Electrochemistry in concerned with the interrelation of electrical and chemical effects. i.e. production of chemical change by electrical energy (electrolysis) and conversion of chemical energy into electrical energy. The reactions involve transfer of electrons and are therefore redox reactions.

Substances that allow the flow of electrons as electric current through them are called electrical conductors. These substances obey ohm’s law and are of two types:

Matallic / electronic Conductor

• Flow of electricity due to movement of electrons

• No chemical change as there is no transfer of matter.

• Faraday’s law is not followed

• Conduction decreases with temperature because kernels start vibrating which interfere in the flow of electrons

• Depends upon nature, structure of metal and number of valence electrons per atom.

Electrolytic conductor

• Flow of electricity due to movement of ions

• Ions are oxidised or reduced at the electrodes, hence involve transfer of matter .

• Faraday’s law is followed

• Conduction increases with temperature because dissociation increases and viscosity decreases

• Depends upon nature of electrolyte (weak or strong), size of ions and their solvation, nature of solvent and viscosity.

Electrochemistry I : Electrolytic cells

The device in which conversion of electrical energy into chemical energy is done is known as electrolytic cell. An electrolytic cell consists of a vessel for electrolytic solution or molten electrolyte in which two metallic electrodes connected to a source of electric current are immersed.

If an electrolytic solution consists of more than two ions then during electrolysis all the ions are not discharged simultaneously but certain ions are liberated at the electrode in preference to the others. This is based on the principle of preferential discharged theory which states that the ion which requires least energy is discharged first. The discharge potential of ${\mathrm{H}}^{+}$ions is lower than ${\mathrm{Na}}^{+}$ ions, similarly discharge potential of ${\mathrm{Cl}}^{-}$ion is lower than that of ${\mathrm{OH}}^{-}$ions. This can be explained by a few examples:

• Electrolysis of ${\mathrm{CuSO}}_4$ solution using Pt electrodes

Possible reactions at :

Cathode

${\mathrm{Cu}}^{2+}(\mathrm{aq})+\mathrm{e}⟶\mathrm{Cu}(\mathrm{s}){\mathrm{E}}_{{\mathrm{cu}}^{2+}\mid \mathrm{cu}}^{\circ }=0.34\mathrm{V}$

$2{\mathrm{H}}_2\mathrm{O}(\mathrm{I})+2{\mathrm{e}}^{-}⟶{\mathrm{H}}_2(\mathrm{g})+2{\mathrm{OH}}^{-}(\mathrm{aq}){\mathrm{E}}_{{\mathrm{H}}_2\mathrm{O}\mid {\mathrm{H}}_2}^{\circ }=0.83\mathrm{V}$

Anode

$2{\mathrm{SO}}_4^{2-}(\mathrm{aq})⟶{\mathrm{S}}_2{\mathrm{O}}_8^{2-}(\mathrm{aq})+2{\mathrm{e}}^{-}{\mathrm{E}}_{{{\mathrm{SO}}_4}^{2-}\mid {\mathrm{S}}_2{\mathrm{O}}_8^{2-}}^{\circ }=-2.9\mathrm{V}$

$2{\mathrm{H}}_2\mathrm{O}(\mathrm{I})⟶{\mathrm{O}}_2(\mathrm{g})+4{\mathrm{H}}^{+}(\mathrm{aq})+4{\mathrm{e}}^{-}{\mathrm{E}}_{{\mathrm{H}}_2\mathrm{O}\mid {\mathrm{O}}^{2+}}^{\circ }=-1.23\mathrm{V}$

$\mathrm{Cu}(\mathrm{s})⟶{\mathrm{Cu}}^{2+}(\mathrm{aq})+2{\mathrm{e}}^{-}{\mathrm{E}}_{\mathrm{cu}\mid {\mathrm{cu}}^{2+}}^{\circ }=-0.34\mathrm{V}$

On comparing electrode discharge potential, it can be said that when ${\mathrm{CuSO}}_4$ is electrolysed using Cu electrodes. $\mathrm{Cu}$ is dissolved at $\mathrm{Cu}$ anode and $\mathrm{Cu}$ is deposited at $\mathrm{Cu}$ cathode

• Aqueous ${\mathrm{H}}_2{\mathrm{SO}}_4$ solution

Possible reactions at :

Cathode

$2{\mathrm{H}}_2\mathrm{O}(\mathrm{I})+2\mathrm{e}-⟶{\mathrm{H}}_2(\mathrm{g})+2{\mathrm{OH}}^{-}(\mathrm{aq})$ ${\mathrm{E}}_{{\mathrm{H}}_2\mathrm{O}{\mathrm{H}}_2}^0=-0.83\mathrm{V}$

