Ebook chemistry basis elements part 1

First Edition, 2009 ISBN 978 93 80168 56 2 © All rights reserved Published by Global Media 1819, Bhagirath Palace, Chandni Chowk, Delhi 110 006 Email globalmedia@dkpd com Table of Contents 1 The Equil[.]

First Edition, 2009

ISBN 978 93 80168 56 2

© All rights reserved

Published by:

Global Media

1819, Bhagirath Palace,

Chandni Chowk, Delhi-110 006 Email: globalmedia@dkpd.com

Table of Contents

1 The Equilibria

2 Chemical Equilibrium

3 Surface Phenomenon

4 Bioinorganic Chemistry

5 Bio-organic Chemistry

6 The Photochemistry

7 Quantum Chemistry

8 Analytical Chemistry

9 Nuclear Chemistry

10 Organic Name Reactions

11 Reagents in Organic Synthesis

12 Kinetics of Reactions

13 The Macromolecules

14 Fast Reactions

15 Conformational Analysis

The Equilibria

1

The Equilibria

The E°cell of an aluminium-air battery is 2.73 volts and it involves a 12 electron process The free energy change (DG°) of the battery in kJ Calculated :

DG = _ nFE° = _ 12 × 96500 × 2.73 J

= _ 3161340 J = _3161.34 Id

For the following reaction—

H 2 (g) + C1 2 (g) 2HCl (g)

DG° is _ 262 kJ Calculate the equilibrium constant (K) for the reaction at 298 K

DG° = _ 262000 J = _2.303 RT log K

Hydrogen-oxygen Fuel Cell

Two half-cells of hydrogen-oxygen fuel cell under basic conditions can be depicted as OH_/O2 (g)/ Pt and OH_/H2 (g)/

Chemistry : Basic Elements

Pt and their standard electrode potentials at 25°C are 0.4009 and _ 0.8279 V respectively Write the half cell reactions and the complete cell reaction Depict the complete cell and the e.m.f of the cell Calculated :

_Pt |H2(g)| OH_| O2(g)| Pt +

At anode, the reaction is H2 + 2OH_ ® 2H2O + 2e_

At cathode, the reaction is 2e_ + O2 + H2O ® 2OH_

The cell reaction is H2 + O2 ® H2O

The e.m.f of the cell is = 0.4009 _ (_0.8279) = 1.2288 V

The reduction potentials of Ag+/Ag and Fe+3/Fe+2 are 0.799 and 0.771 V respectively The

equilibrium constant of the reaction

Ag + Fe +3 Fe +2 + Ag +

Ag | Ag+ || Fe+3 · Fe+2| Pt(+)

E°cell = 0.771 _ 0.799 = 0.028 V

At (_ ), Ag ® Ag+ + e_

At (+), Fe3+ + e_ ® Fe+2

Cell Reaction : Ag + Fe3+ ® Ag+ + Fe2+, Here n = 1

K = 0.33628

The EMF of the cell, Pb | PbCl2 || AgCl | Ag at 300 K is 0.50 V If the temperature coefficient of EMF

is _ 2 × 104 volt deg_1, DH and DS for the cell reaction Calculated :

Pb + 2AgCl ® PbCl 2 + 2Ag

Pb | PbCl2| Cl_ | AgCl | Ag +

The cell reaction is Pb + 2AgCl ® PbCl2 + 2Ag

The Equilibria

DG = _ nFE = _ 2 × 96500 × 0.5 J

= _ 96,500 J or _ 96.5 kJ

DS =

= 2 × 96500 (_2 × 10_4) = _ 38.6 JK_1

DG = DH _ TDS,

_ 96,500 = DH _ 300 (_38.6)

DH = _ 96,500 _ 300 × 38.6

= _ 108,080 J

or, _108.08 kJ

The potential of pentane/oxygen fuel cell given that the standard free energies of formation (in kJ/mole) at 298 K are _ 8.2, _ 237.2 and _ 394.4 for pentane, H 2 O (1) and CO 2 (g) respectively Calculated :

