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CBSE NOTES CLASS 12 CHEMISTRY CHAPTER 3

ELECTROCHEMISTRY

Electrochemistry

Electrochemistry is the branch of chemistry which deals with the study of production of electricity from energy released during spontaneous chemical reactions and the use of electrical energy to bring about non-spontaneous chemical transformations.

Importance of electrochemistry

1. Production of metals like Na, Mg, Ca and Al.

2. Electroplating.

3. Purification of metals.

4. Batteries and cells used in various instruments.

Conductors

Substances that allow electric current to pass through them are known as conductors.

Metallic conductors or electronic conductors

Substances which allow the electric current to pass through them by the movement of electrons are called metallic conductors, e.g., metals.

Electrolytic conductors or electrolytes

Substances which allow the passage of electricity through their fused state or aqueous solution and undergo chemical decomposition, are called electrolytic conductors, e.g., aqueous solution of acids, bases and salts.

Electrolytes are of two types

1. Strong electrolytes - The electrolytes that completely dissociate or ionise into ions are called strong electrolytes. e.g., HCl, NaOH, K2SO4.

2. Weak electrolytes - The electrolytes that dissociate partially (α < 1) are called weak electrolytes, e.g., CH3COOH, H2CO3, NH4OH, H2S, etc.

Electrochemical cell

Galvanic cell or Voltaic cell or electrochemical cell converts the chemical energy of a spontaneous redox reaction into electrical energy.

In the Galvanic cell, the Gibbs energy of the spontaneous redox reaction is converted into electrical work which may be used for running motor or other electrical gadgets like heater, fan, geyser, etc.

Daniel cell

An electrochemical cell using zinc and copper metals as electrodes, is known as Daniell cell. It is represented as

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

It has an electrical potential equal to 1.1V when concentration of Zn2+ and Cu2+ ions is unity (1 mol dm–3).

If an external opposite potential is applied in the galvanic cell and increased slowly, the reaction continues till the opposing voltage reaches the value 1.1V. The reaction stops altogether when the external opposing voltage reaches 1.1 Volt and no current flows through the cell.

Any further increase in the external potential again starts the reaction but in the opposite direction. It now functions as an electrolytic cell, a device for using electrical energy to carry non-spontaneous chemical reactions.

When Eext < 1.1 V

1. Electrons flow from Zn rod to Cu rod hence current flows from Cu to Zn.

2. Zn dissolves at anode and copper deposits at cathode.

When Eext = 1.1 V

1. No flow of electrons or current.

2. No chemical reaction.

When Eext > 1.1 V

1. Electrons flow from Cu to Zn and current flows from Zn to Cu.

2. Zinc is deposited at the zinc electrode and copper dissolves at copper electrode

Cell half reactions

The reaction is a combination of two half reactions whose addition gives the overall cell reaction:

1. Cu2+ + 2e → Cu(s) (reduction half reaction)

2. Zn(s) → Zn2+ + 2e (oxidation half reaction)

The reduction half reaction occurs on the copper electrode while the oxidation half reaction occurs on the zinc electrode.

These two portions of the cell are called half-cells or redox couples.

The copper electrode is called the reduction half-cell and the zinc electrode, the oxidation half-cell.

Difference between an electrolytic cell and a galvanic cell

 Electrochemical cell (Galvanic Cell) Electrolytic cell A Galvanic cell converts chemical energy into electrical energy. An electrolytic cell converts electrical energy into chemical energy. The redox reaction is spontaneous and is responsible for the production of electrical energy. The redox reaction is not spontaneous and electrical energy has to be supplied to initiate the reaction. The two half-cells are set up in different containers, being connected through the salt bridge or porous partition. Both the electrodes are placed in a same container in the solution or molten electrolyte. The anode is negative and cathode is positive electrode. The reaction at the anode is oxidation and that at the cathode is reduction. The anode is positive and cathode is the negative electrode. The reaction at the anode is oxidation and that at the cathode is reduction. The electrons are supplied by the species getting oxidized. They move from anode to the cathode in the external circuit. The external battery supplies the electrons. They enter through the cathode and come out through the anode.

