Electrochemistry is the study of producing electricity from chemical reactions that occur spontaneously as well as the use of electrical energy to cause chemical reactions that don't occur spontaneously.
Galvanic cells convert chemical energy into electrical energy, whereas electrolytic cells turn electrical energy into chemical energy.
Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)��(�)+��2+(��)→��2+(��)+��(�)
Oxidation Half: Zn(s)→Zn2+(aq)+2e−��(�)→��2+(��)+2�−
Reduction Half: Cu2+(aq)+2e−→Cu(s)��2+(��)+2�−→��(�)
Zn�� is the reducing agent, while Cu2+��2+ is the oxidising agent.
Faraday proposed the following two laws to explain the relationship between the amount of substance produced and the current or equivalent weight.
It states that "the amount (mass) of any substance deposited or liberated at any electrode is directly proportional to the amount of electricity passed through the electrolyte."
w∝Qw=ZQ=ZitCharge(Q)=current(i)×time(t)�∝��=��=���Charge(�)=current(�)×����(�)
Z = Electrochemical equivalence
Z=E96500�=�96500
(where, E = Equivalent weight)
"When the same amount of electricity is passed through solutions of different electrolytes connected in series, the weight of substance produced at the electrodes is directly proportional to their equivalent weight," it states.
i.e. Weight of Cu deposited (w1) Weight of Ag deposited (w2)= Equivalent weight of Cu(E1) Equivalent weight of Ag(E2) Weight of Cu deposited (�1) Weight of Ag deposited (�2)= Equivalent weight of Cu(�1) Equivalent weight of Ag(�2)
M(s)⟷OxidationReductionMn+(aq)+ne−�(�)⟷OxidationReduction��+(��)+��−
Mn+(aq)+ne−⟷ReductionOxidationM(s)��+(��)+��−⟷ReductionOxidation�(�)
It's defined as the electrode potential of a given electrode when compared to a standard hydrogen electrode under standard conditions. The standard conditions are as follows:
The half-cell potential values are standard, and they are represented as standard reduction potential values in the Electrochemical Series table at the end.
The difference between the electrode potential of two half cells is called cell potential. Electromotive force occurs when no current is drawn from the cell (EMF).
Ecell=Ecathode+Eanode�����=����ℎ���+������
We use the anode's oxidation potential and the cathode's reduction potential in this equation.
Since the anode is on the left and the cathode is on the right, the following is the result:
=ER+EL=��+��
For a Daniel Cell, therefore:
E∘cell=E∘Cu2+/Cu−E∘Zn/Zn2+=0.34+(0.76)=1.10V�����∘=���2+/��∘−���/��2+∘=0.34+(0.76)=1.10�
The following conventions or notations are used to write the cell diagram by IUPAC regulations.
The Daniel cell is represented as follows:
Zn(s)|Zn2+(C1)||Cu2+(C2)|Cu(s)��(�)|��2+(�1)||��2+(�2)|��(�)
Anodic Chamber: Zn(s)|Zn2+(aq)��(�)|��2+(��)
Cathodic Chamber: Cu2+(aq)|Cu(s)��2+(��)|��(�)
Zn(s)|Zn2+(1M)||Cu2+(1M)|Cu��(�)|��2+(1�)||��2+(1�)|�� , EMF = +1.1 V
Zn(s)|Zn2+(C1)||H+(C2)|H2(Pt)(s)��(�)|��2+(�1)||�+(�2)|�2(��)(�)
The salt bridge keeps the balance of charges and completes the circuit by allowing ions to pass freely over it.
It has a gel inside with an inert electrolyte like Na2SO4��2��4 or KNO3���3 as an electrolyte. Negative ions migrate to the anode and positive ions move across the salt bridge to the cathode, maintaining the equilibrium of charges in the cell.
Diagram of Salt Bridge
ΔG=−nFEcellΔ�=−�������
ΔGΔ� must be negative and cell potential must be positive for a spontaneous cell response to occur.
The standard value of ΔGΔ� will be obtained if we utilise the standard value of cell potential in the following equation.
ΔG∘=−nFE∘cellΔ�∘=−�������∘
At anode: M→Mn++ne−�→��++��−
At cathode: Mn++ne−→M��++��−→�
We employ a variety of salts that are only very little soluble in the metal itself as electrodes.
