1.ElectrochemistryTheory.pdf

17. ELECTROCHEMISTRY 1. INTRODUCTION Electrochemistry is a branch of chemistry which deals with inter-conversion of chemical energies and electrical energy. We’ll discuss electrolytic reactions (reactions that occur when electricity passes through solutions) as well...

17. ELECTROCHEMISTRY 1. INTRODUCTION Electrochemistry is a branch of chemistry which deals with inter-conversion of chemical energies and electrical energy. We’ll discuss electrolytic reactions (reactions that occur when electricity passes through solutions) as well as electromagnetic reaction (reactions that produce electric energy). Some examples of electrolytic reactions are electrolysis, electroplating, electro refining of metals, etc. Some examples of electro genetic reactions are reactions occurring in cells or batteries. 2. TERMINOLOGIES IN ELECTROCHEMISTY Some important terms used in Electrochemistry are as follows: (a) Electrical Conductors: Materials that allow flow of electrons are known as conductors. There are broadly two types of conductors-Electronic/Metallic and Electrolytic/Solution. Table 17.1: Difference between Electronic Conductor and Electrolytic Conductors Electronic Conductors or Metallic Conductors Electrolytic Conductors or Solution Conductors 1. Passage of current by movement of electrons in the Passage of current by ions in molten state or in aqueous solution metallic lattice, e.g., Cu, Ag, etc. of electrolytes, e.g., NaCl (aq) or NaCl (fused). 2. Passage of current brings in only physical changes. Passage of current brings in physical as well as chemical changes. 3. It generally shows no transfer of matter. It involves transfer of matter in the form of ions. 4. It generally shows an increase in resistance during the It generally shows a decrease in resistance due to decrease in passage of current due to increase in temperature. viscosity of the medium and degree of hydration of ions with Thermal motion of metal ions hindering the flow of increase in temperature. electrons increases with increase in temperature. 5. The conducting power of metals is usually high. The conducting power of electrolytic conductors is relatively low. (b) Insulators: Those materials which don’t allow the passage of electrons are known as Insulators. For e.g. wood, wool, plastic, silk, etc. (c) Electrolytes: The substance that in solution or in the molten state, conducts electric current and is simultaneously decomposed is called an electrolyte.The extent or degree of dissociation of different electrolytes in solution is different. Electrolytes can be broadly categorized into two: Strong and Weak Electrolytes. (d) Strong Electrolytes: Substances which are largely dissociated and form a highly conducting liquid in water are strong electrolytes, e.g., All salts (except CdBr2, HgCl2), mineral acids like HCl, H2SO4, HNO3, etc. and bases like NaOH, KOH, etc. are strong electrolytes. The strong electrolytes are almost 100% ionized at normal dilution. 1 7. 2 | Electrochemistr y (e) Weak Electrolytes: Substances which dissociate only to a small extent in aqueous solution forming low conducting liquid are weak electrolytes, e.g., All organic acids (except sulphonic acids), inorganic acids like HCN, H3BO3, etc. and bases like NH3, amines, etc. are weak electrolytes. (f) Electrodes: In order to pass the current through an electrolytic conductor, two rods or plates are always needed which are connected with the terminals of a battery. These rods/plates are called Electrodes. The electrode where oxidation reaction takes place is anode and electrode where reduction takes place is cathode. 3. ELECTROLYSIS The phenomenon in which passage of current through an electrolyte (molten or solution) brings in chemical changes involving electronation (reduction) as well as de-electronation (oxidation) of ions is known as electrolysis. 3.1 Preferential Discharge Theory If an electrolytic solution consists of more than two ions and the electrolysis is done, it is observed that all the ions are not discharged from the electrodes simultaneously but certain ions are liberated from the electrodes in preference to others. This is explained by preferential discharge theory. It states that if more than one type of ions are attracted towards a particular electrode, then the one discharged is the ion which requires least energy. The potential at which the ion is discharged or deposited on the appropriate electrode is termed the discharge potential or deposition potential. The values of discharge potential are different for different ions. Table 17.2: Examples of preferential discharge theory Electrolyte Electrode Cathodic reaction Anodic reaction Aqueous acidified Pt 2Cl− → Cl2 + 2e− CuCl2 solution Cu2+ + 2e− → Cu Molten PbBr2 Pt 2Br − → Br2 + 2e− Pb2+ + 2e− → Pb Sodium chloride Hg 2Cl− → Cl2 + 2e− solution 2Na+ + 2e− → 2Na Silver nitrate solution Pt Ag+ + e− → Ag 1 2OH− → O2 + H2O + 2e− 2 Sodium nitrate Pt 2H+ + 2e− → H2 1 solution 2OH− → O2 + H2O + 2e− 2 Illustration 1: Find the charge in coulomb on 1 g-ion of N3−.  (JEE MAIN) Sol: First determine charge on one ions of this can be calculated as product of number of electron and charge of electron. According to Avogadro’s law one g of ion contains 6.02 × 1023 ions. So, charge on one g-ion of N3− can be calculated by multiplying charge. Charge on one ions of N3− into Avogadro number. 3− −19 Charge on one ions of N =3 × 1.6 × 10 coulomb One g-ion = 6.02 × 1023 ions Thus, charge on one g-ion of N3− =3 × 1.6 × 10 −19 × 6.02 × 1023 = 2.89 × 105 coulomb Illustration 2: Explain the reaction: (a) 2KI + Cl2 → 2KCl + I2 , (b) 2KClO3 + I2 → 2KIO3 + Cl2 Sol: Compound which undergoes oxidation acts as a reducing agent and compound which undergoes reduction acts as an oxidizing agent.  