Electrode potential

A potential difference develops between the electrode and the electrolyte which is called electrode potential.

When the concentrations of all the species involved in a half-cell is unity then the electrode potential is known as standard electrode potential.

The half-cell in which oxidation takes place is called anode and it has a negative potential with respect to the solution.

The half-cell in which reduction takes place is called cathode and it has a positive potential with respect to the solution.

By convention cathode is represented on the RHS and anode on the LHS.

Transport number or transference number

The current flowing through an electrolytic solution is carried by the ions. The fraction of the current carried by an ion is called its transport number or transference number.

Functions of salt bridge

Salt-bridge generally contains solution of strong electrolyte such as KNO3, KCl etc. KCI is preferred because the transport numbers of K+ and Cl- are almost same.

1. It completes the circuit and allows the flow of current.

2. It maintains the electrical neutrality on both sides.

Cell potential or emf

The potential difference between the two electrodes of a galvanic cell is called the cell potential. It is the difference between the electrode potentials (reduction potentials) of the cathode and anode, or, the potential of the half-cell on the right hand side minus the potential of the half-cell on the left hand side.

Ecell = Eright – Eleft

It is also called the cell electromotive force (emf) of the cell when no current is drawn through the cell.

A cell of almost constant emf is called standard cell.

Example – A cell with copper and silver electrodes

Consider a cell consisting of copper and silver electrodes.

Cell reaction

Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)

Half-cell reactions

Cathode reaction (reduction)

2Ag+(aq) + 2e → 2Ag(s)

Anode reaction (oxidation)

Cu(s) → Cu2+ (aq) + 2e

The silver electrode acts as a cathode and copper electrode acts as an anode.

This cell can be represented as

Cu(s) | Cu2+(aq) || 2Ag+(aq) | 2Ag(s)

and emf of the cell is given by,

Ecell = Eright – Eleft = EAg+|Ag – ECu2+|Cu

Reference electrode

An electrode of known potential is called reference electrode. It may be primary reference electrode like hydrogen electrode or secondary reference electrode like calomel electrode.

Standard hydrogen electrode (SHE), also known as normal hydrogen electrode (NHE), consists of platinum wire, carrying platinum foil coated with finely divided platinum black. The wire is sealed into a glass tube, placed in beaker containing 1M HCl. The H2 gas at 1 bar pressure is bubbled through the solution at 298K.

Half-cell is

Pt(s) | H2(g) | H+(aq)

In SHE at the surface of plantinum, either of the following reaction can take place

The electrode potential of SHE has been fixed as zero at all temperatures.

Drawbacks of SHE

1. It is difficult to maintain 1 bar pressure of H2 gas.

2. It is difficult to maintain H+ ion concentration 1M.

3. The platinum electrode is easily poisoned by traces of impurities.

Hence, calomel electrodes are used as reference electrodes.

Calomel electrode consists of mercury in contact with Hg2Cl2 (calomel) paste in a solution of KCl.

Measurement of electrode potential

The potential of individual half-cell cannot be measured. Only the difference between the two half-cell potentials can be measured. This is called the emf of the cell.

Ecell = Eright – Eleft

EMF of Daniel cell

The measured emf of the cell

Pt(s) | H2(g, 1 bar) | H+ (aq, 1 M) || Cu2+(aq, 1 M) | Cu

is 0.34 V. This is also the value for the standard electrode potential of the half-cell corresponding to the reaction

Cu2+ (aq, 1 M) + 2 e → Cu(s)

Similarly, the measured emf of the cell

Pt(s) | H2(g, 1 bar) | H+ (aq, 1 M) || Zn2+ (aq, 1 M) | Zn

is –0.76 V corresponding to the standard electrode potential of the half-cell reaction

Zn2+ (aq, 1 M) + 2e → Zn(s)

Therefore, emf of the Daniel cell will be,

Ecell = Eright – Eleft

= 0.34 – (–0.76) = 1.1 V

For a reaction to take place spontaneously, the standard electrode potential should be positive.

