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Author: Lewis Evans

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2 Table of Contents (18.1) Balancing oxidation–reduction equations(18.2) Galvanic cells (18.3) Standard reduction potentials (18.4) Cell potential, electrical work, and free energy (18.5) Dependence of cell potential on concentration (18.6) Batteries (18.7) Corrosion (18.8) Electrolysis (18.9) Commercial electrolytic processes

3 Oxidation - Loss of electrons Reduction - Gain of electrons Review of Terms Oxidation - Loss of electrons Reduction - Gain of electrons Reducing agent - Electron donor Oxidizing agent - Electron acceptor Copyright © Cengage Learning. All rights reserved

4 Review of Terms (Continued)Electrochemistry: Study of the interchange of chemical and electrical energy Oxidation–reduction (redox) reaction Involves a transfer of electrons from the reducing agent to the oxidizing agent Copyright © Cengage Learning. All rights reserved

5 Overall reaction is split into two half-reactionsHalf-Reaction Method Overall reaction is split into two half-reactions One involves oxidation, and the other involves reduction Unbalanced equation for the oxidation–reduction reaction between cerium(IV) ion and tin(II) ion Reduction Oxidation Copyright © Cengage Learning. All rights reserved

6 Problem-Solving Strategy - Half-Reaction Method (Acidic Solution)Write separate equations for the oxidation and reduction half-reactions For each half-reaction: Balance all the elements except H and O Balance O using H2O Balance H using H+ Balance the charge using electrons Copyright © Cengage Learning. All rights reserved

7 Add the half-reactions, and cancel identical species Problem-Solving Strategy - Half-Reaction Method (Acidic Solution) (Continued) If necessary, multiply one or both balanced half-reactions by an integer Helps equalize the number of electrons transferred in the two half-reactions Add the half-reactions, and cancel identical species Check that the elements and charges are balanced Copyright © Cengage Learning. All rights reserved

8 Interactive Example 18.1 - Balancing Oxidation–Reduction Reactions (Acidic)Potassium dichromate (K2Cr2O7) is a bright orange compound that can be reduced to a blue-violet solution of Cr3+ ions Under certain conditions, K2Cr2O7 reacts with ethanol (C2H5OH) as follows: Balance this equation using the half-reaction method

9 Interactive Example 18.1 - SolutionThe reduction half-reaction is: Chromium is reduced from an oxidation state of +6 in Cr2O72– to one of the +3 in Cr3+ The oxidation half-reaction is: Carbon is oxidized from an oxidation state of –2 in C2H5OH to +4 in CO2

10 Interactive Example 18.1 - Solution (Continued 1)Balancing all elements except hydrogen and oxygen in the first half-reaction Balancing oxygen using H2O Balancing hydrogen using H+

11 Interactive Example 18.1 - Solution (Continued 2)Balancing the charge using electrons Next, we turn to the oxidation half-reaction Balancing carbon

12 Interactive Example 18.1 - Solution (Continued 3)Balancing oxygen using H2O Balancing hydrogen using H+ We then balance the charge by adding 12e– to the right side

13 Interactive Example 18.1 - Solution (Continued 4)In the reduction half-reaction, there are 6 electrons on the left-hand side, and there are 12 electrons on the right-hand side of the oxidation half-reaction Thus, we multiply the reduction half-reaction by 2 to give:

14 Interactive Example 18.1 - Solution (Continued 5)Adding the half-reactions and canceling identical species

15 Interactive Example 18.1 - Solution (Continued 6)Check that elements and charges are balanced

17 Problem-Solving Strategy - Half-Reaction Method (Basic Solution)Use the half-reaction method as specified for acidic solutions to obtain the final balanced equation as if H+ ions were present To both sides of the equation obtained above, add a number of OH– ions that is equal to the number of H+ ions We want to eliminate H+ by forming H2O Copyright © Cengage Learning. All rights reserved

18 Check that elements and charges are balancedProblem-Solving Strategy - Half-Reaction Method (Basic Solution) (Continued) Form H2O on the side containing both H+ and OH– ions, and eliminate the number of H2O molecules that appear on both sides of the equation Check that elements and charges are balanced Copyright © Cengage Learning. All rights reserved

19 Critical Thinking When balancing redox reactions occurring in basic solutions, the text instructs you to first use the half-reaction method as specified for acidic solutions What if you started by adding OH– first instead of H+? What potential problem could there be with this approach?

