gain electrons and are reduced.

!n an electrochemical cell exactly the same reaction occurs, but the redox reagents are

placed in separate pots (or Ghal*-cellsH". The half%cells are connected with a metal

wire so that electrons can flow from one reagent to the other when the circuit is

completed.

e.g. the inc metal : copper ion cell

The salt-bride is used to complete the circuit without using a metal wire : metals

are good reducing agents and would set up more competing redox reactions. A salt bridge can be made by dipping a filter paper in an aqueous solution of potassium

nitrate and allows ions to flow between half cells.

The reaction driving the cell is the same as that above in the test%tube reaction but

electrical energy is produced instead of heat.

n(s) Cu'O#(aq) n'O#(aq) Cu(s)

() (2) (2) ()

!n this circuit, electrons travel from le*t to riht because the inc has lost electrons

and the copper(!!" ions have gained them.

% - %

As reaction occurs, the blue

colour of the +u&# ions fades,

the inc metal dissolves, a pin7

precipitate of copper is seen

and heat energy is released

e-

copper metal

copper(!!" sulphate

solution

i.e. +u&#(aq"

inc(!!" sulphate

solution

i.e. 6n&#(aq"

inc metal

salt

bride



As electrons flow through the wire away from the inc, the metal dissolves, the

copper electrode grows fatter and the blue colour of the copper(!!" ion fades. When

either reagent runs out, the cell becomes GflatH.

8.$. Electro%oti&e *orce (e%*) and electode potentials

+onsider the cell produced if the inc above is replaced with magnesium. As

magnesium is more reactive (i.e. a better reducing agent" than inc, it pushes electronsaround the circuit much more strongly. !f a voltmeter is added in place of the bulb, it

will measure the electro%oti&e *orce (e%*) of the cell.

As electrode potentials are used to show the strength of oxidiing agents and reducing

agents, they can be used to predict the emf of any cell. *lectrode potentials are

actually measured using cellsI this is why they have units of volts.

The half equations for the cell above are4

('" +u&# # &e% +u *o = #).5

(&" ;g&# # &e% ;g *o = %&.-

*quation & will go bac7wards as its *o is more negative, so the cell reaction is

Cu2 6 Cu 62

and e%* (."#) (2."7) 2.71&

1o if the cell is connected up as shown, the voltmeter will read &.-'v. There are

several ways of representing emf, such as Eo

reaction and Eo

cell.

Again, electrons flow away from the mort reactive metal (magnesium" until all of it

dissolves or all of the copper(!!" ions have reacted. !f the voltmeter is replaced with

an electrical supply with a voltage higher than &.-'v, electrons can be pushed bac7 to

the magnesium and the cell is re%charged. Euring this process, the reverse reaction

occurs.

8.7. 6easurin electrode potentials

As a circuit is required to measure a potential difference, single electrode potentials

cannot be measured. !nstead, chemists have chosen a standard electrode againstwhich all other electrodes are then compared. The standard h;droen electrode has

% -5 %

high resistance voltmeter

/

copper metal

copper(!!" sulphate

solution

i.e. +u&#(aq"

magnesium

sulphate solution

i.e. ;g&#(aq"

magnesium

metal



been given an electrode potential of ).)v and is included in circuits to measure other

electrode potentials.

a) ?he circuit used to %easure Eo *or %etal s;ste%s at 24

C

e.g. to measure *o for Cu2 2e- Cu

As the electrode potential of hydrogen has been given a value of ).)v, the voltmeter

reading gives *o

for the copper(!!"Dcopper half equation, in this case, #).5v.The whole apparatus can be placed in a te%perature controlled aterbath at &<°+

for more accurate results.

b) ?he circuit used to %easure Eo *or aseous s;ste%s at 24 C

e.g. to measure *o for Cl2 2e- 2Cl-

% -< %

?he 'tandard ?he Cu2>Cu

!;droen electrode

Electrode (hal* cell)

2! 2e- !2 E

o .& E

o A

!(aq)

1 %old%-"

platinu%

%etal

!2()

1at%

voltmeter

/

copper %etal

copper(II) sulphate

solution 1 %old%-"

i.e. +u&#(aq"

/

!(aq)

1 %old%-"

(e.g. /+l"

platinu%

%etal

!2()1at%

platinu%

%etal

Cl2() 1 at%

Cl-(aq) 1 %old%-"

(e.g. Fa+l"



The voltmeter reading again gives the *o value for +l&D+l%. The platinum electrode

must be in contact with both chlorine and chloride ions dissolved in water. A

temperature controlled waterbath increases the accuracy of results.

c) ?he circuit used to %easure Eo *or aqueous ions at 24

C

e.g. to measure *o for 6nO#

- 4e- 8! 6n2 #!2O

Fotice than in the half%cell under test, the platinum electrode must be in contact with

both reactants (;n5%, /#" and products (;n&#". The voltmeter reading again gives

the electrode potential for the ;n5%D;n&# system.

