• Oxidation and Reduction

• Oxidation -

Oxidation may be defind as “ A process which involves the addition of oxygen

or any other electronegative element, or as a process which involves the removal of hydrogen

or any electropositive element.”

• For example-

(i) 2Mg(s) + O2(g) ⟶ 2Mg O (Addition of oxygen)

(ii) Mg(s) + Cl2 (g) ⟶ MgCl2 (Addition of electronegative element chlorine)

(iii) 2H2S (g) + O2 (g) ⟶ 2S (s) +2H2O (Removal of hydrogen)

(iv) 2KI (aq) + H2O(l) + O3(g) ⟶ 2KOH(aq) + I2(s) + O2

(Removal of electropositive element,potassium)

• In all these reactions, the compound underlined has undergone oxidation.

• Reduction- According to the classical a concept, Reduction may be defined

as “ A process which involves the addition of hydrogen or any other

electropositive element,or removal of oxygen or any other electroneative

element.”

Eg.

• Br2 (g) + H2 S (g) ⟶ 2HBr (g) + S (s) (Addition of Hydrogen)

• 2HgCl 2 (aq) + SnCl2 (aq) ⟶ Hg 2Cl 2(s) +SnCl 4 (aq)

(Addition of electropositive element,mercury)

• CuO (s) +H2(g) ⟶ Cu(s) +H2O (liq) (Removal of oxygen)

• Oxidising agent or Oxidant- An oxidising agent is a

substance which supplies oxygen or any other electronegative element, or

removes hydrogen or any other electropositive elements. An oxidising

agent carryingout oxidation is itself reduced in a chemical reaction.

Mg(s) +F2(g) ⟶ 2 MgF2 (s)

4HCl (aq) + MnO2 (s) ⟶ MnCl2(aq) + Cl2(g) + H2O (liq)

K2Cr2O7(aq) +3SO2 (g)+H2SO4(aq) ⟶ Cr2(SO4)3 (aq) +K2SO4 (aq) +H2O (liq)

• Reducing agent or reductant-

A reducing agent or

reductant may be defined as “ A substance which supplies hydrogen or any

other electropossitive element ,or removes oxygen or any other

electronegative element.A reducing agent after carrying out reduction is

itself oxidised in a chemical reaction”.

For eg.

CuO (s) +C (s) ⟶ Co(g) + Zn (s)

Fe2O3(s) +3CO(g) ⟶ 2 Fe(s) + 3CO2(g)

2KMNO4(aq)+3H2SO4(aq)+5HNO2(aq)⟶K2SO4(aq)+2MnSO4(aq)+5HNO3(aq)+3H2O

• Oxidation Number- “Oxidation number of an element may be defined as the

charge which an atom of the element has in its ion or appears to have when present in the

combined state with other atoms.Oxidation numbers are also called oxidation states.”

• General rules to calculate Oxidation number of elements- The

following rules are helpful in determinig the oxidation state or number of elements:

➢ Oxidation number of an element in the uncombined or free state is zero. For eg.

Oxidation number of Cl in Cl2, P in P4 and S in S8 is Zero.

➢ O.N. of hydrogen in all it’s compounds excepting ionic hydrides is +1. In hydrides, the

O.N. of hydrogen is -1. eg. Im LiH, O.N. of hydrogen is -1.

➢ O.N.of oxygen in all its compounds is -2 excepting OF2 and peroxides, where it is +2 and -1

respectively.

➢ O.N.number of alkali metals (IA sub-group) and alkaline earth metal (IIA-sub group) is +1

and +2 respectively.

➢ O.N. of an ion or radical is equal to the charge present on it eg. O.N. of Ca-atom as Ca2+ is 2

and Cl as Cl- ion is -1.