Chem i str y | 17.3 (a) Cl2 acts as oxidizing agent: 2e− + Cl2 → 2Cl− ; 2I − → I2 + 2e− (b) I2 acts as reducing agent: 2Cl5+ + 10e− → Cl ; I02 → 2I5+ + 10e− 2 MASTERJEE CONCEPTS Misconception: Electrolysis does not mean breaking up of an ionic compound into ions. An ionic compound even on dissolution in water furnishes ions. Note: During electrolysis, oxidation-reduction occurs simultaneously. Oxidation occurs at anode whereas reduction occurs at cathode. Nikhil Khandelwal (JEE 2009 AIR 94) 3.2 Faraday’s Law of Electrolysis The relationship between the quantity of electric charge passed through an electrolyte and the amount of the substance deposited at the electrodes was presented as the ‘laws of electrolysis’ by Faraday in 1834. 3.2.1 Faraday’s First Law When an electric current is passed through an electrolyte, the amount of substance deposited is proportional to the quantity of electric charge passed through the electrolyte. If W be the mass of the substance deposited by passing Q coulomb of charge, then according to the law, we have the relation: W ∝ Q Q = current in amperes × time in seconds = I × t So, W ∝ I × t or W = Z×I×t Where Z is a constant, known as electrochemical equivalent and is characteristic of the substance deposited. When a current of one ampere is passed for one second, i.e., one coulomb (Q = 1), then W = Z. Definition of electrochemical equivalent: Mass of the substance deposited by one coulomb of charge or one ampere current for one second. 3.2.2 Faraday’s Second Law When the same quantity of charge is passed through different electrolytes, then the masses of different substances deposited at the respective electrodes will be in the ratio of their equivalent masses. Again according to first law, W= Z × Q When, Q = 96500 coulomb, W becomes gram equivalent mass (E). E Z1 E1 Thus, E= Z × 96500 or Z = ; = 96500 Z 2 E2 3.2.3 Faraday’s Law for Gaseous Electrolytic Product ItVe For the gases, we use V = 96500 Where, V = Volume of gas evolved at STP at an electrode Ve = Equivalent volume = Volume of gas evolved at an electrode at STP by 1 faraday charge  1 7. 4 | Electrochemistr y Example: A 40.0 amp current flowed through molten iron (III) chloride for 10.0 hours (36,000 s). Determine the mass of iron and the volume of chlorine gas (measured at 25ºC and 1 atm) that is produced during this time. Sol: 1. Write the half-reaction that take place at the anode and at the cathode. Anode (oxidation): 2Cl− → Cl2 (g) + 2e− Cathode (reduction): Fe3+ + 3e− → Fe(s) 2. Calculate the number of moles of electrons. 1.44 × 106 C 40.0amps × 36,000s = 1F 1mole e− 1.44 × 106 C × 14.9F ; 14.9F × = 14.9mole e− = 96, 485C 1F 3. Calculate the moles of iron and of chlorine produced using the number of moles of electrons calculated and the stoichiometry from the balanced half-reactions. According to the equations, three moles of electrons produce one mole of iron and 2 moles of electrons produce 1 mole of chlorine gas. 1mole Fe 1mole Cl2 14.9mole = e− × e− × 4.97mole Fe; 14.9mole= 7.45mole Cl2 − 3mole e 2mole e− 4. Calculate the mass of iron using the molar mass and calculate the volume of chlorine gas using the ideal gas law (PV = nRT). 55.847 gFe (7.45moleCl2 )(0.0821atm L / mole K)(298K) 4.97mole Fe × = 278gFe; = 182L Cl2 1mole Fe 1atm Calculating the Time required To determine the quantity of time required to produce a known quantity of a substance given the amount of current that flowed: (i) Find the quantity of substance produced/consumed in moles. (ii) Write the balanced half-reaction involved. (iii) Calculate the number of moles of electrons required. (iv) Convert the moles of electrons into coulombs. (v) Calculate the time required. Example: How long must a 20.0 amp current flow through a solution of ZnSO 4 in order to produce 25.00 g of Zn metal? Sol: (i) Convert the mass of Zn produced into moles using the molar mass of Zn. 1mole Zn 25.00gZn × 0.3823mole Zn = 65.39gZn (ii) Write the half-reaction for the production of Zn at the cathode. Zn2+ (aq) + 2e− → Zn(s) (iii) Calculate the moles of e- required to produce the moles of Zn using the stoichiometry of the balanced half- reaction. According to the equation, 2 moles of electrons will produce one mole of zinc. 2mole e− 0.3823mole Zn × 0.7646mole e− = 1mole Zn  Chem i str y | 17.5 (iv) Convert the moles of electrons into coulombs of charge using Faraday’s constant. 1F 96, 485C 0.76mole e− × = 0.7646F; 0.7646F × = 73,770C 1molee − 1F (v) Calculate the time using the current and the coulombs of charge. 20.0amps = × t 73,770C; = t 3,688s or 1.03h Calculating the Current required To determine the amount of current necessary to produce a known quantity of substance in a given amount of time: (i) Find the quantity of substance produced/or consumed in moles. (ii) Write the equation for the half-reaction taking place. (iii) Calculate the number of moles of electrons required. (iv) Convert the moles of electrons into coulombs of charge. (v) Calculate the current required. Example: What amount of current is required to produce 400.0 L of hydrogen gas, measured at STP, from the electrolysis of water in 1 hour (3600 s)? Sol: (i) Calculate the number of moles of H2. (Remember, at STP, 1 mole of any gas occupies 22.4 L) 1mole H2 400.0L H2 × = 17.9mole H2 22.4L H2 (ii) Write the equation for the half-reaction that takes place. Hydrogen is produced during the reduction of water at the cathode. The equation for this half-reaction is: 4e− + 4H2O(I) → 2H2 (g) + 4OH− (aq) (iii) Calculate the number of moles of electrons. According to the stoichiometry of the equation, 4 mole of e− are required to produce 2 moles of hydrogen gas, or 2 moles of e− s for every one mole of hydrogen gas. 2mole e− 17.9mole H2 × 35.8mole e− = 1mole H2 (iv) Convert the moles of electrons into coulombs of charge. 