21.-Electrochemistry-Notes.pdf

Electrochemistry General Introduction Redox reactions Oxidation and reduction Property Oxidation Reduction 1 Hydrogen Loss Gain 2 Oxygen...

Electrochemistry General Introduction Redox reactions Oxidation and reduction Property Oxidation Reduction 1 Hydrogen Loss Gain 2 Oxygen Gain Loss 3 Electron Loss Gain 4 Oxidation No. Increase Decrease Oxidising agent or oxidant Reducing agent or reductant 1 Under goes reduction Undergoes oxidation 2 Gain electrons Donates electrons 3 Oxidation no. of its atoms increases Oxidation no. of its atoms decreases For example, consider the reaction Zn(s)  CuSO4 (aq)  ZnSO4 (aq)  Cu(s) Or Zn(s)  Cu2 (aq)  Zn2 (aq)  Cu(s) ……(i) Concept Ladder Zn is oxidized on Zn2+ ions while Cu2+ has been Displacement reactions reduced to Cu. can and can't be Similarly in the reaction redox reactions. Single Cu(s)  2 AgNO3 (aq)  Cu(NO3 )2 (aq)  2 Ag(s) displacement reactions Or Cu(s)  2 Ag  Cu (aq)  2 Ag(s) ……(ii)  2 are redox reactions where as double displacement Electrochemistry Cu has been oxidized to Cu2+ ions while Ag+ has reactions are not redox been reduced to Ag. reactions. 1.  Again in the reaction Zn(s)  H2 SO4 (aq)  ZnSO4 (aq)  H2 (g ) Previous Year’s Question Or Zn(s)  2 H (aq)  Zn2 (aq)  H2 (g) ……(iii) Zn has been oxidized to Zn2+ whereas H+ In acidic medium H2O2 changes ions have been reduced to H2 gas. Cr2O72- to CrO5 which has two The substance which gets reduced (—O—O—) bonds. Oxidation state oxidizes the other substance and is called of Cr in CrO5 is [NEET-2014] oxidizing agent or oxidant while the substance (1) +5 (2) +3 (3) +6 (4) -10 which gets oxidized reduces the other substance and is called reducing agent or reductant. Thus, oxidizing agent is a substance which gains Concept Ladder electrons while reducing agent is a substance According to classical which loses electrons. concept, oxidation is an Redox reaction is defined as the reaction addition of oxygen [or which is considered to be made up of two half electronegative radical/ reactions, one involving oxidation, i.e. loss of element] or removal of hydrogen [or electropositive electrons and the other involving reduction, i.e., radical/element gain of electrons. These are known as oxidation half-reaction and reduction half-reaction. For example, the reaction (i) may be split into two half reactions as under: Rack your Brain All redox reactions are exothermic. Why? Electrochemistry Electrochemistry is defined as the branch of physical chemistry, which deals with the study Definition of conversion of electrical energy from chemical Electrochemistry deals with energy which is produced in a redox reaction relationship between electrical Electrochemistry and chemical energy or how to get redox reaction by using electrical energy, which is otherwise non-spontaneous. 2. 3. Electrochemistry  Electro chemical cell concept came from the study of redox reaction This reaction happens by its own. Therefore its Gibb’s free energy is negative. Definition DG < 0 DG > O The device in which (spontaneous) (Non spontaneous chemical energy is converted Work done by the system in electrochemical into electrical energy by Electrochemistry process Work done = –DG spontaneous redox reaction is Electrical work = Charge × potential called Galvanic cell Or Voltaic Difference cell 4. Galvanic Cell Or Voltaic Cell In the above cell, the zinc ion solution begins to build up a positive charge. Similarly, Concept Ladder as copper ions plate on as copper, the solution builds up a negative charge. The half cell reactions In a galvanic cell, reduction will stop unless positive ions can move from the potential of reduction half zinc half-cell to the copper half cell, and negative cell reaction is always greater than oxidation half ions from the copper half-cell can move to the cell reaction. zinc half-cell. It is necessary that these ion flow Electrochemistry occur without mixing of the zinc ion and copper ion solutions. If copper ion comes in contact with the zinc metal. 5.  