CHEMISTRY MASTER: +2 UNIT 5 PAGE- 2

KOHLRAUSCH'S LAW

The molar conductivity, L_m of electrolytes increases with increase in dilution till a limiting value L_m^ois obtained at infinite dilution. In this limit, the positive and negative ions may be thought of as behaving essentially independent of each other. Based up on his exhaustive experimental studies, Kohlrausch gave a generalisation, which is known as Kohlrausch law. It states that at infinite dilution, when the dissociation of the electrolyte is complete, each ion makes a definite contribution towards the molar conductivity of electrolyte, irrespective of the nature of the other ion with which it is associated.

This implies that the molar conductivity of an electrolyte at infinite dilution can be expressed as the sum of the contributions from its individual ions. If l₊^oand l_-^orepresent the molar conductivities of cation and anion respectively at infinite dilution, then the molar conductivity of electrolyte at infinite dilution L_m^o is given by :

where n₊ and n_- represent the number of positive and negative ions furnished by each formula unit of the electrolyte.

For example :

i) One formula unit NaCl furnishes one Na⁺ and one Cl^- ion, therefore,

ii) One formula unit of BaCl₂ furnishes one Ba²⁺ and two Cl^- ions. Therefore,

iii) One formula unit of Na₂SO₄ furnishes two Na⁺ ions and one SO₄²^- ions, therefore ,

In terms Of Equivalent Conductance , Kohlrausch's Law is stated as :

The equivalent conductance of an electrolyte at infinite dilution is the sum of two values - one depending on the cation and other depending on anion.

Mathematically ,

L⁰_eq = l⁰c + l⁰a

where l⁰c and l⁰a are called the ionic conductances at infinite dilution for cation and anion respectively.

The ionic conductivities at infinite dilution for some ions at 298 K are given in the following Table.

TABLE

Ionic conductivities at Infinite dilution at 298 K

Cation	l⁰(S cm² mol^-¹)	Anion	l⁰(Scm² mol^-¹)
H⁺	349.6	OH^-	199.1
K⁺	73.5	Cl^-	76.3
Na⁺	50.1	Br^-	78.1
Ag⁺	61.9	I^-	76.80
NH₄⁺	53.0	NO₃^-	71.40
Cu²⁺	108.0	CH₃COO^-	40.90
Mg²⁺	106.0	SO₄²^-	160.0

Applications of Kohlrausch’s law

Some of the important applications of Kohlrausch's law are :

i. Calculation of Molar conductivities of weak electrolytes at infinite dilution

L_m⁰for weak electrolytes cannot be obtained directly by the extrapolation of plot of L_m versus ÖC .The limiting molar conductivities of weak electrolytes, L_m⁰ can be easily calculated with the help of Kohlrausch's law.

For example, the value of L_m⁰for acetic acid can be calculated from the knowledge of molar conductivities at infinite dilution of strong electrolytes like CH₃COONa, HCl and NaCl as follows :

ii. Calculation of degree of dissociation of weak electrolytes

Molar conductivity of an electrolyte depends upon its degree of dissociation. Higher the degree of dissociation, higher is the molar conductivity. As dilution increases, the degree of dissociation of weak electrolyte also increases and consequently, molar conductivity of electrolyte increases. At infinite dilution molar conductivity becomes maximum because degree of dissociation approaches unity.

Thus, if

Problems

10. Calculate L^o for CaCl₂ and MgSO₄ from the given data :

l^o (Ca²⁺) = 119.0 S cm²mol^-¹, l^o (Cl^-) = 76.3 S cm²mol^-¹. l^o (Mg²⁺) = 106.0 S cm²mol^-¹ and l^o (SO₄²^-) = 160.0 S cm²mol^-¹.(NCRT 90)

11. L^o for NaCl, HCl and NaAc are 126.4, 425.9 and 91.0 S cm²mol^-¹ respectively. Calculate L^o for HAc.

12. At 25⁰C the equivalent conductance of 0.001 M acetic acid is 49.5 ohm^-¹cm²equiv^-¹. The maximum value of equivalent conductance is 390 ohm^-1cm²equiv^-¹. Calculate the degree of ionisation of 0.001 M acetic acid at 25⁰C.

13. The molar conductivities at infinite dilution of HCl, NaCl and CH₃COONa are 380.5, 109.0 and 78.5 ohm^-¹cm²mol^-¹respectively. Calculate the molar conductivity at infinite dilution of acetic acid.

14. At 18⁰C , the equivalent conductivities at infinite dilutions of NH₄Cl, NaOH and NaCl are 129.8, 217.4 and 108.9 ohm^-¹cm²equiv^-¹respectively. If the equivalent conductivity of 0.01 M solution of NH₄OH is 9.33 ohm^-1cm²equiv^-1. Calculate the degree of dissociation of ammonium hydroxide at this dilution.

15. The molar conductivity at infinite dilution of Al₂(SO₄)₃ is 858 S cm²mol^-¹, while that of sulphate ion is 160 S cm²mol^-¹. Calculate the value of ionic conductance of Al³⁺ ion.

16. Calculate L_m⁰for NH₄OH, given L_m⁰for Ba(OH)₂, BaCl₂ and NH₄Cl as 523.28, 280.0 and 129.8 S cm²mol^-¹respectively.

17. The molar conductance of acetic acid at 298 K at concentrations of 0.1 M and 0.001 M are 5.20 and 49.2 S cm²mol^-¹respectively. Calculate the degree of dissociation of acetic acid at these concentrations. Given l⁰(H⁺) and l⁰(CH₃COO^-) as 349.1 and 40.9S cm²mol^-¹.