1F 96, 485C 35.8mole e− × 35.8F; 35.8F × = 3.45 106 C =× 1mole e − 1F (v) Calculate the current required. I × 3600s =3.45 × 106 C; I =958C / s =958amps MASTERJEE CONCEPTS As one faraday (96500 coulombs) deposits one gram equivalent of the substance, hence electrochemical equivalent can be calculated from the equivalent weight, Eq. wt. of the substance i.e., Z = 96500 Note: Knowing the weight of the substance deposited (W gram) on passing a definite quantity of electricity W (Q coulombs), the equivalent weight of the substance can be calculated, i.e., Eq. wt. = × 96500 Q 1 7. 6 | Electrochemistr y MASTERJEE CONCEPTS Tip: The quantity of electricity actually passed is calculated from the current and time as follows: Quantity of electricity in columbs = Current amperes × time in seconds Thus, knowing the quantity of electricity passed, the amount of substance deposited can be calculated. Faraday’s first law and second law can be combined to give a mathematical relation as follows:- E Q Q M C×t M W = ZQ = ×Q = ×E = × = × F F F z F z z = Electrochemical equivalent; Q = Quantity of electricity passed, E = Eq. wt. of the metal, F = 1 Faraday, M = Atomic mass of the metal; z = Valency of the metal; C = Current passed, t = Time for which current is passed. Saurabh Gupta (JEE 2010 AIR 443) Illustration 3: Electric current of 100 ampere is passed through a molten liquid of sodium chloride for 5 hours. Calculate the volume of chlorine gas liberated at the electrode at NTP.  (JEE MAIN) Sol: Here current and time is given so from this first calculate quantity of electricity passed (charge) and from this calculate the amount of chlorine liberated. Volume of Cl2 liberated at NTP can be determined by multiplying the amount of chlorine liberated by 22.4 L The reaction taking place at anode is: 2Cl− → Cl2 + 2e− Q = I × t = 100 × 5 × 60 × 60 coulomb 71.0 g 71.0 g 2×96500 coulomb 1mole The amount of chlorine liberated by passing 100 × 5 × 60 × 60 coulomb of electric charge 1 = × 100 × 5 × = 60 × 60 9.3264 mole Volume of Cl2 liberated at NTP= 9.3264 × 22.4= 201L 2 × 96500 Illustration 4: How much electric charge is required to oxidize (a) 1 mole of H2O toO2 and (b) 1 mole of FeO to Fe2O3 ? (JEE MAIN) Sol: Charge = No of electrons involved in the reaction x faradays constant So first find out the no of electron reaction involved in the reaction by writing the chemical reaction, balancing it and then calculate the charge. 1 (a) The oxidation reaction is: H2O → O2 + 2H+ + 2e− ; Q= 2 × F = 2 × 96500 = 193000 coulomb 1mole 2 2mole 1 1 (b) The oxidation reaction is: FeO + H2O → Fe2O3 + H+ + e− ; O= F= 96500 coulomb 2 2 Illustration 5: An aqueous solution of sodium chloride on electrolysis gives H2 (g),Cl2 (g) and NaOH according to the reaction. 2Cl− (aq.) + 2H2O → 2OH− (aq.) + H2 (g) + Cl2 (g). A direct current of 25 ampere with a current efficiency 62% is passed through 20 L of NaCl solution (20% by mass). Write down the reactions taking place at the anode and cathode. How long will it take to produce 1 kg of Cl2? What will be the molarity of the solution with respect to hydroxide ion? Assume no loss due to evaporation.   (JEE ADVANCED)  Chem i str y | 17.7 Sol: Time can be calculate by using charge and current relationship. Effective current is determined by using current efficiency. Here it is given that we have to find out the molarity of the solution with respect to hydroxide Ion. Volume is given. We have to find out the no of moles of oxygen. This can be achieved by calculating the no of mole of Cl2 present in 1 kg. Reactions at anode and cathode are: 2Cl− → Cl2 + 2e− (at anode) 1000 2H2O + 2e− → H2 + 2OH− ( at cathode) 1kg of= Cl2 = 14.08 mole 71.0 Charge to produce one mole of Cl2=2 x 96500 Charge to produce 14.08 mole of Cl2=2x96500x14.08 62 Effective current = × 25.0 = 15.5 ampere 100 Charge 2 × 96500 × 14.08 Time = = = 175318.7 second = 48.699 hour Current 15.5 OH- ions produced = 2 × moles of Cl2 = 2x14.08 = 28.16 Mole 28.16 Molarity = = = 1.408M Volume 20 Illustration 6: An acidic solution of Cu2+ salt containing 0.4 g of Cu2+ is electrolyzed until all the copper is deposited. The electrolysis is continued for seven more minutes with volume of solution kept at 100 mL and the current at 1.2 amp. Calculate the gases evolved at NTP during the entire electrolysis. (JEE ADVANCED) 0.4 Cu2+ Sol: 0.4 g of = = 0.0126g − equivalent 31.75 8 At the same time, the oxygen deposited at anode = × 0.0126g = 0.00315g − mole 32 After the complete deposition of copper, the electrolysis will discharge hydrogen at cathode and oxygen at anode. The amount of charge passed = 1.2 × 7 × 60= 504 coulomb 1 8 So, Oxygen liberated= × 504= 0.00523g − equivalent = × 0.00523 = 0.001307g − mole 96500 32 1 0.00523g − equivalent = Hydrogen liberated = × 0.00523 =0.00261g − mole 2 Total gases evolved =(0.00315 + 0.001307 + 0.00261)g = − mole 0.007067g − mole NTP 22400 × 0.007067mL = 158.3mL Volume of gases evolved at= 4. ARRHENIUS THEORY OF ELECTROLYTIC DISSOCIATION In order to explain the properties of electrolytic solutions, Arrhenius put forth a comprehensive theory. The main postulates of the theory are: (a) An electrolyte, when dissolved in water, breaks up into two types of charged particles, one carrying a positive charge and the other a negative charge. These charged particles are called ions. Positively charged ions are termed as cations and negatively charged as anions. A +B− + aq. → A + (aq.) + B− (aq.) 1 7. 