The chemical reaction that takes place in the pentane-oxygen fuel cell is

C2H12 (g) + 8 O2 (g) ® 5 CO2 (g) + 6H2O (1)

DG° for the reaction

= [5(_ 394.4) + 6 (_ 237.2)] _ [_ 8.2]

= _ 1972 _ 1423.2 + 8.2 = _ 3387.0 kJ

The electrode reactions are

C5H12 + 10 H2O ® 5 CO2 + 32 H + + 32 e_

8 O2 + 32 H+ + 32 e_ ® 16 H2O

DG° = _ n FE°, DG° = _ 3387 × 103 J, n = 32

Potential of Hydrogen-electrode

Taking the case of hydrogen electrode, consisting of H2 gas

Chemistry : Basic Elements

in equilibrium with H+ ions, the electrode reaction, written as reduction reaction is

H+ + e_ ® H2 (n = 1)

Applying Nernst equation, the electrode potential of the hydrogen electrode is given by

EH+, H2 = E°H+, H2 _

If H2 gas is at 1 atmospheric pressure, aH2 = 1

\ EH+, H2 = E°H+, H2 -

Replacing the activity of H+ ions by its molar concentration, we have

E°H+, H2 = E°H+, H2 +

Since the standard electrode potential of hydrogen electrode is taken as zero

E°M+, H2 = 0

Hence

EH+, H2 = = 0.0591 log [H+] at 25° C

The following cell is used to measure the mean activity coefficient (Y+) of HCl

Pt | H 2 (g) | HCl (aq) | AgCl (s) | Ag (s)

(i) The cell reaction, (ii) activities of H+ and Cl_ ions express the emf of the cell, (iii) the activities in terms of molality of HCl and the mean ionic activity coefficient, obtain an expression for In Y+ in terms of the emf

The cell is Pt | H2 (g) | HCl (aq) | AgCl (s) | Ag (s) i.e.,

The Equilibria

it consist of hydrogen and silver-silver chloride electrodes in HCl as the electrolyte The cell reaction will

be

H2(g)+AgCl (s) ® Ag (s) + H+ (aH+) + Cl_ (aCl_)

and the EMF of the cell is

Ecel = E°Ag|AgCl _

where a is the activity of HCl as a whole Now as the activity (a) of HCl at any molality m is related to the mean activity coefficiency Y+ by the expression a = m2Y2+ , substituting this value in eq (1) we get

Ecel =

Ecel =

or, Ecell + = (2)

All the quantities on the left hand side of equation (2) are known experimentally Hence to calculate Y±,

the value of E°Ag|AgCl is required To determine E°Ag|AgCl the quantity is plotted

against m and the result is extrapolated to m = 0, when m = 0, Y+ = 1, therefore from equation(2)

equal to the value of the ordinate

The mean ionic activity coefficient of 0.1 molar HCl at 25° C given that the EMF of the cell

Calculated :

Chemistry : Basic Elements

H 2 (1 atm) | HC1 (a), AgCl (s) | Ag

is 0.3524 V at 25° C and that the standard electrode potential of Ag-AgCl is 0.2224 V at 25°C

For the given cell

=

Putting T = 298 K, R = 8.314 JK_1 mol_1 and

F = 96500 C, this equation becomes

Ecell + 0.1183 log m = E°Ag|AgCl _ 0.1183 log Y+

Substituting Ecell = 0.3524 V

E°Ag|AgCl = 0.2224 V

and m = 0.l mol kg_1

0.3524 + 0.1183 log 0.1 = 0.2224 _ 0.1183 log Y±

or Y+ = 0.796

The standard reduction potentials of the electrodes Fe+3 | Fe and Fe+2 | Fe are _ 0.035, _ 0.440 V respectively It is easy to oxidise Fe to Fe+2 or Fe to Fe+3