➢ The sum of oxidation numbers of all the atoms in a given molecule is zero and is determined

by the sum of oxidation number of the individual atoms each multiple by the number of the

atoms present in the molecule. For eg. In K2Cr2O7 molecule.

oxidation number of K = +1, thus for 2 X K = 2x1= +2

oxidation number of Cr = +6 thus for 2X Cr = 2x6 = +12

oxidation number of O = -2 thus for 7X 0 = 7x(-2)= -14

For K2Cr2O7 = 2+12-14 = 0

➢ The sum of the oxidation numbers of the atoms in a polyatomic ion is equal to the charge carried by the ion,For example:

• NO3- : N + (O)3 = +5 +3x(-2)=-1

• SO42- : S+ (O) = +6+4x(-2) = -2

• PO43- : P + (O)4= +5+4x(-2)=-3

➢ In compounds containg ions or radicals e.g. (N3-, NO3-,CN-, OH-,NH4

+, CH3+ , NH3

0, CO0, NO+ etc.)It is

more convenient to use to O.N. of the ions or radicals rather than to assign the oxidation numbers of the

individual atoms e.g. O.N. of Cl- ion in NOCl and CH3Cl is -1 because O.N. of NO+ andCH3+ is +1.

➢ O.N. of non-metal is negative in its compound with metal or with hydrogen e.g. O.N. of N in Ca3N2 and

NH3 is -3.

➢ When the sharing of electron pairs takes places between similar atoms,the half shared electrons are counted

along with outermost shells of each of the two like atoms e.g. in Cl2 molecule the sharing of two electrons

takes place between two Cl-atoms, one electron will be counted with each Cl- atoms .i.e. each Cl-atom will

have 7 electrons. Now since chlorine nucleus has +7 charges and 7 negative charges the net charge with Cl-

atom in Cl2 molecule is zero. Thus the O.N. of each Cl atom in Cl2 molecule is zero. Similarly it can be

shown that the O.N. of each N-atom in N2 molecule is also zero.

➢ When the sharing of electrons takes place between two different atoms, the shared

electron are counted along with the outermost shell electrons of more electronegative

atom. On this basis, it can be shown that O.N. of o-atom in H2O, H2O2 and OF2 is -2, -1

and +2 respectively. For example in H2O molecule (H-O-H).

Charges on the nucleus of O-atom = +8

Charges due to 2 electron present in the inner shell of O-atom = -2

Charges due to 6 electrons in the outer shell of O-atom =-6

Charges due to 2 electrons of 2H-atoms counted with O-atom =-2

Apparent charges on O-atom =+8-2-6-2 =-2.

Hence O.N. of O-atom in H2O molecule =-2

➢ There may actually be a difference between the magnitude of valency

and the O.N Let us consider the following compounds of carbon.

CH4 CH3Cl CH2Cl2 CHCl3 CCl4

Valency 4 4 4 4 4

O.N.of

carbon

-4 -2 0 +2 +4

• Example : Pt in H2[Pt Cl6], (ii)- U in UO2(NO3)2, (iii) N in NH4VO3, (iv) Cr in CrO2Cl2

• Solution: (i) Let the O.N.of Pt in H2[PtCl6] is X. Then

2x(+1)+x+6(-1)=0

x-6+2=0

x=+4

Thus ,O.N.of Pt in H2 [PtCl6]

• Let the O.N.Of UO2(NO3)2 is x. Then

x+2x(-2)+2x(-1)=0

x-4-2=0

x=+6

Let the O.N. of N in NH4 NO3 is x .Then

x+4x(-1)+x(-1)=0

x-4-1=0

x=5

• Difference between Valency and Oxidation Number-

1- Valency is equal to the number of H-atoms or double the number of O-atoms which combine with one atom

of the element.

O.N. is the residual charge which an atom has when the other atoms from themolecule are removed as ions

by counting the shared electrons with more electronegative atoms.

2- It is a number only and does not carry any (+) or (-) sign e.g. in CO2 molecule, the co-valency of C-atom is 4

and that of O-atom is 2.

It indicate the charge which may be positive,negative or zero e.g. in CO2molecule oxidation number of C-

atom is +4 and that of O-atom is -2.

3 - The valency of an element is always a whole number.

The O.N of an element may also have fractional value

e.g. O.N. of S in Na2S4O6 is +2.5.

4- Many elements exhibit constant valency e.g. C,N,O etc.