8.8. ?he de*inition o* electrode potential

The actual definition of electrode potential follows directly from the circuits needed to

measure them. !n general terms,

This definition should then be applied to the actual half cell concerned.

e.g.' for a metal, such as copper

e.g.& for a gas, such as chlorine

% -$ %

/

!(aq)

1 %old%-"

(e.g. /+l"

platinu%

%etal

!2()

1at%platinu%

%etal

6nO#

-(aq"

#

6n2(aq" all at

# 1 %old%-"

!(aq"

an electrode potential is the e%* o* a cell co%posed o* the standard h;droen

electrode connected to the hal*-cell= hen all concentrations are 1%old%-"=

pressures are 1at% and te%perature is 28,.

the electrode potential is the e%* o* a cell co%posed o* the standard h;droen

electrode connected to copper %etal in a solution o* Cu 2 ions hen all

concentrations are 1%old%-"= pressures are 1at% and te%perature is


electrode connected to platinu% %etal in contact ith Cl2 as and a

solution o* Cl- ions ith concentrations o* 1%old%-"= pressures o* 1at%

and a te% erature o* 28,.



e.g. for an aqueous ion, such as manganate(!!"

8.. +nal;sin redox reaents &ia titration

8..1. +nalsin reducin aents usin potassiu% %ananate(/II)Jotassium manganate(!!" is a powerful oxidiing agent and reacts rapidly with many

reducing agents. ther salts are 7nown Be.g. Fa;n5, 2a(;n5"&C and all are purple

in colour due to the manganate(!!" ion4

As can be seen from the half equation, mandanate(!!" ions only wor7 in the presence

of acid : for this reason, dilute sulphuric acid must be added to any reducing agent

before titration commences.

+ t;pical %ananate(/II) anal;sis

'. A sample of steel (5.)g" was dissolved in acid to form &<)cm of >e&# solution.

&. A &<cm sample was ta7en via pipette and placed in a conical flas7 and treated

with excess dilute sulphuric acid.

. A burette was rinsed and then filled with potassium manganate(!!" solution of

concentration ).)< moldm%.

5. The >e&#D/&15 solution was then titrated with swirling until a permanent pin7

colour remained in the flas7.

<. The titre was noted and the analysis repeated until concordant results were

achieved.

% -- %


electrode connected to platinu% %etal in an aqueous solution o* 6nO#-=

! and 6n2 ith concentrations o* 1%old%-"= pressures o* 1at% and a

te% erature o* 28,.

6nO#- 8! 4e- 6n2 #!2O

(purple) (colourless)

,6nO# solution

in burette

reducin aent (e.. 3e2)

+0@

dilute !2'O

#

As the titration proceeds, the

purple colour of the manganate

disappears until the end%point

is reached.

Eespite the difficulties in

reading the burette, the

manganate(!!" is placed in the

burette because a colourless to

pinB end-point is easier to

spot than pin7 to colourless



$. n average, &3.&cm of the manganate(!!" solution was needed for reaction.

6nO#- 8! 43e2 6n2 #!2O 43e"

+alculation4

moles of ;n5% used = &3.& x ).)< = '.5' x ')%

')))

∴moles >e&# = < x '.5' x ')% = -.)< x ')% in &< cm

= ).)-)< in &<) cm

∴mass >e&# = mass >e = moles x rmm

= ).)-)< x <$ = .853g

so K >e in the steel = .853 x ')) = 8.8

5.)

'tandardisation o* %ananate(/II) solutions;anganate(!!" ions are such strong oxidiing agents that they will react with dust in

the air. This means that manganate(!!" solutions Ggo offH and their concentration

must be chec7ed before super%accurate analyses. This is done via titration of a 7nown

mass of sodium ethandioate (Fa&+&5". The ethandioate ions are oxidied to carbon

dioxide4

26nO#- 4C2O#

2- 1$! 26n2 8!2O 1CO2

+ t;pical standardi9ation calculation:

A <.)g mass of sodium ethandioate was dissolved in water and made up to &<)cm.

When a &<cm sample was titrated with manganate(!!" solution, a volume of

'$.3cm was needed for reaction.