For instance, when using AgCl���� with Ag�� , there may be a gap between these two phases, as demonstrated by the following reaction:
AgCl(s)+e−→Ag(s)+Cl−����(�)+�−→��(�)+��−
A silver rod is submerged in a solution comprising AgCl(s)����(�) and Cl−��− ions to create this electrode.
Calomel paste (Hg2Cl2)(��2��2) and an electrolyte that contains Cl−��− ions) are mixed with mercury.
At cathode: Hg2Cl2(s)+2e−→2Hg(l)+2Cl−(aq)��2��2(�)+2�−→2��(�)+2��−(��)
At anode: 2Hg(l)+2Cl−(aq)→Hg2Cl2(s)+2e−2��(�)+2��−(��)→��2��2(�)+2�−
This electrode is also used to establish other potentials as a reference point. In its standard form, it is also known as Standard Calomel Electrode (SCE).
In these electrodes, the same metal is employed in two different oxidation states in the same half cell. For instance, when Fe2+��2+ and Fe3+��3+ are dissolved in the same container and an inert platinum electrode is used to transfer electrons.
Possible reactions include the following:
At anode: Fe2+→Fe3++e−��2+→��3++�−
At cathode: Fe3++e−→Fe2+��3++�−→��2+
The relationship between electrode voltage and ion concentration is established. As a result, the reduction potential increases together with the increase in ion concentration. for a common electrochemical reaction type.
aA+bB−→−ne−cC+dD��+��→��−��+��
The Nernst equation can be written as:
Ecell=E0call−RTnFln[C]c[D]d[A]a[B]b�cell=�call0−����ln[�]�[�]�[�]�[�]�
Ec∈l=E∘cdl−2303nFRTlog[C]c[D]d[A]a[B]b��∈�=����∘−2303����log[�]�[�]�[�]�[�]�
Here, substitute the values of R and F as follows:
Ecell=E0ccll−0.0591nlog[C]c[D]d[A]a[B]b�cell=�����0−0.0591�log[�]�[�]�[�]�[�]�
For a Daniel Cell, at equilibrium
Ecell=0=E0cell−2.303RT2Flog[Zn2+][Cu2+]�cell=0=�cell0−2.303RT2Flog[Zn2+][Cu2+]
Eocdl=2.303RT2Flog[Zn2+][Cu2+]Ecdlo=2.303RT2Flog[Zn2+][Cu2+]
But at equilibrium:
[Zn2+][Cu2+]=Kc[��2+][��2+]=��
Eacell=2.303RT2FlogKcE����a=2.303RT2FlogKc
Eocell=2.303×8.314×2982×96500logKcE����o=2.303×8.314×2982×96500logKc
=0.05912logKc=0.05912logKc
In general:
E∘cell=0.0591nlogKcEcell∘=0.0591nlogKc
logKc=nE∘cell0.0591logKc=n�cell∘0.0591
When two electrodes of the same metal are individually dipped into two solutions of the same electrolyte with variable concentrations, and the solutions are connected by a salt bridge, concentration cells are created. For illustration:
H2|H+(C1)||H+(C2)|H2�2|�+(�1)||�+(�2)|�2
Cu|Cu+2(C1)||Cu2+(C2)|Cu��|��+2(�1)||��2+(�2)|��
These are of two types:
1. Electrode Concentration Cells
H2(P1)|H+(C)||H+(C)|H2(P2)�2(�1)|�+(�)||�+(�)|�2(�2)
Ecell=0−0.059nlogP2P1�cell=0−0.059�log�2�1
Where, P2<P1�2<�1 for spontaneous reaction.