Many elements which have constant valency can have variable

O.N. e.g. C,N,O etc.

• Definition based on electron transfer (Electronic concept)-

• According to this concept oxidation is the process in which an ion (cation or anion), or

an atom or a molecule lose one or more electrons.On the other hand ,reduction is the

process in which an ion , or an atom or a molecule gains one or more electrons.

• Examples of Oxidation:

• Fe2+ ⟶ Fe3+ + e-

• 2Cl- ⟶ Cl2 + 2e-

• Na ⟶ Na++1e-

• H2 ⟶ 2H+ + 2e-

• Sn2+ ⟶ Sn4+ +2e-

• S2- ⟶ S+2e-

• Zn⟶ Zn2+ +2e-

• H2O2 ⟶ O2 + 2H+ +2e-

•

• Examples of Reduction-Na++1e-⟶ Na

MnO4- +8H+ + 5e-⟶ Mn2+ + 4H2O

Cr2O7-- + 14 H+ +6e-⟶2Cr3+ + 7H2O

Cl2+2e-⟶2 Cl-

Definition based on the change in oxidation number

(Oxidation number concept): According to this concept ,oxidation

can be defined as a process in which the oxidation number (O.N) of the effective

element increases while reduction is a process in which the oxidation number of the

effective element decrease.

• Example of oxidation-

Mg0⟶Mg2+ + 2e- oxidation number is increased from (0⟶2)Fe2+⟶Fe3+ + e- oxidation number increased from (+2⟶+3)2S2O3

2- ⟶ S4O62- +2e- oxidation number is increased from (+2⟶+2.5)(S=+2) (s=+2.5)

H2O2⟶ O2+2H++e- O.N.increased from (-1⟶0)2Cl- ⟶Cl2 +2e- increased from (-1⟶0)

• Example of reduction-

Cl2+2e-⟶2Cl- oxidation number decrease from( 0⟶ -1)

F + e- ⟶F- decrease from (o to -1)

[Fe(CN)6]3- +e- ⟶[Fe(CN)6]4- decrease from (+3 to +2)

• Redox Reaction:-• A redox reaction is that chemical reaction in which

oxidation and reduction take place simultaneously i.e.in a

redox reaction one substance is oxidised and the other is

reduced and hence every oxidation is accompanied by

reduction and vice-versa.

• 2FeCl3 + H2S ⟶ 2FeCl2 + S + 2H+ + 2Cl- (redox reaction)

• 2FeCl3 + SnCl2⟶ 2FeCl2 +SnCl4 (redox reaction)

H2S⟶S+2H++2e- (oxidation half reaction)

2FeCl2+2e-⟶2FeCl2+2Cl- (reduction half reaction)

------------------------------------------------------

2FeCl3 + H2S⟶2FeCl2 +S+2H++2Cl- (Redox-reaction)

---------------------------------------------------------

▪ Oxidising agent (oxidant) and Reducing agent (Reductant)-

Oxidising agent (oxidant)-

▪ It oxidises some other substance and itself is reduced.

▪ It loses oxygen or any electronegative element.

▪ It accepts hydrogen or any electropositive element.

▪ It gains one or more electrons.

▪ O.N. of effective element decreases.

▪ Reducing Agent (Reductants)-▪ It reduces some other species and itself is oxidised.

▪ It accepts oxygen or any other electronegative element.

▪ It loses hydrogen or any other electropositive element.

▪ It lose one or more electron.

▪ O.N.of effective element increases.

▪ The different terms used in decreasing oxidation-reduction process are given……..

TermIn terms Of Oxidation number change In terms of electron

transfer

Oxidation Increase Loss of electron

Reduction Decreases Gain of electron

Oxidising agent Decreases Gain of electron

Reducing agent Increases Supplies electrons

Substance oxidised Increases Loss of electrons

Substance reduced Decreases Gains electrons

• Metal tendency⟶Oxidised

(Oxidation) M⟶M+ +ne-

“The potential difference between the electrode(metal) and its ion in electrolyte (under equilibrium) is electrode potential or Oxidation potential.”

