Electrochemistry : Electrochemistry is a branch of science which deals with the production of electricity from energy released during spontaneous chemical reactions and the use of electric energy to bring about non spontaneous chemical transformations.
Conductors : A conductor is a substance which allows the passage of electric current. The two types of conductors are i) metallic conductors or electronic conductors ii) Electrolytic conductors or ionic conductors
In metallic conductors the conduction of electricity is due to the flow of electrons. So they are called electronic conductors. The electronic conductance depends on (i) The nature and structure of the metal (ii) The number of valence electrons per atom. (iii)temperature (it decreases with increase in the temperature)
The composition of the metallic conductors remain unchanged by the passage of electric current. In electrolytic conductors the conduction is due to the movement of ions, hence they are called ionic conductors. The conductivity of electrolytic solution depends upon
(i) The nature of the electrolyte (ii) Size of the ions produced and their solvation. (iii)the nature of the solvent and its viscosity. (iv) Concentration of the electrolyte and (v) Temperature (increases with increase in temperature)
On passing a direct current through a solution of an electrolytic conductor change in the composition takes place.
Resistance : Resistance is the opposition to the flow of current. It is denoted by R and is measured in ohm () l R A l R=ρ A Where l is the length and A is area of cross section is resistivity
Conductance (G) Conductance is inverse of resistance 1 A G = = R ρl But 1 =k ρ where k is conductivity or specific conductance A G = k l
Conductivity Conductivity of a solution of an electrolyte is the conductance of a solution placed between two electrodes each of one square meter area kept at a distance of 1 meter apart. SI unit for conductivity is Sm -1.
(1 S cm -1 = 100 Sm -1 ) and S = ohm -1 conductivity of a solution decreases with decrease in concentration of the solution due to decrease in the number of ions per unit volume of the solution.
Cell constant : For measuring the conductivity of a solution of an electrolyte conductivity cell is used. It has two platinum electrodes each of an area of cross section of A and kept at a distance of l.
Cell constant or k = l G * = = Rk A cell constant R
Molar conductivity : Molar conductivity of a solution at a given concentration is the conductance of the volume V of a solution containing one mole of electrolyte kept between two electrodes with area of cross section A and distance of unit length.
It is represented by m m =kv SI unit for m is Sm 2 mol -1 If m is in Sm 2 mol -1 and k in Sm -1 k λ m = 1000C where C is conc. in mol L -1 When m is in S cm 2 mol -1 and k is in Scm -1 1000k λ m = C
1. Conductivity of 0.01 M NaCl solution is 0.12 Sm -1. Calculate its molar conductivity k 0.12 1000C 1000 0.01-2 2 λ m = = = 1.2 10 Sm / mol
2. The molar conductivity of 0.1M nitric acid is 630 S cm 2 /mol. Calculate its conductivity. 1000k λ m = C 1000k 630 = 0.1 630 0.1 k = = 0.063 Scm 1000-1
Limiting molar conductivity : Limiting molar conductivity is the molar conductivity when concentration of the solution approaches zero or at infinite dilution. (Represented by o m)
Variation of molar conductivity with ion concentration : 1) For a strong electrolyte : For a strong electrolyte m increases slowly with dilution as λ = λ 1 o 2 m- AC m is molar conductivity. m
o m is limiting molar conductivity, C concentration in mol L -1 A is constant for a given solvent and temperature and depend on type of the electrolytes(1-1 NaCl, 2-1 CaCl 2 and 2-2 MgSO 4 etc). A plot of m verses 1 C 2 is a straight line
The intercept of the line on the Y axis gives o m for a strong electrolyte. This is called extrapolation method. For a weak electrolyte : Weak electrolytes have low degree of dissociation at moderate concentrations.
Upon dilution the increase in m is very steep due to increase in the degree of dissociation of a weak electrolyte. The increase in m with dilution is more when concentration approaches zero. Hence o m can not be obtained by extrapolating m to zero concentration.