2. Electrolyte Concentration Cell
At 298 K, the EMF of concentration cell can be written as
Zn|Zn2+(C1)||Zn2+(C2)|Zn��|��2+(�1)||��2+(�2)|��
Ecell=0.0591n1logc2clEcell=0.0591n1logc2cl
Where, C2>C1�2>�1 for spontaneous reaction
1. Electrolysis of Molten Sodium Chloride
2NaCl(l)⇌2Na+(l)+2Cl−(l)2����(�)⇌2��+(�)+2��−(�)
The reactions is taking place at the two electrodes may be shown as follows:
At cathode: 2Na++2e−→2Na2��++2�−→2�� , E∘=−2.71V�∘=−2.71�
At anode: 2Cl−→Cl2+2e−2��−→��2+2�− , E∘=−1.36V�∘=−1.36�
Overall reaction:
2Na+(l)+2Cl−−→−−−−−electrolysis2Na(l)+Cl2(g)2��+(�)+2��−→������������2��(�)+��2(�) OR
2NaCl(l)−→−−−−−electrolysis2Na(l)+Cl2(g)2����(�)→������������2��(�)+��2(�)
Electrolysis of an aqueous solution of Sodium Chloride
NaCl(aq)→Na+(aq)+Cl−(aq)����(��)→��+(��)+��−(��)
H2O(l)⇌H+(aq)+OH−(aq)�2�(�)⇌�+(��)+��−(��)
At cathode:
2Na++2e−→2Na2��++2�−→2�� , E∘=−2.71V�∘=−2.71�
2H2O+2e−→H2+2OH−2�2�+2�−→�2+2��− , E∘=−0.83V�∘=−0.83�
Thus H2�2 gas is generated at cathode value Na+��+ ions and remains in solution.
At anode:
2H2O→O2+4H++4e−2�2�→�2+4�++4�− , E∘=−1.23V�∘=−1.23�
2Cl−→Cl2+2e−2��−→��2+2�− , E∘=−1.36V�∘=−1.36�
As a result, the over-voltage principle is used at the anode to evolve Cl2��2 gas while leaving OH−��− ions in the solution.
When Galvanic cells are joined in series to produce a larger voltage, the arrangement is referred to as a "battery."
As long as active components are available, primary cells can be used indefinitely. The cell stops functioning and cannot be utilised once they are gone. A Dry Cell, a Leclanche Cell, a Mercury Cell, for instance.
This cell's typical potential is 1.5 V, and it cannot be replaced once it has been used. This potential diminishes when the battery is continuously depleted.
These are utilised in hearing aids and other small pieces of equipment.
The cell potential is roughly 1.35 V and remains constant throughout the life of the cell.
There are numerous applications for secondary batteries that allow for multiple recharges. For instance, lead storage batteries and Ni−Cd��−�� batteries.
The cell is connected to a cell with a greater potential, which functions as an electrolytic cell and reverses the processes, to recharge it. Pb(s)��(�) and PbO2(s)���2(�) are regenerated at the pertinent electrodes. The voltage generated by these cells is almost constant.
It is the inverse of resistance and is described as the ease with which electric current flows through a wire.
G=1R�=1�
SI unit is Siemen (S).
1S=1ohm−1(mho)1�=1�ℎ�−1(�ℎ�)
It is the reciprocal of resistivity (ρ)(�) .
κ=1ρ=1R×ℓA=G×ℓA�=1�=1�×ℓ�=�×ℓ�
Now is l=1cm�=1�� and A=1cm2�=1��2 , then κ=G�=�
As a result, the conductance of a solution with a cross-sectional area of 1cm21��2 and a length of 1cm1�� can be used to describe an electrolytic solution's conductivity.
An electrolyte is a substance that conducts electricity when it dissociates into ions in solution when it is dissolved or molten.
Examples of strong electrolytes include HCl,NaOH,KCl���,����,���, while weak electrolytes include CH3COOH,NH4OH��3����,��4��
The conductance of electricity by ions present in solutions is referred to as electrolytic or ionic conductance. The electricity that passes through an electrolyte solution is controlled by the following variables.
It is known that, κ=1R×ℓA�=1R×ℓA
We can determine the value of κ� by measuring l�, A�, and R�. The resistance of the solution R� between two parallel electrodes is computed using the "Wheatstone" bridge method.
It consists of a conductivity cell with an unknown resistance R2�2, a conductivity cell with two fixed resistances R3�3 and R4�4, and a variable resistance R1�1. The bridge is balanced when there is no current flowing through the detector. Then, under these conditions:
R1R2=R3R4�1�2=�3�4 or R2=R1R4R3�2=�1�4�3
It is the combined conductivity of all the ions produced when one mole of an electrolyte is dissolved between two large electrodes spaced one centimetre apart.
