Electro Chemistry

ELECTRO CHEMISTRY Electrochemistry deals with the study of electrical properties of solutions of electrolytes and with the inter-relation of chemical phenomenon and electrical energies. Electrical energy is carried through matter in the form of electric current with the help of suitable source and charge carriers (ions or electrons). CONDUCTORS AND NON CONDUCTORS Substances are devided into two classes (a) (a) Non conductor or insulator (b) Conductor Non - conductor : Substances which do not allow electric current to pass through them are called non-conductors or insulators examples - pure water, ice, glass, rubber etc. (b) Conductor : Substances which allow electric current to flow through them are called conductors. Examples - Metals, Aqueous solution of acids, bases and salts, fused salts and impure water etc. CONDUCTORS ARE OF TWO TYPES (i) (i) Metallic conductors (ii) Electrolytic conductors or electrolytes. Metallic conductors : The conductors which conduct electric current by movement of electrons without undergoing any chemical change are known as metallic conductors. Metals (Cu,Ag,Fe,Al etc), non metals (graphite) and various alloys and minerals are examples. (ii) Electrolytic conductors : Those substances whose water solution conducts the electric current and which are decomposed by the passage of current are called electrolytes. In this case, conduction takes place by movement of ions. electrolytes also conduct electricity in fused state and undergo decomposition by passage the electric current. Substances whose aqueous solution does not conduct electric current are called non-electrolytes. They do not conduct electricity in the fused state also. Solutions of cane suger, glycerine, glucose, urea etc. are the examples of non electrolytes. Strong electrolyte : Electrolytes which are completely ionized in aqueous solution or in their molten state, are called strong electrolytes. Example – all salts, strong acid and strong base Weak electrolyte : Electrolytes which are not completely ionized in aqueous solution or in their molten state, are called weak electrolytes. Examples :- All carbonic acids (except sulphonic acid), CH3COOH, HCN, NH3, amine, etc. Difference between metallic and electrolytic conduction S.No. (i) (ii) Metallic conduction Flow of electricity takes place without the decomposition of the substance No transfer of matter takes place. (iii) The resistance to the flow of current increases with the increase in temperature and hence the increase in temperature decreases the conduction. Electrolytic conduction Flow of electricity takes place by chemical changes at electrodes. Transfer of matter takes place in the form of ions The resistance to the flow of current decreases with the increase in temperature and hence increase in temperature increases the conduction. ELECTROLYSIS The process of decomposition of an electrolyte by the passage of electricity is called electrolysis or electrolytic dissociation. It is carried out in electrolytic cell where electrical energy is converted into chemical energy. For electrolysis to take place two suitable electrodes are immersed in the liquid. The solution of an electrolyte contains ions. When an electric potential is applied between the electrodes, the positively charged ions move towards the cathode and negatively ions move towards the anode, when a cation reaches the cathode, its takes up electron(s) and thus gets its charge neutralised. Thus the gain of electrons (decrease in oxidation number) means reduction takes place at the cathode. Similarly an anion when it reaches the anode, gives up electron(s) and thus gets discharged. Loss of electrons(Increase in oxidation number) means oxidation takes place at anode. (a)  The tendency of an electrode to loose electrons is known as the oxidation potential.  The tendency of an electrode to gain electrons is known as the reduction potential. Electrolysis of fused sodium chloride : When fused sodium chloride is electrolysed, Na+ ions moves towards the cathode and Cl ions moves towards the anode. At cathode Na+ ions accept electrons to form sodium metal. At anode each Cl ion loses an electron to form Cl 2 gas. At anode Cl  e Cl Cl + Cl Cl2 At cathode (b) Na+  +e Na Electrolysis of aqueous solution of NaCl The solution of NaCl contain Na+, Cl and small amounts of H+, OH (due to small dissociation of water)  If more than one types of ions are present at a given electrode, then the one ion is the liberated which requires least energy. The energy required to liberate an ion is provided by the applied potential between electrodes. This potential is called discharge or deposition potential. ELECTRO CHEMISTRY ORDER OF DISCHARGE POTENTIAL Higher be the discharge potential, lower will be the tendency of ion to get discharged at the respective electrode. The decreasing order of discharge potential or increasing order of deposition of some of ions are given below : For cations : K+, Na+, Ca2+, Mg2+, Al3+, Zn2+, H+, Cu+, Ag2+, Ag+ For anions : SO4–, NO3–, OH–, Cl–, Br–, l– At cathode H+ +e H H + H H2 At anode Cl e  Cl Cl + Cl Cl2 Na+ ions move towards the cathode and combine with OH ions furnished by feebly ionised water to form NaOH. 2H2O + 2e  H2 + 2OH Na + + OH  NaOH PRODUCTS OF ELECTROLYSIS OF SOME ELECTROLYTES S.No. Electrolyte Electrode Product obtained at anode Product obtained at cathode (i) Aqueous NaCl Pt or Graphite Cl2 H2 (ii) Fused NaCl Pt or Graphite Na (iii) Aqueous NaOH Pt or Graphite Cl2 O2 (iv) Fused NaOH Pt or Graphite O2 Na Cu H2 Pt or Graphite O2 (vi) Dilute HCl Pt or Graphite H2 (vii) Dilute H2SO4 Pt or Graphite Cl2 O2 (viii) Aqueous AgNO3 Pt of Graphite O2 Ag (v) Aqueous CuSO4 H2 ELECTROLYTIC CONDUCTANCE (a) Resistance (R) : Metallic and electrolytic conductors obey ohm's law according to which the resistance of a (b) conductor is the ratio of the applied potential difference (V) to the current following(I). V R= • R is expressed in ohms. I Conductance (C) : The conductance of a conductor is equal to reciprocal of resistance 1 C= • C is expressed in mho. R Specific resistance () : The resistance (R) of a conductor of uniform cross section is directly proportional to its (c) length() and inversely proportional to its area of cross section (A) l l R  R  A A where  is a constant and called resistivity or specific resistance. When  = 1 , A = 1 , then  = R thus the specific resistance may be defined as the resistance of a conductor of unit length and unit area of cross section.  Unit of  ohm .cm (d) Specific conductance() : It is defined as the reciprocal of specific resistance 1  The above definitions apply to metallic conductors and electrolytes. =  In the case of solution of electrolytes, the resistance offered by the solution to the flow of current is –  Directly proportional to the distance between the electrodes R Inversely proportional to the area of cross section of the electrodes 1 l R R= A A A The conductance C = ρl Cl 1  =  so  = A ρ 2 If  = 1 cm and A = 1 cm then =C Hence specific conductivity of a solution is defined as the conductance of one centimeter cube of the solution of the electrolyte.  =  Cl A l = cell constant A Cell constant (x) is a fixed quantity for a particular cell and is defined as the distance between two parallel electrodes of a cell devided by the area of cross section of the electrodes. Where   = C × cell constant  (e) κ = R = Specific conductivity × resistance C Unit of  ohm–1 cm–1 cell constant = Equivalent Conductance (eq) : It is defined as the product of specific conductance () and the volume (V in mL) of the solution which contain one gram equivalent of the electrolyte.  eq =  × V If concentration of solution is C - gram equivalent per litre then eq = Unit  κ×1000 C  ohm–1 cm2 per gram equivalent Ex. The resistance of a 1N solution of salt is 50 Calculate the equivalent conductance of the solution, if the two platinum electrodes in solution are 2.1cm apart and each having an area of 4.2cm2. Sol. κ= = 1 ρ = 1l    RA 1 2.1 1 × = 50 4.2 100 and λ eq. = κ ×1000 N = 1 1000 × = 10 100 1 ELECTRO CHEMISTRY (f) Molar conductance : (mor ) It is defined as the product of specific conductance () and the volume (V in mL)) in which contains one mole of the electrolyte. m  =  × v and Unit κ ×1000 m =  M ohm–1 cm2 per mole Ex. Which of the following have maximum molar conductivity. (i) 0.08M solution and its specific conductivity is 2 × 10–2  cm–1. (ii) 0.1M solution and its resistivity is 50 cm. Sol. (i) λM = (ii) λM =  κ= κ ×1000 M = 2 ×10–2 × 1000 0.08 = 250 —1 cm2 mol—1 κ ×1000 M 1 ρ 1 1000 × = 200 –1 cm2 mol–1 50 0.1 So, the molar conductivity of 0.08M solution will be greater than 0.1M solution.   M¾ Relation between eq. and M : λM = κ ×1000 M and λ eq. = κ ×1000 N We know that Normality = Valency Factor × Molarity or Ex. Sol. N = n × M so λ eq. = λM V.F. If resistivity of 0.8 M KCl solution is 2.5 x 10 3  cm calculate  m of the solution.  = 2.5 x 10–3  cm K= m = 103 = 4 x 102 2.5 4×102 ×1000×10 = 5 x 105 –1 cm2 mole–1 0.8 FACTORS INFLUENCING THE CONDUCTIVITY OF ELECTROLYTES (a) Inter ionic attraction : If inter ionic attraction between solute is more, then the conductivity will be less. (b) Polarity of solvent : If the solvent is greater polarized then the ionization and conductivity will be more. (c) Viscosity of medium : By increasing the viscosity of medium, the conductivity decreases. (d) Temperature : As the temperature of electrolyte solution is increased, the conductivity increases because K.E. of ions increases and all type of attraction forces decreases and the viscosity of medium decreases. (e) Dilution : (i) The degree of ionisation increases with the increase of dilution of the solution the conductivity is increases due to increasing the number of ions. (ii) Effect of dilution on specific conductance : Specific conductance decreases with the increase of dilution of the solution due to the presence of no. of ions in 1cm3 solution decreases. (iii) Effect of dilution on equivalent/molar conductivity : The equivalent/molar conductivity increases with dilution. This is due to the fact that degree of ionisation increases with dilution.  When the whole of the electrolyte has ionised, further addition of the water bring a small change in the value of equivalent/molar conductance. This stage is called infinite dilution.  The ratio of equivalent conductivity at any dilution to equivalent conductivity at infinite dilution is called conductivity ratio or degree of dissociation of solute - α= Ex. Sol. λ λ The equivalent conductivity of H2SO4 at infinite dilution is 384 cm2 eq–1. If 49 g H2SO4 per litre is present in solution and specific resistance is 18.4 then calculate the degree of dissociation. 49 Equivalent of H2SO4 = = 1N 49 1 1 1000 × κ 1000 ×1 specific conductance = =  λ eq. = = = 55 specific resistance 18.4 N 18.4 Degree of dissociation ( = KOHLRAUSCH LAW : λ Ceq. λ eq. = 55 = 0.14 % = 14% 384 This law states that the equivalent conductivity of any electrolyte at infinite dilution ( eq. ) is the sum of ionic conductances of the cation and anion given by the electrolytes at infinite dilution. λ eq = λ a + λ c consider a salt NaCl, its equivalent conductivity at infinite dilution is the sum of two terms i.e. λ eq = λ Na + λ Cl These terms are known as ionic mobilities of anion and cation. For AxBy type solution :+ – 1 1 λ + - λ a + c Z Z here c = cation & a = anion and Z+, Z– are the charges on the cation and anion respectively. λ eq = Ex. At infinite dilution the equivalent conductance of Al+3 and SO4–2 ion are 189 and 160  cm2eq–1 respectively. Calculate the equivalent and molar conductivity at infinite dilute of Al2(SO4)3. ELECTRO CHEMISTRY Sol. λ eq.Al SO  2  4 3  1  1  = λ Al λ –2 +3 + 3 2 SO4 1 1 ×189 + ×160 = 143 cm2 eq–1 3 2 Molar conductivity = eq × V. F. = 143 × 6 = 858 cm2 mol–1 Application : This law can be used to (a) Determine the equivalent conductivities of weak electrolytes at infinite dilution. (b) Determination of degree of dissociation, since , the degree of dissociation of an electrolyte at the dilution V is = given by  = Ex.   where   = a + c 0m Na+ = 150  –1 cm2 mole–1 ; 0eq Ba2+ = 100  –1 cm2 eq–1 0eq SO 24 = 125  -1 cm2 eq–1 ; 0m Al3+ = 300  –1 cm2 mole–1 0m NH4 = 200  –1 cm2 mole–1 ; 0m , Cl– = 150  –1 cm2 mole–1 Then calculate Sol. (a) 0eq , Al3+ (b) 0eq Al2(SO4)3 (c) 0m (NH4)2SO4 (d) 0m NaCl, BaCl2. 6H2O (e) 0m , (NH4)2 SO4 Al2(SO4)3 . 24H2O (f) 0eq NaCl (a) 0eq Al3+ = 300 = 100 3 (b) 0eq Al2(SO4)3 = 100 + 125 = 225 (c) 0m (NH4)2SO4 = 2 x 200 + 2 x 125 = 650 (d) 0m NaCl.BaCl2.6H2O = 150 + 200 + 3 x 150 = 800 r–1 (e) 0m , (NH4)2 SO4 Al2(SO4)3 . 24H2O 0m (NH4)2 = 400 + 600 + 4 x 250 = 2000 (f) 0eq NaCl = 300  –1 cm2 eq–1 Ex. To calculate 0m or λ 0eq of weak electrolyte 0 λ 0mCH 3COOH = λ mCH COO– + 0m H+ 3 0 0 0 0 0 = ( λ mCH3COO – +  mNa  ) – λ mNa + + λ 0mH + + λ mCl+ – λ mCl – λ 0CH3COOH = λ 0mCH3COONa + 0mHCl – λ 0mNaCl Ex. Calculate 0m of oxalic acid, given that 0eq Na2C2O4 = 400  –1 cm2 eq–1 0m H2SO4 = 700  –1 cm2 mole–1 0eq Na2SO4 = 450  –1 cm2 eq–1 Sol. 0m H2C2O4 = 700 + 800 – 900 = 600  –1 cm2 mole 0eq = 400 + 700 – 450 2 λm = 350 – 50 = 300 2  m = 600 Ex. If conductivity of water used to make saturated solution of AgCl is found to be 3.1 x 10 –5  –1 cm–1 and conductance of the solution of AgCl = 4.5 x 10 –5  –1 cm–1 If 0M AgNO3 = 200  –1 cm2 mole–1 0M NaNO3 = 310  –1 cm2 mole–1 calculate KSP of AgCl Sol. 0M AgCl = 140 Total conductance = 10 –5 S= 140× 4 ×10 –5 ×1000 140 1.4 ×10 –4 14 S = 5.4 x 10–4 S2 = 1 x 10–8 = Ex. To calculate K W of water H2O() + H2O()  H2O+(aq) + OH–(aq) 0  m =  M,H2O = 0M H+ + 0M OH– = K ×1000 - Concentration of water molecules 100% dissociated Ask molarity molarity = [H+] = [OH–] = KW = [H+][OH–] = K ×1000 λ M  K ×1000    0  λM  2 Ka or [H + ][OH – ] Kb = H 2O CONCENTRATION CELLS : A concentration cell consists of two electrodes of the same material, each electrode dipping in a solution of its own ions and the solution being at different concentrations.  