Since the standard reduction potentials are given the standard oxidation potentials will be + 0.036 V and + 0.44 V The standard oxidation potential of Fe | Fe+2 is more positive than that of Fe|Fe+3 electrode So

it is easy to oxidise Fe to Fe+2

(a) Write sell electrode for the following reaction Cu (OH)2 (s) ® Cu 2+ + 2OH _

(b) Write the cell reaction for the following cell Pt | H2 | HCl | Hg 2 Cl 2 | Hg | Pt

The Equilibria

(a) Half-cell reactions are

Cu(OH)2 (s) + 2e - ® Cu + 2OH_

Cu2+ + 2e_ ® Cu (s)

and electrodes are Cu2 + | Cu and Pt | Cu(OH)2 | OH_

(b) The electrode reactions are

2 H+ + 2 e_ ® H2(g)

Hg2Cl2 + 2e_ ® 2Hg (1) + 2Cl_

The cell reaction is

Hg2Cl2 (s) + H2 ® 2 Hg (1) + 2H+ + 2Cl_

The standard electrode potentials of the electrodes Cu2+

| Cu and Ag+ | Ag are 0.337 V and 0.7991 V The concentration of Ag+ in a solution containing 0.06

M of Cu2+ ion such that both the metals can be deposited together ? The activity coefficients are unity and both silver and copper do not dissolve among themselves Assumed :

The individual reactions are

Cu2+ + 2e_ ® Cu (s)

and Ag+ + e_ ® Ag (s)

The electrode potentials given by Nernst equation

= 0.337 _ = 0.337 _ 0.036 = 0.301

E(Ag+ |Ag) = 0.7991 _

Chemistry : Basic Elements

= 108.428

or, [Ag+] = 10_8.428 = 0.37 × 10_8mol dm_3

The Ksp of AgI by forming proper cell Give E°, I _ AgI (s)|Ag = _ 0.151 V and E°Ag+|Ag = 0.7991V Calculated :

The cell can be written as Ag | Ag+; I_ | AgI | Ag

At left electrode Ag (s) ® Ag+ + e_ E° = 0.7991V

At right electrode AgI (s) + e_ ® Ag (s) + I_, E° = _ 0.151 V

AgI (s) ® Ag+ + I_

The standard emf of the cell is E°

= E°R _ E°L = _ 0.151 _ 0.7991 = _ 0.9501 V

We know that log Ksp = = _ 16.1

At 25° C, (¶E°/¶T) p = _1.25 × 10 _3 VK _1 and E° = 1.36 V for the cell

Pt|H 2 (g)|HCl(aq)|Cl 2 |Pt

The enthalpy and entropy for the cell reaction calculated The cell reaction is :

H 2 (g) + Cl 2 (g) = 2HCl (aq)

DS° = nF

DS° = (2 mol) (96,485 C mol_1) (_1.25 × 10_3 VK_1) = _ 241JK_1

We also know that DH° = _ nF

DH° = (2 mol) (96,485C mol_1)

[(1.36 V _ 298.15 K) (1.25 × 10_3 VK_1)]

DH° = _ 191 kJ

The Equilibria

The fact that two electrodes must have the same potential when equilibrium is attained to calculate K for the familiar reaction Utilized :

Zn + Cu +2 Zn +2 + Cu

If (aZn+2) and (aCu+2) represent the activities of ions when equilibrium is attained, the potentials of Zn |

Zn+ 2 and Cu | Cu2+ electrodes, which must then be equal are given by

= + 0.763 _ 0.0296 log (aZn+2)

= _ 0.337 _ 0.0296 log (aCu+2)

Equating these potentials, it is seen that

+ 0.763 _ 0.0296 log (aZn+2)e = _ 0.337 _ 0.0296 log (aCu+2)

= K = 1.7 × 1037

Finely divided metallic lead and tin, shaken with solutions containing stannous and plumbous

perchlorates until the equilibrium in the reaction

Sn (s) + Pb +2 Sn +2 + Pb (s)

reached; the ratio of the concentrations of stannous and plumbous ions at equilibrium i.e.,

(CSn+2/CPb+2) was found to be 2.98 at 25°C The standard oxidation potential of the