18. The resistance of 0.01 M acetic acid solution when measured in a conductivity cell of cell constant 0.366 cm^-¹, is found to be 2220 ohm. Calculate the degree of dissociation of acetic acid at this concentration. Also find the dissociation constant of acetic acid. Given that the value of l⁰(H⁺) and l⁰(CH₃COO^-) as 349.1 and 40.9 S cm²mol^-¹.

GALVANIC CELLS

Galvanic cell is the device in which chemical energy is converted into electrical energy. It is also called electrochemical or voltaic cell. In a Galvanic cell, a redox reaction is carried out in an indirect manner and the decrease in free energy during the chemical process appears as electrical energy. An indirect redox reaction is such that reduction and oxidation processes are carried out in separate vessels. In order to understand this phenomenon, let us consider Zn-CuSO₄ reaction as the basis of the cell reaction.

In its simple form, a zinc strip is dipped in the ZnSO₄solution and the copper strip is dipped in CuSO₄solution taken in separate beakers. The metallic strips which acts as electrodes are connected by conducting wire through a voltmeter. The two solutions are joined by an inverted U-tube known as salt bridge. The U-tube is filled with the solution of some electrolyte such as KCl, KNO₃ or NH₄Cl to which gelatin or agar-agar has been added to convert it into a semi-solid paste. A schematic diagram of this cell has been shown in Fig 3.

Fig 3. Electrochemical Cell.

The deflection of the voltmeter indicates that there is a potential difference between the two electrodes. It has been found that the conventional current flows through the outer circuit from copper to zinc strip. It implies that the electron flow occurs from zinc to copper strip. This cell converts the chemical energy liberated during the reaction Zn(s) + Cu²⁺(aq) ® Zn²⁺ (aq) + Cu(s) to electrical energy and it has an electric potential equal to 1.1 V when activity (nearly equal to molarity) of Cu² and Zn²⁺ ion is unity. Such a device is called galvanic cell or voltaic cell.

If an external opposite potential is applied [Fig a ] and increased slowly, we find that the reaction continues to take place till opposing voltage reaches the value 1.1 V [(Fig b] when the reaction stops altogether and current flows through the cell. Further increase in external potential now again starts the reaction but in opposite direction [Fig c] . It now functions as an electrolytic cell ; a device for using electrical energy to carry non-spontaneous chemical reactions. Both types of cells are quite important.

Fig (a) When E_ext< 1.1 V

(i) Electrons flow from Zn to Cu rod and hence current flows from Cu to Zn.

(ii) Zn dissolves at anode and copper deposits at cathode.

Fig (b) When Eext = 1.1 V

(i) No flow of electrons or current.

(ii) No chemical reaction.

Fig (c) When E_ext > 1.1 V

(i) Electrons flow from Cu to Zn current from Zn to Cu .

(ii) Zinc is deposited at the zinc electrode and copper dissolves at copper electrode.

Working of the cell

i) Zinc undergoes oxidation to form zinc ions.

Zn(s) ® Zn²⁺⁽aq) + 2 e^- (oxidation)

ii) The electrons liberated during oxidation are pushed through

the connecting wires to copper strip.

iii) Copper ions move towards copper strip, pick up the electrons and get reduced to copper atoms which are deposited at the copper strip.

Cu²⁺⁽aq) + 2 e^- ® Cu(s)

The electrode at which oxidation occurs is anode and that at which reduction occurs is cathode. In the above cell, zinc strip is anode and copper strip is cathode. Due to the oxidation process occurring at the anode it becomes a source of electrons and acquires a negative charge in the cell. Similarly, due to reduction process occurring at the cathode it acquire positive charge and becomes a receiver of the electrons. Thus in the electrochemical cell,

Anode constitutes - ve terminal and

Cathode constitutes + ve terminal.

Functions of Salt bridge

In the electrochemical cell a salt bridge serves two important functions.

i) It allows the flow of current by completing the circuit.

ii) It maintains electrical neutrality.

The transference of electrons from anode to cathode leads to a net positive charge around the anode due to increase in the concentration of cations and a net negative charge around the cathode due to deposition of positive ions at the cathode. The positive charge around the anode will prevent electrons to flow from it and the negative charge (due to excess of anions in solution) collected around the cathode will prevent the acceptance of electrons from the anode. The reaction, thus, stops and no current will flow. The salt bridge comes to aid and restores the electro-neutrality of the solutions in the two compartments. It contains a concentrated solution of an inert electrolyte the ions of which are not involved in electrochemical reactions. The anions of the electrolyte in the salt bridge migrates to the anode compartment and cations to the cathode compartment. Thus, the salt bridge prevents the accumulation of charges and maintains the flow of current.

Representations of Galvanic cells

Galvanic cell is a combination of two half-cells, viz., oxidation half-cell and reduction half-cell. If M represents the symbol of the element and Mⁿ⁺ represents its cation in solution, then :

Oxidation half cell is represented as : M | Mⁿ⁺ (C)

Reduction half cell is represented as : Mⁿ⁺(C)| M

In both notations, C represents the concentration of the ions in solution. Conventionally, a cell is represented by writing the cathode on the right hand side and anode on the left hand side. Two vertical lines are put between the two half-cells which indicate the salt bridge.