Kohlrausch s law of independent migration of ions: The limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.
Eg λ m(nacl)= λ o o o Na + λ + - Cl In general for an electrolyte AxBy xa y + + yb x - o o + yλ o x y A B λ m(a B )= xλ y + x - o o o 2 Ca Cl λ mcacl = λ + 2λ 2+ -
The relationship between degree of dissociation() and limiting molar conductivity ( o m) for a weak electrolyte Where λ m molar conductivity α= λ at a given concentration m o m o m is the limiting molar conductivity.
2 Cα Ka = 1-α cλ 2 Ka = m λ o (λ o - λ ) m m m
Calculate the molar conductivity of a solution of MgCl 2 at infinite dilution given that the molar ionic conductivities of λ = 106.1 Scm mol and λ = 76.3 Scm mol o 2-1 o 2-1 2+ - (Mg ) (Cl ) λ = λ + 2λ o o o MgCl 2+ - 2 Mg Cl = 106.1 + 2(76.3) = 258.7 Scm mol 2-1
Problem : The values of limiting molar conductivities ( o m) for NH 4 Cl, NaOH and NaCl are respectively 149.74; 248.1 and 126.4 Scm 2 mol -1. Calculate the limiting molar conductivity of NH 4 OH
o o o o NH OH NH Cl NaOH NaCl λ = λ + λ - λ 4 4 = 149.74+248.1-126.4 = 271.44 Scm 2 mol -1
Faraday s first law of electrolysis: 2 faradys first law The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte either through its aqueous solution or molten state.
If w is 2 faradys the first mass law of the substance deposited and Q is the current passed in coulombs w Q But Q = I t where I is the current strength in ampere and t is time in seconds.
Faraday: 2 faradys first law A Faraday is the charge of one mole of electrons in coloumbs. It is represented by F. 1 Faraday = Avogadro No. Charge of one electron in Coloumbs
2 faradys first law = 6.02 10 23 1.6021 10-19 C = 96487C 96500 C 1 Faraday of electric current is required to reduce 1 mole of cations with unit positive charge eg to reduce 1 mole of silver ions to get 1 mole of silver i.e, 108g of Ag 96500 C of electricity is required
+ - Ag(aq)+ e 2 faradys first law Ag 1mol. 96500C 108g To reduce 1 mole of cations with two unit positive charge 2 Faraday of current is required. Eg to reduce 1 mole of Mg 2+ ions to get 1 mole i.e, 24g of Mg 2F i.e, 2 96500C of
Electricity is required. 2 faradys first law Mg + 2e Mg 2+ - (aq) 1 mole 2 96500C 24g In general to reduce 1 mole of M n+ cation nf of electric current is required M + ne M n+ - (aq) nf
Faraday s 2 faradys first second law law of electrolysis: The amounts of different substances liberated by the same quantity of electricity passing through the electrolytic solution are proportional to their chemical equivalent weights.