The two solutions are separated by a salt bridge. e.g. Ag(s) | Ag+ (a 1) || Ag+ (a 2) | Ag(s) (a1 < a2) a 1 , a2 are concentrations of each half cell At LHS electrode Anode : Ag (s)  Ag+(a1) + e– At RHS electrode Cathode : Ag+(a2) + e–  Ag(s) The net cell reaction is : Ag+ (a2)  Ag+ (a1)  ELECTRO CHEMISTRY The Nernst eq. is a1 0.059 log a n 2 E cell = – (Here n = 1, Temp, 298 K) Likewise, the e.m.f. of the cell consisting of two hydrogen electrodes operating at different pressure P 1 and P 2 (P 1 > P 2 ) and dipping into a solution HCl is : Ecell = P1 0.059 log P 2 2 (at 298 K) Solubility product and EMF (Metal Insoluble Salt Electrode) :  A half cell containing metal M and its sparingly soluble salt MA in a saturated solution. i.e M(s) | MA (satd) or a metal, its sparingly soluble salt in contact with a solution of a soluble salt NaA of the same anion, i.e. M(s) | MA(s) | NaA is set up.   The solubility product of a sparingly doubles salt is a kind of equilibrium constant. MX (s) M+(aq) + X– (aq) Ksp = [M+] [X–] At Anode M (s) At Cathode M+ (aq) + e– e – + MX(s) Overall reaction : MX (s) Cell representation : M | M+M– (satd. sol.) || NaA | MX(s) | M (s) M(s) + X– (aq) M+ (aq) + X– (aq) Eº cell = E°red – E°ox From thermodynamics ° = – nFE° G° = –2.303 RT log Ksp combining both equations – 2.303 RT log Ksp = – nFE° 2.303 RT or E° = log Ksp nF E° = 0.0591 log Ksp n at 25°C Work done by a cell : Ex. (i) Let 'n' faraday charge be taken out of a cell of EMF 'E' ; then work done by the cell will be calculated as : work = Charge × Potential = nFE (ii) Work done by cell = Decrease in free energy so – G = nFE or Wmax = + nFEº where Eº is standard EMF of the cell Calculate the maximum work that can be obtained from the Daniell cell given below Zn(s) | Zn2+ (aq) || Cu2+ (aq) | Cu (s). Given that Eº Zn2  / Zn = – 0.76 V and Eº Cu2  / Cu = + 0.34 V.. Sol. Cell reaction is : Zn(s) + Cu2+ (aq)  Cu(s) + Zn2+ (aq) Here n = 2 Eºcell = Eºcathode – Eºanode (On the basis of reduction potential) = + 0.34 – (0.76) = 1.10 V We know that : Wmax = Gº = – nFEº = – (2 mol) × (96500 C mol) × (1.10 V) = – 212300 C.V. = – 212300 J or Wmax = – 212300 J Ex. Calculate Ksp if (PbSO4) Ecell at 298 K of this electrode is 0.236 V Pb(s) |PbSO4(s)|Na2SO4(aq)||Pb(NO3)|Pb(s) 0.01 M Sol. I Ecell = E0cell – 0.236 = E 0cell 0.059 log 2 0.059 + 2 0.1 M  0.01   0.1    E 0cell = 0.236 – 0.03 = 0.206 – 0.059 log[Ksp] Sol. II Let it be conc. cell Anode Pb(s)  Pb 2  +2e– (a ) Cathode Pb (2s) + 2e–  Pb(s) Pb(2c) 2+ Pb (a) Ecell = E 0cell – 0.236 =  (Pb 2+ ) a  0.059  log  2+ 2  (Pb )c   (Pb 2+ ) a  0.059  log  2+ 2  (Pb )c  (Pb2+)a = 10–9M Ksp = (Pb2+)a (SO 24 )a = 10–11 Let if it is not a conc cell Anode : Pb(s) + SO 24 (aq)  PbSO4(s) + 2e– Cathode : Pb2+(aq) + 2e–  Pb(s) Pb2+(c) + SO 24 (a)  PbSO4(s) 0.059  0  0 0 0 log K sp  E 0cell = E Pb 2+ /Pb – ESO2–4 |PbSO4 |Pb = E Pb2+ |Pb –  E Pb2+ |Pb+ + 2   0.059 0 log Ksp E cell = – 2 0.236 =   1  K SP  0.059 0.059  = log  log Ksp – log  2+ 2–  = log 2 2  0.1× 0.01   (Pb ) c (SO 4 )a   K SP   10 –3  log10–8   Ksp = 10–11 Calomel Electrode :  A calomel cell consists of a platinum electrode dipping into mercury in contact with calomel (dimercury (I) chloride, Hg2Cl2) and potassium chloride solution.   Usually  The the solution is saturated with potassium chloride. cell has E0 = 0.28 V (with respect of SHE) at 25°C ELECTRO CHEMISTRY Standard (normal) calomel electrode when [Cl– ] = 1M = 1N At Anode At Cathode 2Hg()  Hg22 (aq) + 2e– Hg22 + 2e–  2Hg() Hg22 (aq) + 2Cl–  Hg2Cl2(s) Hg2Cl2  Hg22 (aq) + 2Cl–(aq) 2Hg() + 2Cl–(aq)  Hg2Cl2(s) + 2e– Hg2Cl2(s) + 2e–  Hg() + 2Cl-  Cell representation Pt(s) |Hg() | Hg2Cl2(s) | Cl–(aq)|| cathode  Cell 0 EPt Hg |Hg Cl E 0Cl |Hg Cl 2 E Cl– /Hg Cl 2  2 Cl 2 /Hg E M n+ /M(Hg)Pt = SOP 0 = E Cl / Hg Cl 2 2 / Hg 0 = E Mn+ /M(Hg)Pt – representation (Anode) || Cl–(aq) | Hg2Cl2|Hg() | Pt(s) 2 – 2 |Hg( ) = SRP RT n[Cl–] F 1 RT n[ n+ ] M F THERMODYNAMICS OF CELL POTENTIAL We know that : G0 = –nF E 0cell G = – nFEcell G = H – TS From thermodynamics H = E + PV enthalpy function. Substituting G = E + PV – TS By partial differentiation ...... (i)  G=  E+P V+V P–T S–S T according to 1st law of thermodynamics E=q+W dW = –PdV  E =  q – PdV ...... (ii)  q=  E+P V according to 2nd law dq = dq = Tds ...... (iii) T From (i), (ii) and (iii) dG = V  P – S  T at constant pressure, which is actually the condition for all normal cell reaction. dG = –S  T  G S= T ds = (ΔG) T G = – nFEcell S = – S = +nF E cell T E cell = temperature coefficient of cell reaction. T G = H – TS H = – nF E cell + nFT E cell T Kirchoff's equation Cp =  (H) T Cp (of products) – Cp (of reactants) = Cp = nFT  2 E cell –nFE cell –nFE cell + nFT + 2 T T T  2 E cell T 2 ELECTROLYSIS & ELECTROLYTIC CELL ELECTROLYSIS  Electrolyte is a combination of cations and anions which in fused state can conduct electricity.  This is possible due to the movement of ions from which it is made of and electrolyte.  The process of using an electric current to bring about chemical change is called electrolysis.  Electrolysis is a process of oxidation and reduction due to current in the electrolyte.  The product obtained during electrolysis depends on following factors. • The nature of the electrolyte • The concentration of electrolyte •The charge density flowing during electrolysis. • The nature of the electrode ACTIVE v/s INACTIVE ELECTRODES The metal electrodes in the cell that are active because the metals themselves are components of the half reactions.   As the cell operates, the mass of the zinc electrode gradually decreases, and the [Zn2+] in the anode half – cell increases. At the same time, the mass of the copper electrode increases and the [Cu2+] in the cathode half – cell decreases; we say that the Cu2+ "plates out" on the electrode.  For many redox reactions, however, there are no reactants or products capable of serving as electrodes. Inactive electrodes are used, most commonly rods of graphite or platinum, materials that conduct electrons into or out of the cell but cannot take part in the half -reactions.  In a voltaic cell based on the following half reactions, for instance, the species cannot act as electrodes : 2I –(aq)  I2(s) +2e– [anode ; oxidation] MnO 4– (aq) + 8H + (aq) + 5e–  Mn2+ (aq) + 4H2O() [cathode ; reduction] Therefore, each half – cell consists of inactive electrodes immersed in an electrolyte solution that contains all the species involved in that half -reaction. In the anode half-cell, I– ions are oxidized to solid I 2. The electrons released flow into the graphite anode, through the wire, and into the graphite cathode. From there, the electrons are consumed by MnO4– ions as they are reduced to Mn2+ ions. Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 204 ELECTRO CHEMISTRY  Examples of Electrolysis Using inert (pt/graphite) electrodes. Cathode (red) : Pb2+ + 2e–  Pb(s) Anode : 2Br-  Br2 + 2e- E0 = – 1.08 V Ecell = – 0.126 – (0.108) x 10 = – 1.206 V Eext > 1.206 V  E0 = 0.126V Electrolysis of CuSO4 molten Cathode : Cu2+ + 2e–  Cu Anode : 2 SO 24  S2 O 82 + 2e– E0 = +0.34 V E0 = – 2.05 V H2S2O8 – Marshall's acid peroxy disulphuric acid. Ecell = 0.34 – (2.05) = – 1.71 V (negative  not feasible) O O H O S O O S O H O O  Electrolysis of aq CuSO4 Cathode : Cu2+ + 2e–  Cu(s) 2e + 2H2O()  H2(g) + 2OH–(aq) Anode : 2 SO 24  S2 O 82 + 2e– E0 = –2.05 V 2H2O()  O2 + 4H+ + 4e–   E0 = 0.34 V E0 = –0.83V E0 = –1.23 V Electrolysis of aq NaBr solution (initially PH = 7) Cathode : Na+(aq) + e–  Na(s) 2e– + 2H2O()  H2 + 2OH– E0 = – 2 V E0 = – 0.83 V Anode : 2Br-  Br2 + 2e– E 0OX = – 1.08 V 2H2O ()  O2 + 4H+ + 4e– E 0OX = – 1.23 V Electrolysis of aq NaCl Cathode : Na+ + e–  Na 2e– + 2H2O()  H2(g) + 2OHAnode : E0 = – 2V E0 = –0.83 V E 0OX = –1.30 V 2Cl–  Cl2 + 2e– 2H2O()  O2 + 4H+ + 4e– E 0OX = –1.23 V Rate of production of Cl2 is more than rate of production of O2 gas. EE YYPP OOI INNS TTS K According to thermodynamics, oxidation of H2O to produce O2 should take place on anode but experimentally (experiment from chemical kinetics) the rate of oxidation of water is found to be very slow. To increase it's rate, the greater potential difference is applied called over voltage or over potential but because of this oxidation of Cl– ions also become feasible and this takes place on anode. Electrolysis using attackable (reactive) electrodes.  Electrolysis of aq. CuSO4 using Cu electrode. Cathode (reduction) : Anode (oxidation) : Cu2+ + 2e–  Cu E0 = + 0.34 V 2H2O() + 2e–  H2(g) + 2OH– E0 = –0.83 V SO 24  S2 O 82 + 2e– E 0OX = – 2.05 V 2H2O()  O2 + 2H+ + 4e– E0 = 1.23 V Cu(s)  Cu2+ + 2e– E0 = –0.34 V electrolytic refining  AgNO 3(aq) using Cu cathode & Ag anode. Cathode : Anode : Ag+ + e–  Ag(s) 2H2O () + 2e–  H2(g) + 2OH– E0 = 0.8 V E0 = – 0.083 V NO3–  X (No reaction) 2H2O()  O2 + 4H+ + 4e– E0 = – 1.23 V Ag(s)  Ag+(aq) + e– E0 = – 0.80 V FARADAY'S LAWS OF ELECTROLYSIS : Michael Faraday deduced two important law : (a) FARADAY'S FIRST LAW OF ELECROLYSIS This law states that "The amount of a substance deposited or dissolved at an electrode is directly proportional to the charge passing through the electrolytes. If a current of I amperes is passed for t seconds, the quantity of charge Q in coulombs. If W, gram of substances is deposited by Q coulombs of electricity, then W  Q  It W = Z It = E It 96500 Where Z is constant of proportionality and is known as electrochemical equivalent.  When Q = 1 coulomb, W = Z thus electro chemical equivalent may be defined as the weight in grams of an element liberated by the passage of 1 coulomb of electricity. Ex. How many cc of chlorine will be deposited by 100 amp. current flowing for 5 hours through melted NaCl. Sol. Q = It = 100 × 5 × 60 × 60 = 18 × 105 W = ZQ = 18× 35.5 E 18E ×103 = 662.2 g × 18 × 105 = × 105 = 965 96500 96500  Volume of 71 g Cl2 at NTP = 22.4 L  volume of 662.2 g Cl2 at NTP = 22.4 × 662.2 = 208.9 L 71 ELECTRO CHEMISTRY Ex. The time required to coat a metal surface of 80 cm2 with 0.005mm thick layer of silver (density = 10.5 g cm–3) with the passage of 3A current through silver nitrate solution is – Sol.  Volume of layer of silver = 0.005 × 10–1 × 80 = 0.04 cm3  Mass = Density × volume = 10.5 × 0.04 = 0.42 g w= So t= (b) E × It 96500  0.42 = 108 × 3× t 96500 0.42 × 96500 = 125.09 seconds. 108 × 3 FARADAY'S SECOND LAW : This law states that the amounts of different substances deposited at electrodes by passage of the same quantity of electricity are proportional to their chemical equivalent (E). W  E If W1 and W2 be the amounts of two different substances deposited at electrodes and E1 and E2 be the equivalent weights then - W1 E1 = W2 E 2 combining the two laws W  It E W = ItE F Where 1 is proportionality constant and F is called faraday.. F when It = F then W = E Hence faraday (F) is the quantity of charge in coulombs required to deposit one g equivalent of any substance. The Faraday (F) is also the quantity of charge carried by one mole of electrones. F = e × N = 1.6 × 10–19 × 6.023 × 1023 = 96500 coulombs. Ex. The same current if passed through solution of silver nitrate and cupric salt connected in series. If the weight of silver deposited is 1.08 g. Calculate the weight of copper deposited Sol. According to faradays second law W1 E1 = W2 E 2  1.08 108 =  W2 = 0.3175 g W2 31.75 CURRENT EFFICIENCY current efficiency = charge actually used in electricity x 100 charge passed mass actually produced current efficiency = mass that should have been produced x 100 Ex. Calculate volume of the gases liberated at STP if 1 L of 0.2 molar solution of CuSO4 is electrolysed by 5.79 A current for 10000 seconds. Sol. No. of moles of e– = Cathode : 5.79×10000 579 = = 0.6 96500 965 Cu2+ + 2e-  Cu(s) 0.2 mole 0.4 mole 2H2O() + 2e–  H2 + 2OH– 0.2 mole of e–  0.1 mole of H2 at S.T.P.. Anode : 2H2O()  O2 + 4H+ + 4e– 4 mole of e–  1 mole of O2 0.6 mole of e–  0.15 mole of O2 so, total moles = 0.25 mole Total volume = 5.6 Ltr. Ex. The electrochemical equivalent of copper is 0.0003296 g coulomb–1. Calculate the amount of copper deposited by a current of 0.5 ampere flowing through copper sulphate solution for 50 minutes. Sol. According to Faraday's first law, W = 2 It. W = 0.5 × 50 × 60 × 0.003296 = 0.4944 g Ex. An electric current is passed through three cells connected in series containing ZnSO4, acidulated water and CuSO4 respectively. What amount of Zn and H2 are liberated when 6.25 g of Cu is deposited? Eq. wt. of Cu and Zn are 31.70 and 32.6 respectively. Sol.  Eq. of Cu = Eq. of Zn = Eq. of H2 WH 2 6.25 WZn = = 31.70 32.6 1 Ex. The cell consists of three compartments separated by porous barriers. The first contains a cobalt electrode in 5.00 L of 0.100 M cobalt (II) nitrate; the second contains 5.00 L of 0.100 M AgNO3. Assuming that the current within the cell is carried equally by the positive and negative ions, tabulate the concentrations of ions of each type in each compartment of the cell after the passage of 0.100 mole electrons. Given Co2+ + 2e–  Co Eº = – .28 V Ag+ + e–  Ag Eº = 0.80 V + 2+ Spontaneous reaction is : 2Ag + Co Co + 2Ag Eº = 1.08 V Sol. In the left compartment. Cobalt will be oxidized to cobalt (II) ion. In the right compartment, silver ion will be reduced to silver. The passage of 0.100 mol electrons will cause the following quantities of change. Compartment 1 Effect of electrode Positive ion movement 3+ + 0.0500 mol Co 2+ –0.0250 mol Co Compartment 2 Compartment 3 + – 0.100 mol Ag + 0.0250 mol Co2+ + 0.0500 mol K+ – 0.0500 mol K+ Negative ion movement + 0.0500 mol NO3– – 0.0500 mol NO3– Changing the numbers of mol to concentrators in 5.0 L compartments and adding or subtracting yields the following results : ELECTRO CHEMISTRY Final concentrations (M) Co2+ 0.015 Co2+ – 0.210 + NO3– 0.100 NO3 0.20 M NO 3 Co 0.10 M Co – 2+ 5.0 L 0.00500 Ag + 0.0800 K 0.090 0.0900 K+ NO3– 0.0100 0.10 M NO3 0.10 M K – – 0.10 M NO 3 + 0.10 M Ag 5.0 L + Ag 5.0 L ELECTRO CHEMICAL CELL/GALVANAIC CELL/ VOLTAIC CELL Example – Daniel Cell  A cell in which the chemical energy is transformed into electrical energy.  The chemical reaction occuring in a galvanic cell is always a redox reaction.  