The positive value of the standard electrode potential in the first case indicates that Cu2+ ions get reduced more easily than H+ ions. The reverse process cannot occur, that is, hydrogen ions cannot oxidize Cu (or alternatively we can say that hydrogen gas can reduce copper ion) under the standard conditions described above. Thus, Cu does not dissolve in HCl. In nitric acid get gets dissolved as it is oxidised by nitrate ion and not by hydrogen ion.

The negative value of the standard electrode potential in the second case indicates that hydrogen ions can oxidise zinc (or zinc can reduce hydrogen ions). Therefore Zn will dissolve in acid.

Electrochemical series

It is the arrangement of electrodes in the increasing order of their standard reduction potentials.

Standard Electrode Potentials at 298 K (ECS)

 F2(g) + 2e– → 2F– 2.87 Co3+ + e– → Co2+ 1.81 H2O2 + 2H+ + 2e– → 2H2O 1.78 MnO4– + 8H+ + 5e– → Mn2+ + 4H2O 1.51 Au3+ + 3e– → Au(s) 1.40 Cl2(g) + 2e– → 2Cl– 1.36 Cr2O72– + 14H+ + 6e– → 2Cr3+ + 7H2O 1.33 O2(g) + 4H+ + 4e– → 2H2O 1.23 MnO2(s) + 4H+ + 2e– → Mn2+ + 2H2O 1.23 Br2 + 2e– → 2Br– 1.09 NO3– + 4H+ + 3e– → NO(g) + 2H2O 0.97 2Hg2+ + 2e– → Hg22+ 0.92 Ag+ + e– → Ag(s) 0.80 Fe3+ + e– → Fe2+ 0.77 O2(g) + 2H+ + 2e– → H2O2 0.68 I2 + 2e– → 2I– 0.54 Cu+ + e– → Cu(s) 0.52 Cu2+ + 2e– → Cu(s) 0.34 AgCl(s) + e– → Ag(s) + Cl– 0.22 AgBr(s) + e– → Ag(s) + Br– 0.10 2H+ + 2e– → H2(g) 0.00 Pb2+ + 2e– → Pb(s) –0.13 Sn2+ + 2e– → Sn(s) –0.14 Ni2+ + 2e– → Ni(s) –0.25 Fe2+ + 2e– → Fe(s) –0.44 Cr3+ + 3e– → Cr(s) –0.74 Zn2+ + 2e– → Zn(s) –0.76 2H2O + 2e– → H2(g) + 2OH–(aq) –0.83 Al3+ + 3e– → Al(s) –1.66 Mg2+ + 2e– → Mg(s) –2.36 Na+ + e– → Na(s) –2.71 Ca2+ + 2e– → Ca(s) –2.87 K+ + e– → K(s) –2.93 Li+ + e– → Li(s) –3.05
1. A negative Eo means that the redox couple is a stronger reducing agent than the H+/H2 couple.

2. A positive Eo means that the redox couple is a weaker reducing agent than the H+/H2 couple

Applications of electrochemical series (ECS)

1. The lower the value of Eo, the greater the tendency to form cation.

M → Mn+ + ne-

2. Metals placed below hydrogen in ECS replace hydrogen from dilute acids but metals placed above hydrogen cannot replace hydrogen from dilute acids.

Reactions possible

Ca + dil H2SO4 → CaSO4 + H2

Ca + 2H+ → Ca2+ + H2

Reactions not possible

Cu + dil H2SO4 → CuSO4 + H2

Cu + 2H+ → Ca2+ + H2

1. Oxides of metals placed below hydrogen are not reduced by H2 but oxides of iron and metals placed above iron are reduced by H2·
• SnO, PbO, CuO are reduced by H2

• CaO, K2O are not reduced by H2·

2. Reducing character increases down the series.

3. Reactivity metals increases down the series.

4. Determination of emf - emf is the difference of reduction potentials of two half-cells.

emf = Eright – Eleft

Note: If the value of emf is positive, then reaction takes place spontaneously, otherwise not.