$2{\mathrm{SO}}_4{}^{2\cdot }(\mathrm{aq})⟶{\mathrm{S}}_2{\mathrm{O}}_8{}^2\cdot (\mathrm{aq})+2{\mathrm{e}}^{-}$ ${\mathrm{E}}_{{\mathrm{SO}}_4{}^2-\mathrm{s}{\mathbf{s}}^2{\mathrm{e}}^2}^0=-2.01\mathrm{V}$

Anode

$2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})⟶{\mathrm{O}}_2(\mathrm{g})+4{\mathrm{H}}^{+}(\mathrm{aq})+4\mathrm{e}$ ${\mathrm{E}}_{{\mathrm{H}}_2\mathrm{O}{\mathrm{O}}_2}^0=-1.23\mathrm{V}$

On electrolysis, ${\mathrm{H}}_2$ gas is evolved at cathode and ${\mathrm{O}}_2$ gas is evolved at anode. It has been observed that at high current density or for highly concentrated ${\mathrm{H}}_2{\mathrm{SO}}_4$ solution, ${\mathrm{SO}}_4{}^2$ ions get oxidised to ${\mathrm{S}}_2{\mathrm{O}}_8{}^{2\cdot }$.

• Aqueous $\mathrm{NaCl}$ solution

Possible reactions at :

Cathode :

${\mathrm{Na}}^{+}(\mathrm{aq})+{\mathrm{e}}^{-}⟶\mathrm{Na}(\mathrm{s}){\mathrm{E}}_{{\mathrm{Na}}^{+}\mid \mathrm{Na}}^0=-2.71\mathrm{V}$

$2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})+{\mathrm{O}}_2(\mathrm{g})+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2(\mathrm{g})+2{\mathrm{OH}}^{-}(\mathrm{aq}){\mathrm{E}}_{{\mathrm{H}}_2\mathrm{O}{\mathrm{H}}_2}^0=-0.83\mathrm{V}$

Anode:

$2\mathrm{Cl}(\mathrm{aq})⟶{\mathrm{Cl}}_2(\mathrm{g})+2{\mathrm{e}}^{-}$ ${\mathrm{E}}_{{\mathrm{Cl}}^{-}\mid {\mathrm{Cl}}_2}^0=-1.36\mathrm{V}$

$2{\mathrm{H}}_2\mathrm{O}(\mathrm{I})⟶{\mathrm{O}}_2(\mathrm{g})+4{\mathrm{H}}^{+}+4{\mathrm{e}}^{-}$ ${\mathrm{E}}_{{\mathrm{H}}_2\mathrm{O}{\mathrm{O}}_2}^0=-1.23\mathrm{V}$

During electrolysis using Pt electrodes ${\mathrm{H}}_2(\mathrm{g})$ at cathode and ${\mathrm{O}}_2(\mathrm{g})$ at anode are evolved. But when $\mathrm{Hg}$ electrodes are used instead of $\mathrm{Pt},{\mathrm{Na}}^{+}(\mathrm{aq})$ gets deposited which further reacts with $\mathrm{Hg}$ to form sodium - amalgam.

Quantitative analysis of Electrolysis - Faraday’s Laws

Faraday’s first law of electrolysis

Mass of substance deposited or liberated at any electrode is directly proportional to the quantity of electricity passed through the electrolyte.

$m=zlt$

$z=$ electrochemical equivalent $={\displaystyle \frac{\text{ atomic mass }}{\mathrm{nF}}}$

$F=$ Faraday constant $=96500\mathrm{C}{\mathrm{moL}}^{-1}$

$I=$ Current in Ampere

$t=$ time in seconds

Electrochemical equivalent

if $I=IA,T=IS$

$m=z$

It is the mass of substance deposited when a current of $1\mathrm{A}$ is passed for 1 second.

Faraday’s second law of electrolysis

When the same quantity of electricity is passed through solutions of different electrolytes connected is series, masses of substances produced at electrodes is directly proportional to their equivalent masses. ${\displaystyle \frac{m_1}{m_2}}={\displaystyle \frac{E_1}{E_2}}$

Solved Examples:

Question 1. During electrolysis of aq. ${\mathrm{CuSO}}_4$, a current of $2\mathrm{A}$ is passed for exactly 2 hours. Write the reactions involved at cathode and anode. What is the amount of the product being formed at the respective electrodes at ${25}^{\circ }\mathrm{C}$ and $1\mathrm{atm}$ pressure?