20 Balance this equation using the half-reaction methodInteractive Example Balancing Oxidation–Reduction Reactions (Basic) Silver is sometimes found in nature as large nuggets; more often it is found mixed with other metals and their ores An aqueous solution containing cyanide ion is often used to extract the silver using the following reaction that occurs in basic solution: Balance this equation using the half-reaction method

21 Interactive Example 18.2 - SolutionBalance the equation as if H+ ions were present Balance the oxidation half-reaction Balance carbon and nitrogen Balance the charge

22 Interactive Example 18.2 - Solution (Continued 1)Balance the reduction half-reaction: Balance oxygen Balance hydrogen Balance the charge

23 Interactive Example 18.2 - Solution (Continued 2)Multiply the balanced oxidation half-reaction by 4 Add the half-reactions, and cancel identical species

24 Interactive Example 18.2 - Solution (Continued 3)Add OH– ions to both sides of the balanced equation to eliminate the H+ ions We need to add 4OH– to each side 4H2O(l)

25 Interactive Example 18.2 - Solution (Continued 4)Eliminate as many H2O molecules as possible Check that elements and charges are balanced

26 Device in which chemical energy is changed to electrical energy Galvanic Cell Device in which chemical energy is changed to electrical energy Uses a spontaneous redox reaction to produce a current that can be used to do work Reaction occurs at the interface between the electrode and the solution where electron transfer is facilitated Copyright © Cengage Learning. All rights reserved

28 Galvanic Cell - ComponentsA salt bridge or a porous disk is used to permit ions to flow without extensive mixing of the solutions Salt bridge: Contains a strong electrolyte in a U-tube Porous disk: Contains tiny passages that allow hindered flow of ions Copyright © Cengage Learning. All rights reserved

30 Cell Potential (Ecell)Pull or driving force on electrons Termed as the electromotive force (emf) of the cell Volt (V): Unit of electrical potential Defined as 1 joule of work per coulomb of charge transferred Copyright © Cengage Learning. All rights reserved

31 Measuring Cell PotentialUse a voltmeter Voltmeter: Draws current via a known resistance Maximum cell potential can be ascertained by measuring it under zero current Potentiometer: Variable voltage device, which is powered by an external source, inserted in opposition to the cell potential Copyright © Cengage Learning. All rights reserved

33 Standard Reduction PotentialsE°values corresponding to reduction half-reactions with all solutes at 1 M and all gases at 1 atm All half-reactions are given as reduction processes in standard tables

34 Table 18.1 - Standard Reduction Potentials at 25°C (298 K) for Many Common Half-Reactions

35 Table Standard Reduction Potentials at 25°C (298 K) for Many Common Half-Reactions (Continued)

36 Obtaining a Balanced Oxidation–Reduction Reaction - ManipulationsWhen a half-reaction is reversed, the sign of E° is reversed When a half-reaction is multiplied by an integer, E° remains the same Standard reduction potential is an intensive property Copyright © Cengage Learning. All rights reserved

37 Standard Reduction Potentials - ExampleRedox reaction Half-reactions (1) (2) To balance the cell reaction and calculate the standard cell potential, reverse reaction (2) Copyright © Cengage Learning. All rights reserved

38 Standard Reduction Potentials - Example (Continued 1)Each Cu atom produces two electrons but each Fe3+ ion accepts only one electron Therefore, reaction (1) must be multiplied by 2 Copyright © Cengage Learning. All rights reserved

40 Critical Thinking What if you want to “plate out” copper metal from an aqueous Cu2+ solution? Use Table 18.1 to determine several metals you can place in the solution to plate copper metal from the solution Defend your choices Why can Zn not be plated out from an aqueous solution of Zn2+ using the choices in Table 18.1?