Examples :

Zn | Zn²⁺ || Cu²⁺ | Cu Net Cell reaction:

Zn(s) + Cu²⁺(aq) ® Zn²⁺(aq) + Cu(s)

Cu|Cu²⁺(1M) ||Ag⁺(1M) |Ag

Cu(s)+ 2 Ag⁺(aq) ® 2 Ag(s) + Cu²⁺(aq)

Ni|Ni²⁺(1M) ||Ag⁺(1M) |Ag

Ni(s) + 2 Ag⁺(aq) ® 2 Ag(s) + Ni²⁺(aq)

ELECTRODE POTENTIAL

When a strip of metal (M) is brought in contact with the solution containing its own ions (Mⁿ⁺), then either of the following three processes can take place :-

1) The metal ion Mⁿ⁺may collide with metallic strip and bounce back without any change.

ii) The metal ion Mⁿ⁺may collide with the strip , gain electrons and get converted into the metal atom i.e., the ion is reduced.

Mⁿ⁺ + n e^- ® M

iii) The metal atom on the strip loses 'n' electrons and enter the solution as Mⁿ⁺ ion, i.e., metal is oxidised.

M ® Mⁿ⁺ + n e^-

The above changes have been shown in Fig 4.

Fig 4 Electrode equilibrium

If the metal has a relatively high tendency to get oxidised , its atoms would start losing electrons, change into positive ions and pass into the solution. The electrons lost accumulate in the metal strip and cause it to develop negative charge. The negative charge developed on the strip does not allow metal atoms to continue losing electrons, but it would re-attract the metal ions from the solution in an attempt to neutralise its charge. Ultimately, a state of equilibrium will be established between metal and its ions at the interface.

Similarly, if the metal ions have relatively greater tendency to get reduced , they will accept electrons from the metal strip and consequently, a net positive charge is developed on the metal strip. Ultimately, a similar equilibrium is established between the metal ions and metal atoms at the interface.

The equilibrium between the metal and its ions leads to the separation of charges. This results in a potential difference set up between the metal and solution of its ions is known as half Cell Electrode Potential. In fact, it is the measure of the tendency of an electrode in a half cell to lose or gain electrons.

According to the present convention, the half-cell reactions and their potentials are represented by reduction potentials. It may be noted that :

i) Reduction potential(tendency to gain electrons) and oxidation potential(tendency to lose electrons) of an electrode are numerically equal but have opposite signs.

ii) Reduction potential increases with increase in the concentration of the ions and decrease in concentration of the ions in solution.

iii) The reduction potential of the electrode, when the concentration of the ions in solution is 1 mol L^-¹ and temperature is 298 K is called standard reduction potential and is represented as E^o_red.

EMF OF THE CELL

The electrochemical cell consists of two half-cells. The electrodes in these half-cells have different electrode potentials. When the circuit is completed the loss of electron occurs at the electrode having lower potentials, whereas the gain of electrons occurs at the electrode with higher reduction potential. The difference in the electrode potentials of the two electrodes of the cell is termed as electromotive force (EMF) or Cell voltage.

Mathematically, it can be expressed as :

EMF = E_Red(cathode) - E_Red(anode)

or simply as :

EMF = E_cathode - E_anode

Since in the representation of a cell, the cathode is written on the right hand side and the anode on the left hand side, therefore EMF of a cell is also sometimes written as :

EMF = E_Right - E_Left

EMF = E_R- E_L

EMF of the cell may be defined as the potential difference between the two terminals of the cell when either no or very little current is drawn from it. It is measured with the help of a potentiometer or vaccum tube voltmeter.

STANDARD ELECRODE POTENTIALS

Absolute value of electrode potential of an electrode cannot be determined because a half-cell by itself cannot cause the movement of charges(flow of electrons). It is due to the fact that once equilibrium is reached between the electrode and solution in a half cell , no further displacement of charges can occur unless and until it is connected to another half cell with different electrode potential. This difficulty is overcome by finding the electrode potentials of various electrodes relative to reference electrode whose electrode potential is arbitrarily fixed. The common reference electrode used for this purpose is Standard Hydrogen Electrode (SHE) or Normal Hydrogen Electrode (NHE), whose electrode potential is arbitrarily taken to be zero.

NORMAL HYDROGEN ELECTRODE

Normal hydrogen electrode consists of a platinum wire sealed into a glass tube carrying a platinum foil at one end. The platinum foil is coated with finely divided platinum. The electrode is placed in a beaker containing H⁺ ions. Hydrogen gas at 1 atmosphere pressure is continuously bubbled through the solution at temperature of 298 K. The oxidation or reduction in the NHE takes place at the platinum foil. Hence it can act as anode as well as cathode and may be represented as :

Pt, ½ H₂ (1atm)½H⁺(1M) or

H⁺(1M) ½ ½ H₂ (1atm), Pt respectively.

If NHE acts as anode, then oxidation will take place at it as :

H₂(g) ⇌ 2 H⁺(aq) + 2 e^-

If NHE acts as cathode then reduction will take place at it as

2 H⁺(aq) + 2 e^-⇌ H₂(g)

Fig 5 The normal hydrogen electrode

The electrode potentials of other electrodes are determined by coupling them with NHE. The electrode potential of an electrode determined relative to the standard hydrogen electrode under standard conditions is called standard Electrode Potential. It is represented by E⁰. The standard conditions are 1M concentration of ions at 298 K temperature and 1 atmospheric pressure.