Metal tendency-----------Reduced(reduction) Mn+ + ne- ⟶ M

from solution from metal

“The potential difference between the electrode(metal) and its ion in electrolyte (under equilibrium) is electrode potential or Reduction potential.”

• Eelectrode Potential -

When a metal rod is placed in

asolution of its own ions, the rod becomes more electropositive orelectronegative with respect to the solution.Thus, a potentialdifference is established between the metal and the solution. Thisdifference of potential is known as electrode potential.It is denotedby ‘E’.

“OR”

“ The tendency of an electrode to lose or gain electrons when it is incontact with its own ions in solution, is called electrode potential.”

Since the tendency to gain electrons means also the tendency to getreduced, this tendency is called reduction potential.

• Similarly ,the tendency to lose electrons means the tendency to get oxidised.Hence ,this tendency is called oxidation potential.

• Example - When zinc rod is placed in the solution ZnSO4 , zinc rod becomeselectronegative and a potential difference is established between the metal and zinc ions.Thepotential difference so produced is called the potential of zinc electrode.

• Factors affecting the Electrode Potential-

1-Nature of metal and its ion-

Zn⟶Zn2+

Fe⟶Fe2+

Fe⟶Fe3+

2-Concentration of ions-

In case of solution concentrarion term is defined in term of molar while incase of gas concentration is defined interm of partial pressure.

3- Temperature- (25+273)= 298 K (250c)

• Standard electrode potential-

concentration [Mn+]= 1m

Presuure in case of gas= 1atm/1bar

Temperature= 298K (250C)

• The standard electrode potential can be defined as the electrode potential when all the reactants and products are at their unit concentration at 250C and 1 atm (atmospheric pressure), It is represented by E0.

• Standard electrode potential= SEP (already given)

• Oxidation M⟶Mn+ + ne-

Em/mn+⟶ oxidation potential

E0m/m

n+ = standard oxidation potential

Reduction Xn+ + ne- ⟶ X

• Exn+/x = reduction potential

• E0xn+/x = standard electrode potential

According to IUPAC

Zn⟶Zn2++2e-

E0Zn/Zn

2+ =+0.76 v (SOP)

Zn2++2e- ⟶Zn

E0Zn

2+/Zn = -0.76 v (SRP)

• Acoording to IUPAC

SRP⟶SEP or Reduction potential = Electrode potential

Standard Reduction Potential = -Standard oxidation potential

• Standard electrode potential-

concentration [Mn+]= 1m

Presuure in case of gas= 1atm/1bar

Temperature= 298K (250C)

• The standard electrode potential can be defined as the electrode potential when all the reactants and products are at their unit concentration at 250C and 1 atm (atmospheric pressure), It is represented by E0.

EMF cell = ERP - ERP

ECell = E cathode - E anode

(RP) (RP)

OR

• Ecell = E right - E left

• (RP) (RP)

• ECell can calculate at any temperature or any

concentration by NERNST equation.

• E0Cell = Standard EMF of cell at Temperatur is 250c,

concentration is 1 molar and partial pressure is 1 atm. then

• E0Cell = E 0cathode - E0

anode

(RP) (RP)

• E 0cell = E0right - E0

left

• Question-

• Ni/Ni2+ (1M) II Ag+ (1M)/Ag

Left-anode Right-cathode

oxidation Reduction

Given: E0Ni/Ni

2+ = -0.25 v or E0Ni

2+/Ni = -(-0.25) = +0.25 v (reduction potential)

E0Ag

+/Ag = 0.80 v

calculate standard electrode potential.