During the chemical process, the reduction in free energy will obtain as a result in the form of electrical energy. Galvanic cell is made up of two half cells i.e., anodic and cathodic. The cell reaction is of redox kind. Oxidation takes place at anode and reduction at cathode. It is also known as voltaic cell. It may be represented as shown in fig. Zinc rod immersed in ZnSO4 behaves as anode and copper rod immersed in CuSO4 behaves as cathode. Oxidation takes place at anode : Zn(s)  Zn2+ + 2e– (loss of electron : oxidation) Reduction takes place at cathode : Cu2+ + 2e–  Cu(s) (gain of electron ; reduction) Over all process : Zn (s) + Cu2+  Cu (s) + Zn2+ In galvanic cell like Daniell cell ; electrons flow from anode (zinc rod) to the cathode (copper rod) through external circuit ; zinc dissolves as Zn2+ ; Cu2+ ion in the cathode cell picks up two electron and become deposited at cathode. e I  e Salt Bridge Zn Cu ClZn Inert electrolyte KCl, KNO3, NH4Cl etc. with Gelatin or Agar - Agar Cu +2 +2 K+ ZnSO4 CuSO 4 Zn Zn +2e (Oxidation) Anode -ve terminal +2  Cu+2 +2e   Cu (Reduction) Cathode +ve terminal  Salt bridge : 1. It allows to flow of current by completing the circuit. 2. It maintains electrical neutrality of electrolytes in two half cells. Characteristics of electrolyte used in salt bridge : 1. The electrolyte should be inert 2. The cations and anions of the electrolyte used should be of the same ionic mobility. ELECTROLYTIC CELL This cell converts electric energy into chemical energy. The entire assembly except that of the external battery is known as the electrolytic cell. e – – + e – –ve ion Anode (+) Cathode (–) +ve ion Electrolyte Electrolytic cell Cell reaction : Zn + Cu+2  Zn+2 + Cu Representation of Galvanic cell. Zn/Zn+2 // Cu+2/ Cu (C1) (C2) Anode Cathode salt bridge Representation of electrochemical cell : Ex. (i) Zn (s) | Zn2+ (aq) | Cu2+ (aq.) / Cu (s) (ii) Pt/H2 (g) (1 atm)|H+ (aq) | Cu2+ (aq.) | Cu (s). 1. First of all, the anode is written. 2. After anode, the electrolyte of the anode should be written with its concentration in the bracket. 3. A slash ( / ) is put between anode and electrolyte to denote the surface barrier between the two as they exists in different phases. 4. Then we denote salt bridge by two vertical parallel lines. (II). 5. Then, we write the electrolyte of cathode (negative electrode) followed by its concentration in bracket. 6. Finally, after a slash, we write the cathode. 7. In case of gas, the gas is to be indicated after the electrode in case of anode and before the electrode in case of cathode. Ex. Pt, H2/H+ or H+ / H2 , Pt ELECTRO CHEMISTRY ELECTRODE POTENTIAL When a strip of metal is brought in contact with the solution containing its own ions then the strip of metal gets positively charged or negatively charged and results into a potential being developed between the metallic strip and its solution which is known as electrode potential. At anode M M+n + ne— (Oxidation Potential) At cathode M+n ne— M (Reduction Potential) The value of electrode potential depends upon : (1) the nature of electrode (2) the concentration of solution (3) the temperature STANDARD ELECTRODE POTENTIAL (E0) : If the concentration of ions is unity, temperature is 25°C and pressure is 1 atm (standard conditions), the potential of the electrode is called standard electrode potential. The given value of electrode potential be regarded as reduction potential unless it is specifically mentioned that it is an oxidation potential. ELECTRO MOTIVE FORCE OF CELL OR CELLVOLTAGE : The difference in the electrode potentials of the two electrodes of the cell is termed as electro motive force [EMF] or cell voltage. Ecell = Ered (cathode ) – Ered (anode) or Ecell = Eoxi.(anode) – Eoxi.(Cathode) or Ecell = Eoxi.(anode) + Ered(cathode) Ex. Write short hand notation for the following reaction, Sn2+ (aq) + 2Ag+ (aq)  Sn4+ (aq) + 2Ag(s). Sol. The cell consists of a platinum wire anode dipping into an Sn+2 solution and a silver cathode dipping into an Ag+ solution therefore Pt(s) | Sn2+(aq) , Sn4+ (aq) || Ag+ (aq) | Ag(s). ELECTRO CHEMICAL SERIES Standard Aqueous Electrode potentials at 298K 'The Electrochemical series' Element Li K Ba Ca Na Mg Al Mn Zn Cr Fe Cd Ni Sn Pb H2 Cu I2 + Li + e  Li + Ba Ca +2 +2 + 2e  Ba – 2.90 – + 2e  Ca – 2.87 – Na + e  Na +2 – 2.71 – + 2e  Mg +3 – Al + 3e  A+F2l Mn Zn +2 +2 – + 2e  Mn – + 2e  Zn – 2.37 – 1.66 – 1.18 – 0.76 +3 – – 0.74 +2 – – 0.44 – – 0.40 Cr + 3e  Cr Fe Cd + 2e  Fe +2 + 2e  Cd +2 – – 0.25 Sn +2 – – 0.14 Pb +2 – – 0.13 Ni + 2e  Ni + 2e  Sn + 2e  Pb + – 2H + 2e  H2 Cu +2 0 – + 2e  Cu – – I2 + 2e  2I +2 Hg 2 Ag + 0.54 + 2e  2Hg – Ag + e  Ag Hg +2 0.34 0.79 0.8 – + 2e  Hg 0.85 Br2 – – Br2 + 2e  2Br 1.08 Cl2 – – 1.36 Pt Au F2 Cl2 + 2e  2Cl Pt +2 Au – + 2e  Pt +3 + 3e  Au – – F2 + 2e  2F Calculate E0cell of (at 298 K), Zn(s) / ZnSO4(aq) || CuSO4(aq) / Cu(s) given that E0Zn/Zn2+(aq) = 0.76 V E0Cu(s) / Cu2+(aq) = – 0.34 V Sol. 0 potential E , Volts – 3.05 – + Mg Standard electrode Reduction – 2.93 – K +e K Hg Hg Ex. Electrode Reduction Reaction E0cell = (S.R.P)cathode – (S.R.P)anode = 0.34 – (– 0.76) = 1.1 V 1.2 1.5 2.87 ELECTRO CHEMISTRY Ex. Given the cell Ag AgCl(s) | NaCl (0.05 M) || Ag NO 3 (0.30) | Ag (a) Write half reaction occurring at the anode. (b) Write half reaction occurring at the cathode. (c) Write the net ionic equation of the reaction. (d) calculate E°cell at 25°C. (e) Does the cell reaction go spontaneous as written ? 0 (Given E°AgCl,Cl = + 0.22 volt) ; E Ag / Ag = + 0.80 volt) Sol. (a) LHS electrode is anode and half reaction is oxidation. Ag+ + Cl–  AgCl(s) + e– ... (i) (b) RHS electrode is cathode and half reaction is reduction. ... (ii) Ag + e  Ag(s) (c) From equation (i) and (ii) cell reaction is : Cl– (0.05 M) + Ag+ (0.30 M)  AgCl(s) (d) E°cell = E°right – E°left = (0.80 – 0.22 volt = 0.58 volt (e) Yes, the e.m.f. value is positive, the reaction will be spontaneous as written in the cell reaction. Ex. An aqueous solution containing Na+ , Sn2+, Cl– & SO42– ions, all at unit concentration, is electrolysed between a silver anode and a platinum cathode. What changes occur at the electrodes when current is passed through 0 the cell ? Given : E Ag |Ag = 0.799 V,, 0 0 ESn E 0Cl |Cr – = 1.36 V , ES0 O2– |SO2– = 2 V, ESn 2+ 4+ |Sn = – 0.14 V, |Sn 2+ = 0.13 V 2 2 8 4 Sol. (A) Sn2+ is reduced and Cl– is oxidized (B) Ag is oxidized and Sn2+ is reduced (C) Sn2+ is reduced and Sn2+ is oxidized (D) H+ is reduced and Sn2+ is oxidised At anode either Ag can oxidised to Ag+ or Sn2+ to Sn4+ or Cl– to Cl2 or SO 24 to S 2O82 Their respective oxidation potential values are –0.799 V, 0.13 V, –1.36 V and –2 V. From these values, it is evident that Sn2+ would be oxidised first, followed by Ag at anode. At cathode, either Na + can get reduced to Na. or Sn2+ to Sn or H+ to H2. The reduction potential value for Na+ is highly negative while for Sn2+ | Sn is –0.14 V and for H+ + e– 1    1/2 H 2  E H + |H 2 = 0.059log –7  is –0.413 V. Thus Sn 2+ will get reduced at cathode 10   followed by H+  (C) IMPORTANT POINTS ABOUT SERIES : 1. Electrode whose standard reduction potential (SRP) is less, act as anode, and other one which has high reduction potential acts as cathode. 2. Metals near the top of the series are strongly electropositive. 3. Metals near the top of the series can displace more electronegative metal below them from their salt. For example : 2AgNO3 + Cu Cu(NO3)2 + 2Ag CuSO4 + Ag Reaction is not observed 4. Metal above hydrogen can displace H2 from dilute acid For example Na + H2SO4 Na2SO4 + H2 Cu + H2SO4 Reaction is not observed 5. Hydroxides of metal in the upper part of series are strongly basic while hydroxides of a metal in lower part are weakly basic. 