Measurement of Standard Electrode Potentials

In order to find the standard electrode potential of an electrode, the electrode in the standard conditions is connected to normal hydrogen electrode to constitute a cell. The potential difference between the electrodes is determined with the help of potentiometer or vaccum tube voltmeter.

E⁰_cell= E⁰_cathode- E⁰_anode

Knowing E⁰_cell, and the electrode potential of one of the electrode(NHE) that of the other can be calculated.

For example, in order to find out the standard electrode potential of zinc electrode containing 1M concentration of Zn²⁺ ions , it is connected to NHE (Fig 6) .

Fig 6 Measurement of standard electrode potential

of Zinc electrode

The voltmeter reading shows the potential difference (E⁰_cell) of 0.76 V. The electron flow is found to be from zinc electrode to NHE. From this, it follows that, Zn acts as anode while NHE acts as cathode. The net cell reaction is :

Zn(s) + 2 H⁺(aq) ® Zn²⁺(aq) + H₂(g)

Now, E⁰cell = E⁰cathode - E⁰anode

0.76V = E⁰ _H⁺_/½_H2 - E⁰_Zn²⁺_/Zn

= 0 - E⁰Zn²⁺/Zn

E^o_Zn²⁺_{/ Zn} = - 0.76 V

When standard copper electrode is coupled with NHE (Fig 7) the voltmeter reading shows a potential difference (E⁰_cell)of 0.34 V. The electron flow is found to occur from NHE towards copper electrode. Therefore NHE in this cell acts as anode and copper electrode acts as cathode.

Fig 7 Measurement of Standard electrode

potential of Cu electrode

The cell reaction is :

Cu²⁺(aq) + H₂(g)® Cu(s) + 2 H⁺(aq)

E⁰_cell= E⁰_cathode- E⁰_anode

E⁰_cell= E⁰_Cu²⁺_/Cu- E⁰_H⁺_/_½_H2

0.34V = E⁰_Cu²⁺_{/ Cu}- 0

E⁰_Cu²⁺_/Cu= 0.34 V

The reduction potentials of other electrodes including metals as well as non-metals can be determined in a similar way. For example, the standard electrode potential of chlorine can be determined by using electrode consisting of chlorine gas at one atmospheric pressure and 298 K in equilibrium with 1 M solution of chloride ions. For metals of variable valence, the redox potential of one ion in equilibrium with other ion of different charge can also be determined. For example, redox potential of Fe³⁺(aq), Fe²⁺(aq)|Pt can be determined by using the cell,

Pt, ½ H₂(g) ½H⁺(1M)║Fe³⁺(1M),Fe²⁺(1 M) ½Pt

Note

It is not always convenient to use NHE electrode as a reference electrode because of the following difficulties.

i) Unity concentration of H⁺ ions is difficult to be maintained.

ii) 1 atmospheric pressure of H₂ gas cannot be maintained

uniformly.

iii) The hydrogen electrode gets poisoned even if traces of

impurities are present.

To overcome these difficulties some other electrodes like calomel electrode or silver-silver chloride are used as reference electrodes. The standard electrode potentials of these electrodes are given below :

i) Calomel electrode : Hg-Hg₂Cl₂| KCl(aq) : E_RED= 0.2422V

ii) Silver-Silver chloride electrode :

Ag-AgCl(s) | KCl(aq) : E_RED= 0.2225V

ELECTROCHEMICAL SERIES

This is a list in which some half-cell reactions and their reduction potentials are properly arranged in the increasing order from top to bottom. The electrochemical series is also called activity series. The table is so arranged that oxidising agents are listed in the increasing order of their strength and reduction potential. The electrochemical series is given in the following Table.

Applications of Electrochemical series

The applications of the electrochemical series are given below

1. Relative strengths of oxidising agents

With the help of the series , we can predict the relative oxidising or reducing strengths of substances. In the electrochemical series the substances are arranged in the increasing order of reduction potentials, i.e., increasing tendency for reduction to occur or as power as oxidising agent or decreasing tendency for oxidation to occur or power as reducing agent. For example, zinc comes above iron in the table and has less reduction potential and therefore, zinc is stronger reducing agent than iron. But copper comes below iron in the table and has more reduction potential and therefore iron is stronger reducing agent than copper or copper is stronger oxidising agent than iron. Thus, the substances which have lower reduction potentials are stronger reducing agents while those which have higher reduction potentials are stronger oxidising agents.

Standard Reduction Electrode Potentials at 298 K

2. Calculation of the E.M.F of the cell

The E.M.F of the cell which is the difference between the reduction potential of the cathode and anode and is determined by the following steps:

Step I : Write the two half-cell reactions in such a way that the reaction taking place at the left hand electrode is written as an oxidation reaction and that taking place at the right electrode is written as reduction.

Step II : Multiplying one of the equations if necessary by suitable number to equate the number of electrons in the two equations. However, it may be noted that electrode potential, E⁰ are not multiplied.