41 Interactive Example 18.3 - Galvanic CellsConsider a galvanic cell based on the following reaction: The half-reactions are: Give the balanced cell reaction, and calculate E°for the cell

42 Interactive Example 18.3 - SolutionThe half-reaction involving magnesium must be reversed and since this is the oxidation process, it is the anode Also, since the two half-reactions involve different numbers of electrons, they must be multiplied by integers

43 Interactive Example 18.3 - Solution (Continued)

44 Describe electrochemical cells Anode components are listed on the left Line Notations Describe electrochemical cells Anode components are listed on the left Cathode components are listed on the right Anode and cathode components are separated by double vertical lines The lines indicate a salt bridge or a porous disk Copyright © Cengage Learning. All rights reserved

45 Line Notations (Continued)Phase difference (boundary) is indicated by a single vertical line Example Mg(s)|Mg2+(aq)||Al3+(aq)|Al(s) Mg → Mg2+ + 2e– (anode) Al3+ + 3e– → Al (cathode) Copyright © Cengage Learning. All rights reserved

46 Components in the Description of a Galvanic CellCell potential and the balanced cell reaction Cell potential is always positive for a galvanic cell = E°(cathode) – E°(anode) Direction of electron flow Obtained by inspecting the half-reactions and using the direction that gives a positive Designation of the anode and cathode Copyright © Cengage Learning. All rights reserved

47 Components in the Description of a Galvanic Cell (Continued)Nature of each electrode and the ions present in each compartment If none of the substances participating in the half–reaction are conducting solids, a chemically inert conductor is required Copyright © Cengage Learning. All rights reserved

48 Example 17-4 - Description of a Galvanic CellDescribe completely the galvanic cell based on the following half-reactions under standard conditions: Ag+ + e– → Ag E° = 0.80 V (1) Fe3+ + e– → Fe2+ E° = 0.77 V (2)

49 Since a positive value is required, reaction (2) must run in reverseExample Solution Since a positive value is required, reaction (2) must run in reverse Ag+ + e– → Ag E°(cathode) = 0.80 V Fe2+ → Fe3+ + e– –E° (anode) = – 0.77 V Ag+ (aq) + Fe2+(aq) → Fe3+(aq) + Ag(s) = 0.03 V

50 Example 17-4 - Solution (Continued 1)Ag+ receives electrons, and Fe2+ loses electrons in the cell reaction The electrons will flow from the compartment containing Fe2+ to the compartment containing Ag+ Oxidation occurs in the compartment containing Fe2+ (electrons flow from Fe2+ to Ag+) Hence this compartment functions as the anode Reduction occurs in the compartment containing Ag+, so this compartment functions as the cathode

51 Example 17-4 - Solution (Continued 2)The electrode in the Ag/Ag+ compartment is silver metal, and an inert conductor, such as platinum, must be used in the Fe2+/Fe3+ compartment Appropriate counterions are assumed to be present Line notation Pt(s)| Fe2+(aq), Fe3+(aq)|| Ag+(aq)|Ag(s)

52 Work and Cell PotentialCell potential (E) and work (w) have opposite signs In any real, spontaneous process some energy is always wasted Actual work realized is always less than the calculated maximum Copyright © Cengage Learning. All rights reserved

53 Work and Cell Potential (Continued)Actual work done is given by the following equation: E - Actual potential difference at which current flows q - Quantity of charge in coulombs transferred Faraday (F): Charge on 1 mole of electrons F = 96,485 C/mol e– Copyright © Cengage Learning. All rights reserved

54 Cell Potential and Free EnergyChange in free energy equals the maximum useful work that can be obtained from a process wmax = ΔG For a galvanic cell wmax = – qEmax = ΔG Since q = nF Therefore, ΔG = – qEmax = – nFEmax Copyright © Cengage Learning. All rights reserved

55 Maximum Cell PotentialDirectly related to the free energy difference between the reactants and the products in the cell ΔG° = –nFE° Galvanic cell will run in the direction that provides a positive Ecell value Positive Ecell value implies negative ΔG value, which is the condition for spontaneity Copyright © Cengage Learning. All rights reserved

56 Interactive Example 18.5 - Calculating ΔG°for a Cell ReactionUsing the data in Table 18.1, calculate ΔG°for the following reaction: Is this reaction spontaneous?