• Solution:

E0Cell = E0

cathode - E0anode

(RP) (RP)= 0.80-(-025)= 0.80+0.25= 1.05 v

• Question: Calculate standard cell potential of DaniellCell-

Zn/Zn2+ (1M) II Cu2+ (1M)/Cu

• Given: E0Zn

2+/Zn= -0.76 v (RP)

E0Cu

2+/Cu = +0.34 v (RP)

Solution: E0Cell = E0

cathode - E0anode

= 0.34-(-0.76)

= 0.34+0.76

= 1.10 v

• Caculate standard cell potential

Zn/Zn2+ (1M) II Pb2+ (1M)/Pb

Given: E0Zn/Zn

2+ = +0.76 v

E0Pb

2+ /Pb = +0.76 v

E0Zn/Zn

2+ = -0.76 (RP)

E0 cell = E0 cathode - E0 anode

= +0.76 – (-0.76)

=+ 1.52 v

Effect of Metal Ion Concentration on the Magnitude of Electrode Potential –(Nerst Equation)-

We know that the standard oxidation potential of Zn/Zn2+

electrode in which the Zn rod is dipped in 1M solution of ZnSO4 is +0.76 volt.

If the concentration of the solution is not 1M, then the Oxidation potential

corresponding to the reaction:

Zn⟶ Zn2+ +2e-

Is not equal to +0.76 volt but is givenby:

E= E0zn / zn

2+ - log

Where n= No.of electrons in the electrode reaction =2

• Now since the concentration of the elements and solid =1, [Zn] = 1 and hence:

E= E0zn / zn

2+ - log [Zn +]

In general :

E= E0 - log [Zn +]

This is Nerst equation.

This is the equation for a half-cell oxidation reaction and hence:

Eox = E0– log [ion]

The value of Ered for the half –cell reduction is given by reversing the sign of Eox , i.e.

• Ered = - Eox= -{ E0- log [Ion]}

Ered = - E0 + log [Ion]

Effect Of metal Ion concentration on the Magnitude of e.m.f. of a given cell-

We have seen that the standard e.m.f. of Daniellcell represented as:

Zn/Zn2+ (1M) II Cu2+ (1M)/Cu

• In which the concentration of each of the metal ions is unity is give

E0Cell = E0

cathode - E0anode

= 0.34-(-0.76)

= 0.34+0.76

= 1.10volt

If the concentration of each of the metal ions is different as in the cell:

Zn/Zn2+ (0.1M) II cu2+ (0.1M)/ cu

The e.m.f. is given by:

Ecell = E0cell - log [ ]

Where n = No.of electrons involved in the redox reaction

Zn⟶Zn2+(O.1M) + 2e (oxidation)

Reduction half-reaction (cathode reaction)

Cu2+(0.01M)+2e ⟶ Cu

Cell reaction is represented as Zn+ Cu2+(0.01M) ⟶ Zn2+(0.1M) +Cu (n=2)

=

Now since concentration of the elements and solids =1, [Cu] or [Zn] and hence

= =

ECell = E0cell - log

= 1.10-0.0295x1

= 1.0705 volt.

Electro chemical series or e.m.f. Series –

“ The list of elements

or ions arranged in the decreasing order of their standard reduction

potential values is called electro-chemical or e.m.f. series or activity

series.

❖ From this series the following points should keep in mind while studying

the use of this series.

❖ The forword reaction shown in the central column of the series is a

reduction reaction and hence the standard electrode potential values are

standard reduction potential value.

❖ The species shown at the left hand side act as oxidising agents while those

given at the right hand side of the reduction reaction act as reducing agents.

❖ The standard Electrode potential values are decreasing from top to bottom

in the series i.e. possitive values (e.g. (F2/2F-, E0=+2.85 V e.g. (F2/2F-,

E0=+2.85 V)⟶Zero (2H+/H, E0 =∓0.00V) ⟶ negative E0 values

• e.g. (Li+/Li, E0=-3.04 V). Thus the species lying above hydroge have positive E0values while those lying below hydrogen have negative E0 .

• From top to bottom the oxidising power of oxidising agents is decrease (i.e. F2 to Li+

ion ) while the reducing power of reducing agents is increase in the same direction (i.e.from F- ion to Li metal). Thus F2 is the strongest oxidising agent while Li+ ion is theweakest oxidising agent. similarly F- ion is the weakest reducing agent while Li metal isthe strongest reducing agent.

• Aplication Of Electrochemical series:

• To compare the oxising power and reducing power of metals or non-metals-

The magnitude of E0 value of ametal or a non-metal gives a measure of itstendency to gain electrons to get reduced and hence to act as an oxidising agent or loseelectrons to get oxidised and hence to act as a reducing agent.