Step III : The E.M.F. of the cell (E⁰cell )is equal to the difference between the standard reduction potential of the right electrode (E⁰_R ) and standard electrode potential of the left electrode (E⁰_L ). Thus,

E⁰_cell= E⁰_R- E⁰_L

Step IV : If the E.M.F of the cell is +ve, the reaction is feasible in the given direction. But if E.M.F of the cell is- ve, the cell reaction is not feasible in the given direction. The reaction must be occurring in the reverse direction.

3. Predicting the feasibility of a redox reaction

The electrochemical series helps in finding out whether a given redox reaction is feasible or not in the given direction from the E⁰values of the two electrodes. In general , a redox reaction is feasible only if the species that has higher reduction potential is reduced i.e., accepts electrons and the species which has lower reduction potential is oxidised, i.e., loses the electrons. Otherwise, a redox reaction is not feasible. In other words, the species to release electrons must have lower reduction potential as compared to the species which is to accept electrons.

Illustration

Consider the redox reaction :

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

E⁰ value of Cu = +0.34 V and E⁰ value of Ag=+0.80 V. Since the reduction potential of Ag is more than that of Cu, this means that silver has greater tendency to get reduced in comparison to copper. Thus, the reaction

Ag⁺(aq) + e^- ® Ag(s)

occurs more readily than the reaction :

Cu²⁺(aq) + 2 e^- ® Cu(s)

Similarly, since the reduction potential of copper is less than that of Ag, this means that copper will be oxidised in comparison to Ag. Thus, the reaction:

Cu(s) ® Cu²⁺(aq) + 2 e^-

occurs more readily than

Ag(s) ® Ag⁺ + e^-

Therefore , silver will be reduced and copper will be oxidised and the above reaction is not feasible. Rather the reverse reaction:

Cu(s) + Ag⁺(aq) ® Cu²⁺(aq) + 2 Ag(s)

can occur.

4. To Predict whether a metal can liberate hydrogen from acid or not

The electrochemical series can also be used to predict whether a metal can liberate hydrogen from acid or not. While the metals like zinc, magnesium and nickel can liberate hydrogen from acids like HCl, H₂SO₄ etc. metals like copper and silver cannot do so. In general, only those metals can liberate hydrogen from the acid which have negative values of reduction potentials i.e., - E⁰values.

M ® M⁺(aq) + e^-

(monovalent)

H⁺(aq) + e^-® ½ H₂

Problems

19. What happens when, if copper sulphate solution is stored in a zinc vessel and why ?

20. Can 1 M solution of Ferrous sulphate be stored in a nickel vessel ?

21. Standard reduction potentials E⁰for the electrode reaction in water at 298 K for a few systems are given below :

Electrode E⁰(volts)

Zn²⁺; Zn -0.76

Cu²⁺; Cu +0.34

Ag⁺; Ag +0.80

From the data , calculate the standard cell potentials

for the following galvanic cells.

i) Zn½ Zn²⁺(1M)½½ Cu²⁺ (1M)½Cu

ii) Ag½ Ag⁺(1M)½½ Cu²⁺ (1M)½Cu

22. Using standard electrode potentials , predict the reactions, if any, that occurs between the following :

a) Fe³⁺(aq) and I^-(aq)

b) Ag⁺(aq) and Cu(s)

c) Fe²⁺(aq) and Br₂⁰ (aq)

d) Ag(s) and Fe³⁺(aq)

e) Br^-(aq) and Fe³⁺(aq)

23. Arrange the following metals in the order in which they displace each other from their salt solutions. Al, Cu, Fe, Mg, Zn.

24. Electrode potentials of the metals in their respective solutions are provided. Arrange them in their increasing order of reducing power :

K⁺½K = -2.93 V : Ag⁺½Ag= +0.80 V

Hg⁺½Hg = +0.79 V : Mg²⁺½Mg = -2.37 V

Cr³⁺½Cr = -0.74 V

28. The following reaction :

Zn(s) + Co²⁺(aq) ® Zn²⁺(aq) + Co(s)

occurs in a cell. Write the electrode reactions and

compute the standard E.M.F of the cell. Given that :

E⁰ _Zn_®_Zn²⁺= 0.76 V.

E⁰ _Co_®_Co²⁺ = 0.28 V

29. Depict the electrochemical cell in which the reaction is

Zn(s) + 2 Ag⁺(aq) ® Zn²⁺(aq) + 2 Ag(s)

Also show :

a) Cathode and anode reactions.

b) Movement of ions and electrons.

c) electrode reactions.

30. Calculate standard cell potentials of Galvanic cells in which reactions are as follows :

a) 2 Cr(s) +3 Cd²⁺(aq) ® 2Cr³⁺(aq) + 3 Cd(s)

b) Fe²⁺(aq) + Ag⁺(aq) ® Fe³⁺(aq) + Ag(s)

31. The Zinc/silver oxide cell is used in hearing aids and electronic watches.

Zn ® Zn²⁺ + 2 e^- : E⁰= 0.76V

Ag₂O + H₂O + 2 e^-® 2 Ag + 2 OH^- : E⁰=0.344 V

a) What is oxidised and reduced ?

b) Find E⁰of the cell and DG in Joules.

32. Predict the products of electrolysis of each of the following :

(i) An aqueous solution of AgNO₃ with silver electrodes.

(ii) An aqueous solution of AgNO₃ with inert electrodes.

(iii) A dilute aqueous solution of H₂SO₄.

(iv) An aqueous acidified solution of cupric chloride using platinum electrodes.