Sn | Sn+2 electrode at 25°C, E°Pb is 0.126 volt at 25°C

If the ratio of the concentrations is equal to the ratio of the activities in terms of molalities, as is probably the case if the solutions are dilute,

Chemistry : Basic Elements

\ K = = 2.98

The oxidation and reduction processes in the above reaction is

Sn (s) Sn+2 + 2e-

and Pb+2 + 2e_ Pb (s)

so that the complete reaction as written in the question above, takes place in the reversible cell

Sn | Sn+2 || Pb+2| Pb

for the passage of two faradays i.e., n is 2

In this case E°cell = E°Sn _ E°Pb

E°cell = E°Sn _ 0.126

E°Sn = +0.140 volt

The standard potential of the Sn | Sn+2 electrode is thus + 0.140 volt at 25° C

Nature of the Electrode Process

In an alkali-chlorine cell a saturated (about 6 N) solution of sodium chloride is electrolyzed, at

ordinary temperatures, between a steel cathode (hydrogen overvoltage 0.2) and a graphite anode

(oxygen overvoltage 0.6 volt; chlorine overvoltage negligible) The nature of the electrode process Explained :

At Cathode : The cations present in the solution are H+ and Na+; the concentrations are 10_7 g (for a

neutral solution) and 6 g ion per liter, respectively The standard oxidation potentials are 0.0 and + 2.71V respectively; hence the reversible potentials in the given electrolyte are :

ENa = + 2.71 _ 0.059 log 6 = + 2.66 volt

The Equilibria

Since hydrogen overvoltage is 0.2, the potential for the discharge of hydrogen ions and the evolution of hydrogen gas is + 0.41 + 0.2 = 0.61 volt, this is much below required for Na+ ion discharge, only by raising the potential to 2.66 volt the discharge of Na+ ions become possible The removal of H+ ions by discharge leaves an excess of OH_ ions in the solution and this accounts for the formation of NaOH

At Anode : The anions present are OH_ and Cl_; the concentrations being 10_7 and 6 g ion per litre as for

H+ and Na+ respectively The standard oxidation potentials of O2 and C12 are _ 0.40 and _1.36

respectively and hence the reversible potentials in the given electrolyte are

EO = _ 0.40 + 0.059 log 10_7 = _ 0.81 volt

ECl = _ 1.36 + 0.059 log 6 = _1.31 volt

Allowing for the overvoltage (0.6 volt) the oxygen evolution potential resulting from the discharge of

OH_ ions is _ 0.81 _ 0.6 = _ 1.41 volt and hence discharge of chloride ions and the formation of chlorine gas, will take place in preference By increasing the anode potential, oxygen evolution would tend to occur

Four types of fuel cells They are superior for :

Depending on the kind of fuel used, the types of fuel cells are—

1 Hydrogen-oxygen fuel cell

2 Hydrocarbon-oxygen fuel cell

3 Carbon monoxide-oxygen fuel cell

4 Solid coal-oxygen fuel cell

Superiority of Fuel Cells :

1 They possess very high efficiency (75.90%) In heat engines efficiency is around 40% or less

2 The individual cells can be stacked and connected in series to generate higher voltage

Chemistry : Basic Elements

3 They are also very light

4 The fuel cells do not create pollution problems

5 These cells play an important role in manned space flights

The temperature dependence of the emf of an electrochemical cell can often be written in the form :

E = (a + bT + cT2 + dT3) volt

where a, b, c and d are constants A certain commercially suitable battery was found to have a = 1.19237, b = _ 1.537 × 10_4, c = 2.73 × 10_8 and d = 1.78 × 10_11 DG, DH and DS for this cell calculated at 27° C if n = 3

DG = _ n EF, where n = 3,

DG = _ 3(23.06) [1.19237 _ (1.537 × 10_4) (300)

+ (2.73 × 10_8)(300)2 + (1.78 × 10_4)(300)3]

= _ 79.52 k cal

DS = nF = nF (0 + b + 2cT + 3dT2)

DS = 3 (23.06) [_ 1.537 × 10_4 + 2(2.73 × 10_8)T]

+ 3 (1.78 × 10_11) T2

DS = _ 9.17cal deg_1

DH = DG + TDS

DH = _79.52 + 300(_ 9.17 × 10_3)

DH = _ 82.27 k cal

A voltaic cell is constructed using Al and Al+3 in one half cell and Ag and Ag+ in the other half-cell (a) What total reaction will occur ?