NERNST EQUATION

The relationship between the concentration of ions and electrode potential is given by Nernst equation.

For the reaction,

M → Mn+ + ne-,

The half reaction electrode potential is given by,

For a general electrochemical cell,

aA + bB → cC + dD;

Where,

R is gas constant (8.314 JK–1 mol–1),

F is Faraday constant (96487 C mol–1)

(Therefore, nF is the total amount of charge transferred per mole),

T is temperature in kelvin and [Mn+] is the concentration of the species, Mn+.

Note: Concentration of pure solids and liquids is taken as unity.

For Daniel cell, the electrode potential for any given concentration of Cu2+ and Zn2+ ions, we can write,

For Cathode:

For Anode:

Therefore, the cell potential will be,

${\mathrm{E}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}={\mathrm{E}}_{{\mathrm{C}\mathrm{u}}^{2+}|\mathrm{C}\mathrm{u}}{\mathrm{E}}_{{\mathrm{Z}\mathrm{n}}^{2+}|\mathrm{Z}\mathrm{n}}$

Now at equilibrium

$\frac{\left[{\mathrm{Z}\mathrm{n}}^{2+}\right]}{\left[{\mathrm{C}\mathrm{u}}^{2+}\right]}={\mathrm{K}}_{\mathrm{c}},$

Therefore,

Relationship between free energy change and equilibrium constant

Electrical work done is equal to electrical potential multiplied by total charge passed.

If we want to obtain maximum work from a galvanic cell then charge has to be passed reversibly. The reversible work done by a galvanic cell is equal to decrease in its Gibbs energy.

If the emf of the cell is Ecell and nF is the amount of charge passed and ΔrG is the Gibbs energy of the reaction, then

ΔrG = – nFEcell

Ecell is an intensive parameter but ΔrG is an extensive thermodynamic property and the value depends on n.

For reaction,

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

ΔrG = – 2FEcell

And for the reaction

2Zn(s) + 2Cu2+(aq) → 2Zn2+(aq) + 2Cu(s)

ΔrG = – 4FEcell

If the concentration of all the reacting species is unity, then Ecell = Eocell

At equilibrium, Ecell becomes zero, so Standard Gibbs energy of the reaction,

ΔrGo = – nFEocell = – RT ln Kc

Concentration cell

A concentration cell is a limited form of a galvanic cell that has two equivalent half-cellsof the same composition differing only in concentrations.

We can calculate the potential developed by such a cell using the Nernst Equation.

Resistance of a cell

The resistance R of a cell depends

1. Directly on the length of column (distance between the electrodes) of solution, l i.e.

2. Inversely on area of cross-section A of the electrode plates, i.e.

3. On the nature of of the electrolyte

On combining the above factors, we get

The proportionality constant ρ (Rho) is called resistivity of electrolyte.

• R can be calculated using wheat stone bridge.

When the Wheatstone bridge is balanced

Conductance of a cell

It is the ease of flow of electric current through the conductor. It is reciprocal of resistance (R).

Where,

The SI unit of conductance is siemens, represented by the symbol ‘S’ and is equal to ohm–1 (also known as mho) or Ω–1.

Conductivity of a cell ($\mathrm{\kappa }$)

The inverse of resistivity is called conductivity or specific conductance. It is represented by the symbol, κ (Greek, kappa).

Where,

The SI unit is S m-1

Unit of cell constant is cm-1 or m-1.

Molar conductivity of a solution (Λm)

Molar conductivity of a solution at a given concentration is defined as,

where M = molarity and c is concentration in mole per m3

The unit of molar conductivity is Ω-1 cm2 mol-1 or S cm2 mol-1.

1 S m2 mol–1 = 104 S cm2 mol–1

or 1 S cm2 mol–1 = 10–4 S m2 mol–1.

Factors affecting conductivity

1. Nature of electrolyte - The strong electrolytes like KNO3, KCl, NaOH etc. are completely ionised in aqueous solution and have high values of molar conductivity.