6. The activity of non metals increases from top to bottom. 7. The metals which come below copper, form unstable oxides i.e., these are decomposed on heating. NERNST EQUATION With the help of Nernst equation, we can calculate the electrode potential of electrode or EMF of cell. E cell = E 0 – RT [Product] log e nF [Reactant] Where - E0 = standard electrode potential R = gas constant T = temperature (in K) F = Faraday (96500 coulomb mol–1) n = no. of e– gained or loosed in balanced equation. E cell = E 0 – or 0.0591 [Product] 0.0591 log10 logQ = E° – n [Reactant] n Let, in the cell : at Anode : +n M1 M1 + ne– for this reaction – E oxi = E at Cathode : 0 oxi  M1+n  0.0591 – log10 n  M1  ....(1) M2  0.0591 log10 n  M +n 2   ....(2) +n M2 + ne— M2 for this reaction – E red. = E 0red. – Note : Concentration of solid taken as unity. so [M1] = [M2] = 1 We know that EMF of cell is EMF = Eoxi (Anode) + Ered. (Cathode) By adding equation (1) & (2) we get : EMF = E0oxi – 0.0591 1 0.0591 log log  M1+n  + E 0red. – n  M +n n 2   ELECTRO CHEMISTRY E 0 oxi + E 0red.  – 0.0591  log  M1+n  – log  M 2+n    n   EMF = E cell = E 0cell – Ex.  M1+n  0.0591 log  +n  n  M 2  The 0.1M copper sulphate solution in which copper electrode is dipped at 25°C. Calculate the electrode potential of copper electrode [Given E0 Cu+2 / Cu = 0.34V] Sol. Cu+2 + 2e— Cu E red = E red0 – 0.0591 [Product] log n [Reactant] here n=2 so E = 0.34 – 0.0591 log10 2 = 0.34 – 0.03 = 0.31 Volts Ex. The EMF of the cell Cr / Cr+3 (0.1M) // Fe+2 (0.01M) / Fe (Given E° Cr+3 / Cr = – 0.75V E° Fe+2 / Fe = – 0.45 V) Sol. Half cell reactions are : At Anode  [Cr Cr+3 + 3e-— ] × 2 At Cathode  [Fe+2 + 2e— Fe ] × 3 ——————————— 2Cr + 3Fe+2 2Cr+3 + 3Fe over all reaction E°cell = oxidation pot. + Reduction pot. = 0.75 + (–0.45) = 0.30 Ecell = E0 – 0.0591 [Product] log n [Reactant] 2 = 0.30 – Cr +3  0.0591 log  3 6  Fe +2  = 0.30 – 0.0591  0.12 = 0.30 – 0.24 = 0.26 Volt. log 6 6 0.013 DIFFERENT TYPES OF ELECTRODES : 1. Metal - Metal ion electrode M/Mn+ Mn+ + ne–  M(s) E = E° + 2. 0.0591 log [Mn+] n Gas - ion Electrode Pt / H2 (P atm) / H+ (XM) as a reduction electrode H+ (aq) + e–  1 H (P atm) 2 2 1 E = E° – 0.0591 log 3. PH22 [H + ] Oxidation - reduction Electrode Pt / Fe2+ , Fe3+ as a reduction electrode Fe3+ + e–  Fe2+ [Fe 2+ ] E = E° – 0.0591 log [Fe3+ ] 4. Metal - Metal insoluble salt electrode Ex. Ag/AgCl, Cl– as a reduction electrode AgCl (s) + e–  Ag (s) + Cl– 0 – E Cl– /AgCl/Ag = E Cl – /AgCl/Ag – 0.0591 log[Cl ] Note : 1. 2. The value of electrode potential does not depend on stoichiometry of half reactions. In electrolytic cell, cathode is negative terminal whereas in Galvanic cell cathode is positive terminal. 3. Work obtained by electro chemical cell = –free energy of cell G = –nFEcell 4. G° = –nFE°cell Cell reaction is spontaneous, if G = – ve or Ecell = +ve 5. Cell reaction will be in equilibrium Ex. G = 0 or –nFEcell; = 0 In nernst equation 0.0591 [Product] E cell = E 0cell – log10 n [Reactant] 6. or Ecell = 0 If Ecell = 0, then E 0cell = 0.0591 [Product] log10 n [Reactant] 216 ELECTRO CHEMISTRY or E 0cell = 0.0591 log10 K n or E 0cell = RT log e K nF put the value of E0Cell in G° = – nFE°cell we get ΔG 0 = –nF RT log e K  G° = – RT loge K nF Ex. Calculate R.P. of hydrogen electrode at 298K which is prepared with the help of aq. solution of acetic acid with 0.1 M conc at 1 atm pressure Ka = 1.8 x 10 –5. Sol. [H+] = Ka × c = 1.8×10 –5 ×10 –1 = 1.8×10 –6 2H+ + 2e–  H2 ERedn= E 0red – Ex. PH2 0.059 log [H + ]2 2 ERedn= – 1   0.059 log  –6  2  1.8×10  ERedn = – 0.059 x 5.74 = –0.169 V 2 (E 0Reduction = 0) = – 0.059 [6 – log (1.8)] 2 Which is stronger oxidizing agent (i) K2Cr2O7 in solution in which [Cr2O72–] = 0.1 M, [Cr3+] = 10–2 M and [H+] = 10–1 M (ii) KMnO4 in a solution in which [MnO 4–] = 10–1M, [Mn2+] = 10–2M, [H+] = 10–2 M E 0 Cr2O72 – / Cr 3 = 1.33 V Sol. E 0MnO4 / Mn 2 = 1.51 V (i) 14H+ + Cr2O72–  2Cr+3 + 7H2O + 6e– ERedn = 1.33 –  10 –4 ×10  0.059 0.059 log  x 11 –14  = 1.33 – 10 6 6   ERedn = 1.33 – 0.649 = 1.330 – 0.108 = 1.222 V 6 (ii) 5e– + 8H+ + Mn O4  Mn2+ + 4H2O  10  0.059 log  –16 –1  = 1.51 – 0.059 x 3 = 1.51 – 0.18 = 1.33 V 5  10 ×10  –2 ERedn = 1.51 – EReduction is more so, good oxidising agent Ex. Calculate Ecell of Cl2 (g) Cl – (aq) Cr2O72– , Cr +3  in H 2SO 4  = 0.05M 0.1atm 10 –2 M 0.01M 0.1 M Pt(s) 0 given that E Cr O2 2 7 / Cr  3 E0 Cl– | Cl2 = –1.36 V = 1.33 V Sol. 6e– + 14H+ + Cr2O72–  2Cr+3 + 7H2O [2Cl–  Cl2 + 2e–] x 3 14H+ + 6Cl– + Cr2O72–  3Cl2 + 2Cr+3 + 7H2O E 0cell = 1.33 – (+1.36) = – 0.03 3+ 2 Ecell = – 0.03 – 3 [Cr ] [PCl ] 0.059× 23 0.059 log + 14 – 6 2 2– = – 0.03 – 6 6 [H ] [Cl ] [Cr2O 7 ] Ecell = – 0.26 V Ex. The E°cell for the reaction Fe + Zn2+ Zn + Fe2+ , is – 0.32 volt at 25°C. What will be the equilibrium concentration of Fe2+ , when a piece of iron is placed in a 1 M Zn2+ solution ? Sol. We have the Nernst equation at equilibrium at 25°C 0.0591 log K n E° = ... (i) Since E°Cell for the given reaction is negative, therefore, the reverse reaction is feasible for which E°cell will be + 0.32 V, Thus for Zn + Fe2+ Now , E° = [Zn 2+ ] 0.0591 log or [Fe2+ ] n log [Fe2+] = – 10.829 [Fe ] = 1.483 x 10 2+ –11 Fe + Zn2+ ; E°Cell + 0.32 V 0.32 = 1 0.0591 log [Fe2+ ] 2 Taking antilog, M REFERENCE ELECTRODE An electrode used to calculate the electrode potential of other electrodes. (a) Standard Hydrogen Electrode (SHE) : It consist of a platinum electrode over which H2 gas (1 atm pressure) is bubbled and the electrode is immersed in a solution that is 1 M in H+ at 25°C. 2H+ + 2e–  H2(g) (1 atm) (1M) the potential of this electrode at 25°C is taken as Zero volt. (b) Calomel Electrode : Hg (l) / Hg2Cl2 (s)/Cl– (aq) It is prepared by a Pt wire in contact with a paste of Hg and Hg2Cl2 present in a KCl solution. reaction 1 Hg2Cl2(s) + e–  Hg (l) + Cl– 2 E = E° – 0.0591 log [Cl–]. ELECTRO CHEMISTRY HEAT OF REACTION INSIDE THE CELL Let n Faraday charge flows out of a cell of e.m.f. E, then –G = nFE ........(i) Gibbs Helmoholtz equation (from thermodynamics) may be given as,  ΔG  G = H + T   T  p .......(ii) From Eqs. (i) and (ii), we have   (–nFE)   E  –nFE = H + T  = H – nFT     T  p  T  p   E  H = – nFE + nFT    T  p ENTROPY CHANGE INSIDE THE CELL We known that G = H – TS or G = H – TS ...(i) where G = Free energy change , H = Enthalpy change and S = entropy change. According to Gibbs Helmoholtz equation,  ΔG  G = H + T   T  p .......(ii) From Eqs. (i) and (ii), we have  ΔG  –TS = T   T  p or  ΔG  or S = –   T  p  E  S = nF    T  p  E  where   is called temperature coefficient of e.m.f. of the cell.  