Question 2. I $\mathrm{L}$ buffer solution $(0.3\mathrm{M}\mathrm{NaH}{\mathrm{HO}}_4$ and $0.3\mathrm{M}{\mathrm{Na}}_2{\mathrm{HPO}}_4)$ was taken in each of the two compartments of an electrolytic cell, with Pt electrodes. The only reaction that can occur is electrolysis of ${\mathrm{H}}_2\mathrm{O}$. The current of $0.965\mathrm{A}$ is passed for exactly $2\mathrm{hrs}$. What is the $\mathrm{pH}$ at cathode and anode? (pKa for ${\mathrm{NaH}}_2{\mathrm{PO}}_4=7.2$ )

Electrochemistry II : Electrolytic Conductance

• Conductance- It is the reciprocal of resistance, i.e. $G={\displaystyle \frac{1}{R}}$ Its units are ${\mathrm{hm}}^{-1}$ or ${\mathrm{\Omega }}^{-1}$ or mho or Siemen(S).

• Specific resistance or Resistivity- If ’ $I$ ’ is the length of a conductor and ’ $A$ ’ is its area of cross section, $R\propto {\displaystyle \frac{1}{A}}$ or $R=\rho {\displaystyle \frac{1}{A}}$ where $\rho$ is called specific resistance or resistivity.

• The units of resistivity are $\mathrm{\Omega }\mathrm{cm}$ or $\mathrm{\Omega }\mathrm{m}$.

$(1\mathrm{\Omega }\mathrm{m}=100\mathrm{\Omega }\mathrm{cm}$ or $\mathrm{\Omega }\mathrm{cm}=0.01\mathrm{\Omega }\mathrm{m}$ ).

• Specific conductivity (or simply called conductivity)- It is the reciprocal of specific resistance,

i.e., $\kappa ={\displaystyle \frac{1}{A}}$ Hence, from $R=\rho {\displaystyle \frac{1}{A}},{\displaystyle \frac{1}{G}}={\displaystyle \frac{1}{\kappa }}{\displaystyle \frac{1}{A}}$ or $\kappa =G\times {\displaystyle \frac{1}{A}}$.

If $I=1\mathrm{cm}$ and $a=1{\mathrm{cm}}^2,\kappa =G$. Hence, conductivity is the conductance of $1{\mathrm{cm}}^3$ of the conductor. Units of $\kappa =0{\mathrm{hm}}^{-1}{\mathrm{cm}}^{-1}$ or $\mathrm{SI}$ units are ${\mathrm{\Omega }}^{-1}{\mathrm{m}}^{-1}$ or $\mathrm{S}\mathrm{m}{\mathrm{m}}^{-1}(1{\mathrm{cm}}^{-1}=100{\mathrm{Sm}}^{-1})$.

• Molar conductivity $\left({\mathrm{\Lambda }}_m\right)$ - of a solution is the conductance of all the ions produced from one mole of the electrolyte dissolved in a given volume of the solution when the electrodes are one $\mathrm{cm}$ apart and the area of the electrodes is so large that the whole of the solution is contained between them.

${\mathrm{\Lambda }}_{\mathrm{m}}={\displaystyle \frac{\kappa \times 1000}{\text{ Molarity }}}$

Sl unit $=\mathrm{S}{\mathrm{m}}^2{\mathrm{mol}}^{-1}$

• Equivalent Conductivity $\left({\mathrm{\Lambda }}_{\mathrm{eq}}\right)$ - Of a solution is defined as the conductance of all the ions produced from one gram equivalent of the electrolyte dissolved in a given volume of the solution when the distance between the electrodes is one $\mathrm{cm}$ and the area of the electrodes is so large that whole of the solution is contained between them.

${\mathrm{\Lambda }}_{\mathrm{eq}}={\displaystyle \frac{\kappa \times 1000}{\text{ Normality }}}$

Unit $=S{\mathrm{m}}^2$ eqvt ${}^{-1}$

Variation of Conductance, Conductivity, Equivalent and Molar conductivities with Dilution

• Effect of dilution- Conductance increases (because total no. of ions increase), specific conductivity decreases (because no. of ions/cc decrease), equivalent and molar conductivity increase with dilution (because it is product of $\kappa$ and $V$ and $V$ increases much more than decrease in the value of $\kappa$ ).

• Variation of molar conductivity with concentration- For a strong electrolyte, it is given by Debye - Huckel - Onsager equation. viz. ${\mathrm{\Lambda }}_m={\mathrm{\Lambda }}_m{}^{\alpha }-A\sqrt{c}$ where $A$ is a constant depending upon the nature of the solvent and temperature, ${\mathrm{\Lambda }}_{\mathrm{m}}{}^{\circ }$ is the molar conductivity at infinite dilution (called limiting molar conductivity). Thus, a plot of ${\mathrm{\Lambda }}_{\mathrm{m}}\mathrm{v}.\sqrt{c}$ will be linear with slope $=-A$. However, some deviation is observed at higher concentration as shown in the fig. (1) below. Further, for a weak electrolyte, ${\mathrm{\Lambda }}_{\mathrm{m}}$ is much less and the increase with dilution is slow in the beginning and then very steep at large dilutions.