8 | Electrochemistr y (b) The process of splitting of the molecules into ions of an electrolyte is called ionization. The fraction of the total number of molecules present in solution as ions is known as degree of ionizations or degree of dissociation. It is denoted by ‘ α ’ Number of molecules dissociated int o ions (c) α= Total number of molecules (d) Ions present in solution constantly re-unite to form neutral molecules and, thus, there is a state of dynamic equilibrium between the ionized and non-ionized molecules, i.e. AB  A + + B− [A + ][B− ] (e) Applying the law of mass action to the above equilibrium = K. K is known as ionization constant. The [AB] electrolytes having high value of K are termed strong electrolytes and those having low value of K as weak electrolytes. (f) When an electric current is passed through the electrolytic solution, the positive ions (cations) move towards cathode and the negative ions (anions) move towards anode and get discharged, i.e., electrolysis occurs. The ions are discharged always in equivalent amounts, no matter what their relative speeds are. (g) The electrolytic solution is always neutral in nature as the total charge on one set of ions is always equal to the total charge on the other set of ions. However, it is not necessary that the number of two sets of ions must be equal always. AB  A + + B− (Both ions are equal) NaCl  Na+ + Cl− (Both ions are equal) AB2  A2+ + 2B− (Anions are double that of cations) 2+ − BaCl2  Ba + 2Cl (Anions are double that of cations) A2B  2A + + B2− (Cations are double that of anions) Na2SO 4  2Na+ + SO24− (Cations are double that of anions) (h) The properties of electrolytes in solution are the properties of ions present in solution. For example, acidic solution always contains H+ ions while basic solution contains OH− ions and characteristic properties of solutions are those of H+ ions and OH− ions respectively. Limitations of Arrhenius Theory (i) You cannot apply Ostwald’s dilutions law which is based on Arrhenius theory to strong electrolytes. (ii) Strong electrolytes conduct electricity in a fused state, i.e., in the absence of water. This is in contradiction of Arrhenius theory which states that the presence of solvent is imperative for ionization. (iii) Arrhenius theory assumes independent existence of ions but fails to account for the factors which influence the mobility of the ions. 4.1 Factors Affecting Degree of Ionization (a) Nature of solute: When the ionizable parts of a molecule of a substance are held more by covalent bonding than by electrovalent bonding, less ions are furnished in solution. (b) Nature of solvent: The main function of the solvent is to weaken the electrostatic forces of attraction between the two ions and separate them. (c) Dilution: The extent of ionization of an electrolyte is inversely proportional to the concentration of its solution. Thus, degree of ionization increases with the increase of dilution of the solution, i.e., decreasing the concentration of the solution.  Chem i str y | 17.9 (d) Temperature: The degree of ionization increases with the increase in temperature. This is due to the fact that at higher temperatures molecular speed is greater than before which overcomes the forces of attraction between the ions. 5. ELECTRICAL CONDUCTANCE The conductance is the property of the conductor (metallic as well as electrolytic) which facilitates the flow of electricity through it. It is equal to the reciprocal of resistance, i.e. 1 1 Conductance = = ... (i) Resistance R It is expressed in the unit called reciprocal ohm ( ohm−1 or mho) or Siemens. 5.1 Specific Conductance or Conductivity The resistance of any conductor varies directly with its length (l) and inversely with its cross-sectional area (a), l l i.e. R ∝ or R = ρ  …..... (ii) a a Where, ρ is called the specific resistance. If l = 1 cm and a = 1cm2 , then R = ρ ... (iii) The specific resistance is, thus, defined as the resistance of one centimeter cube of a conductor. The reciprocal of specific resistance is termed the specific conductance or it is the conductance of one centimeter cube of a conductor. It is denoted by the symbol κ, Thus, 1 κ= , κ= kappa − The specific conductance... (iv) ρ Specific conductance is also called conductivity. a. 1 l.1 l l  From eq. (ii), we= have ρ R or = ; κ= × C  = cell constant  l ρ a R a a  or Specific conductance = conductance × cell constant 5.2 Equivalent Conductance Equivalent conductance is defined as the conductance of all the ions produced by one gram-equivalent of an electrolyte in a given solution. It is denoted by Λ. In general Λ = κ × V ... (v) Where, V is the volume in mL containing 1 g-equivalent of the electrolyte. In case the concentration of the solution is c g-equivalent per liter, then the volume containing 1 g-equivalent of the electrolyte will be 1000/c. 1000 So, equivalent conductance Λ = κ × ... (vi) c 1000 Λ = κ× ; where, N = normality. The unit of equivalent conductance is ohm−1 cm2 eq−1. N 5.3 Molar Conductance The molar conductance is defined as the conductance of all the ions produced by ionization of 1 g-mole of an electrolyte when present in V ml of solution. It is denoted by µ. Molar conductance µ = κ × V ... (vii) 1 7. 