Types of Conductors Electronic conductors or Metallic conductors Electrolytic Conductors or Solution Conductors 1 It allows current to pass in metallic lattice by the It allows current to pass in molten state or in movement of electron. e.g., Cu, Ag, etc. aq. solution of electrolytes by movement of ions. e.g., NaCl(aq) or NaCl (fused). 2 There is only physical change occur during There is physical as well as chemical changes passage of current. occurs during passage of current. 3 It generally shows no transfer of matter. It involves transfer of matter in form of ions. 4 During the passage of current, resistance During the passage of current, resistance increases due to increase in temperature. generally decreases as there is decreases in Thermal motion of metal ions which results in degree of hydration of ions and viscosity of the hindrance in the flow of electrons increases with medium with increase in temperature. increase in temperature. 5 The conducting power of metal is usually high. The conducting power of electrolytic conductors is relatively low. Electrolytes and Electrolysis (1) Definition : When current is passed through Rack your Brain the aqueous solution of a substance and the substance decomposed of into its ions An aqueous solution and then the substance is termed as electrolyte copper sulfate is electrolysed using platinum electrodes in and this process is known as electrolytic another case. Will the products decomposition or electrolysis. of electrolysis be same of Examples of electrolytes are solutions of different? bases, acids, fused salts and salts in water etc. Electrolytes may be strong or weak. Concept Ladder Solutions of alcohol, glycerine, cane sugar etc., are examples of non-electrolytes. Higher is the value of Strong and Weak Electrolytes dissociation constant Electrochemistry (i) The extent or degree of dissociation of greater is the degree of different electrolytes in solution is different. dissociation and stronger is the electrolyte. 6. (ii) Strong electrolytes are the substances which are largely dissociated and form a Rack your Brain highly conducting liquid in water. e.g., All salt (except CdBr2, HgCl2), mineral acids like Mineral acids have covalent HCl, H2SO4, NHO3, etc., and bases like KOH, bond, still they are strong NaOH, etc., are strong electrolytes. They are electrolytes. Why? almost 100% ionized at normal dilution. (iii) Weak electrolytes are the substances which are low conducting liquid formed to a small extent in aqueous solution, e.g., all organic acids (except sulfonic acids), inorganic acids like H3BO3, HCN etc., and bases like ammonia, amines etc., are weak electrolytes. Concept Ladder Factors Influencing Degree of Dissociation The degree of dissociation (a) of an electrolyte in Dilution of solute follows solution is given by: ostwald's law of dilution for weak electrolytes. Mole dissoicated at any time  Total mole present at t = 0 or dissolved initially The variation of a of an electrolyte is directed by (i) Nature of solute : All ionic compounds (strong electrolytes) have a ~ 1 at normal dilution. Most of the polar covalent compounds (weak electrolytes) have a b → degree of dissociation of a > degree of dissociation of b Q.5 Calculate solubility product of BaSO4 If its conductivity and molar conductance at saturation are 3.06 × 10–6 S cm–1 and 153 S cm2 mol–1 respectively. A.5 BaSO4  Ba 2  SO24 s s KSP = s2 K  1000  3.06  106  1000  s= =  1   153 S cm mol  2 0 m Ksp = (2 × 10–5)2 = 2 × 10–5 mol/lit = 4 × 10–10 Q.6 The equivalent conductances of sodium chloride, hydrochloric acid and sodium acetate at infinite dilution are 126.45, 426.16 and 91.0 ohm–1 cm2 equiv–1, respectively at 25°C. Calculate the equivalent conductance of acetic acid at infinite dilution. A.6 According to Kohlrausch’s law  CH COONa  CH COO  Na  91.0.....1. 