57 Interactive Example 18.5 - SolutionThe half-reactions are: We can calculate ΔG° from the equation ΔG° = –nFE°

58 Interactive Example 18.5 - Solution (Continued 1)Since two electrons are transferred per atom in the reaction, 2 moles of electrons are required per mole of reactants and products Thus, n = 2 mol e–, F = 96,485 C/mol e–, and E° = 0.78 V = 0.78 J/C

59 Interactive Example 18.5 - Solution (Continued 2)The process is spontaneous, as indicated by both the negative sign of ΔG° and the positive sign of This reaction is used industrially to deposit copper metal from solutions resulting from the dissolving of copper ores

60 Interactive Example 18.7 - The Effects of Concentration on EFor the cell reaction Predict whether Ecell is larger or smaller than for the following case: [Al3+] = 2.0 M, [Mn2+] = 1.0 M [Al3+] = 1.0 M, [Mn2+] = 3.0 M Copyright © Cengage Learning. All rights reserved

61 Interactive Example 18.7 - SolutionA product concentration has been raised above 1.0 M This will oppose the cell reaction and will cause Ecell to be less than (Ecell < 0.48 V) A reactant concentration has been increased above 1.0 M Ecell will be greater than (Ecell > 0.48 V) Copyright © Cengage Learning. All rights reserved

62 Concentration Cells Cells in which both compartments have the same components at different concentrations Cell potential is a factor of difference in concentration Voltages are small

63 Example 18.8 - Concentration CellsDetermine the direction of electron flow, and designate the anode and cathode for the cell represented in the figure on the right:

64 Example Solution The concentrations of Fe2+ ion in the two compartments can (eventually) be equalized by transferring electrons from the left compartment to the right This will cause Fe2+ to be formed in the left compartment, and iron metal will be deposited (by reducing Fe2+ ions to Fe) on the right electrode

65 Example 18.8 - Solution (Continued)Since electron flow is from left to right, oxidation occurs in the left compartment (the anode) and reduction occurs in the right (the cathode)

66 Given in a form that is valid at 25°CNernst Equation Explains the relationship between cell potential and concentrations of cell components Given in a form that is valid at 25°C Resulting potential calculated from this equation is the maximum potential before any current flow Copyright © Cengage Learning. All rights reserved

67 Nernst Equation and EquilibriumA cell will spontaneously discharge until it reaches equilibrium Q = K (equilibrium constant) Ecell = 0

68 Nernst Equation and Equilibrium (Continued)Dead battery Battery in which the cell reaction has reached equilibrium At equilibrium: Both cell compartments have the same free energy ΔG = 0 Cell can no longer work

69 What if you are told that E°= 0 for an electrolytic cell?Critical Thinking What if you are told that E°= 0 for an electrolytic cell? Does this mean the cell is “dead”? What if E = 0? Explain your answer in each case Copyright © Cengage Learning. All rights reserved

70 Example 18.9 - The Nernst EquationDescribe the cell based on the following half-reactions: Where, T = 25°C [VO2+] = 2.0 M [H+] = 0.50 M [VO2+] = 1.0 ×10–2 M [Zn2+] = 1.0 × 10–1 M Copyright © Cengage Learning. All rights reserved

71 Example Solution The balanced cell reaction is obtained by reversing reaction (2) and multiplying reaction (1) by 2 2 × reaction (1) Reaction (2) reversed Copyright © Cengage Learning. All rights reserved

72 Example 18.9 - Solution (Continued 1)Since the cell contains components at concentrations other than 1 M, we must use the Nernst equation, where n = 2 (since two electrons are transferred), to calculate the cell potential At 25°C we can use the following equation: Copyright © Cengage Learning. All rights reserved

73 Example 18.9 - Solution (Continued 2)E is the cell potential at any condition, and E° is the cell potential under standard conditions (1.0 M or 1 atm, and 25°C). E equals zero when the cell is in equilibrium ("dead" battery). E is equal to zero for a concentration cell. Copyright © Cengage Learning. All rights reserved

74 Consider the cell described below:Exercise Consider the cell described below: Zn|Zn2+(1.00 M)||Cu2+(1.00 M)|Cu Calculate the cell potential after the reaction has operated long enough for the [Zn2+] to have changed by 0.20 mol/L Assume T = 25°C 1.09 V Copyright © Cengage Learning. All rights reserved

75 Determining Ion ConcentrationCell potentials can help determine concentration of an ion pH meter - Device used to measure concentration using observed potential Composed of: Standard electrode of known potential Glass electrode: Changes potentials based on H+ ion concentration in a solution Potentiometer

76 Figure 18.12 - A Glass ElectrodeElectrical potential Depends on the difference in [H+] between the reference solution and the solution into which the electrode is dipped Varies with the pH of the solution being tested