• Reasons for increase of ${\mathrm{\Lambda }}_{\mathrm{m}}$ with dilution- Molar conductivity of a strong electrolyte increases with dilution because interionic attractions decrease with dilution. Small deviations at higher concentration are due to large interionic attractions. Molar conductivity of a weak electrolyte increases with dilution because dissociation increases with dilution.

Fig : Variation of ${\mathrm{\Lambda }}_{\mathrm{m}}$ with $\sqrt{\mathrm{C}}$ for strong and weak electrolyte

• Inability to determine ${\mathrm{\Lambda }}_{\mathrm{m}}{}^0$, experimentally for a weak electrolyte- Molar conductivity at infinite dilution $({\mathrm{\Lambda }}_{\mathrm{m}}{}^{\circ }$ or ${\mathrm{\Lambda }}_{\mathrm{m}}{}^{\circ }$ ) for a strong electrolyte can be found by extrapolation to zero concentration but that of weak electrolyte cannot be thus found.

  • Moreover, at infinite dilution though the dissociation is complete, concentration of ions per unit volume is so low that conductivity cannot be measured accurately.
    
    

• Kohlrausch’s Law-

It states that “Equivalent conductivity of any electrolyte at infinite dilution is the sum of the equivalent conductivities of its cations and anions at infinite dilution”.

ie., ${\mathrm{\Lambda }}_{eq}{}^0\equiv {\lambda }_{\mathrm{c}}{}^0+{\lambda }_{\mathrm{a}}{}^0$

or “molar conductivity of an electrolyte at infinite dilution is the sum of the ionic conductivities of the cations and the anions at infinite dilution each multiplied by the number of ions present per formula unit

${\mathrm{\Lambda }}_m{}^0=v_a{\lambda }_a{}^0+v_c{\lambda }_c{}^0$

Applications of Kohlrausch’s Law

a). In calculation of ${\mathrm{\Lambda }}_{\mathrm{m}}{}^0$ or ${\mathrm{\Lambda }}_{\mathrm{eq}}{}^0$ for weak electrolytes, e.g., ${\mathrm{\Lambda }}^{\circ }{}_{{\mathrm{CH}}_3\mathrm{COOH}}={\lambda }_{{\mathrm{CH}}_3{\mathrm{COO}}^{-}}^0+{\lambda }^0{\mathrm{H}}^{+}=$ ${\mathrm{\Lambda }}_{{\mathrm{CH}}_{\mathrm{S}}\mathrm{CONa}\mathrm{a}}^{\circ }+{\mathrm{\Lambda }}_{\mathrm{HCl}}^{\circ }-{\mathrm{\Lambda }}_{\mathrm{NaCl}}^{\circ }$

b). In calculation of degree of dissociation, i.e., $\alpha ={\mathrm{\Lambda }}_m{}^c/{\mathrm{\Lambda }}_m^0$ c). In calculation of dissociation constant by using value of $\alpha$ calculated above ${\mathrm{K}}_{\mathrm{c}}={\displaystyle \frac{c{\alpha }^2}{1-\alpha }}$

d). In calculation of solubility of sparingly soluble salts using the relation, solubility $=1000\kappa /{\mathrm{\Lambda }}_m^{\circ }$.

Solved Examples

Question 3. An aqueous solution of ’ $A$ ’ is slowly added to an aqueous solution of ’ $B$ ‘. The variation in conductivity of these reactions is given below

Match the list I & II,

Question 4. When $4\mathrm{A}$ of current is passed through $0.1\mathrm{M}{\mathrm{Fe}}^{34}(\mathrm{aq})$ solution $(\mathrm{V}=1\mathrm{L})$ for $1\mathrm{hr}$. it is partly reduced to $\mathrm{Fe}(\mathrm{s})$ and partly to ${\mathrm{Fe}}^{2+}(\mathrm{aq})$. How much of ${\mathrm{Fe}}^{2+}(\mathrm{aq})$ is present in solution.

Question 5. What mass of peroxysulphuric acid is formed from electrolytic oxidation of sulphuric acid if the volumes of ${\mathrm{H}}_2$ and ${\mathrm{O}}_2$ gases collected from electrodes at NTP $(0^{\circ }\mathrm{C},1\mathrm{atm})$ are $6.72\mathrm{L}$ and $2.24\mathrm{L}$, respectively. Also, write the reactions occurring at two electrodes.

Question 6. The resistance of a conductivity cell filled with a solution of an electrolyte of concentration $0.1\mathrm{M}$ is $100\mathrm{\Omega }$. The conductivity of this solution is $1.25\mathrm{S}{\mathrm{m}}^{-1}$. Resistance of the same cell when filled with $0.2\mathrm{M}$ of the same solution is $500\mathrm{\Omega }$. What is the molar conductivity of 0.02 M solution of the electrolyte.