3 3  HCl  H  Cl   426.16......2  NaCl  Na  Cl   126.45.....3 By adding eqn (1) and (2) and subtracting (3) CH COO  Na  H  Cl   Na  Cl   91.0  426.16  126.45 3 CH COO  H   CH COOH  390.7 ohm1cm2equiv 1 3 3 Electrochemistry 18. FACTORS AFFECTING ELECTROLYTIC CONDUCTANCE Rack your Brain Inter ionic attraction :- inter ionic attraction between ions of solute Conductivity of solution is more, then the conductance will be less. dependes upon the polarity of Polarity of solvent :- solvent. Would the solution of If solvent has high-dielectric constant then benzene and toluene conduct electrolysis? the ionization and conductance will be higher. Viscosity of medium :- On increasing the viscosity of medium, the conductance decreases. Temperature :- As the temperature of electrolytic solution is increased, the conductance Concept Ladder increases Transport number is the Hydrated size : Due to hydration of ions fraction of current carried conductance decreases. by ion. Dilution :- Transport number = (i) On increasing the dilution conductance Current carried by ion (G) increases. Total current carried For strong electrolyte on dilution interionic force of attraction decreases therefore conductance increases. For weak electrolyte with dilution degree of dissociation (a) increases therefore conductance increases. (ii) On dilution specific conductance Rack your Brain decreases because on dilution number of ions in 1 ml solution decreases. Is there any relationship between (iii) On dilution equivalent and molar molar conductance and dilution? conductance increases because Electrochemistry with dilution normality or molarity decreases. 19.  Representation of a cell Concept Ladder In a galvanic cell, cathode is positive with respect to anode. Net reaction Q.7 Give the cell reaction from the cell notation Zn(s) | Zn2+ (aq) || Fe3+ (aq)|, Fe2+(aq)| Pt A.7 The half cell reactions are Anode Zn(s) → Zn2+ (aq) + 2e– Cathode Fe3+(aq) + e- → Fe2+(aq) and the cell reaction is : Zn(s) + 2Fe3+(aq) → Zn2+(aq) + 2Fe2+(aq) Electrode potential Concept Ladder Metal acquires either a negative or positive Decreasing order of metal- charge w.r.t. the solution when it is placed in a lic character is given below: solution of its ions. By this a definite potential is K > Na > Ba > Ca > Mg > Al developed between the metal and the solution. > Zn > Fe > Ni > Sn This difference in potential is known as electrode H > Cu > Hg > Ag > Au > Pt potential. It depends on the concentration of ions, nature of electrode and temperature. Condition in which the circuit stops working Electrochemistry 20.  No current will flow. In this situation salt bridge is introduced. Salt bridge construction : Rack your Brain 1. It is inverted U–shaped tube 2. Contains inert electrolyte like KCl, KNO3, On what criteria the sign NH4,Cl etc. in agar–agar gel & gelatin. convention of electrodes is decided ? 3. Ions of electrolyte do not react with ions of electrode solution. Salt bridge functions : Concept Ladder 1. To complete circuit. Another function of salt 2. To maintain electrical neutrality. bridge is that it prevents liquid-liquid junction Standard Reduction Potential : potential i.e., the poten- Conditions: tial difference that arises Electrochemistry between the two solutions when they are directly in contact with each other 21.  Concentration = 1 M Rack your Brain For gases pressure = 1 atm Temperature = 298 K Why current drops to zero when salt bridge is pulled out? According to the IUPAC conventions, Standard electrode potential = Standard reduction potential Cell potential Or emf of the cell (Ecell) Concept Ladder The difference between electrode potential Electrode an which of two half cells is known as cell potential oxidation occurs is called Ecell  EOP A  ERP c anode (-ve pole) Electrode on which Electrochemistry Ecell  ERP C  ER P A reduction occurs is called Q.8 For the cell reaction 2Ce4+ + Co → 2Ce3+ cathode (+ve pole) 22.  + Co2+, E0cell is 1.89 V E0CO2+ |CO is 0.28 V, what is the value of ECe4+ |Ce3+ 0 A.8 E0cell  ECe 0 4 |Ce3   E0CO2  |CO 1.89 = E0 4+ 3+ – (–.28) Rack your Brain Ce |Ce E0Ce4+ |Ce3+ = 1.61 V Which electrode of a galvanic cell corresponds to the higher potential energy. Q.9 Which of the following metal displaces hydrogen from H2SO4 solution or acidified water? (1) Mg (2) Al (3) Fe (4) All of these A.9 (4) Previous Year’s Question Relationship between Gibb’s free energy change and emf of cell The pressure of H2 required to Work done = Decrease in free energy change make the potential of H2 elctrode Charge on 1 mole e– = NA  e is zero in pure water at 298K is = 6.023 × 1023 × 1.6 × 10–19 (1) 10-10 atm (2) 10-4 atm = 96500 = 1 Faraday (3) 10-14 atm (4) 10-12 atm [1 Faraday = charge on 1 mole of electron] Charge of n moles of electron Concept Ladder q = nF Electron flow from anode Work done = charge × cell potential to cathode in the external DG = –nFEcell circuit. Inner circuit is Reactions DG E completed by the flow of ions through the salt Electrochemistry Spontaneous (–) (+) bridge. Non- spontaneous (+) (–) Equilibrium 0 0 23.  Electrochemical Series The elements arranged in the decreasing order of reduction potential, the series attained is known as Electrochemical series Electrochemistry 24.  Q.10 If E0 for Au+ + e- → Au (s) is 1.69 V & E0 for 3e- + Au+3 → Au is 1.40 V, then E0 will be (1) 0.19 V (2) 2.945 V (3) 1.255 V (4) none A.10 (3) Au+ + e– → Au(s) E° = 1.69 V..(1); DG0 1 Au3+ + 3e– → Au(s) E° = 1.40 V..(2); DG0 G 3  G 2  G1 2 –2 × F × E° = –3 × F × 1.40 + 1 × 1.69 × F E° = 1.255 V Concept Ladder Nernst equation: When the concentration is not equal to 1 Molar. ERP ↑ = Reduction tendency ↑ Then we use Nernst equation EOP ↑ = Oxidation tendency ↑ If nothing is given According to thermodynamics, E°RP ↑ = Reduction tendency ↑ DG = DG° + RT ln Q E°OP ↑ = Oxidation tendency ↑ (\–DG = nFE and –DG° = nFE°) –nFE = –nFE° + RT lnQ Ecell = E0 − 2.303RT log Q cell nF where E0 = standard electrode potential, R = gas constant, T = temperature (in K) F = Faraday (96500 coulomb mol–1), Previous Year’s Question n = number of moles of e– gained lost or If the E0Cell for a given reaction transferred in balanced equation. has a negative value, which of At 25°C, above equation may be written as the following gives the correct relationship for the value of DG0 Ecell = E0cell − 0.0591 log Q n and Keq? [NEET 2016] 0.0591 [P] (1) DG > 0; Keq < 1 0 Ecell = Ecell − log 0 n [R] (2) DG0 > 0; Keq > 1 Electrochemistry (3) DG0 < 0; Keq > 1 (4) DG0 < 0; Keq < 1 25.  Q.11 Calculate the EMF of a Daniel cell when the concentration of ZnSO4 and CuSO4 are 0.001 M and 0.1 M respectively. The standard EMF of the cell is 1.1V A.11 E = 1.159 V Zn(S) + CuSO4 → ZnSO4 + Cu 0.0591  Zn2  0.0591 10 3 Ecell  E0cell  log  2   1.1  log 7  1.159 V 2  Cu  2 10 Q.12 Calculate E0 and E for the cell Sn | Sn2+ (1M) || Pb2+(10-3M) | Pb, E0 (Sn2+| Sn) = -0.14V, E0 (Pb2+| Pb) = -0.13V. Is cell reaction is feasible? A.12 No, Ecell = -0.078V 0.059  Sn2+  0.059 Ecell = -E0Sn2+ /Sn + EPb 0 - log  2+  = +0.14 - 0.13 - log1013 2 Pb 2 2+ /Pb   0.059 Previous Year’s Question Take  0.01    0.078 V 2 Applications of Nernst Equation Consider the half-cell reduction reaction (i) Calculation of electrode potential (ERP or EOP) Mn2+ + 2e- → Mn, E0 = -1.18V Mn2+ → Mn3+ + e-, E0 = -1.51V 0.0591 1 The E0 for the reaction, ERP = E RP  0 log n [M  n ] 3Mn2+ → Mn0 + 2Mn+3 If [M+n] increases, then ERP increases and possibility of the forward reaction are respectively? (ii) Calculation of electrode potential and pH of [NEET 2016] hydrogen electrode - (1) -4.18 V and yes 2H+ + 2e– → H2(g) (2) +0.33 V and Yes (3) +2.69 V and No 0.0591 PH ERP = E RP  0 log 2 2 (4) -2.69 V and No 2 [H ] 0.0591 1 \ E 0RP = 0 ; ERP = E RP  0 log  2 2 [H ] Rack your Brain [ PH2 is taken 1 atm] Electrochemistry Why Cu2+ dispropostionates in ERP = 0.0591 log [H ] + aqueous solution? ERP = –0.0591pH EOP = +0.0591 pH 26. (iii) Calculation of equilibrium constant (Keq) Previous Year’s Question and DG° From Nernst equation – In the electrochemical cell; Zn | ZnSO4 (0.01M) || CuSO4(1.0M) 0.0591 [P] Ecell = E Cell − log | Cu the emf of this Daniell cell n [R ] is E1 when the concentration of At equilibrium, Ecell = 0 and [P] = K ZnSO4 is changed to 1.0M and eq [R] that of CuSO4 changed to 0.01 0.0591 M the emf changes to E0. From E Cell = logK eq n the followings, which one is the 2.303RT relationship between E1 and E2? E Cell = log K eq nF [NEET 2017]  (1) E1 < E2 nF E Cell = 2.303 RT logKeq (2) E1 > E2 DG° = –2.303 RT logKeq (3) E2 = 0 ≠ E2 (4) E1 = E2 Q.13 Calculate the equilibrium constant for the reaction Fe  2  Ce4  Fe3  Ce3  ,[E0Ce4 /Ce3  1.44V; EFe 0 3 /Fe2  0.68V] 2.303 RT = 0.06 at 25° C, log 4.68 = 0.67 F A.13 Kc = 4.68 × 1012 Fe2+ + Ce4+  Fe3+ + Ce3+ 0.06 E0  1.44  0.68  0.76 V  log KC 1 Kc  4.64  1012 CONCENTRATION CELL ( E°Cell = 0) Rack your Brain Electrode Gas concentration cell : Pt, H2 (P1) | H+ (C) | H2 (P2 ), Pt Will change in tempature affect At 25° C, E0Cell? [For spontaneity of such cell reaction, p1 >p2] Electrochemistry 27.  Q.14 Pt | Cl2 (g, P2 ) | Cl– (aq,C) || Cl– (aq, C) | Cl2 (g, P1) | Pt EMF of cell is positive if (1) P1 > P2 (2) P2 > P1 (3) P1 = P2 (4) We cannot predict A.14 For spontaneous reaction P1 > P2 Anode half cell 0.0591 P.....(1) EOX  EOx 0  log 2 2 C Rack your Brain Cathode half cell Why Pt electrodes are used with 0.0591 C Ered  Ered   log......  2  the combination of the gas when 2 P1 oxidised or reduced part is a gas? Ecell  EOX  Ered 0.0591 P C  EOX   Ered   log 2. 2 C P1 0.0591 P Ecell  log 1 2 P2 Electrolyte concentration cells Zn(s) | ZnSO4 (C1) || ZnSO4 (C2 ) | Zn(s) Concept Ladder 2.303RT C  E log  1  2F A potential difference  C2  develops across the bound- ary of two solutions which [For spontaneity of such cell reaction,C2> is known has liquid junc- C1] tion potential of diffusion potential. Electrochemistry 28. Electrolytic cell Concept Ladder NaCl(molten) → Na+ + Cl– The minimum voltage Reactions at Anode (oxidation) required for discharge of ions is called discharge 2Cl– → Cl2 (g) + 2e– potential Cathode (reduction) 2Na+ + 2e– → 2Na(l) Q.15 By electrolysis of aq. MgSO4 with the help of inert electrode calculate ratio of moler of substance deposited at cathode and anode (1) 1 : 4 (2) 1 : 2 (3) 2 : 1 (4) 4 : 1 A.15 At cathode At anode 1 2H2O + 2e– → H2 + 2H– H2O → O2  2H  2e 2 nf =2 nf =2 Electrochemistry gm eq. H2 = mol H2 × 2 gm eq. H2 = mol × 4 mol H2 4 2 = = mol O2 2 1 29.  Q.16 Aq. Na2SO4 is electrolysed then after electrolysis pH will be (1) increased (2) decreased (3) unchanged (4) None of these A.16 (3) Quantitative aspects of Electrolysis Fe+2 + 2e–1 → Fe Previous Year’s Question 2 mole of e– → 1 mol Fe On electrolysis of dil. sulphuric 1 mole of e – → 1/2 mol Fe acid using platinum (Pt) electrode, the product obtained Molecular weight at anode will be : Equivalent weight of Fe2+ = 2 [NEET–2020] 1 Faraday will discharge 1 Equivalent. (1) Hydrogen gas Faraday’s law (2) Oxygen gas (3) H2S gas (a) First law of electrolysis : (4) SO2 gas Amount of substance liberated or deposited at an electrode is proportional to value of charge passed (utilized) through the solution. Rack your Brain W ∝Q W = ZQ where Z = electrochemical equivalent Electrolysis of water is often done with a small amount of then W = Z (when Q = 1 coulomb) sulphuric acid added to water. Amount of substance deposited or Why? liberated by 1 coulomb charge is called electrochemical equivalent. Let I ampere current is passed till 't' seconds Q = It Previous Year’s Question W = ZIt. 1 Faraday = 96500 coulomb = Charge on one The weight of silver (at. wt = mole electrons 108) displaced by a quantity of electricity which displaces Let 'E' is equivalent weight then 'E' gram 5600mL O8O2 at STP will be will be liberated by 96500 coulombs. [NEET–2014] Electrochemistry E (1) 5.4 g (2) 10.8 g \ 1 Coulomb will liberate gram; 96500 (3) 54.0 g (4) 108.0 g 30.  E EIt Z = 96500 \W = 96500 W It = = number of g eq = number of faraday’s E 96500 Q.17 How long a current of 2 A has to be passed through a solution of AgNO3 to coat a metal surface of 80 cm2 with 5 mm thick layer? Density of silver = 10.8 g/cm3. A.17 M M d= ⇒ 10.8 = 80 ×5 × 10-4 ⇒ M = 10.8 × 400 × 10–4 V E ×I× t 108 × 2 × t W= ⇒ 10.8 × 400 ×10–4 = ⇒ t = 193 s 96500 96500 (b) Second law of electrolysis: The weight of substances deposited or Concept Ladder dissolved at anode or cathode are in ratio of their equivalent weights, when Amount of substance deposited same amount of charge is passed through different electrolyte solutions connected in  Eq. wt Q    series.  