T  p SOME COMMERCIAL BATTERIES Any battery or cell that we use as a source of electrical energy is basically an electrochemical cell where oxidising and reducing agents are made to react by using a suitable device. In principle, any redox reaction can be used as the basis of an electrochemical cell, but there are limitations to the use of most reactions as the basis of practical batteries. A battery should be reasonably right and compact and its voltage should not vary appreciably during the use. There are mainly two types of cells : (i) primary cells and (ii) secondary cells. In primary cells, the reaction occurs only once and the battery then becomes dead over a period of time and cannot be used again. (For example, dry cell, mercury cell.) Contrary to this, secondary cells can be recharged by passing a current through them so that they can be used again and again. (For example, lead storage battery, nickel- cadmium storage cell.) PRIMARY CELLS The most familiar type of battery is the dry cell which is a compact of Leclanche cell known after its discover Leclanche pitch seal (fig.): In this cell, the anode consists of a zinc container and and carbon. The space between the electrodes is filled with graphite (carbon) cathode (with metal cap) MnO2 + C a moist paste of NH4Cl and ZnCl2. The electrode reactions paste of NH4Cl, ZnCl2 the cathode is a graphite rod surrounded by powdered MnO2 are complex, but they can be written approximately as follows. zinc anode Anode Zn(s) Zn+2 + 2e– Cathode MnO2 + NH4+ + e–  MnO(OH) + NH3 In the cathode reaction, manganese is reduced from the +4 Fig. : A Dry Cell oxidation state to the +3 state. Ammonia is not liberated as a gas but combines with Zn2+ to form Zn(NH3)42+ ion. SECONDARY CELLS The most important secondary cell is the lead storage battery. It consists of a lead anode and a grid of lead packed with lead dioxide as cathode. A solution of sulphuric acid (38 percent by mass or having density 1.30 g cm–3) is used as an electrolyte. When the cell discharges, it operates as a voltaic cell. The lead is oxidised to Pb+2 ion and lead plates acquire a negative charge. At Anode : Pb Pb+2 + 2e– .....(i) +2 Pb ions combine with sulphate ions of solution to form insoluble lead sulphate PbSO4 , when begins to coat lead electrode Pb+2 + SO4 2– PbSO4 (Precipitation) .....(ii) The electrons given by Pb are used at cathode (PbO2 electrode) At Cathode : PbO2 + 4H+ + 2e– Pb2+ + 2H2O .....(iii) Pb+2 + SO4–2 PbSO4 (Precipitation) .....(iv) Over all cell reaction add equation (i), (ii), (iii) & (iv) Pb + PbO2 + 4H+ + 2SO42– 2PbSO4 + 2H2O Ecell is 2.041 volt (A 12 V lead storage battery is generally used which consist of 6 cells each producing 2 volt) When a potential slightly greater than the potential of battery is applied, the battery can be recharged. During recharging the cell is operated like an electrolytic cell i.e. now electrical energy is supplied to it from an external source. The electrodes reactions are reverse of those that occur during discharge : 2PbSO4 + 2H2O Pb + PbO2 + 2H2SO4 It is clear that H2SO4 is used up during the discharge & produced during recharging reaction of cell. Another secondary cell is the nickel-cadmium storage cell which has a longer life than the lead storage cell but it is more expensive to manufacture. ELECTRO CHEMISTRY FUEL CELLS It is possible to make batteries in which the reactants are fed Water anode (-) (+) cathode continuously to the electrodes. Electrical cells that are designated to convert the free energy from the combustion of fuels such as hydrogen, carbon monoxide or methane aqueous electrolyte directly into electrical energy are called FUEL CELLS. One of the most successful fuel cells uses the reaction of hydrogen with oxygen to form water (fig.). This cell has been used for O2 H2 electric power in the Apollo space programme. The water vapour produced was condensed and added to the drinking water supply for the astronauts. Fig. : A simple fuel cell In the cell (fig.) hydrogen and oxygen are bubbled through a porous carbon electrode into concentrated aqueous sodium hydroxide. Catalysts are incorporated in the electrode. The electrode reactions are : Anode 2[H2(g) + 2OH– (aq)  2H2O() + 2e– ] Cathode O2(g) + 2H2O() + 4e– 4OH– (aq) ———————————————————————— Overall reaction : 2H2(g) + O2(g)  2H2O() This cells runs continuously as long as the reactants are supplied. CORROSION Corrosion is basically an electrochemical phenomenon. The rusting of iron, the tarnishing of silver, development of a green coating on copper and bronze are some of the examples of corrosion. Corrosion causes enormous damage to building, bridges, ships and to many other articles made of iron. In corrosion, a metal is oxidised by loss of electrons to oxygen and forms metal oxide. Corrosion of iron(which is commonly known as rusting) occurs in presence of water and oxygen (air). Rust Fe2O3 Air O2(g) cathode Z n e Z  n S Z 4O n 2+ Z 2n +e C  -(v O x eA iedoar tn em tio d a)n l l- nA I N e H g t4 r aC A -le S al tB ir gd e etc g a.tor iw ly th e K G eC I la,t iK n N o rO 3, C S K u +O 4 C - eu 2u + e  C u ev atdh cto d uer etio m )ina l Fe2 water e C 2+ C u R ( C + anode O2 + 4H+ + 4e-  2H2O Fe  Fe2+ + 2eIron Fig. : The rusting of Iron (Iron in contact with water forms the anode and in contact with air forms the 2+ cathode. At the anode iron is oxidised to Fe ,and at the cathode oxygen is reduced to water Although the chemistry of corrosion is complex but it is understood that at one spot of an iron object oxidation occurs and that spot behaves as an anode. Anode Fe(s) Fe2+ (aq) + 2e– ; ( E 0 Fe 2  / Fe  0 .44 V ) Electrons released at anode move through the metal and go to another spot on the metal and reduce the oxygen in presence of H+ ions (which is believed to be available from H2CO3 formed due to dissolution of CO2 in water). This spot works as a cathode : Cathode O2(g) + 4H+ (aq) + 4e–  2H2O () ; (E0 = 1.23 V)  Here Fe2+ ions move through water on the surface of the iron object. (If water present is saline, it will help more in carrying the current in the miniature cell thus formed and will enhance corrosion). The overall reaction of the miniature cell is the sum of the cathode and anode reactions as follows : 2Fe(s) + O2(g) + 4H+(aq) 2Fe2+(aq) + 2H2O() ; E0Cell = 1.67 V) 2+ The Fe ions are further oxidised by atmospheric oxygen to Fe3+ (as Fe2O3) and comes out as rust in the form of a hydrated iron (III) oxide expressed as Fe2O3 . xH2O. 4Fe2+(aq) + O2(g) + 4H2O()  2Fe2O3(s) + 8H+ H+ ions produced in the above reaction help further in rusting. Impurities present in iron also enhance rusting by setting a number of miniature cells. Very pure iron does not rust quickly. PREVENTION FROM CORROSION Corrosion of a metal is prevented by applying protective coating (such as grease, paint or metal coatings) on its (metals) surface. In case of iron this coating is done in two ways : (i) electrolysis (Cr, Ni and Cd coating) (ii) dipping iron objects in a molten metal (Zn and Sn coating). Use of zinc coating to protect iron is called Galvanisation. In galvanising iron, zinc being more reactive than iron serves as an anode and is oxidised 0 (E0 2+ = –0.76V and E 2+ = –0.44V ). Zn /Zn Fe /Fe Zn Zn Water drop 2+ 2+ Zn + 2e - O2(g) O2 + 2H+ + 2e2H2O Iron cathode Zinc anode Fig. : Galvanised iron : Cathodic protection of iron in contact with zinc is provided by galvanization. Here it is important to note that iron does not rust even after Magnesium anode 2+ Mg Mg + 2e zinc coating is broken which is not true