Electrochemistry III : Electrochemical Cells

Difference between electrolytic and galvanic cell

Note: The role of salt bridge is to complete the internal circuit and maintain electroneutrality. Even in the absence of salt bridge, the chemical reactions at respective electrodes continue to occur, provided the internal circuit is complete.

Type of Electrodes:

Whether a given electrode acts as anode or cathode depends upon the other electrode with which it is coupled within an electrochemical cell.

In electrochemical series, different half cells have been arranged in the order of increasing standard reduction potential with respect to standard hydrogen electrode.

The measure of ability of an electrode to undergo oxidation is referred to as it oxidation potential.

The measure of ability of an electrode to undergo reduction is referred to as its reduction potential.

A cell can be constructed by coupling two electrodes.

$(\bullet )$ Metal - Metal ion electrode ${\mathrm{M}}^{n+}/\mathrm{M}$ : metal in contact with solution of its ions

${\mathrm{M}}_{(\mathrm{aq})}^{\mathrm{n}+}+\mathrm{ne}\to \mathrm{M}(\mathrm{s})$

Highly reactive metal are taken in the form of amalgam and electrode is referred to as metal (amalgam) - metal ion electrode ${\mathrm{M}}^{n+}\mid \mathrm{M}(\mathrm{Hg})$

$(\bullet )$ Gas - ion electrode : ${\mathrm{H}}^{+}\mid {\mathrm{H}}_2,\mathrm{Pt}$

$(\bullet )$ Redox electrode : insert metal in contact with the solution of ions of an element in two different oxidation states.

$\mathrm{Pt}\mid {\mathrm{Fe}}^{3+},{\mathrm{Fe}}^{2+}$

The only organic electrode known as an example of redox electrode is (Quinhydrone) electrode : an equimolar mixture of quinone and hydroquinone

$\mathrm{Pt}\mid {\mathrm{H}}_2\mathrm{Q},\mathrm{Q},{\mathrm{H}}^{+}$

$(\bullet )$ Metal - Insoluble salt - ion electrode : metal coated with insoluble salt and in contact with solution of its anion. Three most important examples of this category are:

(i) Calomel electrode $\mathrm{Hg}\mid {\mathrm{Hg}}_2{\mathrm{Cl}}_2\mid {\mathrm{Cl}}^{-}$

Paste of $\mathrm{Hg}\mathrm{\&}{\mathrm{Hg}}_2{\mathrm{Cl}}_2$ is known as calomel

(ii) Silver - Silver chloride electrode $\mathrm{Ag}\mid \mathrm{AgCl}\mid Cl{}^{-}$

(iii) mercury - mercurous sulphate electrode $\mathrm{Hg}\mid {\mathrm{Hg}}_2{\mathrm{SO}}_4\mid {\mathrm{SO}}_4{}^{2-}$

Effect of Electrolytic Concentration on Electrode Potential and EMF of a cell : Nernst equation for Electrode Potential:

(i) Write electrode reaction as reduction reaction

${\mathrm{M}}^{n+}+n{\mathrm{e}}^{+}⟶\mathrm{M}$

(ii) Apply Nernst equation,

$E_{\text{electrode }}=E_{\text{electrode }}^0-{\displaystyle \frac{2\cdot 303RT}{nF}}\log {\displaystyle \frac{1}{\left[{\mathrm{M}}^{n+}\right]}}$

At $298\mathrm{K}$,

$E_{\text{electode }}=E_{\text{electrode }}^0-{\displaystyle \frac{0.0591}{n}}\log {\displaystyle \frac{1}{\left[{\mathrm{M}}^{n+}\right]}}$

Nernst equation for EMF of a cell

(i) Write cell reaction

$\text{ e.g., }aA+bB⟶xX+yY$

(ii) Apply Nernst equation,

$E_{\text{cell }}=E_{\text{cell }}^0-{\displaystyle \frac{2.303RT}{nF}}\log {\displaystyle \frac{[\mathrm{X}{]}^{\mathrm{x}}[\mathrm{Y}{]}^{\mathrm{y}}}{[\mathrm{A}{]}^{\mathrm{a}}[\mathrm{B}{]}^{\mathrm{b}}}}$

At $298\mathrm{K}$,

$E_{\text{cell }}=E_{\text{cell }}^0-{\displaystyle \frac{0.0591}{n}}\log {\displaystyle \frac{[X{]}^x[Y{]}^y}{[A{]}^2[B{]}^b}}$

Molar concentration of pure solids, pure liquids and gases at one bar pressure are taken as 1

$(\bullet )$ Relationship between standard EMF ( $E_{\text{cell }}^0)$ and Equilibrium constant $(K)$