(b) What half reaction will occur at each electrode?

(c) What is the anode and which is cathode ?

The Equilibria

(d) How many volts will the cell produce if [Al+3] and [Ag+] are 1.0 M :

(a) The reduction potential for Ag is positive, relative to hydrogen and the reduction potential for

aluminium is negative Therefore silver will be reduced and Al will be oxidized The balanced equation for the reaction is

Al + 3Ag+ Al3+ + 3Ag

(b) The half reactions are

Al ® Al3+ + 3e_

3Ag+ + 3e_ ® Ag

(c) Since the oxidation occurs at the Al electrode this is called the anode Reduction occurs at the Ag electrode and is called the cathode

(d) The voltage or potential difference, produced by the cell is just the total difference between the standard potentials for Al and Ag The standard reduction potential for Al is _1.66 V and for Ag it is + 0.80 V The total difference is 2.46 This is the voltage produced by the cell

With the help of electrochemical series Which substance can be used to oxidize fluorides to fluorine Shown :

From the values of standard potentials in electrochemical series

F2 + 2e_ 2F_ E° = 2.87 V

This half cell has the highest reduction potential This implies that fluorides cannot be oxidized

chemically by any substance listed in electrochemical series They can be oxidized only electrolytically

Tin is (II) stable towards disproportionation in non-complexing media ? Given

E° Sn+2Sn = _ 0.15 V,

Chemistry : Basic Elements

E° Sn+4 , Sn+2 = 0.15 V

The disproportionation reaction is

2 Sn+2(aq) Sn+4 (aq) + Sn (s)

which results from the following half-reactions

(i) Sn+2 (aq) + 2e_ Sn (s) E°el = _ 0.15 V

(ii) Sn+2(aq) Sn+4 (aq) + 2e_ E°el = _ 0.15 V

Adding (i) and (ii) 2Sn+2 (aq) Sn+4 (aq) + Sn (s) E° = _ 0.30 V

Since E° is negative, the disproportionation reaction is not spontaneous Hence tin (II) is stable

The galvanic cell for each of the following reactions and write down the corresponding expression for the cell potential Constructed :

(1) Zn(s) + H2SO4 (aq) ZnSO4 (aq) + H2(g)

(2) Fe(s) + Cl2(g) FeCl2(aq)

(1) Zn is oxidized to Zn2+ and H+ is reduced to H2 Thus we have

RHC, 2H+(aq) + 2e_ H2(g) (1)

LHC, Zn2+ (aq) + 2e_ Zn (s) (2)

Subtracting eq (2) from eq (1) we have

Zn (s) + 2H + (aq) H2(g) + Zn2+ (aq)

The cell would be

Zn (s) | Zn2 + (aq) || H+ (aq) | H2 (g) | Pt

and the cell potential will be given by

Ecell =

where E°cell = E°(H+,H2 | Pt) _ E° (Zn2+ |Zn)

The Equilibria

= _E°(Zn2+ |Zn)

(2) Fe (s) + Cl2(g) FeCl2 (aq)

Fe is oxidized to Fe2+ and Cl2 is reduced to Cl_ Thus we have

RHC, Cl2 (g) + 2e_ 2 Cl_ (aq) (1)

LHC, Fe2+ (aq) + 2e_ Fe (s) (2)

Subtracting eq (2) from eq (1), we have

Fe (s) + Cl2 (g) Fe2+(aq) + 2Cl_ (aq)

The cell would be Fe | FeCl2 (aq) | Cl2 (g) | Pt and the cell potential is given by