96500  W1 E1 i.e. = W2 E2 Q.18 One ampere current for how much time should be passed through Aq. Soln of 2 molar, 200 ml AgNO3 , so that 1 mol of Ag can be deposited A.18 AgNO3  Ag + w i.t W 1 t 1 t = molarity    103  1  E 96500 E 9600 96500 Electrochemistry T= 96.5 sec 31.  Q.19 When 1 M 2L Aq. AgNO3 and 1 M, 2L aq NaCl are connected in units If 10.8 gm Ag deposited on cathode, then calculated volume of Cl2 (at NTP) collected at anode A.19 gm aq. Of Ag = gm aq. Of Cl2 Previous Year’s Question w VSTP   nf E 22.4 When 0.1 mole MnO42- is oxidised the quantity of electricity concept (Z) = Eq.wt 96500 required to completely oxidise AgNO3 + e– = Ag = 1 MnO42- to MnO4- is [NEET–2014] 10.8 VSTP (1) 96500 C (2) 2×96500C  2 108 22.4 (3) 9650C (4) 96.50C 1 2NaCl → Cl2 + 2e– = nf = 2 Deniell cell Deniell cell converts chemical energy liberated during the redox reaction to electrical Concept Ladder energy and has an electrical potential equal to Electrochemical equivalent 1.1V when the concentration or activity of Zn and 2+ Eq.wt Cu2+ ions is unity (1 mol dm-3). (Z) = 96500 Zn(s) + Cu2+ (aq)   Zn2+(aq) + Cu(s) Electrochemistry 32. Primary Batteries (Dry cells and alkaline batteries) Cathode, reductions: Rack your Brain 2NH4  aq  2e  2NH3  g   H2  g  What is the most common Anode Oxidation : Primary Battery? Zn  s   Zn2  aq  2e The pressure will be builded up and the cell is reptured by the formation of two gases at the cathode. Concept Ladder Mecury cell suitable for low Zn2  aq  2NH3  g   2Cl   aq  Zn NH3 2 Cl 2  s  current devices like hearing acids, watches, etc. con- 2MnO2  s   H2  g   Mn2O3  s   H2O l  sists of zinc-mercury amal- gam as anode and a paste Electrochemistry of HgO and carbon as the cathode. 33.  SECONDARY OR RECHARGABLE BATTERIES Concept Ladder The lead storage battery – Discharging reaction : The discharging process of the storage cell is based on Pb(s)  PbO2 (s)  2 H2 SO4  PbSO4 (s)  2 H2 O the principles of electro- Cathode reduction : chemical cell, whereas the charging process is based 2MnO2  s   H2  g   Mn2O3  s   H2O l  upon the principles of elec- Anode Oxidation trolytic cells. PbO2  s   4H  aq  SO42  aq  2e  PbSO4  s   2H2 O l  Net Cell reaction Pb  s   SO24  aq  PbSO4  s   2e Charging reaction is reverse of discharging Rack your Brain reaction with the reaction on anode will be Electrochemistry What is end life of the battery? written on cathode and vice versa 34. Nickel - Cadmium ("Ni - Cad") batteries Cathode (Reduction) : NiO2 + 2H2O + 2e– → Ni(OH)2 + 2OH– Anode (Oxidation): Concept Ladder Cd + 2OH– → Cd(OH)2 + 2e– Discharge potential of Net cell reaction : negative ions is as follows NiO2(OH) + Cd + H2O → 2Ni(OH)2 + Cd(OH)2 SO24  NO3  OH  Br   I Electrochemistry 35.  Fuel Cell Concept Ladder Cathode, reduction : O2(g) + 2 H2O(l) + 4e– → 4OH– (aq) Efficiency of a Fuel Cell E° = 1.23 V G nFE Anode, Oxidation :   H H H2 (g) → 2H+ (aq) + 2e– E° = 0 V Corrosion Cells and Reactions Previous Year’s Question A corrosion system can be regarded as a short- circuited electrochemical cell in which the anodic A device that converts energy process is something like of combustion of fuels like Fe(s) → Fe2+(aq) + 2e– hydrogen and methane, directly into electrical energy is known as and the cathodic steps can be given as [2015, Cancelled] O2 + 2H2O + 4e– → 4OH– (1) Dynamo Electrochemistry 2H+ + 2e– → H2 (g) (2) Ni-Cd cell M2+ + 2e– → M(s) (3) Fuel cell (4) Electrolytic cell 36. Control of Corrosion Sacrificial coatings When coating of more active metal is applied on another metal, negative charge is supplied to the metal. Cathodic Production By maintaining a continuous supply of negative electrical charge on a metal for Concept Ladder inhabitation for dissolution of positive ions. In a galvanic cell, cathode is positive with respect to anode. Electrochemistry 37.  