For the cell reaction at equilibrium, e.g.,

$\mathrm{Zn}(\mathrm{s})+{\mathrm{Cu}}^{2+}(\mathrm{aq})\rightleftharpoons {\mathrm{Zn}}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})$

$E_{\text{cell }}=0$. Hence,

$0=E_{\text{cell }}^0-{\displaystyle \frac{2.303RT}{nF}}\log K_{\mathrm{c}}$

or $E_{\text{cell }}^{\circ }={\displaystyle \frac{2.303RT}{nF}}\log K_c$

$={\displaystyle \frac{0.0591}{n}}\log K_{\mathrm{c}}\text{ at }298\mathrm{K}$

where $K_c$ is equilibrium constant.

$(\bullet )$ On calculation, if $\log K_{\mathrm{c}}=x$, then $K_c={10}^x$

$(\bullet )$ In terms of natural logarithm, at $298\mathrm{K}$,

$E_{\text{cell }}^0={\displaystyle \frac{0.0257}{n}}\ln K_{\mathrm{c}}(\therefore \ln K_{\mathrm{c}}=2.303\log K_{\mathrm{c}})$

On calculation if $\mathrm{In}{K}_{\mathrm{c}}=x$, then $K_{\mathrm{c}}=e^{\mathrm{x}}$.

$(\bullet )$ Relation between Gibbs energy change $\left(\mathrm{\Delta }{G}^0\right)$ and $\mathrm{EMF}\left(E_{\text{cell }}^0\right)$

$-\mathrm{\Delta }G=nF{E}_{\text{cell }}$ or $\mathrm{\Delta }{G}^0=(-)nF{E}_{\text{cell }}^0$

Relation between the cell reaction and the Gibb’s energy change : $\mathrm{\Delta }{G}^0=-nF{E}^0$

$E_{\text{cel }}^0$ is an intensive property and thus its value does not depend upon the size of the cell and the extent to which the cell reaction is carried out. If the overall reaction

${\mathrm{Fe}}^{3+}+{\mathrm{e}}^{-}\rightleftharpoons {\mathrm{Fe}}^{2+}$

is written as

$2{\mathrm{Fe}}^{3+}+2{\mathrm{e}}^{-}\rightleftharpoons 2{\mathrm{Fe}}^{2+}$

Then $E^0$ remains the same but $\mathrm{\Delta }{G}^0$ changes as there is a change in the number of electrons involved in the two reactions.

Putting $E_{\text{cell }}^0={\displaystyle \frac{RT}{nF}}\ln K$

$\mathrm{\Delta }{G}^0=-RT$ in $K=-2.303RT\log K$.

Further, decrease in Gibb’s energy = Maximum electrical work done. Hence, maximum electrical work done $=-nF{E}_{\text{cell }}$

Concentration cell- If two half cells are of the same type differing only in the concentration of ions, it is called electrolyte concentration cell,

e.g., $\mathrm{Zn}\left|{\mathrm{Zn}}^{2+}\left({\mathrm{c}}_1\right)|{\mathrm{Zn}}^{2+}\left({\mathrm{c}}_2\right)\right|\mathrm{Zn}$,

For such a cell, $E_{\text{cell }}={\displaystyle \frac{0.0591}{n}}\log {\displaystyle \frac{c_2}{c_1}}$ where $c_2>c_1$.

Commercial cell / batteries- These are of three types

(i) Primary cells which cannot be recharged.

(ii) Secondary cells which can be recharged

(iii) Fuel cells in which redox reaction is combustion of a fuel (e.g. ${\mathrm{H}}_2,{\mathrm{CH}}_4$ )

I Primary cells

(a) Dry cell :

Anode $=$ zinc cylinder,

Cathode $=$ Graphite rod surrounded by ${\mathrm{MnO}}_2+\mathrm{C}+{\mathrm{NH}}_4\mathrm{Cl}+{\mathrm{ZnCl}}_2$ acting as electrolyte.

Anode reaction : $\mathrm{Zn}⟶{\mathrm{Zn}}^{2+}+2\mathrm{e}$,

Cathode reaction : $2{\mathrm{MnO}}_2+2{\mathrm{NH}}_4^{+}+2\mathrm{e}⟶{\mathrm{Mn}}_2{\mathrm{O}}_3+2{\mathrm{NH}}_3+{\mathrm{H}}_{2}0.E_{\text{cell }}=1.25-1.50\mathrm{V}$.

They do not have long life because ${\mathrm{NH}}_4\mathrm{Cl}$ corrodes zinc container

(b) Mercury Cell :

Anode $=\mathrm{Zn}$ container, Cathode $=$ Carbon rod.

Electrolyte $=$ Paste of $\mathrm{HgO}+\mathrm{KOH}$.