77 Ion-Selective ElectrodesElectrodes that are sensitive to the concentration of any specific ion Examples Using a crystal of lanthanum(III) fluoride (LaF3) in an electrode to measure [F –] Using solid sliver sulfide (Ag2S) to measure [Ag+] and [S2–]

78 Table 18.2 - Some Ions Whose Concentrations Can Be Detected by Ion-Selective Electrodes

79 Calculation of Equilibrium Constants for Redox ReactionsE°and ΔG°have a quantitative relationship At equilibrium, Ecell = 0 and Q = K

80 Interactive Example 18.10 - Equilibrium Constants from Cell PotentialsFor the oxidation–reduction reaction The appropriate half-reactions are Balance the redox reaction, and calculate E°and K (at 25°C)

81 Interactive Example 18.10 - SolutionTo obtain the balanced reaction, we must reverse reaction (2), multiply it by 2, and add it to reaction (1)

82 Interactive Example 18.10 - Solution (Continued 1)In this reaction, 2 moles of electrons are transferred for every unit of reaction For every 2 moles of Cr2+ reacting with 1 mole of S4O62– to form 2 moles of Cr3+ and 2 moles of S2O32– Thus n = 2

83 Interactive Example 18.10 - Solution (Continued 2)The value of K is found by taking the antilog of 22.6 K = = 4 ×1022 This very large equilibrium constant is not unusual for a redox reaction

84 Group of galvanic cells that are connected in seriesBattery Group of galvanic cells that are connected in series Potentials of each individual cell add up to give the total battery potential Essential source of portable power in today’s society Copyright © Cengage Learning. All rights reserved

85 Nickel–cadmium battery Lithium-ion batteryTypes of Battery Lead storage battery Dry cell battery Silver cell Mercury cell Nickel–cadmium battery Lithium-ion battery Copyright © Cengage Learning. All rights reserved

86 Components Electrode reactions Lead Storage Battery Anode - LeadCathode - Lead coated with lead dioxide Electrolyte solution - Sulfuric acid Electrode reactions

88 Dry Cell Battery - Acid VersionComponents Zinc inner case that acts as the anode Carbon rod that is in contact with a moist plate of solid MnO2 Solid NH4Cl and carbon that acts as the cathode

89 Dry Cell Battery - Alkaline VersionComponents Solid NH4Cl is replaced with NaOH or KOH Lasts for a longer period of time Zinc corrodes less rapidly under basic conditions

90 Figure 18.14 - A Common Dry Cell Battery

91 Silver Cell and Mercury Cell BatteriesContains a Zn anode and a cathode that uses Ag2O as the oxidizing agent Mercury cell Contains a Zn anode and a cathode that uses HgO as the oxidizing agent

93 Nickel–Cadmium and Lithium-Ion BatteriesNickel–cadmium battery Lithium-ion batteries Li+ ions migrate from the cathode to the anode and enter the interior as the battery is charged Charge-balancing electrons travel to the anode through the external circuit in the charger The opposite process occurs on discharge

94 Galvanic cells for which reactants are continuously supplied Fuel Cells Galvanic cells for which reactants are continuously supplied The image depicts a hydrogen–oxygen fuel cell Anode reaction 2H2 + 4OH– → 4H2O + 4e– Cathode reaction 4e– + O2 + 2H2O → 4OH–

95 Can be viewed as the method of returning metals to their natural stateProcess of Corrosion Can be viewed as the method of returning metals to their natural state Natural state - Ores from which metals are originally obtained Involves the oxidation of a metal Spontaneous process Copyright © Cengage Learning. All rights reserved

96 Process of Corrosion (Continued)Some metals tend to develop a thin oxide coating to protect against further oxidation Aluminum forms a thin layer of aluminum oxide (Al2(OH)6) Copper forms an external layer of greenish copper carbonate called patina Silver sulfide forms silver tarnish Copyright © Cengage Learning. All rights reserved

98 Application of paint or metal platingCorrosion Prevention Application of paint or metal plating Galvanizing: Process in which steel is coated with zinc to prevent corrosion Alloying Cathodic protection Protects steel in buried fuel tanks and pipelines Copyright © Cengage Learning. All rights reserved

100 Electrolysis Forcing a current through a cell to produce a chemical change for which the cell potential is negative Electrolytic cell: Device that uses electrical energy to produce a chemical change Copyright © Cengage Learning. All rights reserved

101 Figure 18.19 - A Galvanic Cell and an Electrolytic Cell

102 Stoichiometry of ElectrolysisRefers to determining how much chemical change occurs with the flow of a given current for a specified time Steps to solve a stoichiometry problem Copyright © Cengage Learning. All rights reserved

103 Interactive Example 18.11 - ElectroplatingHow long must a current of 5.00 A be applied to a solution of Ag+ to produce 10.5 g silver metal?

