Ebook chemistry basis elements part 1

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 standa

10 Organic Name Reactions

11 Reagents in Organic Synthesis

12 Kinetics of Reactions

13 The Macromolecules

14 Fast Reactions

15 Conformational Analysis

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

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

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

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

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 _

or, Ecell + = (2)

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

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

(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

E (Cu2+ | Cu) = E° _

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

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

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

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 :

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

(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

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

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)

Ecell =

with E°cell = E° (Cl_ |Cl2| Pt) _ E° (Fe2+ | Fe)

In acid solution the following are true under the standard conditions

(a) H 2 S will react with oxygen to give H 2 O and Sulphur

(b) H 2 S will not react in the corresponding reaction with Selenium and Tellurium

(c) H2Se will react with Sulphur giving to H 2 S and Se but not react with Tellurium

The hydrides H 2 O, H 2 S, H 2 Se and H 2 Te 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+ + O2 + 2e_ H2O (Reduction)

E° (H+ |O2| Pt) > E° (H+ | H2S | S)

(b) Similarly we have

E°(H+|H2S|S) > E° (H+ | H2Se | Se)

E° (H+ | H2S | S) > E° (H+ | H2Te | Te)

(c) Similarly we have

E° (H+ | H2S | S) > E° (H+ | H2Se | Se)

E° (H+ | H2Se | Se) > E° (H+ | H2Te | Te)

Arranging the above reduction potentials in the, decreasing order, we get

E° (H+ | O2 | Pt) > E° (H+ | H2S | Pt)

> E° (H+ | H2Se | Se) > E°(H+ | H2Te | Te)

The above-order implies that the tendency for the reduction reaction

2H+ + Sn + 2e_ H2X

to take place is greatest for oxygen and least for Te The tendency for the reverse reaction i.e., the

oxidation of hydride to the corresponding element will be greatest for H2Te and least for H2O

Determined at 298 K for the cell

Pt | Q·QH 2 , H + |1 M KCl, Hg 2 Cl 2 (S) | Hg | Pt

(a) its emf when pH = 5.0

(b) the pH when Ecell = 0

The Equilibria

(c) the positive electrode when pH = 7.5

For the given cell, we have

Half cell Reaction Potential

RHC Ecalomel = E°calomel

(1 mol dm_3 calomel)

LHC Q + H+ + e_ QH2 EQ,QH2, H+|Pt =

(quin hydrone)

The emf of the cell is Ecell = E°calomel _ EQ,QH2,H+|Pt

= E°calomel _ E°Q,QH2, h+ | Pt +at 298 K

A + B 1 C + D, 2A + 2B 2 2C + 2D)

How are E° and K for the two reactions related ?

In aqueous solution Cobalt(III) ion is able to oxidize under

Chemistry : Basic Elements

Co 3 + + e_ ® Co 2 +, E° = 1.842V The formation constant for [Co(NH 3 ) 6 ] 3 + is 5 × 10 3 and for [Co(NH 3 ) 6 ] 2 +, 1 × 10 5 An aqueous solution of [Co(NH 3 ) 6 ] 3 + in 1 M NH3 does not oxidize water Shown :

are + 0.137 V and + 0.518 V The arrangement of the cell for the reaction :

CuCl ® u + + Cl _

The standard cell EMF The cell reaction spontaneous The solubility product of CuCl Calculated :

The electrode reaction for the half cell Cu | CuCl | Cl_ is

CuCl + e_ ® Cu + Cl_ (Reduction)

At the Cu | Cu+

Cu ® Cu+ + e_ (Oxidation)

_

Overall cell reaction: CuCl ® Cu+ + Cl_

The concerned cell is

Chemistry : Basic Elements

Cu (s) | Cu+ || Cl_ | CuCl (s) | Cu+(s)

Standard EMF of such a cell = E°rhs _ E°Ihs

= 0.137 _ 0.518 = _ 0.381 V

The cell reaction is not spontaneous under standard conditions

The equilibrium constant for the cell reaction, K is given by the relation

In K = _ 0.381 Here n = l

0.0592 log K = _ 0.381 log K =

log K = _ 6.4358 = 7_.5642

K = 3.666 × 10_7

Solubility product = 3.666 × 10_7

Chemical Equilibrium

2

Chemical Equilibrium

The average bond energies of S 8 , H 2 and H 2 S are 264, 436 and 338 kJ mol _l respectively The

enthalpy of the reaction, estimated :

S 8 + 8H 2 8H 2 S

DH of the above reaction = (Total bond energies of the reactants)

_ (Total bond energy of the product)