Cell reaction : $\mathrm{Zn}+\mathrm{HgO}⟶\mathrm{ZnO}+\mathrm{Hg}$. $E_{\text{cel }}=1.35\mathrm{V}$.

These are used in hearing aids and watches.

II Secondary Cells

a) Lead stroage battery :

Anode $=$ Lead, Cathode $=$ grid of $\mathrm{Pb}$ packed with ${\mathrm{PbO}}_2$

Electrolyte $={\mathrm{H}}_2{\mathrm{SO}}_4(38\mathrm{\%}).E_{\text{een }}=2\mathrm{V}$

During discharge, Anode reaction : $\mathrm{Pb}+{\mathrm{SO}}_4^{2\cdot }⟶{\mathrm{PbSO}}_4+2{\mathrm{e}}^{-}(\mathrm{ox})$

Cathode reaction: ${\mathrm{PbO}}_2+{\mathrm{SO}}_4{}^2+4{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}⟶{\mathrm{PbSO}}_4+2{\mathrm{H}}_2\mathrm{O}$ (red)

During recharging, Cathode reaction : ${\mathrm{PbSO}}_4+2\mathrm{e}⟶\mathrm{Pb}+{\mathrm{SO}}_4{}^{2-}$ (red)

Anode reaction : ${\mathrm{PbSO}}_4+2{\mathrm{H}}_2\mathrm{O}⟶{\mathrm{PbO}}_2+{\mathrm{SO}}_4{}^{2-}+4{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}(\mathrm{ox})$

Note : Nature of an electrode of a secondary cell is opposite during discharge and recharge. The negative electrode which acts as anode during discharge but becomes cathode during recharge. Similarly, the positive electrode acts as cathode during discharge but becomes anode during recharge.

b) Ni-Cd storage cell :

Anode $=\mathrm{Cd}$, Cathode $={\mathrm{NiO}}_2$,

Electrolyte $=\mathrm{KOH}$ sol. $E_{\text{ceu }}=1.4\mathrm{V}$.

Anode reaction : $\mathrm{Cd}+2{\mathrm{OH}}^{-}⟶\mathrm{Cd}(\mathrm{OH}{)}_2+2{\mathrm{e}}^{-}$

Cathode reaction: ${\mathrm{NiO}}_2+2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-}⟶\mathrm{Ni}(\mathrm{OH}{)}_2+2{\mathrm{OH}}^{-}$

III Fuel Cells,

e.g., ${\mathrm{H}}_2-{\mathrm{O}}_2$ fuel cell,

Electrolyte $=$ conc. $\mathrm{KOH}$ sol. $E_{\text{cell }}=1.23\mathrm{V}$ (theoretical)

Anode reaction : $2{\mathrm{H}}_2+4{\mathrm{OH}}^{-}⟶4{\mathrm{H}}_2\mathrm{O}+4{\mathrm{e}}^{-}$,

Cathode reaction: ${\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{O}+4{\mathrm{e}}^{-}⟶4{\mathrm{OH}}^{-}$,

Efficiency of fuel cell $={\displaystyle \frac{\mathrm{\Delta }G}{\mathrm{\Delta }H}}\times 100={\displaystyle \frac{-n{\mathrm{EF}}_{\text{cell }}}{\mathrm{\Delta }\mathrm{H}}}\times 100={\displaystyle \frac{-2\times 96500\times 1.231\mathrm{J}}{-285800\mathrm{J}}}\times 100=83\mathrm{\%}$

Corrosion

Corrosion is the process of oxidation which results in change of metal surface into salts like oxides, sulphides, carbonates etc. due to attack of atmospheric gases.

Rust- Chemically, it is hydrated ferric oxide, ${\mathrm{Fe}}_2{\mathrm{O}}_3\bullet \times {\mathrm{H}}_2\mathrm{O}$

Theory of rusting: Rusting of iron can be explained on the basis of Electrochemical theory as follows :-

$\underset{\text{(layer on the surface)}}{{\mathrm{H}}_2\mathrm{O}}+\underset{\text{(from air)}}{{\mathrm{CO}}_2}⟶{\mathrm{H}}_2{\mathrm{CO}}_3$

${\mathrm{H}}_2{\mathrm{CO}}_3\rightleftharpoons 2{\mathrm{H}}^{+}+{\mathrm{CO}}_3{}^{2-}$

At Anode:

$2\mathrm{Fe}⟶2{\mathrm{Fe}}^{2+}+2\mathrm{e}$

At Cathode:

${\mathrm{H}}^{+}+{\mathrm{e}}^{-}⟶\mathrm{H}$

$4\mathrm{H}+\underset{\text{ (Dissolved) }}{{\mathrm{O}}_2}2{\mathrm{H}}_2\mathrm{O}$

or $4{\mathrm{H}}^{+}+{\mathrm{O}}_2+4{\mathrm{e}}^{-}⟶2{\mathrm{H}}_2\mathrm{O}$

Further, ferrous ions formed at the anode react with the dissolved ${\mathrm{O}}_2$ to from ${\mathrm{Fe}}_2{\mathrm{O}}_3$.