NERNST EQUATION FOR ELECTRODE AND CELL POTENTIAL

The tendency of metal to lose electrons or tendency of its ions to gain electrons depends upon the concentration of the ions in solution. The tendency to lose or gain electrons is expressed in terms of electrode potentials. The value of electrode potential therefore changes with the variation in the concentration of the ions. The quantitative relationship between the concentration of ions and electrode potentials is given by Nernst Equation. For a general electrode reaction:

Mⁿ⁺+ n e^- ® M

the Nernst equation can be written as :

Substituting the values of R ( = 8.314 J K^-¹ mol^-¹),

T ( = 298 K) and F ( = 96,500 C) , the Nernst equation at 298 K becomes :

In general for any electrode:

It may be noted that the concentration of solid phase is taken to be unity.

Calculation of cell potential using Nernst equation

Consider a Daniel cell in which the concentration of the two solution may not be 1M. The cell is :

Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)

The E.M.F of the cell is :

E_cell= E(Cu²⁺|Cu) - E(Zn²⁺|Zn) …….. (2)

The electrode potential for the two electrodes can be calculated by using Nernst Equation as :

Substituting these values in Equation (2) we get :

Now,

and the concentrations of solids is taken as unity so that [Zn(s)] = 1 ; [Cu(s)] = 1.

Therefore,

At 298 K,

In the Daniel cell, the valencies of zinc and copper are the same, i.e., n = 2.

Consider an example in which the valencies of the two metals used in two half-cells are not the same.

Consider the cell:

Cu|Cu²⁺(aq) ||Ag⁺(aq) |Ag

The E.M.F of the cell :

E_cell= E_cathode- E_anode ………(3)

During the reaction, two electrons are released by one copper atom but one electron is accepted by Ag⁺according to the reaction :

Cu(s) ® Cu²⁺(aq) + 2 e^- ( oxidation half cell)

Ag⁺(aq) + e^- ® Ag(s) (reduction half cell)

To balance the loss and gain of electrons, the reduction half-cell is multiplied by 2, so that the net reactions are :

Cu(s) ® Cu²⁺(aq) + 2 e^-

2 Ag⁺(aq) + 2 e^- ® 2 Ag(s)

The electrode potential of two electrodes may be written according to Nernst equation as :

Where E⁰(Ag⁺IAg) remain unchanged.

Substituting in Equation(3), we get :

In general , for an electrochemical cell reaction :

The Nernst Equation may be written as :

at 298 K.

The values of a, b,c, d and n are obtained from the balanced cell reactions.

Concentration Cell

A cell in which both electrodes are of the same type but solutions of electrolyte in which they dip have different concentrations is called concentration cell. For example, in the concentration cell [Fig] :

Cu½CuSO₄(C₁) ½½ CuSO₄(C₂)½Cu

The standard potentials of the two electrodes will cancel and the cell potential is given by :

A copper ion concentration cell

Problems

33. Calculate the E.M.F of the cell in which the reaction is :

Mg(s) + 2 Ag⁺(aq) ® Mg²⁺(aq) + 2 Ag(s)

when [Mg²⁺] = 0.13M and [Ag⁺] = 1.0 x 10^-⁴ M

34. Write Nernst equation and calculate the E.M.F of the following cells at 298 K.

i) Mg(s)½ Mg²⁺(0.001M)½½ Cu²⁺ (0.0001M)½Cu(s)

ii) Fe(s)½ Fe²⁺(0.001M)½½ H⁺ (1M) H₂(1atm),Pt

iii) Sn(s)½ Sn²⁺(0.050M)½½ H⁺ (0.02M) H₂(1atm),Pt

35. A Galvanic cell consists of a metallic zinc plate immersed in 0.1 M Zn(NO₃)₂ solution and a metallic plate of lead in 0.02 M Pb(NO₃)₂ solution. Calculate the E.M.F of the cell. Write equations for cell reactions and represent the cell.

Nernst equation and Equilibrium constant

The Galvanic cell does not continue working indefinitely and stops working after some time. In fact as the cell reaction progresses, there is a fall in concentration of cations around cathode due to reduction and at the same time there is an increase in the concentration of the metal cations around the anode due to oxidation. Consequently, electrode potential of cathode decreases and that of the anode increases with passage of time. Ultimately, a stage is reached when potential difference (E_cathode- E_anode) becomes zero and there is no flow of electrons. For example

in a cell :

Zn½ Zn²⁺(aq)½½ Cu²⁺ (aq)½Cu

the flow of electrons from zinc to copper stops when E_Zn²⁺/_Zn becomes equal to E Cu²⁺/Cu. In this situation the concentration of Zn²⁺(aq) and that of Cu²⁺(aq) will be equilibrium concentrations because the cell reaction attains equilibrium:

Zn(s) + Cu²⁺(aq) == Zn²⁺(aq) + Cu(s)

Considering the concentration of Zn(s) and Cu(s) to be unity, the equilibrium constant for the above reaction K_C is given by :

Applying Nernst equation for cell potentials :

Since E_cell at equilibrium is zero,

Significance of Kc

The value of Kc gives the extent of the cell reaction. For example, the value of Kc for Zn-CuSO₄cell at 298 K is 2 x 10³⁷, which shows that the reaction has proceeded to completion.