1 0 | Electrochemistr y Where, V is the volume in mL containing 1 g-mole of the electrolyte. If c is the concentration of the solution in 1000 g-mole per liter, then µ = κ × Its unit is ohm−1 cm2 mol−1. c Molar conductance Molecular mass Equivalent conductance = ; where, n = n Equivalent mass Illustration 7: 1.0 N solution of a salt surrounding two platinum electrodes 2.1 cm apart and 4.2 sq. cm in area was found to offer a resistance of 50 ohm. Calculate the equivalent conductivity of the solution. (JEE MAIN) Sol: As Equivalent conductivity = κ × V In order to find equivalent conductivity we have to calculate specific conductance. l 1 Specific conductance ( κ ) is given as κ =. a R l 1 Given, l = 2.1 cm, a = 4.2 sq. cm, R = 50 ohm. Specific conductance, κ =. a R 2.1 1 Or = κ × = 0.01 ohm−1 cm−1 ; Equivalent conductivity = κ × V 4.2 50 V = The volume containing 1 g-equivalent = 1000 mL = 0.01 × 1000 = 10 ohm−1 cm2 eq−1 So, Equivalent conductivity Illustration 8: The specific conductivity of 0.02M KCl solution at 25ºC is 2.768 × 10−3 ohm−1 cm−1. The resistance of this solution at 25ºC when measured with a particular cell was 250.2 ohm. The resistance of 0.01 M CuSO 4 solution at 25ºC measured with the same cell was 8331 ohm. Calculate the molar conductivity of the copper sulphate solution. (JEE ADVANCED) 1000 Sol: Molar conductivity is given = by Sp. cond. × so first we have to calculate specific conductivity of the C solution. Sp.conductivity is given as a product of cell constant and conductance. Now cell constant is not provided; we can calculate it from the conductance and Sp.conductivity of KCl solution. Sp. cond. of KCl 2.768 × 10−3 Cell constant = = = 2.768 × 10−3 × 250.2 Conductance of KCl 1 / 250.2 For 0.01 M CuSO4 solution 1 Sp. Conductivity = Cell constant × Conductance = 2.768 × 10−3 × 250.2 × 8331 1000 2.768 × 10−3 × 250.2 1000 = Sp. cond. Molar conductance = × = × 8.312 ohm−1 cm2 mol−1 C 8331 1 / 100 6. KOHLRAUSCH’S LAW At infinite dilution, when dissociation is complete, each ion makes a definite contribution towards molar conductance of the electrolyte, irrespective of the nature of the ion with which it is associated and the value of molar conductions of its constituent ions, i.e., Λ = λ + + λ − λC and λa are called the ionic conductance of cation and anion at infinite dilution respectively. The ionic conductance are proportional to their ionic mobilities. Thus, at infinite dilution, λC = kuC and λa = kua, where, uC and ua are ionic mobilities of cation and anion respectively at infinite dilution. The value of k is equal to 96500 C, i.e., one Faraday. Thus, assuming that increase in equivalent conductance with dilution is due to increase in the degree of dissociation of the electrolyte, it is evident that the electrolyte achieves the degree of dissociation as unity when it is completely ionized at infinite dilution. Therefore, at any other dilution, the equivalent conductance is proportional to the degree of dissociation. Thus,  Chem i str y | 17.11 Λ Equivalent conductance at a given concentration Degree of dissociation = α = Λ∞ Equivalent conductance at inf inite dilution Ionic Mobility, µ: It is the distance travelled by an ions per second under a potential gradient of 1 volt per meter. 1. For an, µ = λ º /F 2. Ionic mobility of an ion depends on its charge, size, viscosity of solvent, temperature, etc. 3. For aqueous solution, greater the charge or smaller the size of gaseous ion, greater will be the size of aqueous ion. When such a big ion moves in solution, it experiences greater resistance by the size of solvent particles. This results in a decrease in its conductance as well as ionic mobility. Following are the increasing order of ionic mobilities of some ions: Li+ < Na+ < K + < Rb+ < Cs+ ; F− < Cl− < Br − < I − ; Al3+ < Mg2+ < Na+ 4. The size of gaseous H+ ion is smallest among all the ions and hence its ionic mobility should be minimum but among all the ions, it is maximum. The ion with second highest ionic mobility is OH–. The very high ionic mobilities of these ions are due to interchange of hydrogen bonds and covalent bonds, by which migration of charge occurs without any large displacement in the ions (Grotthus mechanism). Applications of Kohlrausch’s Law: 0 0 (a) Determining Λm of a weak electrolyte: In order to calculate Λm of a weak electrolyte say CH3COOH, we 0 determine experimentally Λm values of the following three strong electrolytes: (i) A strong electrolyte containing same cation as in the test electrolyte, say HCl (ii) A strong electrolyte containing same anion as in the test electrolyte, say CH3COONa (iii) A strong electrolyte containing same anion of (a) and cation of (b) i.e. NaCl. 0 0 0 0 0 Λm of CH3COOH is then given as: Λm (CH3COOH) = Λm (HCl) + Λm (CH3COONa) − Λm (NaCl) 0 Proof: Λm (HCl) = λH0 + λ ... (i) Cl− 0 Λm (CH3COONa) = λ0 +λ ... (ii) CH3COO− Na+ 0 Λm (NaCl) = λ0 + λ0 − ... (iii) Na+ Cl Adding equation (I) and equation (II) and subtracting (III) from them: 0 0 0 Λ(HCl) + Λ(CH − Λ(NaCl) = λ0 + + λ0 = Λ0(CH 3COONa) (H ) (CH3COO− ) 3COOH) No. of molecules ionised Λm (b) Determination = of degree of dissociation (α): α = Total number of molecules dissolved Λ0 m 0 1000 κ (c) Determination of solubility of sparingly soluble salt: Λm = , C Where C is the molarity of solution and hence the solubility. 0 0 (d) Determination of ionic product of water: From Kohlrausch’s law, we determine Λm of H2O where Λm is the molar conductance of water at infinite dilution when one mole of water is completely ionized to give one mole 0 of H+ and one mole of OH− ions i.e. Λm (H2O) = λ0 + + λ0 H OH− κ × 1000 Again using the following Λm = , where C=molar concentration i.e. mole L−1 or mole dm−3 C  1 7. 1 2 | Electrochemistr y κ ⇒ Λm = , where C = concentration in mole m−3 C 0 0 κ κ Assuming that Λm differs very little from Λm ; Λm = ⇒C= C Λm0 Specific conductance ( κ ) of pure water is determined experimentally. Thereafter, molar concentration of dissociated water is determined using the above equation. K w is then calculated as: K w = C2 7. THEORY OF WEAK ELECTROLYTES (i) Electrolytes that are not completely ionized when dissolved in a polar medium like water are called weak electrolytes. There exists equilibrium between ions and unionized molecules. AB  A + + B− (ii) The Concept of chemical equilibrium and law of mass action can be applied to ionic equilibrium also. AB  A⁺ + B⁻ t = 0 C 0 0 [A + ][B− ] Cα × Cα Cα 2 teq. C – Cα Cα; K Cα = = ; K= ... (i) [AB] C(1 − α ) 1−α For weak electrolytes, α p2 , oxidation occurs at LHS electrode and reduction occurs at RHS electrode. 