;oles Fa&+&5 used = mass = <.) = ).)-' in &<)cm

rmm '5

= .-' x ')% in &<cm

∴moles ;n5% = .-' x ')% x & = ).)'58&

<

∴conc. ;n5% = moles x '))) = ).)'58& x '))) = .888 %old%-"

volume '$.3

8..2. +nal;sin oxidi9in aents ith potassiu% iodide and sodiu% thiosulphate

!n these analyses, the oxidiing agent is first treated with a large excess of potassium

iodide (@!" solution. The iodide ions are oxidied to iodine and the reaction mixture

turns brown. The amount of iodine is then found by titration with sodium

thiosulphate solution (Fa&1&"4

% -3 %

2'2O

"

2- I2 '

#O

$

2- 2I-

brown colourless

0a2'

2O

" solution

in burette

oxidisin aent (e.. Cu2)and

excess ,I produces I2

Euring the titration, the brown colour of

the iodine slowly fades to a yellow colour.'tarch solution is then added before the

end%point is reached as a blue-blacB to

colourless change is easier to spot.



+ t;pical iodide>thiosulphate anal;sis

'. A brass coin (.)g" was dissolved in acid to form &<)cm of +u&# solution.

&. A &<cm sample was ta7en via pipette and placed in a conical flas7 and treatedwith excess potassium iodide to form a brown solution of iodine.

&+u&# # 5!% → &+u! # !&

. A burette was rinsed and then filled with sodium thiosulphate solution of

concentration ).& moldm%.

5. The iodine solution was then titrated with swirling until a yellow colour remained.

After adding starch, the titration continued until the blue blac7 colour disappeared.

<. The titre was noted and the analysis repeated until concordant results were

achieved.

$. n average, '3.5cm of the thiosulphate solution was needed for reaction.

+alculation4 moles of 1&&% used = '3.5 x ).& = .$3 x ')%

')))

2'2O"2- I2 '#O$

2- 2I-

∴moles !& formed = .$3 x ')% ÷ & = '.35 x ')% in &< cm

∴moles +u&# formed = '.35 x ')% x & = .$3 x ')% in &< cm

= ).)$3 in &<) cm

∴mass +u&# = mass +u = moles x rmm

= ).)$3 x $.< = &.-g

so K +u in the coin = &.- x ')) = 77.

.)

0ote: The thiosulphate ion can actually be oxidied in two different ways depending

on the strength of the oxidiing agent. This is shown by the following half equations4

2'O#2- 1! 8e- '2O"

2- 4!2O Eo .47 /

'#O$2- 2e- 2'2O"

2- Eo . /

As *o

!&D!% is only #).<5 , iodine will only react to form 15$&%. 1tronger oxidiing

agents li7e chlorine (*o

+l&D+l% = '.$ " ma7es both 15$&% and 15

&% simultaneously in

different reactions and can therefore not be analysed by titration with thiosulphate.

% -8 %



8.1. ?he rustin o* iron

Rust is hydrated iron(!!!" oxide, >e&.n/&. The rusting of iron metal is anelectrochemical process and occurs when the surface is in contact with water which

contains dissolved oxygen.

When iron metal is impure, heated or mechanically Gwor7edH, areas of the metallic

lattice are wea7ened and some iron(!!" ion start to dissolve in the water : this 7ic7s

off the rusting process4

@iara% o* the rustin process:

1. At areas of wea7ness, iron(!!" ions dissolve, leaving behind electrons in the metal4

>e → >e&# # &e% (oxidation"

2. *lectrons travel through the metal to another point on the surface where oxygen is

reduced to hydroxide ions4

& # &/& # 5e% → 5/% (reduction"

". The iron(!!" hydroxide now present in the water is further oxidied to rust

&>e(/"& # L& → >e& # &/&

As more and more rust forms it fla7es away from the surface, exposing fresh metal

and so deep pitting eventually results where the iron(!!" ions dissolve.

Rust pre&entionThere are two general methods used to prevent iron from rusting4

a) Co&erae o* the iron sur*ace

!f water and oxygen are not in contact with the metal then no rusting can occur.

Jainting, oiling and GtinningH all wor7 in this way. As tin is unreactive, it does not

readily corrode.

>e&# # &e% >e *o = %).55

1n&# # &e% 1n *o = %).'5

b) 'acri*icial protection

% 3) %

e- e- e- e-

O!- O2

1. 3e2 ions dissolve

". >urther oxidation to 3e"

iron

%etal

ater ith

dissol&ed O2

2. ox;en

reduced



!f a bloc7 of a reactive metal (magnesium or inc" is placed in contact with the

iron, the reactive metal loses electrons to the iron, preventing the formation of >e &#

ions. The reactive metal dissolves as ions instead, GsacrificedH for the iron. The

reactive metal must therefore be replaced readily.

;g&# # &e% ;g *o = %&.-

6n&# # &e% 6n *o = %).-$

>e&# # &e% >e *o = %).55

% 3' %

Documents

1.Redox Equilibria