Q.20 Compare the conductance in following Colour of KI solution containing starch turns blue when Cl2 water is added. Explain. A.20 Chlorine placed below iodine in electrochemical series having more reduction potential and thus shows reduction whereas I– undergoes oxidation. The I2 so formed get absorbed in starch to give blue colour. 2I   2e Cl 2  2e  2Cl  Cl   2I   I2  2Cl  Q.21 How many moles of electrons are needed for the reduction of 20 mL of 0.5M solution of KMnO4 in acid medium ? A.21 Moles of KMnO4 = M × V (L) = 0.5 × 20 × 10–3 = 10–2 [ Mn7+ + 5e- → Mn2+]  1 mol KMnO4 required = 5 mol e- \ 10–2 mol KMnO4 required = 5 × 10–2 mol e- Q.22 Which cells were used in te Apollo space program? What was the product used for? A.22 H2—O2 fuel cell. The product H2O was used for drinking by the astronauts. Q.23 An aqueous solution of NaCl is electrolysed with inert electrodes. Write the equations for the reactions taking place at cathode and anode. What happens if NaNO3(aq.) is used instead of NaCl ? A.23 For NaCl(aq.)anode : 2Cl– → Cl2 + 2e– cathode : 2H+ + 2e– → H2 Electrochemistry For NaNO3 (aq.) anode : 2OH– → H2O + 1/2 O2 + 2e– cathode : 2H+ + 2e– → H2 38. Q.24  The value of  for NH4Cl, NaOH and NaCl are 129.8, 248.1 and 126.4 ohm–1 cm2  mol–1 respectively. Calculate  for NH4OH solution. A.24  NH4OH   NH4 Cl   NaOH   NaCl  129.8 + 248.1 - 126.4 = 251.5 ohm-1cm2mol-1 Q.25 Given the standard electrode potentials ; K+/K = –2.93 V, Ag+/Ag = 0.80 V, Hg2+/Hg = 0.79 V, Mg2+/Mg = –2.37 V, Cr3+/Cr = –0.74V. Arrange these metals in their increasing order of reducing power. A.25 More is E°RP, hence more is the oxidising power or more is the tendency to get reduced or lesser is reducing power. Ag < Hg < Cr < Mg < K Q.26 Iron does not rust even if zinc coating on its surface is broken but the same is not true when coating is of tin. A.26 Zn is more reactive than Fe, this implies that if a crack appears on the surface of Fe coated with Zn even then Zn will take part in the redox reaction and not Fe. In other words, Zn will be corroded in preference to Fe, but same is not in the case with Sn. It is less reactive than Fe, when a crack appears on the surface of Fe coated with Sn, then Fe will take part in the redox reaction and not Sn. Therefore, Fe will be corroded under these circumstances. Q.27 How will show that Faraday's second law of electrolysis is simply corollary of the first law. A.27 According to Faraday's first law of electrolysis. w = Z × Q If same quantity of electricity is passed through two electrolytes, i.e., Q1 = Q2 = Q, then In case of Electrochemistry first electrolyte, w1 = Z1 × Q and In case of second electrolyte, w2 = Z2 × Q 39.  w1 Z1 E1 / 96500 E1 On diving = = = w2 Z2 E2 / 96500 E2 Where E1 and E2 are their equivalent masses Q.28 Calculate the equilibrium constant for the reaction at 298 K Zn  s   Cu  aq   Zn  aq  Cu  s    2 2  Given E0zn2  /Zn  0.76V and E0Cu2  /Cu  0.34V nE0Cell A.29 We know that log = 0.0591 E0cell  Ecathode 0  Eanode 0  = [(+0.34 V) – (– 0.76 V)] = 1.10 V, n =2, 2   1.10 V  log Kc =  37.29  Kc = Antilog 37.29 = 1.95 × 1037 0.0591 V  Q.30 Given that, Co3+ + e–   Co2+ E° = + 1.82V 2H2O   O2 + 4H+ + 4e– ; E° = –1.23V. Explain why Co3+ is not stable in aqueous solutions A.30 4 [Co+3 + e–   Co2+] ; E° = + 1.82V 2H2O   O2 + 4H+ + 4e– ; E° = – 1.23 V E° for first reaction is positive, hence cell reaction is spontaneous. This implies Co3+ ions will take part in the reaction. Therefore, Co3+ is not stable. Electrochemistry 40. Q.31 The measured e.m.f. at 25°C for the cell reaction, Zn (s) + Cu2+ (1.0M)   Cu (s) + Zn2+ (0.1 M) is 1.3 volt Calculate E° for the cell reaction. A.31 Using Nernst equation (at 298 K), 0.0591 V Zn2  aq  Ecell  E  0 cell log 2 Cu2  aq  Substituting the values 1.3V  E0  0.0591V log 0.1  1.3V  E0  0.02955 V log 101 cell cell 2 1.0 1.3V  Ecell  0.02955 V log 10  Ecell  1.3 V  0.02955  1.27 V 0 0 Q.32 The resistance of a 0.01 N solution of an electrolyte was found to 210 ohm at 298 K using a conductivity cell with a cell constant of 0.88 cm–1. Calculate