The weak electrolytes are ionised to a lesser extent in aqueous solution and have lower values of molar conductivity.

2. Concentration of the solution - The concentrated solutions of strong electrolytes have significant interionic attractions, which reduce the speed of ions and lower the value of Λm.

As the dilution decreases such attractions also decrease which increases the value of Λm.

3. Temperature - The increase of temperature decreases inter-ionic attractions and increases kinetic energy of ions and their speed. Thus, Λm and Λeq increase with temperature.

4. Size of the ions produced and their solvation

5. The nature of the solvent and its viscosity. Higher the viscosity, lower is conductivity

Variation of conductivity and molar conductivity with concentration

Conductivity of strong electrolytes

For strong electrolytes, Λm increases slowly with dilution and is given by,

If we plot Λm against $\sqrt{\mathrm{C}}$, we obtain a straight line with intercept equal to ${\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{o}}$ and slope equal to ‘–A’. The value of ‘A’ for a given solvent and temperature depends on the type of electrolyte i.e., the charges on the cation and anion produced on the dissociation of the electrolyte in the solution.

The limiting value, ${\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{o}}$ (the molar conductivity at zero concentration or at infinite dilution) can be obtained extrapolating the graph.

Note: Depending upon the type of ions involved, NaCl, CaCl2, MgSO4 are known as 1-1, 2-1 and 2-2 electrolytes respectively.

All electrolytes of a particular type have the same value for ‘A’.

Conductivity of weak electrolytes

For weak electrolytes like acetic acid, Λm increases steeply on dilution, especially near lower concentrations. This is because they have lower degree of dissociation at higher concentrations and hence for such electrolytes, the change in Λm with dilution is due to increase in the degree of dissociation and consequently the number of ions in total volume of solution that contains 1 mol of electrolyte.

Therefore, ${\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{o}}$ cannot be obtained by extrapolation of ${{\mathrm{\Lambda }}^{\mathrm{c}}}_{\mathrm{m}}$ to zero concentration.

At infinite dilution (i.e., concentration c → zero) electrolyte dissociates completely (α = 1), but at such low concentration the conductivity of the solution is so low that it cannot be measured accurately.

Therefore, ${\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{o}}$ for weak electrolytes is obtained by using Kohlrausch’s law of independent migration of ions.

Kohlrausch’s law of independent migration of ions

The limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.

where:

is the limiting molar ionic conductivity of ion i.

is the number of ions i in the formula unit of the electrolyte

For example

• If λoNa+ and λoCl– are limiting molar conductivity of the sodium and chloride ions respectively, then the limiting molar conductivity for sodium chloride is given by

Limiting molar conductivity for Na2SO4

Limiting molar conductivity for BaCl2

Applications of Kohlrausch’s law

1. Determination of molar conductivities of weak electrolytes at infinite dilution

Consider the example of dissociation of CH3COOH, as follows,

By Kohlrausch’s law,

Also,

Adding equation (ii) and (iii) and subtracting equation (i), we get,

Since CH3COOK, HCl and KCl all are strong electrolytes, we can find there molar conductivities at infinite dilution by graphical method.

2. Determination of degree of dissociation (α) of an electrolyte at a given dilution.

For weak electrolytes like CH3COOH, if α is the degree of dissociation of the electrolyte

Where, ${{\mathrm{\Lambda }}^{\mathrm{c}}}_{\mathrm{m}}$ is the molar conductivity at concentration c and ${\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{o}}$ is the molar conductivity at infinite dilution.

3. Determination of dissociation constant Kc

${\mathrm{K}}_{\mathrm{c}}=\frac{\mathrm{c}{\mathrm{\alpha }}^{2}}{1-\mathrm{\alpha }}$

4. Finding solubility of sparingly solutble salts

5. Calculation of ionic product of water

Ionic conductance of H+ and OH- ions at infinite dilution has been found to be,

Therefore, by Kohlrausch’s law,

The specific conductance, κ of water has been found to be κ = 5.54 × 10-8 S cm-1

Using the formula,

$=\left(1.01×{10}^{-7}\right)×\left(1.01×{10}^{-7}\right)$

$=1.02×{10}^{-14}$

ELECTROLYSIS

It is the process of decomposition of an electrolyte when electric current is passed through either its aqueous solution or molten state.