0.0591 (p ) Ecell = log 1 at 25º C 2 (p2 ) In the amalgam cells, two amalgams of the same metal at two different concentrations are immersed in the same electrolytic solution. M(HgC1 ) | M n+ | Zn(HgC2 )  Chem i str y | 17.23 0.0591 C The e.m.f of the cell is given by the expression Ecell = log 1 at 25º C n C2 (b) Electrolyte concentration cells: In these cells, electrodes are identical but these are immersed in a solution of the same electrolyte of different concentrations. The source of electrical energy in the cell is the tendency of the electrolyte to diffuse from a solution of higher concentration to that of lower concentration. With the expiry of time, the two concentrations tend to become equal. Thus, at the beginning, the e.m.f of the cell is at its maximum and it gradually falls to zero. Such a cell is represented in the following manner: ( C2 is greater than C1 ). Zn | Zn2+ (C1 ) || Zn2+ (C2 ) | Zn M | Mn+ (C1 ) || Mn+ (C2 ) | M or Anode Cathode 0.0591 C The e.m.f of the cell is given by the following expression: Ecell = log 2(RHS) at 25ºC n C1(LHS) The concentration cells are used to determine the solubility of sparingly soluble salts, valency of the cation of the electrolyte and transition point of the two allotropic forms of a metal used as electrodes, etc. Example: Find the standard cell potential for an electrochemical cell with the following cell reaction. Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) Sol: (i) Write the half-reactions for each process. Zn(s) → Zn2+ (aq) + 2e− ; Cu2+ (aq) + 2e− → Cu(s) º (ii) Look up the standard potentials for the reduction half-reaction. Ereduction of Cu2+ = +0.339 V (iii) Look up the standard reduction potential for the reverse of the oxidation reaction and change the sign. º Ereduction of Zn2+ = −0.762 V,Eoxidation º of Zn = −( −0.762 V) = +0.762 V (iv) Add the cell potentials together to get the overall standard cell potential. º º Oxidation: Zn(s) → Zn2+ (aq) + 2e− Eoxidation = −Ereduction = −( −0.762 V) = +0.762 V º Reduction: Cu2+ (aq) + 2e− → Cu(s) Ereduction = +0.339 V º Overall: Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) Ecell = +1.101 V Example: Predict the cell potential for the following reaction when the pressure of the oxygen gas is 2.50 atm, the hydrogen ion concentration is 0.10 M and the bromide ion concentration is 0.25 M. O2 (g) + 4H+ (aq) + 4Br − (aq) → 2H2O(l ) + 2Br2 (l ) º Sol: (i) Calculate the standard cell potential for the reaction, Ecell, using the tabled values: º º Oxidation: 4Br (aq) → 2Br2 (l ) + 4e Eoxidation = −Ereduction = −( +1.077 V) = −1.077 V − − + − º Reduction: O2 (g) + 4H (aq) + 4e → 2H2O(l ) Ereduction = +1.229 V Overall: O2 (g) + 4H+ (aq) + 4Br − (aq) → 2H2O(l ) + 2Br2 (l ) Eº = +0.152 V cell (ii) Determine the new cell potential resulting from the changed conditions. (iii) Calculate the value for the reaction quotient, Q. (Note: We calculate Q using molar concentrations for solutions and pressures for gases. Water and bromine are both liquids, therefore they are not included in the calculation of Q.)  1 7. 2 4 | Electrochemistr y 1 1 Q = ; Q Q 1.02 × 106 ;= + 4 − 4 PO [H ] [Br ] (2.50atm)(0.10 M)4 (0.25M)4 2 (iv) Calculate the number of moles of electrons transferred in the balanced equation, n. n = 4 moles of electrons (v) Substitute values into the Nernst equation and solve for the non-standard cell potential, Ecell. +0.152 V − (0.0257 / 4)ln(1.02 × 106 ),Ecell = Ecell = 0.063V Illustration 11: Reaction → 2Ag + Cd2+. The standard electrode potentials for Ag+ → Ag and Cd2+ → Cd couples are 0.80 volt and −0.40 volt respectively (i) What is the standard potential E for this reaction? (ii) For the electrochemical cell, in which this reaction takes place which electrode is negative electrode?  (JEE MAIN) Sol: First write down the two half-cell. Standard potential E for the cell is given by standard potential of reducing electrode+ standard potential of oxidising electrode. The electrode having less electrode potential act as negative electrode. (i) The half reactions are: 2Ag+ + 2e− → 2Ag Re duction (Cathode) Eº = 0.80 volt (Reduction potential); Cd → Cd2+ + 2e− Ag+ /Ag Oxidation ( Anode ) Eº = −0.40 volt (Reduction potential) or Eº = +0.40 volt Cd2 + /Cd Cd/Cd2 + Eº = Eº + Eº = 0.40 + 0.80 = 1.20 volt Cd/Cd2 + Ag+ /Ag Illustration 12: The standard oxidation potential of zinc is 0.76 volt and of silver is − 0.80 volt. Calculate the e.m.f of the cell: Zn | Zn(NO3 )2 || AgNO3 | Ag At 25ºC. (JEE MAIN) 0.25M 0.1M Sol: First calculate the standard potential for reaction which is calculated as º º º Ecell = Eoxidation − Ereduction º After calculating Ecell e.m.f of the cell can be easily calculated using following equation º 0.0591 [Products] E= cell Ecell − log n [Reactants] º The cell reaction is Zn + 2Ag+ → 2Ag + Zn2+ ; Eoxidation of Zn = 0.76 volt º º º º Ereduction of Ag = 0.80 volt ; Ecell = Eoxidation of Zn + Ereduction of Ag = 0.76 + 0.80 = 1.56 volt º 0.0591 [Products] We know that, E= cell Ecell − log n [Reactants] º 0.0591 0.25 0.0591 = Ecell − log 1.56 − = × 1.3979 = (1.56 − 0.0413) volt = 1.5187 volt. n 0.1 × 0.1 2  Chem i str y | 17.25 Illustration 13: Calculate the e.m.f of the cell. Mg(s) | Mg2+ (0.2M) || Ag+ (1 × 10 −3 ) | Ag Eº +0.8 volt, Eº 2+ = −2.37 volt = Ag+ /Ag Mg /Mg What will be the effect on e.m.f if concentration of Mg2+ ion is decreased to 0.1 M?  (JEE MAIN) Sol: First calculate the standard potential for reaction which is calculated as º º º Ecell = Eoxidation − Ereduction º After calculating Ecell e.m.f of the cell can be easily calculated using following equation º 0.0591 [Products] E= cell Ecell − log n [Reactants] º º Ecell = ECathode − EºAnode = 0.80 − ( −2.37) = 3.17 volt º 0.0591 Mg2+ Cell reaction, Mg + 2Ag+ → 2Ag + Mg2+ ; E= cell Ecell − log n [Ag+ ]2 0.0591 0.2 = 3.17 − log 3.0134 volt when Mg2+ = 0.1M 3.17 − 0.1566 = = 2 [1 × 10−3 ]2 º 0.0591 0.1 E= cell Ecell − log = (3.17 − 0.1477) volt = 3.0223 volt. 2 (1 × 10−3 )2 Illustrations 14: To find the standard potential of M3+ / M electrode, the following cell is constituted: Pt | M | M3+ (0.0018mol−1L) || Ag+ (0.01mol−1L) Ag The e.m.f of this cell is found to be 0.42 volt. Calculate the standard potential of the half reaction M3+ + 3e− → M. Eº = 0.80 volt.  (JEE MAIN) Ag+ /Ag Sol: Here e.m.f is given we have to calculate standard potential of anode. So first we have to calculate standard potential of the cell and subtract it from the provided standard potential of cathode. Standard potential of cell can be determined using following expression, Nernst equation º 0.0591 [M3+ ] E= cell Ecell − log 3 [Ag+ ]3 The cell reaction is M + 3Ag+ → 3Ag + M3+ º 0.0591 [M3+ ] Applying Nernst equation, E= cell Ecell − log 3 [Ag+ ]3 º 0.0591 (0.0018) º º 0.42 = Ecell − log Ecell = − 0.064 ; Ecell =(0.42 + 0.064) =0.484 volt 3 (0.01)3 º º Ecell = ECathode − EºAnode or = EºAnode ECathode º º − ECell =(0.80 − 0.484) =0.32 volt. 1 7. 2 6 | Electrochemistr y PROBLEM-SOLVING TACTICS (a) Related to electrolysis: Electrolysis comprises of passing an electric current through either a molten salt or an ionic solution. Thus the ions are “forced” to undergo either oxidation (at the anode) or reduction (at the cathode). Most electrolysis problems are really stoichiometry problems with the addition of some amount of electric current. The quantities of substances produced or consumed by the electrolysis process is dependent upon the following: (i) Electric current measured in amperes or amps (ii) Time measured in seconds (iii) The number of electrons required to produce or consume 1 mole of the substance (b) To calculate amps, time, coulombs, faradays and moles of electrons: Three equations related these quantities: (i) Amperes × time = Coulombs (ii) 96,485 coulombs = 1 Faraday (iii) 1 Faraday = 1 mole of electrons The through process for interconverting amperes and moles of electrons is: Amps and time Coulombs Faradays Moles of electrons Use of these equations are illustrated in the following sections. (c) To calculate the quantity of substance produced or consumed: To determine the quantity of substance either produced or consumed during electrolysis, given the time a known current flowed: (i) Write the balanced half-reactions involved. (ii) Calculate the number of moles of electrons that were transferred. (iii) Calculate the number of moles of substance that was produced/consumed at the electrode. (iv) Convert the moles of substance to desired units of measure. (d) Determination of standard cell potentials: A cell’s standard state potential is the potential of the cell under standard state conditions, and it is approximated with concentrations of 1 mole per liter (1 M) and pressures of 1 atmosphere at 25ºC. (i) To calculate the standard cell potential for a reaction. (ii) Write the oxidation and reduction half-reactions for the cell. º (iii) Look up the reduction potential, Ereduction, for the reduction half-reaction in a table of reduction potentials. (iv) Look up the reduction potential for the reverse of the oxidation half-reaction and reverse the sign to º º obtain the oxidation potential. For the oxidation half-reaction, Eoxidation = −Ereduction. (v) Add the potentials of the half-cells to get the overall standard cell potential. º º º Ecell −Ereduction = + Eoxidation (e) For determining non-standard state cell potentials: To determine the cell potential when the conditions are other than standard state (concentrations not 1 molar and/or pressures not 1 atmosphere): (i) Determine the standard state cell potential. (ii) Determine the new cell potential resulting from the changed conditions. (iii) Determine Q, the reaction quotient.  Chem i str y | 17.27 (iv) Determine n, the number of electrons transferred in the reaction “n”. (v) Determine Ecell, the cell potential at the non-standard state conditions using the Nernst equation. º E= cell Ecell − (RT / nF)lnQ º Ecell = cell potential at non-standard state conditions; Ecell = standard state cell potential R = constant (8.31 J/mole K); T = absolute temperature (Kelvin scale) F = Faraday’s constant (96,485 C/mole e− ) n = Number of moles of electrons transferred in the balanced equation for the reaction occurring in the cell; [C]c [D]d Q = Reaction quotient for the reaction. aA + bB → cC + dD, Q = [A]a [B]b If the temperature of the cell remains at 25ºC, the equation simplifies to: º º E= cell Ecell − (0.0257 / n)lnQ or in terms of log10 ; E= cell Ecell − (0.0592 / n)logQ POINTS TO REMEMBER S.No. Description 1 Electrolyte Any substance which dissolves in water to form a solution that will conduct an electric current (ionic substances).Electrolytes may be classified as strong (NaCl, HCl, NaOH) or weak (NH4OH, CH3COOH, HF). Solutions that do not conduct electricity at all are called non-electrolytes. 2 Strong and Weak Strong electrolyte - Solutions in which the substance dissolved (solute) is present Electrolytes entirely as ions. Weak electrolyte - A solute that yields a relatively low concentration of ions in solution. 