Chemical equivalent weight 2 faradys first law Atomic mass = Number of electrons required to reduce the cation Eg + - (aq) Ag + e Ag Eq weight of Ag = At mass 1
2 faradys first law 2+ - (aq) Mg + 2e Mg Eq weight of Mg = 3+ - (aq) At wt 2 Al + 3e Al Eq weight of Al = At wt 3
Problems: 2 faradys first law 1) A solution of Ni(NO 3 ) 2 is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of nickel is deposited at the cathode? (Mol mass of Ni = 58.7)
Solution: Q = I t 2 faradys first law = 5 20 60 = 6000C 2+ - Ni + 2e Ni 2 96500C 193000C 58.7g
For 193000C of electricity mass of nickel 2 faradys first law obtained = 58.7g For 6000C of electricity 6000 58.7 = 1.812g 193000
2) How 2 faradys long first law it will take for the deposition of 0.2g of silver when silver nitrate solution is electrolysed using 0.5 ampere of current (Mol mass of Ag = 108)
+ - Ag + e Ag 2 faradys first law 96500C 108g For 108g of silver to be deposited current required is 96500C. For 0.2g of Ag 0.2 96500 But Q = I t = 178.7C = Q 108 Q 178.7 t = = = 357.4 sec I 0.5
Electrochemical Cell or Galvanic cell It is a device in which chemical energy of spontaneous redox reaction is converted into electrical energy. An electrochemical cell consists of two half cells or electrodes
Electrode An electrode is a metal in contact with a solution containing its own ions e.g Zn (s) / Zn 2+ ( aq) or Cu(s)/Cu 2+ ( aq) Non metallic electrodes may also be prepared in presence of a inert metal like Platinum which provides surface for the
oxidation or reduction and for the conduction of electrons. Eg Pt (s) H 2 (g) H + (aq) or Pt (s) Br 2 (aq) Br - (aq)
Electrode Potential The potential difference developed between the electrode (metal) and the electrolyte (solution containing its own ions) when both the metal and the solution are in equilibrium is called electrode potential. It is expressed in terms of volt. When coupled with SHE, if reduction takes place at the given electrode the electrode
Nernst Equation : n+ - M + ne reduction M (aq) (s) Oxidised state reduced state RT nf o E n+ = E n+ - ln (M / M) (M / M) RT o = E n+ - ln (M / M) n+ nf [M ] [reduced state] [oxidised state] [M]
2.303RT [M] nf [M ] o = E n+ - log (M / M) 10 n+ Substituting the values R= 8.314JK -1 mol -1 T =298K F = 96500C and [M] = 1 0.059 1 n M o E n+ = E n+ - log (M / M) (M / M) 10 n+
where E is the electrode potential (reduction) E o standard electrode potential or standard reduction potential (SRP) n=number of electrons exchanged.
Standard Electrode potential : Standard electrode potential is the electrode potential when the concentrations of all the species involved is unity (1M) and if a gas is involved its pressure should be 1 bar.
Standard Hydrogen Electrode SHE
It consists of a platinum electrode coated with platinum black. The electrode is dipped in 1M HCl. Pure hydrogen gas is bubbled through it under a pressure of 1 bar. S.H.E is represented as Pt(s) H 2 (g)(1bar) H + (aq)(1m) The reduction reaction taking place is
1 H ( aq) e H2( g) 2 S.H.E is assigned an electrode potential of 0.0 V at all temperatures.
Electrochemical cell Daniel cell To prepare Daniel cell get a zinc electrode by dipping zinc rod in 1M ZnSO 4 solution. Get a copper electrode by dipping a copper plate in 1 M CuSO 4 solution. Couple these two electrodes using a salt bridge to get Daniel cell
Reactions taking place At anode Zn Zn 2e 2 oxdn At cathode Cu 2e Cu 2 redn 2 2 ( s) ( aq) ( aq) ( s) Net cell reaction Zn Cu Zn Cu
Zinc rod is the ve pole and copper plate is the +ve pole. Flow of electrons is from Zinc to copper but the flow of current is from copper to zinc. Daniel cell can be represented as Zn/ Zn 2+ (aq) Cu 2+ (aq) /Cu
EMF of Daniel cell E o cell = E o R - E o L = E o Cu - E o Zn = 0.34-(-0.76) = 1.10V If an external opposite potential is applied in the Daniel cell and E ext is increased slow i) If E ext < 1.10 V
Zinc rod is the ve pole, copper is the positive pole, electrons flow from zinc to copper and the current flows from copper to zinc. 2) If E ext = 1.10V: No reaction takes place and there is no flow of electrons or current. 3) If E ext > 1.10V
All the reactions taking place are reversed. Zinc is deposited at the zinc electrode and copper dissolves at copper electrode. Electrons flow from copper to zinc and current flows from zinc to copper. Electrochemical cell becomes an electrolytic cell.