Electrochemical cell and Gibb’s energy of the Reaction

A Galvanic cell is a source of electrical energy which can be used for different kinds of work. Unlike heat energy, electrical energy can be quantitatively converted into work. The electrical work or electrical energy is equal to the product of E.M.F of the cell and the electrical charge that flows through the external wire. It may be put as :

W_Ele= (EMF) x (Electrical charge flowing through the wire)

If n is the number of moles of electrons flowing through the wire and F is 1 Faraday of electricity , then,

Charge flowing through the wire = n F

If E_cell is the EMF of the cell, then

W_Ele = n F E_cell.

But the Gibb’s energy change (DG) for a process occurring at constant T and P is the measure of useful work that can be obtained. Mathematically,

DG = W_{non-expansion}.

For electrical systems, W_{non-expansion} can be written as W_Elect, because work involved is electrical work . In electrochemical cells the work is done by the system and by conventions it is taken to be negative. Thus , electrical work performed by the cell can be written as - W_Elect.

Now equating Gibb’s energy change(DG) to electrrical work done by the cell(- W_Elect), we have

DG = - W_E_lect = - n F E_cell

For the sake of comparison, standard cell potentials are used. Therefore, standard Gibb’s energy change is given by :

DG⁰ = - n F E⁰_cell

Significance of the above equation lies in the fact that it can help in predicting the feasibility of the cell reaction. For cell reaction to be spontaneous, DG must be negative, and for DG to be negative, the value of E must be positive.

Problems

36. For the Cell

Ni(s)½ Ni²⁺(aq)½½ Ag⁺ (aq)½Ag(s)

Calculate the equilibrium constant at 25⁰C. How much

maximum work would be obtained by the operation of

the cell ?

37. Calculate the equilibrium constant for the following

reaction :

Cu(s)+ 2 Ag⁺(aq) ® 2 Ag(s) + Cu²⁺(aq)

; E⁰_cell = 0.46 V

38. Write Nernst Equation for the reaction :

Cu(s)+ 2 Ag⁺(aq) ® 2 Ag(s) + Cu²⁺(aq)

when it takes place in a Daniel cell.

39. The standard electrode potential for Daniel cell is 1.10 V. Calculate the Gibb’s energy for the reaction :

Zn(s) + Cu²⁺(aq) ® Zn²⁺ (aq) + Cu(s)

Some commercial cells or batteries

Any cell or battery that we employ as a source of electrical energy is basically an electrochemical cell. The term battery

is used to represent the arrangement of two or more galvanic cells connected in series. Theoretically, the basis of the electrochemical cell is any of redox reactions. However, in practice, the redox reaction used should give an arrangement which fulfills the following requirements :

i) It should be light and compact.

ii) Its voltage should not vary appreciably during its use.

iii) It should provide power for a longer period and

iv) It should be rechargeable.

The Galvanic cells can be broadly classified into two categories, viz.,Primary cells and Secondary Cells.

1. PRIMARY CELLS

This type of cells become dead over a period of time and the chemical reaction stops. They cannot be recharged or used again. Some common examples are dry cell, mercury cell, etc.

a) Dry Cell

It is a compact form of Leclanche cell

known after its discoverer. In this cell, the anode consists of a zinc container and cathode is a graphite rod surrounded by powdered MnO₂ and carbon . The space between the electrodes is filled with a moist paste of NH₄Cl and ZnCl₂.

Fig 8 A Dry Cell

The electrode reactions are complex, but they can be written approximately as follows :

Anode : Zn(s) ® Zn²⁺ + 2 e^-

Cathode : MnO₂ + NH₄⁺ + e^-® MnO(OH) + NH₃

(Oxidation state of Mn changes from +4 to +3)

In the cathode reaction, manganese is reduced from +4 oxidation state to +3 state. Ammonia is not liberated as a gas, but combines with Zn²⁺ to form[ Zn(NH₃)₄]²⁺ ion. Dry cells do not have an indefinite life as acidic NH₄Cl corrodes the zinc container even when not in use. Dry cells have a potential of approximately 1.25 to 1.5 V.

b) Mercury cells

It is a miniature cell which has found use in small electrical circuits(such as hearing aides, watches and camera). In mercury cell , the anode is zinc-mercury amalgam and cathode is a paste of HgO and carbon. The electrolyte is a paste of KOH and ZnO.

Commonly used mercury cell. The reducing agent is zinc and oxidising agent is mercury (II) oxide.

The reaction of the cell is as follows :

Anode : Zn(Hg) + 2 OH^-® ZnO(s) + H₂O(ℓ) + 2e^-

Cathode: HgO(s) + H₂O + 2e^-® Hg(ℓ) + 2OH^-

Overall reaction : Zn(Hg) + HgO(s) ® ZnO(s) + Hg(ℓ)

The cell potential is approximately 1.35 V and remains constant during its life time as the overall reaction does not involve any ion in solution whose concentration can change during its life time.

2. Secondary cells

This type of cells can be recharged by passing direct current through them and can be used again and again. Some examples are lead storage battery, nickel-cadmium storage cell, etc.

a) LEAD STORAGE BATTERY

This is the most commonly used battery in automobiles. Each battery consists of a number of voltaic cells connected in series. Three to six such cells are generally combined to get 6 to 12 volt battery. In each cell , the anode is a grid of lead packed with finely divided spongy lead and the cathode is a grid of lead packed with PbO₂. The electrolyte is an aqueous solution of sulphuric acid(38% by mass) having a density 1.30 g ml^-1.sulphuric acid.