Gain of electrons (reduction)

M + e- ⟶ M-

oxidising agent Reducing agent

Loss of electrons (oxidation)

M ⟶ M+ + e-

Reducing agent oxidising agent

Thus the oxidising or reducing property of a given element in solution is measured by the magnitude of its E0 value.

Oxidising agent:

The species having +E0 i.e. ,the species lying above hydrogen in theelectrochemical series show a strong tendency to gain electrons to undergo reduction andhence they are strong oxidiasing agents.

As already said,with the decrease of +E0 value of a given species, the tendency of thatspecies to gain electrons to undergo reduction decrease and hence the oxidising power ofthat species also decrease.

• Examples: The metal like Au, Pt, Hg, Cu etc. which have positive E0 values are strong oxidising agents. The oxidising power of these metals is in the order :

Au >Pt > Hg > Ag> Cu, since Au has the maximum E0 value while Cu has the minimum E0 value as shown below:

Au3+ + 3e- ⟶ Au, E0= +1.42 V

Pt2+ + 2e- ⟶ Pt, E0= +1.20 V E0

values decrease Oxidising

power decrease

Hg2+ + 2e- ⟶ Hg, E0= +0.85 V

Ag+ + e - ⟶ Ag, E0= +0.80 V

Cu2 + 2e- ⟶ Cu , E0= +0.34 V

• (ii)- The non-metals like F2, Cl2, Br2,I2 and O2 which have +E0 values are all strong agents. F2 which has the maximum E0 value is the strongest oxidising agent while O2

which has the lowest E0 value is the weakest oxidising agent.Thus the oxidising power of these non-metals is in the order:

F2 > Cl2 > Br2 >I2 > O2

Since E0 values are also decreasing in the same trend as shown below:

F2 +2e- ⟶ 2F- , E0 = +1.42 V

Cl2 +2e- ⟶ 2Cl -, E0 = +1.20 V E0 values decreasing

Br2 +2e- ⟶ 2Br -, E0 = +0.85 V

I2 +e - ⟶ 2I- E0 = +0.80 V Oxidising power

decreasing

O2 +2H - +2e- ⟶ 2OH -,E0 = +0.34 V

• As per electrochemical series, E0 values decrease from F2 to Li+ ion , the tendency

of these species to gain electrons to undergo reduction decreases and hence their

oxidising power decreases from F2 to Li+ ion . F2 is the strongest oxidising agent

while Li+ ion is the weakest oxidising agent.

Reducing agent- The species having - E0 values i.e. the species lying below

hydrogen in the electrochemical series they are strong reducing agent. As already

said, with the decrease of E0 of a given species (i.e. as E0 values becomes more

negative), the tendency of that species to lose electrons to undergo oxidation

increases and hence the reducing power of that species also increase.

Examples: (i)- To apply the above rule we can consider the reducing property of alkali

metals (group IA) and alkaline earth metals (group IIA). E0 values of these metals

are given below:

Alkali metals (IA group)

Li+ + e- ⟶ Li, E0= -3.04V

Na+ + e- ⟶ Na, E0= -2.71V

K+ + e- ⟶ K, E0= -2.92V

Rb++ e- ⟶ Rb, E0= -2.92V

Cs+ + e- ⟶ Cs , E0= -2.92V

E0 values of alkali metals show that these metals have high negative values and hence are strong reducing agent .Since Li has the minimum E0 value (i.e. Li has maximum negative E0 value) It is the strongest reducing agent of all the alkali metals.

Alkali earth metals (IIA group)

Be++ + 2e- ⟶ Be, E0= -1.70V

Mg+++ 2e- ⟶ Mg, E0= -2.38V

Ca+++ 2e- ⟶ Ca, E0= -2.76V

Sr++ + 2e- ⟶ Sr , E0= -2.89V

Ba++ + 2e- ⟶ Ba, E0= -2.92V

Similarly all the alkaline earth metal have high negative E0 values and hence ,likealkali metals behaves as a strong reducing agent.Since E0 values are decreasingfrom Be to Ba, the reducing power of these metals is also increasing in the samedirection, i.e. Be ,with maximun E0 is the weakest reducing agent while Ba whichhas minimum E0 value is the strongest reducing agent.