in case of tin coating over iron (E0 Sn2  / Sn  0.14 V) . Now if the coating is broken, iron is exposed and iron being more active than both copper and tin, is corroded. Here iron corrodes more rapidly than it does in the absence of tin. But tin protects copper in the same way as zinc protects iron ( E0 Cu2  / Cu  0.34 V ). Instead of coating more reactive metals on iron, the use of such metals is made as sacrificial anode. This method of preventing iron from corrosion is called CATHODIC PROTECTION. In this method, a plate of reactive metal (Zn or Mg) is buried beside the iron pipe or tank and connected Iron pipe (cathode) O2 + 2H2O + 4e 4OH Fig. : Cathodic protection by connecting a block of magnesium to an iron pipe burried underground. to it by wires. Here iron becomes cathode and more reactive metal becomes anode. The reactive metal anode is sacrificed to protect the iron. Since these reactive metal plates are oxidised quickly, they are replaced from time to time which is easy to do. Ex. During the discharge of a lead storage battery the density of H2SO4 falls from 1 g/cc to 2 g/C, H2SO4 of density of 1 g/C. C is X% by weight and that of density of 2 g/c.c is Y% by weight. The battery holds V litre of acid before discharging. Calculate the total charge released at anode of the battery. The reactions occurring during discharging are. At anode : Pb + SO 24  PbSO4 + 2e– At cathode : PbO2 + 4H+ + SO42– +2e–  PbSO4 + 2H2O ELECTRO CHEMISTRY Sol. Mass of acid solution before discharge of lead storage battery (LSB) = (V x 103 x 1 ) g = (1000 x V 1 )g X   Mass of H2SO4 before discharge of LSB = 1000× Vρ1 × g = (10 x V1 X)g 100   Net reaction during discharging : Pb + PbO2 + 2H2SO4  PbSO4 + 2H2O From the reaction, it is evident that the moles of electron exchanged (lost at anode and gain at cathode) is equal to the moles of H2SO4 consumed or moles of H2O produced. Let the moles of H2SO4 produced be x, then Mass of H2O produced during discharge of LSB = (18x) g Mass of H2SO4 consumed during discharge of LSB = (98x) g  Mass of H2SO4 after discharge of LSB = [(10 V1 X)] – 98x] g Mass of acid solution after discharge of LSB = [(1000 V1 )– 98x + 18x] = [(1000 V1 ) – 80x]g Mass of H 2SO4 after discharge % of H2SO4 after discharge of LSB = Mass of acid solution after discharge × 100 [(1000× Vρ1 ) – 98x] Y = [(1000× Vρ ) – 80x] x 100 1 x can be calculated as all other quantities are known.  Total charge released at cathode, Q = nF = xF. Ex. A lead storage cell is discharged which causes the H 2SO4 electrolyte to change from a concentration of 34.6% by weight (density 1.261 g ml –1 at 25°C) to one of 27% by weight. The original volume of electrolyte is one litre. Calculate the total charge released at anode of the battery. Note that the water is produced by the cell reaction as H2SO4 is used up. Over all reaction is. Pb(s) + PbO2(s) + 2H2SO4()  2PbSO4(s) +2H2O() Sol. Before the discharge of lead storage battery, Mass of solution = 1000 x 1.261 = 1261 g 1261×34.6 = 436.3 g. 100 Mass of water = 1261 – 436.3 = 824.7 g After the discharge of lead storage battery, Let the mass of H2O produce as a result of net reaction during discharge (Pb + PbO2 + 2H2SO4  2PbSO 4 + 2H2O) is x g Mass of H 2SO4 = x  Moles of H2O produced = 18 = moles of H2SO4 consumed Mass of H2SO4 consumed = x x 98 18 98 x +x 18 Mass of H 2SO 4 left % by the mass of H2SO4 after discharge = x 100 = 27 Mass of solution after discharge Now, mass of solution after discharge = 1261 – 98x 18 = x 100 = 27 98x 1261  x 18 436.3   x = 22.59 g E = E° – (Products) .0592 log (Reactants) n  E = E° – (Anodicion concentration) .0592 log (Cathodic ion concentration) n ELECTRO CHEMISTRY CELL Electrolytic cell (1) Electrochemical cell Electrolytic cell : Converts electrical energy into chemical energy + – Anode (+) Pt Cathode (–) Pt Electrolyte : Molten NaCl NaCl (2) Cathode : Na +(aq.) + e Na(s) Anode : – Cl(aq.)  Cl2(g) + e 1 2 Deposition of material at any electrode follow faraday's law of electrolysis. Faraday's Ist Law : w = Z it w= where M × it n  factor  9 65 00 w = mass deposite (gm) M = molar mass i = current (Amp.) t = time (sec.) Faraday's second law : (–) Cell (3) (+) (1) At any electrode for material deposited. w1 w 2 w 3 = = E1 E 2 E3 (–) (+) (2) (–) (+) (3) Note : Order of discharge potential. Cathode : Au+3 > Ag+ > Cu+2 > Zn+2 > H2O > Al+3 > Mg+2 > Na+ > Li+ – – – – Anode : SO 2– 4 < NO 3 < H 2 O < Cl < Br < I PRODUCTS OF ELECTROLYSIS OF SOME ELECTROLYTES S.No. (4) Electrolyte Electrode (i) Aqueous NaCl Pt or Graphite Cl2 H2 (ii) Fused NaCl Pt or Graphite Pt or Graphite Cl2 O2 Na (iii) Aqueous NaOH (iv) Fused NaOH Pt or Graphite O2 Na (v) Pt or Graphite O2 Cu (vi) Dilute HCl Pt or Graphite Pt or Graphite Cl2 O2 H2 (vii) Dilute H2SO4 (viii) Aqueous AgNO3 Pt of Graphite O2 Ag Aqueous CuSO4 Cu Zn Inert electrolyte KCl, KNO3, NH4Cl etc. with Gelatin or Agar - Agar +2 K+ CuSO 4 Zn Zn +2e (Oxidation) Anode -ve terminal +2  Cu+2 +2e   Cu (Reduction) Cathode +ve terminal E°Cell = SRPcathode – SRPAnode = SRPcathode + SOPat anode Half cell reaction : Anode: 2 Zn(s)  Zn (aq.) + 2e Cathode : 2 Cu (aq.) + 2e Cus 2 2 Zn(s) + Cu (aq.)  Zn (aq.)  Cu(s) [Zn +2 ] Q= ;n=2 [Cu +2 ] Nearest equation : Cu +2 ZnSO4 Cell reaction : e Salt Bridge Cl- (7) H2 I  Zn (6) H2 Electrochemical Cell : converts chemical energy into electrical energy. e (5) Product obtained at cathode Product obtained at anode ECell = E°Cell – 0.059 logQ at 298 K n  ELECTRO CHEMISTRY (8) Max electrical work done = nFE° = –G° electrical work done = nFE = –G Different type of electrodes/half cell Type Example Metal - Metal ion M/M Gas - ion Pt / H2 (P atm) Half-cell reaction n+ M Metal insoluble salt Anion , Fe 3+ Fe  M(s) 3+ E  Eo  0.0591 log[M n  ] n E  E o  0.0591 log – + e  Fe AgCl (s) + e – Ag/AgCl, Cl 2+ – Cl (aq)/Hg/Hg 2 Cl2 2 [Fe 2 ] [Fe 3 ] E Cl / AgCl/ Ag  E 0Cl / AgCl /Ag  Ag (s) + Cl Hg 2 Cl2 (s) + 2e PH [H ] E  E o  0.0591 log – – Calomel electrode (9) 2+ e– 1   H2 (P atm) 2 / H (XM) Pt / Fe +n H+ (aq) + e– + Oxidation - reduction n+ Half-cell reaction – 0.0591 log [Cl ] – – E= E° –0.0591 log [Cl ] - 2Hg(I) + 2Cl (aq.) Gibb's Helmhaltz equation :   ΔG  G = H + T   T   ΔG  H = –nFE + nFT   T  P CONDUCTION IN ELECTROLYTES Symbol Unit Specific Change with concentraction Formula Factors Conductance Specific Conductivity Molar Conductivity C k m –1  Conductance of volume within electrode Decrease with in concentration C 1 R (i) Nature of electrolyte (ii) Concentration of electrolyte (iii) Type of cell. –1 –1  cm Conductance of unit volume Decrease with in concentration k = C × cell constant (i) Nature of electrolyte (ii) Concentration of electrolyte –1 2 –1  cm mol Conductance of that volume which contain exactly one mole Increase with in concentration k = m = K × V V = Volume of solution contain 1 mole of electrolyte (i) Nature of electrolyte (ii) Concentration of electrolyte (10) KOHLRAUSEH'S LAW : Λ m (A x By ) = xλ + + yλ – Λ m (K 2SO 4 ) = 2λ + + λ – Λ m (Na 3 PO 4 ) = 3λ + + λ – Λ m [Fe 2 (SO 4 )3 ] = 2λ + + 3λ – (11) FORMULA l A 1000 (2) m = k × M (1) R =  × (3) λ eq. = k × 1000 N (4) for strong electrolyte m = m – b C 