3 Dissociation The separation of ions that occurs when an ionic substance dissolves: CaCl2(s) + H2O → Ca+2(aq) + 2Cl-(aq). 4 Electrochemical Cells A system of electrodes and electrolytes in which a spontaneous or non-spontaneous redox reaction occurs. 5 Components of a. Electrode: An electrical conductor (metal strip) used to establish contact with a non- Electrochemical cells metallic part of the circuit (usually an electrolyte). b. Anode: The electrode at which oxidation occurs. c. Cathode: The electrode at which reduction occurs. d. Electrolyte: A liquid, paste, or gel that serves to conduct charge by moving ions in the cell. e. Half-cell: A single electrode immersed in a solution of its ions. f. Salt bridge: A device (porous disk or bridge i.e. U-tube containing inert electrolytic solution, KCl, NH4NO3, etc.) placed between the cells which maintains electrical neutrality by allowing ions to migrate between the cells. g. External circuit: The part of the cell where charge is conducted as a current of moving electrons. h. Standard Electrode Reduction Potential E : The measurement, in volts, of the tendency for a half reaction to occur as a reduction half reaction. 1 7. 2 8 | Electrochemistr y S.No. Description 6 Voltaic/Galvanic Cells Redox reactions are spontaneous and chemical energy is transformed into electrical energy. The cell potential E is positive and the anode is the negative electrode. i.e. batteries Zn(s) Zn+2(1M) Cu+2(1M) Cu(s) anode  cathode 7 Electrolytic Cells Cell in which an external electric current is required to drive a non-spontaneous redox reaction. The cell potential (Eo) is negative and the anode is the positive electrode. i.e. electrolysis, electroplating, etc. Cu(s) Cu+2(1M) Cu+2(1M) Cu(s) anode  cathode 8 Quick Comparison of Type of redox reaction cell Galvanic/Voltaic Electrolytic Electrolytic Cells potential (Eºcell) Electron Spontaneous Non-spontaneous flow (Eºcell is positive) (Eºcell is negative) Site of oxidation Creates one Requires one Site of reduction Anode Anode Positive electrode Cathode Cathode Negative electrode Cathode Anode Flow of electrons Anode Cathode Anode to cathode Anode to cathode (negative to positive) (positive to negative) Batteries Electrolysis, electroplating 9 Faraday’s First law of The amount of electrolyte discharged at an electrode is directly proportional to the Electrolysis quantity of electricity passed: W ∝ Q where, = I. t I = Current strength in ampere T = time in seconds ⇒ W = ZQ = Zit Z is a constant called electrochemical equivalence (ECE) 10 Electrochemical It is the amount of an electrolyte discharged on passing one coulomb of electricity. Equivalent 11 Faraday’s Constant It is the charge possessed by 1.0 mole of electrons and it is equal to 96500 coulombs (approx.). In terms of faraday’s constant the number of gram equivalent of electrolyte discharged at an electrode is equal to the number of faraday’s passed.  Q  ⇒ W=E   where, E = Equivalent weight  96500  12 Faraday’s Second Law Second Law: If same quantity of electricity is passed through different cells connected in series, same number of gram equivalent of electrolytes are discharged at each of the electrodes: W1 E1 ⇒ = W2 E2 Where, W1 and W2 are the weights of electrolytes discharged at two different electrodes in two different cells connected in series and E1 and E2 are their respective equivalent weights.  Chem i str y | 17.29 S.No. Description 13 Nernst Equation  2.303 RT [Reduced form] Ehalf–cell = E half–cell – log nF [Oxidised form] At 298 K, the Nernst equation can be written as, Ehalf–cell = E half–cell – 0.0591log [Reduced form] n [Oxidised form] Solved Examples JEE Main/Boards Pt(H2 ) H+ Ag+ Ag Eo cell = 0.7991 V 1bar = a 1= a 1 Example 1: Zn and iron can replace Cu in a solution but Pt and Au cannot. Why? Sol: Since platinum electrode has zero reduction Sol: Both Zn and iron have more EOP than Cu, whereas potential the standard electrode potential will be equal Pt and Au have less EOP to the standard electrode potential of the cell. Ecell = EOP  + ERP Example 2: Which of the following metals cannot H/H+ Ag+ /Ag be obtained by the electrolysis of their aqueous salt   solution and why? Or 0.7991=0 + ERP ∴ ERP = 0.7991 V Ag+ /Ag Ag+ /Ag Al, Na, Cu, Ag. Example 5: Standard reduction potential of the Ag+/Ag Sol: Al and Na cannot be obtained because they have electrode at 298 K is 0.799 V. Given that for AgI, Ksp = 8.7 higher EOP than H and thus, reduction of Cu2+ and Ag+ × 10–17, evaluate the potential of the Ag+/Ag electrode in will give Cu and Ag. a saturated solution of AgI. Also calculate the standard reduction potential of I– / AgI/Ag electrode. Example 3: Calculate the no. of electron lost or gained during electrolysis of 2 g Cl– from NaCl (aq) to give Cl2 Sol: Here solubility product is given from this calculate at anode. the concentration of silver ions. Now substituting this Sol: First calculate the equivalent of Cl- used during the value in Nernst equation determine E + Ag /Ag reaction and on multiplying it with avogadro number will give us the no of electron lost during the reaction As we have found out E , EI/AgI/Ag can find out by Ag+ /Ag ∵ 2Cl– → Cl2 + 2e⁻ using the value of solubility product. Eq. of Cl– used = 2/35.5 ∵ 1 eq. of an element involves 1 faraday charge or N E = E +(0.059/1) log [Ag+]  … (i) Ag+ /Ag Ag+ /Ag electrons ∴ (2/35.5) eq. of an element involves Also, K sp = [Ag+ ][I – ] AgI N× 2 6.023 × 1023 × 2 ∵ [Ag]+ = [I–] (for a saturated solution) = electrons = 35.5 35.5 = 3.4 × 1022 electrons ∴ [Ag+] = (K sp=) (8.7 × 10 –17 ) AgI Example 4: Evaluate the E + and E 2+ from the = 9.32 × 10–19 … (ii) Ag /Ag Zn /Zn given values: ∴ By Eq. (i),