Example

One of the simplest electrolytic cell consists of two copper strips dipping in an aqueous solution of copper sulphate.

If a DC voltage is applied to the two electrodes, then Cu2+ ions discharge at the cathode (negatively charged)

Cu2+(aq) + 2e → Cu (s)

At the anode, copper is converted into Cu2+ ions,

Cu(s) → Cu2+(aq) + 2e

Important points regarding electrolytic cell

1. In electrolytic cell both oxidation and reduction takes place in the same cell

2. Anode is positively charged and cathode is negatively charged, in electrolytic cell.

3. During electrolysis of molten electrolyte, cations are liberated at cathode, while anions at the anode.

4. When two or more ions compete at the electrodes, the ion with higher reduction potential gets liberated at the cathode while the ion with lower reduction potential gets liberated at the anode.

5. For metals to be deposited on the cathode during electrolysis, the voltage required is almost the same as the standard electrode potential. However for liberation of gases, some extra voltage is required on top of the theoretical value of the standard electrode potential. The extra voltage thus required is called over voltage or bubble voltage.

How to predict the products of electrolysis?

When an aqueous solution of an electrolyte is electrolysed, and if the cation has higher reduction potential than water (-0.83V), cation is liberated at the cathode (e.g. in the electrolysis of copper and silver salts) otherwise H2 gas is liberated due to reduction of water (e.g., in the electrolysis of K, Na, Ca salts, etc.)

Similarly if anion has higher oxidation potential than water (-1.23V), anion is liberated (e.g., Br-), otherwise O2 gas is liberated due to oxidation of water.

Electrolysis of molten NaCl

If we use molten NaCl as electrolyte, the products of electrolysis are sodium metal and Cl2 gas. Here we have only one cation (Na+) which is reduced at the cathode

Na+ + e → Na

and one anion (Cl) which is oxidised at the anode

Cl→ ½ Cl2 + e

Electrolysis of aqueous NaCl

During the electrolysis of aqueous sodium chloride solution, the products are NaOH, Cl2 and H2. In this case besides Na+ and Cl ions we also have H+ and OH ions along with the solvent molecules, H2O.

At the cathode, there is competition between the following reduction reactions:

Na+ (aq) + e → Na (s) ${{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}$ = – 2.71 V

H+ (aq) + e → ½ H2 (g) ${{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}$ = 0.00 V

The reaction with higher value of ${{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}$ is preferred and, therefore, the reaction at the cathode during electrolysis will be,

H+ (aq) + e → ½ H2 (g)

But H+ (aq) is produced by the dissociation of H2O, i.e.,

H2O (l ) → H+ (aq) + OH (aq)

Therefore, the net reaction at the cathode is

H2O (l ) + e– → ½ H2(g) + OH

At the anode the following oxidation reactions compete,

Cl (aq) → ½ Cl2 (g) + e, ${{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}$ = 1.36V

2H2O (l) → O2 (g) + 4H+(aq) + 4e , ${{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}$ = 1.23V

The reaction at anode with lower value of ${{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}$ is preferred, therefore, water should get oxidised in preference to Cl (aq). However, on account of overpotential of oxygen (higher extra potential is required to convert to O2), first reaction is preferred.

Thus, the reactions will be,

In the solution: NaCl (aq) + H2O → Na+ (aq) + Cl (aq)

At cathode: H2O(l ) + e → ½ H2(g) + OH (aq)

At anode: Cl (aq) → ½ Cl2(g) + e

Net reaction, will therefore be,

NaCl(aq) + H2O(l) → Na+(aq) + OH(aq) + ½ H2(g) + ½ Cl2(g)

Note: The standard electrode potentials are replaced by electrode potentials given by Nernst equation to take into account the concentration effects.