Fig 9 . Lead Storage Battery

The following reactions take place in the lead storage cell :

At anode : The lead loses two electrons and is oxidised to Pb²⁺ ions.

Pb(s) ® Pb²⁺(aq) + 2e^-

Pb²⁺(aq) + SO₄²^-(aq) ® PbSO₄(s)

The overall anode reaction may be written as :

Pb(s) + SO₄²^-(aq) ® PbSO₄(s) + 2e^-

At cathode: The PbO₂ is reduced as :

PbO₂ (s)+ 4 H⁺ + 2e^-® Pb²⁺(aq) + 2 H₂O

Pb²⁺(aq) + SO₄²^- ® PbSO₄(s)

The overall reaction is :

PbO₂ (s)+ 4 H⁺ + SO₄²^-+ 2e^- ® PbSO₄(s) + 2 H₂O

Thus, the complete electrode reactions and overall cell reaction are :

Anode : Pb (s) + SO₄²^-(aq) ® PbSO₄(s) + 2e^-

Cathode : PbO₂(s) +4 H⁺ +SO₄²^-+ 2e^- ® PbSO₄(s) + 2 H₂O

Overall : Pb(s) + PbO₂ (s)+2 H₂SO₄ ® 2PbSO₄(s) +2H₂O

he cell may be represented as :

Pb(s)½ H₂SO₄(aq) ½PbO₂ ½Pb

It is clear from the above reactions that during the working of the cell, PbSO₄ is formed at each electrode and sulphuric acid is used up. As a result, the concentration of H₂SO₄decreases and the density of the solution also decreases. When the density of H₂SO₄falls below 1.2gml^-¹the battery needs recharging.

Recharging the battery

The cell can be charged by passing electric current of a suitable voltage in the opposite direction. The electrode reaction gets reversed. As a result , the flow of electrons get reversed and lead is deposited on anode and PbO₂ on the cathode. The density of sulphuric acid also increases. This reaction may be written as :

2 PbSO₄(s) + 2 H₂O(ℓ) ® Pb(s) + PbO₂(s) + 2 H₂SO₄

b) Nickel-cadmium storage cells

Nickel-cadmium cell is an important secondary cell which has longer life than the lead storage cell but more expensive to manufacture.

A rechargable nickel-cadmium cell in a jelly roll arrangement and separated by a layer soaked in moist sodium or potassium hydroxide.

It consists of a cadmium anode and the cathode is comprised of a metal grid containing nickel(IV) oxide. These are immersed in KOH solution. The reaction occurring are :

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

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

Net: Cd(s) + NiO₂(s) + 2H₂O(ℓl) ® Cd(OH)₂(s)+ Ni(OH)₂(s)

The cell is also called nicad cell and has voltage = 1.4 V.

As is evident that , there are no gaseous products , the product formed adhere to the electrodes and can be reconverted by charging process. The cell is becoming more popular and finds use in electronic watches and calculators.

FUEL CELLS

In recent years , scientists have designed the cells which convert chemical energy of a fuel directly into electrical energy. Such cells are called fuel cells. These are voltaic cells in which the fuels such as hydrogen, carbon monoxide etc are used to generate electrical energy without the intervention of thermal devices like boiler, turbines, etc.

The conventional method is the conversion of chemical energy of a fuel to liberate heat. The heat energy so produced is used to generate steam for spinning the turbines which are coupled to electrical generators.

Fuel cells are designed in such a way that the materials to be oxidised and reduced at the electrodes are stored outside the cell and are constantly supplied to the electrodes. One of the most successful fuel cell uses the reaction of hydrogen and oxygen to form water and is known as H₂- O₂ Fuel Cell. The cell was used for providing electrical power in Appolo space programme . The water vapours produced during the reaction were condensed and added to drinking water supply for astronauts.The experimental arrangement is shown in Fig10.

The cell consists of porous carbon electrodes which are impregnated with catalyst (Pt, Ag or CoO).

Fuel cell using H₂ and O₂ produces electricity

Hydrogen and oxygen are bubbled through the electrodes into electrolyte which is an aqueous solution of NaOH or KOH .

The electrode reactions are :

Anode : [ H₂(g) + 2 OH^-(aq) ® 2 H₂O(ℓ) + 2 e^-] x 2

Cathode : O₂(g) + 2 H₂O( ℓ ) +4 e^-® 4 OH^-(aq)

Net reaction : 2 H₂(g) + O₂(g) ® 2 H₂O(ℓ)

The cell runs continuously as long as the gases hydrogen and oxygen are supplied at the temperature 525 K and 50 atm. pressure.

Advantages of Fuel cells

Some important advantages of fuel cells are given below :

1) Pollution free working : There is no harmful or objectionable products formed in fuel cells. Hence they do not make any pollution problems.

2) High efficiency : The efficiency of fuel cells is approximately 70 - 75 % , which is much higher than the conventional cells.

3) Continuous source of energy : Unlike conventional batteries, energy can be obtained from the fuel cell continuously so long as the supply of fuel is maintained.

+2 UNIT 5 PAGE- 2

Ionic conductivities at Infinite dilution at 298 K

EMF OF THE CELL

Standard Reduction Electrode Potentials at 298 K

PLUS TWO

QUESTIONS