104 Interactive Example 18.11 - SolutionIn this case, we must use the steps given earlier in reverse

105 Interactive Example 18.11 - Solution (Continued 1)Each Ag+ ion requires one electron to become a silver atom Thus 9.73 ×10 –2 mole of electrons is required, and we can calculate the quantity of charge carried by these electrons

106 Interactive Example 18.11 - Solution (Continued 2)The 5.00 A (5.00 C/s) of current must produce 9.39 × 103 C of charge Thus,

107 Electrolysis of Water - Nonspontaneous Reaction4H2O

108 Interactive Example 18.12 - Relative Oxidizing AbilitiesAn acidic solution contains the ions Ce4+, VO2+, and Fe3+ Using the E°values listed in Table 18.1, give the order of oxidizing ability of these species, and predict which one will be reduced at the cathode of an electrolytic cell at the lowest voltage

109 Interactive Example 18.12 - SolutionThe half-reactions and E°values are as follows: The order of oxidizing ability is therefore: Ce4+ > VO2+ > Fe3+ The Ce4+ ion will be reduced at the lowest voltage in an electrolytic cell

110 Production of AluminumAluminum is extracted from bauxite Not pure aluminum oxide Bauxite is treated with aqueous sodium hydroxide to obtain pure hydrated alumina Alumina dissolves in the basic solution Other metals remain as solids Extraction of purified alumina 2CO2(g) + 2AlO2–(aq) + (n + 1)H2O(l) → 2HCO3–(aq) + Al2O3 · nH2O(s)

111 Production of Aluminum (Continued)Purified alumina is mixed with cryolite and melted Aluminum ion is reduced to aluminum metal in an electrolytic cell Alumina reacts with the cryolite anion Al2O3 + 4AlF63– → 3Al2OF62– + 6F– Overall cell reaction 2Al2O3 + 3C → 4Al + 3CO2 Aluminum produced is 99.5% pure

113 Electrorefining of Metals - ExampleImpure copper that is derived from the reduction of copper ore is cast into large slabs that serve as anodes for electrolytic cells Electrolyte - Aqueous copper sulfate Cathodes - Thin sheets of ultrapure copper

114 Protects metals that corrode easily Metal Plating Protects metals that corrode easily Can be done by making the object as the cathode in a tank containing ions of the plating metal Example - Electroplating of a spoon

115 Electrolysis of Sodium Chloride (Brine)Process used for the production of sodium metal Steps Mix solid NaCl with solid CaCl2 to lower the melting point Electrolyze the mixture in a Downs cell

117 Electrolysis of Sodium Chloride (Brine) (continued)The sodium is drained off in the Downs cell Cooled and cast into blocks Stored in an inert solvent, such as mineral oil, to prevent its oxidation Produces hydrogen and chlorine gas Leaves a solution containing dissolved NaOH and NaCl

118 Contamination of the sodium hydroxide by NaClFigure Mercury Cell for Production of Chlorine and Sodium Hydroxide Contamination of the sodium hydroxide by NaCl Can be eliminated by electrolyzing the brine in a mercury cell Reaction of resulting sodium metal 2Na(s) + 2H2O(l) → Na+(aq) + 2OH–(aq) + H2(g)

119 Currently carried out in diaphragm cellsChlor–Alkali Process Involves: Recovering pure NaOH from an aqueous solution Pumping the re-generated mercury into the electrolysis cell Currently carried out in diaphragm cells Cathode and anode are separated by a diaphragm Allows passage of water molecules, Na+ ions, and Cl– ions Blocks passage of OH– ions

120 Chlor–Alkali Process (Continued)Disadvantage The aqueous effluent from the cathode compartment contains a mixture of NaOH and untreated NaCl Needs to be purified to derive pure NaOH Advancement in the chlor–alkali industry Uses a membrane to separate the anode and cathode compartments in brine electrolysis cells Membrane is impermeable to anions

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