Cell potential : Cell potential is the potential difference between the two electrodes of the galvanic cell. Cell electromotive force(emf) : It is the difference between the electrode potential of the cathode and anode when
no current is drawn through the cell. E cell = E right E left = E - E Of the electrode undergoing reduction Of the electrode undergoing oxidation E o cell = E o R E o L e.g E o Daniel cell = E o R E o L
E o - E o = 0.34 -(-0.76)= 1.10V 2 + 2+ Cu /Cu Zn /Cu Use SHE in the determination of SRP of a given electrode Construct a standard electrode of the given metal by dipping the pure metal in 1M solution of its own ion at 25 o C Couple this standard electrode with SHE using a salt bridge to get galvanic cell.
Measure the emf of the cell using suitable instrument like potentiometer. E o = E o R E o L One of the electrodes of the cell is SHE and its electrode potential is 0.0V. So the electrode potential of the given electrode will be the emf of the cell in magnitude.
If reduction takes place at the given electrode its E o will be +ve but if oxidation takes place at the given electrode is E o will be ve.
Calculation of the emf of the cell for any given concentration of the electrolytes of the electrodes: Consider Daniel cell with given concentrations of the electrolyte EMF or E E E cell 2 2 / Cu / Cu Zn Zn
o 2.303RT 1 o 2.303RT 1 E 2 log E 2 log Cu / Cu 2 Zn / Zn 2 nf Cu nf Zn o o E 2 E 2 Cu / Cu Zn / Zn 2.303RT nf log Zn Cu 2 10 2 Substituting R=8.314JK -1 mol, T=298K F=96500C;n=2 o o = E 2+ - E 2+ - log Cu / Cu Zn / Zn 0.059 Zn 2 Cu 2+ 2+
E o cell 0.059 log 2 Zn Cu 2 2 0.059 1.10 log 2 Zn Cu 2 2 In Daniel cell when [Zn 2+ ] = [Cu 2+ ] E cell =1.10V
if [Zn 2+ ]>[Cu 2+ ] if [Zn 2+ ]<[Cu 2+ ] E cell <1.10v E cell <1.10v For a general electrochemical reaction of the type ne aa bb cc dd E E c C D RT ln nf A B o cell cell a b d
When Daniel cell is discharged completely E cell =0 and Zn Cu 2 K 2 e the equilibrium constant o 2.303RT Zn E = 0 = E - log nf Cu 2+ cell cell 2+
E o cell 2.303RT nf log Zn Cu 2 2 E o cell 2.303RT nf log K c E o cell 0.059 log K n c
Problem 1. Calculate the emf of the cell in which the following reaction takes place. Ni 2 Ag (0.002 M ) Ni (0.160 M ) 2Ag 2 ( s) ( s) Given that E o cell = 1.05V E cell E o cell 0.059 [ Ni ][ Ag ] log 2 [ ][ ] 2 2 ( s) 10 2 Ni( s) Ag
But [M] for any element is taken as unity o 0.059 Ni E = E - log 2 Ag 2+ cell cell 10 + 2 0.059 0.160 = 1.05 - log 2 0.002 2 = 0.914V
2. Calculate the equilibrium constant for the reaction at 298K 2 ( ) 2 2 ( s) ( aq) ( s) Cu Ag aq Cu Ag Given that E o Ag + /Ag = 0.80V and E o (Cu 2+ /Cu) = 0.34V Solution : E o cell log K 0.059 log K n c o ne cell 0.059 c
E E E o o o cell 2 Ag Ag Cu Cu =0.80-0.34=0.46V ( / ) ( / ) 20.46 log K c 15.59 0.059 Taking the antilog K c =3.92 10 15
Relationship between cell potential and Gibbs Energy: r G o nfe o cell G o r Is the Gibb s energy n = no. of electrons exchanged in the net cell reaction F= 96500C
E o cell is standard emf of the cell. But G o r = -RT lnk -nfe o = - RT lnk E o RT ln K nf 2.303RT log K nf o E 0.059 log n K
Problems: The cell in which the following reaction occurs 2Fe 2I 2Fe I 3 2 ( aq) ( aq) ( aq) 2( s) Has E o cell = 0.236V at 298K. Calculate the standard Gibb s energy and the equilibrium constant for the cell reaction.