specific conductance and equivalent conductance of solution A.32 Given for 0.01 N solution. R = 210 ohm   0.88 cm1 A 1  1  K  R  A  K  210  0.88  4.19  10 mho cm 3 1 k  1000 4.19  103  1000 eq   eq   eq  419 mho cm2 eq1 N 0.01 Electrochemistry 41. Electrochemistry 42. Chapter Summary Š An electrochemical cell contains two electrodes (metallic conductors) in contact with an electrolyte. An electrode and its electrolyte comprise an Electrode Compartment. Electrochemical Cells can be classified as: (i) Electrolytic Cells in which external source of current drives a non-spontaneous reaction (ii) Galvanic Cells which generates electricity as a result of a spontaneous cell reaction Š A voltic cell is a cell in which a spontaneous reaction generates an electric current. In a galvanic cell, cathode is positive with respect to anode. The salt-bridge having solution of strong ionic salts like KNO3, KCl, NaNO3, , NaCl etc., which is soaked in a colloidal soln of agar-agar gel which permits the movement of ions of salts only. REPRESENTATION OF A CELL (IUPAC CONVENTIONS) Anodic half-cell is written on left and cathodic half-cell on R.H.S. Zn(s) | ZnSO4 (sol) || CuSO4 (sol) | Cu(s) The net reaction that occurs in the galvanic cell, it is called the cell reaction.. Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) Š A metal and the solution develops definite potential difference between them. This potential difference is called electrode potential. (i) Oxidation potential : When electrode is negatively charged w.r.t. solution, i.e., it acts as anode. Oxidation occurs. M → Mn+ + ne– (ii) Reduction potential : When electrode is positively charged w.r.t. solution. i.e., it acts as cathode. Reduction occurs. Mn+ + ne– → M Š The magnitude of potential depends on the given below factors : (i) Nature of the electrode (ii) Concentration of the ion in solution (iii) Temperature Š Concept for Electromotive force (EMF) of a ECell = R.P. (Cathode) – R.P. (Anode) = R.P. (Cathode) + O.P (Anode) Š Electrolytic Cell : This cell changes EE to chemical energy. Electrolytic cell is defined to be the entire setup except external battery. Electrochemistry Electrolysis of molten sodium Chloride NaCl(molten) → Na+ + Cl– 43.  rxns at anode (oxidation) : cathode (reduction) 2Cl → Cl (g) + 2e : – 2 – 2Na+ + 2e– → 2Na(l) Š Relationship Between DG & Electrode Potential Reactions DG E 1 Spontaneous (-) (+) 2 Non-spontaneous (+) (-) 3 Equilibrium 0 0 Š Nernst Equation : aA + bB → cC + dD, c d 0.0591 C  D Ecell  Ecell 0  log   a  b n  A  B Š Concentration Cell : For such cell, E°Cell = 0. Electrode Gas concentration cell : For spontaneity of such cell reaction, p1>p2 Electrolyte concentration cells: For spontaneity of such cell reaction, C2>C1 Š Commercial Voltaic Cells : After even recharging Primary batteries cannot be returned to their original state, so the battery is "dead" and must be discarded when the reactants are consumed,. Secondary batteries are also called rechargeable or storage batteries. The batteries can be recharged by reversing its reactions. Š The discharging process of the storage cell is based on the principles of electrochemical cell, whereas the charging process is based upon the principles of electrolytic cells. Š Faraday’s laws of Electrolysis : Š First law of electrolysis : Amount of substance liberated or deposited at an electrode is proportional to value of charge passed (utilized) through the solution. Š Second law of electrolysis : The weight of substances deposited or dissolved at anode or cathode are in ratio of their equivalent weights, when same amount of charge is passed through different electrolyte solutions connected in series. Š i.e. w1 /w2 = E1 /E2 Š Conductance Inverse of resistance is called as conductance and inverse of specific resistance is called as specific conductance. Electrochemistry 44.  E0cell Š Equivalent Conductance E0CO2+ |CO Molar Conductance E0Ce4+ |Ce3+ Relation between E 0 Š Kohlrausch's Law of Independent Migration of Ions E0CO2+ |CO cell Electrochemistry 45.