C p is always greater than C v and find the relation between C p and C v

If the volume of the system is kept constant when the heat is added to a system then no work is done by the system Thus the heat absorbed by the system is used up completely to increase the internal energy of the system Again if the pressure of the system is kept constant when the heat is supplied to the system; then some work of expansion is also done by the system in addition to the increase in internal energy Thus if at constant pressure, the temperature of the system is to be raised through the same value as at

constant volume, then some extra heat is required for doing the work of expansion Hence C p > C v

m =

(1) For cooling m will be positive (because dT and dP both will be negative)

(2) For heating m will be negative (because dT is positive while dP is negative)

(3) If m = 0, the gas gets neither heated up nor cooled on adiabatic expansion because m = 0, only if dT =

0 for any value of dP

Two moles of gas at 1 bar and 298 K are compressed at constant temperature by use of a constant pressure of 5 bar If the compression is driven by a 1000 kg mass, it will fall in the earths gravitational field to a certain extent :

where a = 17.23 JK _1 mol _1 , b = 13.61 × 10 _3 JK _2 mol _1 and g = 42.55 × 10 _7 JK _3 mol _1 The change

in molar internal energy when oxygen is heated from 298 to 500 K :

Chemistry : Basic Elements

DU_ =

= 17.23(500 _ 298) + (13.61 × 10_3/2)

(5002 _ 2982) _ (42.55 × 10_7/3)

(5003 _ 2983)

= 4437 J mole-1

The Joule-Thomson coefficient for a real gas is not zero in the limit of zero pressure :

For a real gas the compressibility factor at low pressures can always be represented by the viral equation

Every gas has a definite temperature (at a particular pressure) at which m = 0 Below this temperature m

is positive and above this temperature m is negative This temperature (at a particular pressure) at which

m = 0, i.e., the gas neither cooled down nor heated upon adiabatic expansion and below which

Chemical Equilibrium

m is positive adiabatic expansion and above which m is called the inversion temperature

Putting these values in equation (4), we get

Chemistry : Basic Elements

In a triatomic gas, there will be nine degrees of freedom, three translational, three rotational and three

vibrational, contribution to the energy by each degree of translational and rotational motion is ½RT per mol, whereas that of vibrational mode is RT per mol of gas Thus,

b = 0.0318 dm 3 mol _1 The value of DH for the isothermal

(T = 300 K) compression of 1 mol of oxygen from the pressure of 1 to 200 atm, calculated :

Chemistry : Basic Elements

Heat of neutralisation of HCN and NaOH is _12.13 kJ

mol _1 , DH° ionization of HCN :

DH° = DH°ioniz +DH°neutr

\ _12.13 = DH°ioniz _55.90

DH°ioniz = + 43.77 kJ mol_1

Various properties into Intensive and Extensive variables, listed :

Temperature, Heat capacity, Pressure, Dielectric constant, Density, Boiling point, Viscosity,

Concentration, Refractive index, Enthalpy, Entropy, Gibbs free energy, Molar enthalpy, Chemical potential, Molality, Volume, Mass, Specific heat No of moles Free energy per mole

Intensive Variables : Temperature, Pressure, Dielectric constant, Density, Boiling point, Viscosity,

Concentration, Refractive index, Molar enthalpy Chemical potential, Molality, Specific heat, Free

energy per mole

Extensive Variable : Heat capacity, Enthalpy, Entropy, Gibbs free energy, Volume, Mass, No of moles

Matched :

(a) Isothermal process (1) when P = constant

(b) Adiabatic process (2) when heat capacity of body

= constant

(c) Isobaric process (3) when dE = dH = 0

(d) Polytropic process (4) when T = constant

(e) Quasistatic process (5) when both E and H = constant

(f) Isochoric process (6) when q = constant

(g) Cyclic process (7) when V = constant

Mind = f (anger, suspicion, greed)

Hence (1) Mind is a state function

(2) Differential of mind is an exact differential

Shown mathematically that the magnitude of the work involved in a reversible expansion of an ideal gas from volume V 1 to V 2 is larger than the corresponding work involved in an irreversible expansion against a constant pressure of p 2

Since, in expansion V2 > V1 and p1 > p2, therefore

Wrev _ Wirrev = positive

i.e., the magnitude of the work involved in a reversible expansion is larger than the corresponding work

involved in an irreversible expansion

Chemistry : Basic Elements

The enthalpy of fusion, DH f , of ice at _ 10°C, from the following data, calculated

DH f = 6.02 kJ mol _1 at 0°C,

Chemistry : Basic Elements

Combining equations (3), (4) and (5) we get

Differentiating the above equation with respect to temperature at constant pressure and composition of other components

This equation gives the variation of the chemical potential of a constituent in a mixture with respect to

temperature at constant pressure and composition of the system S_ i is the partial molal entropy of ith

component of the mixture

The entropy of mixing of 1 mole of N 2 and 2 moles of O 2 , assuming the gases to be ideal calculated