In electrochemical series from top to bottom F- ion is the weakest reducing agentwhile Li, metal is the strongest reducing agent.

Electropositive Character of Metals- Metals like K , Ca ,Na etc. which lieat the bottom of the series readily lose their outer-most shell electrons toform metal cations and hence are strongly electropositive (i.e. weaklyelectronegative) those lying at the top are not able to lose the outer-mostshell electrons to form the cations and hence are weakly electropositive(strongly electronegative).

• To predict whether a given metal will displace another metal from aqueous solution of its salt - A metal with lower standard reduction potential will displace another metal with higher standard reduction potential from the aqueous solution of its salt and the metal with higher standard reduction potential gets precipitated. This means that a metal will displace another metal from the aqueous Solution of its salt that lies above it in the electrochemical series.

• Examples: (i)- If a piece of Fe is placed in a solution of CuSO4 some of Fe goes in to solution as Fe++ (aq.) ions and Cu metal gets precipitated , i.e. Fe displaces Cu from CuSO4 solution as shown below:

Fe + CuSO4 ⟶ FeSO4 + Cu

Fe (s) + Cu2+ (aq.) ⟶ Fe2+ (aq.) + Cu(s)

• The displacement of Cu from CuSO4 solution by Fe is because of the fact

that E0Fe

2+/Fe = ( -0.44v ) is less than E0

Cu2+

/Cu = (+0.34V) Why the

displacement of Cu from CuSO4 solution by Fe is possible or why the

reaction.

• Fe (s) + Cu2+ (aq.) ⟶ Fe2+ (aq.) + Cu(s)

Reducing agent oxidising agent oxidising agent reducing agent

Examples: Zn does not displace Mg from a solution of MgSO4 ,i.e. the

reaction shown below is not possible to occur.

Zn+ MgSO4 ⟶ ZnSO4 + Mg

Zn (s) + Mg2+ (aq.) ⟶ Zn2+ (aq.) + Mg(s) ppt.

• For above reaction to occur it is essential that E0Zn

2+/Zn should be less than

• E0Mg

2+/Mg . But actually the former is greater than latter (E0

Zn2+

/Zn = -0.76V,

E0Mg

2+/Mg = -2.34V). This means that the reverse of the above reaction will

proceed spontaneously, i.e. Mg will displace as shown below:

Mg + ZnSO4 ⟶ MgSO4 + Zn

Mg (s) + Zn2+ (aq.) ⟶ Mg2+ (aq.) + Zn(s) ppt.

In the reaction :

Zn (s) + Mg2+ (aq.) ⟶ Zn2+ (aq.) + Mg(s) ppt.

Reducing agent Oxidising agent Oxidising agent Reducing agent

• Since the reducing agent viz, Zn lies above the oxidising agent Mg2+ in the electrochemical series, the reaction is not possible to occur in the direction shown.However ,the reverse reaction takes place spontaneously because the reducing agent viz. Mg lies below the oxidising agent namely Zn2+ ions in the electrochemical series.

Zn2+ (aq.) + Mg(s) ppt. ⟶ Zn (s) + Mg2+ (aq.)

Oxidising agent Reducing agent Reducing agent Oxidising agent

Examples 3- The reaction :

Zn (s) + Fe2+ (aq.) ⟶ Zn2+ (aq.) + Fe(s) is predictable ,since E0Zn

2+/Zn (=-

0.76V) is less than E0Fe

2+/Fe =(-0.44).

This reaction is used for galvanising Fe to Zn to prevent rusting of Fe. Zn coating on Fe prevents the oxidation of Fe to Fe2+ by air . If the galvanised Fe is scratched and some iron is oxidised to Fe2+ by air , Fe2+ thus produced is immadiately reduced by Zn to Fe and thus the rusting of Fe is prevented.