Mathematically, it can be expressed as,
Λm=κ×V,Λm=κ×1000CΛm=�×V,Λm=�×1000C
where, V is the volume of solution in cm3��3 containing 1 mole of electrolyte and C is the molar concentration.
Units: Λm=κ×1000C=Scm−1molcm−3Λm=�×1000C=Scm−1molcm−3
=ohm−1cm2mol−1or Scm2mol−1=ohm−1cm2mol−1or Scm2mol−1
It measures the electrical conductivity of a comparable electrolyte sandwiched between two substantial electrodes spaced a centimetre apart.
Mathematically, it can be written as,
Λeq=κ×v=Λeq=�×v=
Λeq=κ×1000NΛeq=�×1000N
Where, v is the volume of solution in cm3��3 containing 1 equivalent of electrolyte and N is normality.
=Scm−1equivalentcm−3=Ohm−1cm2equivalent−1Scm2equivalent−1=Scm−1equivalentcm−3=Ohm−1cm2equivalent−1Scm2equivalent−1
Conductivity decreases when concentration decreases because there are fewer ions per unit volume carrying the current in the solution. Molar conductivity increases as concentration drops. One mole of electrolyte solution's total volume V has increased as a result, which is why. Increases in a solution's volume more than make up for the loss in κ� caused by dilution.
Molar conductivity at infinite dilution, also known as limiting molar conductivity, is the value of molar conductivity as the concentration approaches zero. Extrapolation of the ΛmΛ� versus c√� curve can be used to determine the molar conductivity at infinite dilution in the case of a strong electrolyte. However, because the curve becomes nearly parallel to the y-axis as concentration approaches zero, extrapolation of the curve cannot be used to derive the value of molar conductivity of a weak electrolyte at infinite dilution.
Debye, Huckel, and Onsagar found the mathematical link between ΛmΛ� and Λ∘mΛ�∘ for a strong electrolyte. In its simplest form, the equation is as follows:
Λm=Λ∞m−bc1/2Λm=Λm∞−bc1/2
It claims that an electrolyte limiting molar conductivity can be represented as the sum of the anion and cation's individual contributions. The limiting molar conductivity of an electrolyte that produces v+�+ cations and v−�− anions upon dissociation is given by:
Λ∞m=v+λ∘++v−λ∘−Λm∞=v+�+∘+v−�−∘
Applications of Kohlrausch’s Law
The molar conductivity of acetic acid at infinite dilution, for instance, can be calculated using the molar conductivities of strong electrolytes like HCl��� , CH3COONa��3����� , and NaCl���� at infinite dilution, as shown below.
Λom(CH3−COOH)=Λom(CH3−cooNa)+Λom(HCl)−Λ∘m(NaCl)Λm(CH3−COOH)o=Λm(CH3−cooNa)o+Λm(HCl)o−Λm(NaCl)∘
Degree of dissociation α=ΛcmΛ∘m�=Λ��Λ�∘
K=cα21−αK=c�21−�
α=ΛcmΛ∞m�=ΛmcΛm∞
K=c(Λcm/Λ∞m)21−Λcm/Λ∞m=C(Λcm)2Λ∞m(Λ∗m−Λcm)K=c(Λmc/Λm∞)21−Λm�/Λm∞=C(Λmc)2Λm∞(Λm∗−Λmc)
Use of ΔGΔ� in Relating EMF values of Half Cell Reactions
Consider the following three half-cell responses:
Fe2++2e−→Fe��2++2�−→�� , E1�1
Fe3++3e−→Fe��3++3�−→�� , E2�2
Fe3++e−→Fe2+��3++�−→��2+ , E3�3
Subtracting the first reaction from the second reaction results in the third reaction. On the other hand, EMF readings do not adhere to the same pattern.