$4{\mathrm{Fe}}^{2+}+\underset{\text{ (Dissolved) }}{{\mathrm{O}}_2+4{\mathrm{H}}_2\mathrm{O}}⟶{\mathrm{Fe}}_2{\mathrm{O}}_3+8{\mathrm{H}}^{+}$

${\mathrm{Fe}}_2{\mathrm{O}}_3+{\mathrm{xH}}_2\mathrm{O}⟶{\mathrm{Fe}}_2{\mathrm{O}}_3.{\mathrm{H}}_2\mathrm{O}$

Factors which enhance corrosion-

(i) Presence of impurities in the metal (pure metals do not corrode)

(ii) Presence of moisture (e.g., in rainy season)

(iii) Presence of electrolytes (e.g., saline water)

Prevention of corrosion-

Corrosion can be prevented by the following methods:

(i) Barrier protection by oil/grease layer, paints or electroplating.

(ii) Sacrificial protection by coating the metal with more electropositive metal e.g., Zn (called galvanisation)

(iii) Electrical protection by connecting the iron pipe to a more electropositive metal (like $\mathrm{Mg}$ ) with a wire.

Question 6. The following electrochemical cell has been set up

Pt (1) | ${\mathrm{Fe}}^{3+},{\mathrm{Fe}}^{2+}(\mathrm{a}=1)||{\mathrm{Ce}}^{4+},{\mathrm{Ce}}^{3+}(a=1)|\mathrm{Pt}(2)$

Predict the direction of flow of current

(given : ${\mathrm{E}}_{{\mathrm{Ce}}^{4+1}1{\mathrm{Ce}}^{3+}}^0=1.61\mathrm{V};{\mathrm{E}}_{{\mathrm{Fe}}^{3+}{\mathrm{Fe}}^{2+}}^0=0.77\mathrm{V}$ )

Question 7. Which of the following reactions are possible:

a) Reduction of ${\mathrm{Fe}}^{3+}$ to ${\mathrm{Fe}}^{2+}$ by $\mathrm{Fe}$

b) Reduction of ${\mathrm{Fe}}^{3+}$ to ${\mathrm{Fe}}^{2+}$ by ${\mathrm{Mn}}^{2+}$

given $E_{{\mathrm{Fe}}^{2+},{\mathrm{Fe}}^{3+}{\mathrm{Pt}}^{-1}}^0=0.77\mathrm{I}\mathrm{V},E_{{\mathrm{Fe}}^{2+}{\mathrm{Fe}}^{\mathrm{Fe}}}^0=-0.440\mathrm{V}$

$E_{{\mathrm{Mn}}_4;{\mathrm{Mn}}^{2+},{\mathrm{H}}^{+}\mid \mathrm{Pt}}^0=1.51\mathrm{V}$

Question 8. Compute ${\mathrm{E}}^0$ for the reaction

${\mathrm{Fe}}^{3+}+{\mathrm{e}}^{-}\rightleftharpoons {\mathrm{Fe}}^{2+}$

Given that

$\begin{array}{ll}{\mathrm{Fe}}^{3+}+3{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Fe}& E_{{\mathrm{Fe}}^{3+}\mid \mathrm{Fe}}^0=-0.036\mathrm{V}\\ {\mathrm{Fe}}^{2+}+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Fe}& E_{{\mathrm{Fe}}^{2+}{}_{\mathrm{Fe}}}^0=-0.440\mathrm{V}\end{array}$

Question 9. A silver electrode is immersed in saturated ${\mathrm{Ag}}_2{\mathrm{SO}}_4(\mathrm{aq})$.

The potential difference between silver and standard hydrogen electrode is found to be $0.711\mathrm{V}$. Determine ${\mathrm{K}}_{\mathrm{sp}}\left({\mathrm{Ag}}_2{\mathrm{SO}}_4\right)$.

(Given $E_{{\mathrm{Ag}}^{+}\mid \mathrm{Ag}}^{\circ }=0.799\mathrm{V}$ )

Question 10. During the discharge of a lead storage battery, density of ${\mathrm{H}}_2{\mathrm{SO}}_4$ fell from 1.29 to $1.14\mathrm{g}/\mathrm{mL}$. The initial solution of sulphuric acid is $39$ by weight and later it becomes $20$ by weight. The battery holds $3\mathrm{L}$ of the acid and volume remains practically constant during discharge. Calculate ampere-hour for which battery must have been used.