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 © All rights reserved Published by: Global Media 1819, Bhagirath Palace, Chandni Chowk, Delhi-110 006 Email: globalmedia@dkpd.com Table of Contents The Equilibria Chemical Equilibrium Surface Phenomenon Bioinorganic Chemistry Bio-organic Chemistry The Photochemistry Quantum Chemistry Analytical Chemistry 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 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— H2(g) + C12 (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 log K = = 45.918 K = 8.279 × 1045 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_ + The cell reaction is H2 + O2 + H2O ® 2OH_ 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 = log K = = = = _ 0.47329 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 _ × 104 volt deg_1, DH and DS for the cell reaction Calculated : Pb + 2AgCl ® PbCl2 + 2Ag Pb | PbCl2| Cl_ | AgCl | Ag + The cell reaction is Pb + 2AgCl ® PbCl2 + 2Ag The Equilibria DG = _ nFE = _ × 96500 × 0.5 J = _ 96,500 J or _ 96.5 kJ DS = = × 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, H2O (1) and CO2 (g) respectively Calculated : The chemical reaction that takes place in the pentane-oxygen fuel cell is C2H12 (g) + O2 (g) ® CO2 (g) + 6H2O (1) DG° for the reaction = [5(_ 394.4) + (_ 237.2)] _ [_ 8.2] = _ 1972 _ 1423.2 + 8.2 = _ 3387.0 kJ The electrode reactions are C5H12 + 10 H2O ® CO2 + 32 H + + 32 e_ O2 + 32 H+ + 32 e_ ® 16 H2O DG° = _ n FE°, DG° = _ 3387 × 103 J, n = 32 E° = = 1.0968 volts 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 atmospheric pressure, aH2 = \ 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 = 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 | H2 (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 _ = (1) 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) E°Ag|AgCl = , i.e., 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 H2 (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 = or, = 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 log Y± = = _ 0.0989 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) ® Cu2+ + 2OH_ (b) Write the cell reaction for the following cell Pt | H2 | HCl | Hg2Cl2 | 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 H+ + e_ ® H2(g) Hg2Cl2 + 2e_ ® 2Hg (1) + 2Cl_ The cell reaction is Hg2Cl2 (s) + H2 ® 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 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 E (Cu2+ | Cu) = E° _ = 0.337 _ 0.036 = 0.301 = 0.337 _ E(Ag+ |Ag) = 0.7991 _ = = 8.428 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 Ksp = 10_16.1 = 7.94 × 10_17 At 25° C, (ảE/ảT)p = _1.25 ì 10_3 VK_1 and E = 1.36 V for the cell Pt|H2(g)|HCl(aq)|Cl2|Pt The enthalpy and entropy for the cell reaction calculated The cell reaction is : H2 (g) + Cl2 (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+ and Cu | Cu2+ electrodes, which must then be equal are given by EZn = = + 0.763 _ 0.0296 log (aZn+2) ECu = = _ 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 In this case E°cell = E°Sn _ E°Pb E°cell = E°Sn _ 0.126 = = 0.0140 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 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 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 : EH = _ 0.059 log 10_7 = + 0.41 volt ENa = + 2.71 _ 0.059 log = + 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 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 = _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— Hydrogen-oxygen fuel cell Hydrocarbon-oxygen fuel cell Carbon monoxide-oxygen fuel cell Solid coal-oxygen fuel cell Superiority of Fuel Cells : They possess very high efficiency (75.90%) In heat engines efficiency is around 40% or less The individual cells can be stacked and connected in series to generate higher voltage Chemistry : Basic Elements They are also very light The fuel cells not create pollution problems 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 = DG = _ n EF, where n = 3, F = 96,487 coulombs is equal to 23.06 kcal volt_1 equiv_1 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 = (23.06) [_ 1.537 × 10_4 + 2(2.73 × 10_8)T] + (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 Sn+2(aq) Sn+4 (aq) + Sn (s) which results from the following half-reactions (i) Sn+2 (aq) + 2e_ (ii) Sn+2(aq) Sn (s) E°el = _ 0.15 V 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) (2) Fe(s) + Cl2(g) ZnSO4 (aq) + H2(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_ LHC, Fe2+ (aq) + 2e_ Cl_ (aq) (1) 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 Ecell = with E°cell = E° (Cl_ |Cl2| Pt) _ E° (Fe2+ | Fe) In acid solution the following are true under the standard conditions (a) H2S will react with oxygen to give H2O and Sulphur (b) H2S will not react in the corresponding reaction with Selenium and Tellurium (c) H2Se will react with Sulphur giving to H2S and Se but not react with Tellurium The hydrides H2O, H2S, H2Se and H2Te in order of their tendency to lose electrons to form the corresponding elements Arranged (a) The reaction is H2S O2 ® H2O + S The two partial reactions are 2H+ + H2S O2 + 2e_ H2O (Reduction) 2H+ + S + 2e_ Chemistry : Basic Elements (Oxidation) A cell corresponding to the given reaction can be constructed with the oxidation reaction at anode and reduction reaction at cathode The cell potential would be E°cell = E°RHC _ E°LHC = E° (H+ | O2 | Pt) _ E° (H+ | H2S | S) Since the cell reaction is spontaneous it follows that ... 0 .15 1 V AgI (s) ® Ag+ + I_ The standard emf of the cell is E° = E°R _ E°L = _ 0 .15 1 _ 0.79 91 = _ 0.95 01 V We know that log Ksp = = _ 16 .1 Ksp = 10 _16 .1 = 7.94 × 10 _17 At 25° C, (ảE/ảT)p = _1. 25... = (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° = _ 19 1 kJ The Equilibria The... kcal volt _1 equiv _1 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 = (23.06) [_ 1. 537 × 10 _4 +