That is: E3≠E2−E1�3≠�2−�1 . But the ΔGΔ� values can be related according to the reactions:
ΔG3Δ�3 = ΔG2−ΔG1Δ�2−Δ�1
−n3FE3=−n2FE2+n1FE1−n3FE3=−n2FE2+n1FE1
−E3=−3E2+2E1−E3=−3E2+2E1
⇒E3=3E2−2E1⇒�3=3�2−2�1
1. Using the data given below find out the strongest reducing agent.
E0Cr2O72−/Cr3+=1.33V�0Cr2O72−/Cr3+=1.33V
E0Cl2/Cl−=1.36V�0Cl2/Cl−=1.36V
E0MnO4−/Mn2+=1.51V�0MnO4−/Mn2+=1.51V
E0Cr3+/Cr=−0.74V�0Cr3+/Cr=−0.74V
Correct Answer: Option (B)
Explanation:
According to the electrochemical series and standard reduction potential of metal, the stronger the reducing agent is, the larger the negative value of the standard reduction potential.
The standard reduction potential of chromium has the highest value among the available alternatives, making it the most potent reducing agent. As a result, choice B is the right one.
2. E0cell�0���� for some half-cell reactions are given below. On the basis of these mark the correct answer
(A) In dilute sulphuric acid solution, hydrogen will be reduced at the cathode.
(B) In concentrated sulphuric acid solution, water will be oxidized at the anode.
(C) In dilute sulphuric acid solution, water will be oxidized at the anode.
(D) In dilute sulphuric acid solution, SO2−4SO42− ion will be oxidized to tetrathionate ion at the anode
Correct Answer: Option (i, iii)
Explanation: In the electrolysis of sulphuric acid, the following reactions occur
2SO2−4(aq)→S2O2−8(aq)+2e−2SO42−(��)→S2O82−(��)+2�−; E0cell=1.96V�����0=1.96V
2H2O(l)→O2(g)+4H+(aq)+4e−2H2O(�)→O2(�)+4H+(��)+4�−;E0cell=1.23V�����0=1.23V
The reaction will lower the value of E0cell�0���� is preferred at anode so the second reaction is feasible.
H++e−→12H2H++�−→12H2; E0cell=0.00V�����0=0.00V
At the cathode, reduction of water occurs. Therefore, in dilute sulphuric acid solution, hydrogen will be reduced at the cathode.
3. Λ0m(H2O)Λ0�(H2O) is equal to _____
Correct Answer: Option (i, iii)
Explanation:
Kohlrausch law states that limiting molar conductivity of any salt species is equal to the sum of the limiting molar conductivity of cations and anions of the electrolyte
Λ0m(H2O)Λ0�(�2�)==Λ0m(HCl)+Λ0m(NaOH)−Λ0m(NaCl)Λ0�(���)+Λ0�(����)−Λ0�(����)
Λ0m(H+)+Λ0m(OH−)=Λ0m(H+)+Λ0m(Cl−)+Λ0m(Na+)+Λ0m(OH)−Λ0m(Na+)−Λ0m(Cl−)Λ0�(�+)+Λ0�(��−)=Λ0�(�+)+Λ0�(��−)+Λ0�(��+)+Λ0�(��)−Λ0�(��+)−Λ0�(��−)
Λ0m(H2O)Λ0�(�2�) = Λ0m(HNO3)+Λ0m(NaOH)−Λ0m(NaNO3)Λ0�(���3)+Λ0�(����)−Λ0�(����3)
Λ0m(H+)+Λ0m(OH−)=Λ0m(H+)+Λ0m(NO3−)+Λ0m(Na+)+Λ0m(OH−)−Λ0m(Na+)−Λ0m(NO3−)Λ0�(�+)+Λ0�(��−)=Λ0�(�+)+Λ0�(��3−)+Λ0�(��+)+Λ0�(��−)−Λ0�(��+)−Λ0�(��3−)
Therefore (i, iii) options are correct.
This chapter will elaborately explain what an electrochemical cell is and how it can be used to separate constituent atoms and ions of salts in a solution. The sole aim of this chapter is to deliver the knowledge related to inorganic chemistry where elements can be stacked in an electrochemical series.
In this series, elements and ions can be listed in terms of their electrode potential. This series indicates that an ion can be released by its parent compound at an electrode. This chapter will also cover the different types of cells used in electrolysis.
It covers how a preferential discharge of ions can be attained by using the specific chemical properties of the constituent ions in a compound. Favourable factors can also be created resulting in the development of a proper electrolytic environment to get a desirable product.