Electrolysis of H2SO4

During the electrolysis of sulphuric acid, the following processes are possible at the anode:

2H2O(l ) → O2(g) + 4H+(aq) +4e, ${{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}$ = 1.23V

2SO4 2– (aq) → S2O82– (aq) + 2e, ${{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}$ = 1.96V

For dilute sulphuric acid, the first reaction is preferred but at higher concentrations of H2SO4, second reaction is preferred.

Effect of nature of electrode on electrolysis

If the electrode is inert (e.g., platinum or gold), it does not participate in the chemical reaction and acts only as source or sink for electrons.

On the other hand, if the electrode is reactive, it participates in the electrode reaction.

Thus, the products of electrolysis may be different for reactive and inert electrodes.

Discharge potential is defined as the minimum potential that must be applied across the electrodes to bring about the electrolysis and subsequent discharge of the ion on the electrode.

FARADAY’S LAWS OF ELECTROLYSIS

1. Faraday’s first law of electrolysis

The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte (solution or melt).

Or

The amount of the substance deposited or liberated at cathode is directly proportional to the quantity of electricity passed through electrolyte.

W ∝ I × t

Or W = I t Z = Q Z

Here, I is the current in amp, t is time in sec, Q is quantity of charge (coulomb) and Z is a constant known as electrochemical equivalent.

When I = 1 amp, t = 1 sec then Q = 1 coulomb, then W = Z.

Thus, electrochemical equivalent is the amount of the substance deposited or liberated by passing 1A current for 1 sec (i.e. 1 coulomb, I × t = Q)

2. Faraday’s second law of electrolysis

The amounts of different substances liberated by the same quantity of electricity passing through the electrolytic solution are proportional to their chemical equivalent weights or

Hence,

The amount of electricity (or charge) required for oxidation or reduction depends on the stoichiometry of the electrode reaction.

For example, in the reaction,

Ag+(aq) + e → Ag(s)

One mole of the electron is required for the reduction of one mole of silver ions.

Now, the charge on one electron is 1.6021× 10–19 C. Therefore, the charge on one mole of electrons

= NA × 1.6021 × 10–19 C

= 6.02 × 1023 mol–1×1.6021×10–19 C

= 96487 C mol–1 ≈ 96500 C mol–1

This quantity of electricity is called Faraday and is represented by the symbol F.

For the electrode reactions

Mg2+ (l) + 2e → Mg(s)

One mole of Mg2+ requires 2 moles of electrons (2F).

And for

Al3+ (l) + 3e → Al(s)

One mole of Al3+ requires 3 moles of electrons (3F).

The charge passed through the electrolytic cell during electrolysis is equal to the product of current in amperes and time in seconds.

In commercial production of metals, current as high as 50,000 amperes is used that amounts to about 0.518 F per second.

BATTERIES

These are source of electrical energy which may have one or more cells connected in series.

For a good quality battery it should be reasonably light, compact and its voltage should not vary appreciably during its use.

Primary batteries

In primary batteries, the reaction occurs only once and after use over a period of time battery becomes dead and cannot be reused.

(i) Dry cell or Leclanche cell

The cell consists of a zinc container that also acts as anode and the cathode is a carbon (graphite) rod surrounded by powdered manganese dioxide and carbon.

The space between the electrodes is filled by a moist paste of ammonium chloride (NH4Cl) and zinc chloride (ZnCl2).

The electrode reactions can be written approximately as,

Anode: Zn(s) → Zn2+ + 2e

Cathode: MnO2 + NH4+ + e → MnO(OH) + NH3

In the reaction at cathode, manganese is reduced from the +4 oxidation state to the +3 state.

Ammonia produced in the reaction forms a complex with Zn2+ to give [Zn(NH3)4]2+.

The cell has a potential of nearly 1.5V.

(ii) Mercury cell

Mercury cell, suitable for low current devices like hearing aids, watches, etc. consists of zinc – mercury amalgam as anode and a paste of HgO and carbon as the cathode.

The electrolyte is a paste of KOH and ZnO.