Solution : n = 2.G o = -nfe o Ecell = - 2 96500 0.236 = - 45548 J 0.059 log K n 0.059 0.236 log K 2
20.236 log K 8 0.059 Taking the antilog K = 10 8
Electrolysis: Electrolysis is the chemical reactions taking place when a direct current is passed through the aqueous solution or molten state of an electrolyte. The product of electrolysis depends on
(i) Nature of the electrolyte (Aqueous solution or molten state) (ii) Types of the electrodes (iii)concentration of the solution 1) eg When aqueous solution of copper sulphate is electrolysed using copper electrodes
At anode Cu Cu 2e oxdn 2 ( s) ( aq) A Cu e Cu 2 redn t cathode ( aq) 2 ( s) Thus copper from anode dissolves and an equivalent amount of pure copper is deposited on cathode. This technique is used in electrolytic refining of crude copper.
2) When molten sodium chloride is electrolysed using inert electrodes At anode 2Cl oxdn Cl 2e redn At cathode Na e Na 2 Thus chlorine gas is liberated at anode and Sodium metal is formed at cathode.
3) When aqueous solution of NaCl is electrolysed. NaCl Na Cl H O H OH 2 At cathode there is a competition between the following reduction reactions
Na e Na E 2.71V o aq ( s) cell 1 o H aq e H2( g ) E cell 0.0V 2 At cathode the reaction with higher value of E o is preferred. Therefore the reaction taking place at cathode is 1 H aq e H 2 2( g)
But H + ions are produced by dissociation of H 2 O. So the net reaction taking place at cathode is 1 H2 O( l) e H2( g) OH 2 The possible reactions at anode are - 1 - o Cl aq Cl 2 aq + e E cell = 1.36V 2 + - o 2H O O + 4H + 4e E = 1.23V 2 (l) 2 cell
At anode the reaction with lower E o is preferred. But on account of over potential of oxygen oxidation of Cl - ions takes place. Thus when aqueous solution of NaCl is electrolysed hydrogen is liberated at cathode, chlorine at anode and NaOH solution is formed
Batteries A battery may have one or more than one galvanic cells connected in series. A battery should be of practical use, reasonably light and its voltage should not change very much during the use.
Batteries are divided into two types; (i) Primary battery in which reaction occurs only once and cannot be recharged. Eg Dry cell or Leclanche cell and Mercury cell (ii) Secondary battery which can be recharged by passing current through it in opposite direction,so that it can be
Reused. Eg Lead storage battery and Nickel cadmium cell. Dry cell or Leclanche cell It consists of a zinc container as an anode. A graphite rod surrounded by a mixture of manganese dioxide and carbon powder is cathode.
The space between the electrodes is filled with electrolyte a moist paste of ammonium chloride and zinc chloride Reaction taking place At anode Zn Zn + 2e (s) 2+ - + - At cathode MnO + NH + e MnO(OH)+ NH 2 4 3 NH 3 produced in the reaction forms a
complex with Zn 2+ to form [Zn(NH 3 ) 4 ] 2+. Zinc container is the ve pole and the graphite rod in the middle is the +ve pole. Cell potential is 1.5V Lead storage Battery : It consists of lead anode and a grid of lead packed with lead dioxide (PbO 2 ) as cathode.
Electrolyte is 38% solution of sulphuric acid. The reactions taking place when the battery is in use are Anode Pb SO PbSO 2e 2 ( s) 4 ( aq) 4( s) Cathode PbO ( s) SO 4H 2e PbSO 2 H O 2 2 4 ( aq) ( aq) 4 ( s) 2 ( l)
The overall reaction is Pb PbO 2 H SO 2 PbSO 2 H O ( s) 2 ( s) 2 4 ( aq) 4 ( s) 2 ( l) On charging the battery the entire reaction is reversed.