The electrode reactions for the cell are,

Anode: Zn(Hg) + 2OH→ ZnO(s) + H2O + 2e

Cathode: HgO + H2O + 2e → Hg(l) + 2OH

The overall reaction:

Zn(Hg) + HgO(s) → ZnO(s) + Hg(l)

• Cell potential for mercury cell is 1.35 V

Secondary batteries

These cells can be recharged and can be used again and again

(i) Lead storage battery

The lead storage battery

Used in automobiles and invertors.

Anode-Spongy lead

Cathode-Grid of lead packed with PbO2

Electrolyte - 38% H2SO4 by mass

The cell reactions when the battery is in use or discharge reaction

Anode: Pb(s) + SO42–(aq) → PbSO4(s) + 2e

Cathode: PbO2(s) + SO42–(aq) + 4H+(aq) + 2e → PbSO4 (s) + 2H2O (l)

Overall cell reaction consisting of cathode and anode reactions is:

Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l)

When recharged the cell reactions are reversed.

(ii) Nickel cadmium storage cell

Has longer life than the lead storage cell but more expensive to manufacture.

Anode-Cadmium

Cathode-Metal grid containing NiO2

Electrolyte - KOH solution

Reactions during discharge

Anode: Cd(s) + 2OH-(aq) → Cd(OH)2(s) + 2e-

Cathode: NiO2(s) + 2H2O(l) + 2e- → Ni(OH)2(s) + 2OH-(aq)

Overall reaction,

Cd (s) + 2Ni(OH)3 (s) → CdO (s) + 2Ni(OH)2 (s) + H2O (l )

Cell potential 1.4 V

FUEL CELLS

Galvanic cells that are designed to convert the energy of combustion of fuels like hydrogen, methane, methanol, etc. directly into electrical energy are called fuel cells.

Hydrogen oxygen fuel cell

Fuel cell using O2 and H2

Electrodes - Made of porous graphite impregnated with catalyst (Pt, Ag or a metal oxide).

Electrolyte - Aqueous solution of KOH or NaOH

Oxygen and hydrogen are continuously fed into the cell.

Reactions

Cathode: O2(g) + 2H2O(l) + 4e→ 4OH(aq)

Anode: 2H2(g) + 4OH(aq) → 4H2O(l) + 4e

Overall reaction:

2H2(g) + O2(g) → 2 H2O(l )

Efficiency of cell - 70%, Efficiency of thermal plants – 40%

Cell potential 1.23 V

CORROSION

Slow formation of undesirable compounds such as oxides, sulphides or carbonates at the surface of metals by reaction with moisture and other atmospheric gases is known as corrosion.

Factos affecting corrosion

1. Reactivity of metals

2. Presence of moisture and atmospheric gases like CO2, SO2, etc.

3. Presence of impurities

4. Presence of electrolyte

Electrochemical theory of rusting of iron

An electrochemical cell, also known as corrosion cell, is developed at the surface of iron.

At a particular spot of an object made of iron, oxidation takes place and that spot behaves as anode.

Anode - Pure iron

Anode reaction: 2Fe(s) → 2 Fe2+ + 4e, EoFe2+|Fe = – 0.44 V

Electrons released at anodic spot move through the metal and go to another spot on the metal and reduce oxygen in presence of H+.

H+ is believed to be available from H2CO3 formed due to dissolution of carbon dioxide from air into water. Hydrogen ion in water may also be available due to dissolution of other acidic oxides from the atmosphere.

This spot behaves as cathode.

Cathode - Impure iron surface

Cathode reaction: O2(g) + 4H+(aq) + 4e → 2H2O(l ), EoH+|O2|H2O = 1.23V

The overall reaction:

2Fe(s) + O2(g) + 4H+(aq) → 2Fe2+(aq) + 2H2O(l), Eocell =1.67 V

Rusting of iron can be prevented by the following methods

1. Barrier protection through coating of paints or electroplating.

2. Through galvanisation or coating of surface with tin metal.

3. By the use of antirust solutions (bis phenol).

4. By cathodic protection in which a metal is protected from corrosion by connecting it to another metal that is more easily oxidised.