Fuel cells Galvanic cells that are designed to convert the energy of combustion of fuels like hydrogen, methane etc directly into electrical energy are called fuel cells. Eg Hydrogen-Oxygen Fuel cell.
In this hydrogen and oxygen gases are bubbled through porous carbon electrodes into concentrated aqueous sodium hydroxide solution. Catalyst like finely divided platinum or palladium is incorporated into the electrodes for increasing the rate of electrode reaction
Reaction taking place are Cathode O + 2H O + 4e Overall reaction is 4OH - - 2 (g) 2 (l) (aq) Anode 2H + 4OH 4H O + 4e 2H + O - - 2(g) (aq) 2 (l) 2H O 2 (g) 2 (g) 2 (l) This is the most common cell used in Apollo space programme to get electricity
So that the bi-product water formed is used as drinking water after distillation for the astronauts. Fuel cells are highly efficient (about 70%) pollution free and is continuous source of energy as long as fuel is supplied.
Corrosion: When a metal is exposed to the atmosphere it is slowly attacked by the constituents of the environment as a result of which the metal is slowly lost in the form of its compound. This is called corrosion.
e.g tarnishing of silver, development of a bluish green coating on copper etc. corrosion for iron is called rusting. Rusting of iron is an electrochemical phenomenon. Air and moisture are essential for corrosion. Presence of electrolytes help rusting. A tiny
electrochemical cell is set up on the surface of iron where a small droplets of water sit on. Reaction taking place are At Anode 2Fe 2Fe + 4e (s) 2+ - - At Cathode O + 4H +(aq)+ 4e 2H O 2 (g) 2 (l) H + are produced from H 2 CO 3 formed due to dissolution of carbon dioxide from air into water
The Fe 2+ ions are further oxidised by atmospheric oxygen to ferric ion which are ultimately converted to hydrated ferric oxide called rust(fe 2 O 3.xH 2 O) Rusting may be prevented by barrier protection like painting, metal plating etc. Underground iron pipes are
prevented from rusting by sacrificial or cathodic protection by connecting the iron pipe with pieces of more electropositive metals like Mg. Mg gets corroded first keeping iron pipe safe. This is corroded magnesium pieces are to replaced by new pieces timely.
Problems: 1. The resistance of a conductivity cell containing 0.001 M KCl solution at 298K is 1500. What is the cell constant if the conductivity of 0.001M KCl solution at 298K is 0.146 10-3 Scm -1?
Cell constant G*= Rk =resistance conductivity =0.146 10-3 Scm -1 1500S -1 = 0.219 cm -1
2. A conductivity cell when filled with 0.01M KCl has a resistance of 747.5 ohm at 25 o C. When the same cell was filled with an aqueous solution of 0.05M CaCl 2 solution the resistance was 876 ohm. Calculate
(i) Conductivity of the solution (ii) Molar conductivity of the solution (given conductivity of 0.01M KCl = 0.14114 sm -1 )
Cell constant G* = Rk = 747.5 0.14114 =0.105.5m -1-1 cell constant 105.5m Conductivity k = = = 0.1204Sm R 876 ohm -1 k 0.1204 Molar conductivity λ m = = = 0.00241sm mol 1000C 1000 0.05 2-1
3. The electrical resistance of a column of 0.05M NaOH solution of diameter 1cm and length 50cm is 5.55 10 3 ohm. Calculate its (i) resistivity (ii) conductivity (iii) molar conductivity
Cell constant G o l = a l = 50 cm Diameter = 1 cm radius = 0.5 cm Area of cross section A = r 2 = 3.14 (0.5) 2 = 0.785 cm 3 50 G = = 63.694 cm 0.785 * -1
1 1 Resistivity ρ = = = 87.135 Ω -2 k 1.148 10 1000k Molarconductivityλ m = C -2 1000 1.148 10 = 0.05 = 229.6 S cm mol 2-1