Electrochemistry Revision Notes PDF

DR/Mohamed Arafa Fi Eljaib 01008925892 Unit four (To illustrate an oxidation-reduction reaction) Dip a zinc plate in a blue solution of copper (II) sulphate. Observation: The zinc ions dissolve in t...

DR/Mohamed Arafa Fi Eljaib 01008925892 Unit four (To illustrate an oxidation-reduction reaction) Dip a zinc plate in a blue solution of copper (II) sulphate. Observation: The zinc ions dissolve in the solution. The blue colour of the solution gradually disappears till it becomes colorless. Reddish brown copper metal deposits on the zinc plate. Conclusion: A spontaneous redox reaction takes place a 𝑍𝑛(𝑠) + 𝐶𝑈𝑠𝑜4(𝑎𝑞.) → 𝑍𝑛𝑠𝑜4(𝑎𝑞.) + 𝐶𝑢(𝑠) The previous reaction consists of two half reactions: 0 +2 Oxidation reaction: 𝑍𝑛(𝑠) → 𝑍𝑛(𝑎𝑞.) + 2𝑒 − +2 0 Reduction reaction: 𝐶𝑢(𝑎𝑞.) + 2𝑒 − → 𝐶𝑢(𝑠) ↓ Galvanic Cells EX. Daniel cell Anode>Oxidation(reducing agent Cathode>reduction>oxidizingagent >+veincrease>electronwith product >+vedecrease> electron with reactant 0 +2 − +2 0 𝑍𝑛(𝑠) → 𝑍𝑛(𝑎𝑞.) + 2𝑒 𝐶𝑢(𝑎𝑞.) + 2𝑒 − → 𝐶𝑢(𝑠) ↓ Mass of anode decrease Mass of cathode increase +2 Conc of 𝑍𝑛 →increase Conc of 𝑐𝑢+2 decrease Anion move to Anode Cation move to cathode Anode positive Cathode negative the flow of electric current stops when: 1-All the zinc plate dissolves in the Zn ½ cell. 2-all copper ions (𝑐𝑢+2 ) disappear 3-remove the salt bridge which does not react with the ions in the two ½ cells or with the electrodes. 1 DR/Mohamed Arafa Fi Eljaib 01008925892 Role of salt bridge: It connects the solutions of the two half cells together. It neutralize the excess +ve and -ve ions in the two half cells. It allows the electric current to flow continuously. Measuring the Electrode Potential Since the potential of the standard hydrogen electrode = zero Volt The potential measured is that of the unknown electrode. Standard Hydrogen Electrode (SHE) It consists of: A platinum sheet (1𝐶𝑚2 ) covered by a layer of spongy black platinum (Which has high tendency to absorb 𝐻2 gas) Dipped in a one molar solution (1 mole/liter) of a strong acid as hydrochloric acid. Hydrogen gas is passed on the surface of the Pt sheet at a constant pressure of one atmosphere. Note: The standard hydrogen electrode has zero potential, yet the potential can change if the concentration of the acid solution (1 molar) or the pressure of hydrogen gas (1 atm.) is changed. emf = Oxidation potential of Anode + Reduction potential of Cathode emf or Ecell= Higher oxidation potential- Lower oxidation potential emf or Ecell= Higher reduction potential - Lower reduction potential Solved Problems: 1) Two elements (A) and (B) having oxidation potentials of 0.76 and - 0.34 Volts respectively. Both elements are divalent. a) Write the cell expression of a galvanic cell formed of the two elements. b) Calculate the emf of the cell. c) Does the cell produce a spontaneous current or not? Give reason. Solution Since the oxidation potential of A>B ∴ A is anode while B is cathode a) Anode Cathode (oxidation) (reduction) +2 A/𝐴 𝐵+2 /B b) 𝐸𝑐𝑒𝑙𝑙 = Higher oxidation potential - Lower oxidation potential = 0.76 - (-0.34) = + 1.1 Volt c) The cell produces a spontaneous current, as the E.M.F. is a +ve value. 2 DR/Mohamed Arafa Fi Eljaib 01008925892 2) A galvanic cell utilizes the following reaction 𝐻2 + 𝐶𝑢+2 → 2𝐻+ + 𝐶𝑢 ↓ a) Write the symbolic representation. b) Which element is the oxidizing agent and which is the reducing agent. Give reason for your choices. c) Calculate the emf of the cell, knowing that the ox. pot. Of copper is- 0.34 V. a) Anode Cathode (oxidation) (reduction) + 𝐻2 /2𝐻 𝐶𝑢+2 /𝐶𝑢 b) Oxidizing agent: Copper ion 𝐶𝑢+2 because it is reduced. Reducing agent: Hydrogen𝐻2 , because it is oxidized. c) It is known that the potential of hydrogen is zero. Ecell = Higher oxidation potential- Lower oxidation potential = zero - (-0.34) = + 0.34 Volt The Electromotive Series (Electrochemical series or Chemical Activity Series) 1) Elements at the top of the E.C.S. they are easily oxidized ∴they are strong reducing agents. 2) Elements at the top of the E.C.S. can replace those elements below them. 3) As the distance between the two elements in the E.C.S. increases, the highly reactive element will easily replace the less reactive one. Golden rule: In any galvanic cell the element with higher oxidation potential is always oxidized & always act as anode, and the element with higher reduction potential is always reduced & always act as cathode. Practical Galvanic Cells Primary Cells: "Galvanic cells that cannot be recharged as their electrode reactions cannot be reversed." 1) The Mercury Cell Ecell 1.35= Volt Advantages: small volume & dry Uses: ear phones, clocks and cameras. - Anode: (-ve) Made of Zinc (Zn) - Cathode: (+ve) Made of Mercury (ii) oxide (HgO) - Electrolyte: Potassium hydroxide (KOH) 3 DR/Mohamed Arafa Fi Eljaib 01008925892 2-The Fuel Cell 2𝐻2(𝑔) + 𝑂2(𝑔) → 2𝐻2 𝑜 + 𝐸𝑛𝑒𝑟𝑔𝑦 Uses: It is used in space ships. Why? Because its gaseous fuel (mixture of hydrogen and oxygen) is the same fuel used in launching rockets. The electrolyte is hydrated potassium hydroxide (KOH). The electromotive force produced Ecell =1.23 Volt Cell Reactions: At Anode: (oxidation) − 2𝐻2(𝑔) + 4𝑂𝐻(𝑎𝑞.) → 4𝐻2 𝑜(𝑔) + 4𝑒 − At Cathode: (reduction) − 𝑂2(𝑔) + 2𝐻2 𝑂(𝑔) + 4𝑒 − → 4𝑂𝐻(𝑎𝑞.) Notes:The fuel cell is not consumed& does not store energy like any other galvanic cell. Why? - Because it is continuously supplied with fuel from an external source. The fuel cell operates at a high temperature ∴ the water produced evaporates and can later be condensed to be used as drinking water for astronauts. Secondary Cells:can be reversed." recharging process takes place by connecting the galvanic cell to an external battery of slightly higher potential and reversing the direction of the current. Lead- Acid Battery(Car Battery or Lead Accumulator) six cells connected in series. Each cell produces Ecell = 2 Volt. The emf of the battery is 2x6 = 12 Volt. Disadvantages: Large size. &contain liquid electrolyte. Composition: Outer vessel made of solid rubber or plastic (polystyrene) which is not affected by acids. - Anode: (-ve) A network of lead plates packed with pure spongy lead (Pb). - Cathode: (+ve) A network of lead plates packed with lead dioxide (𝑷𝒃𝑶𝟐 ) - Electrolyte: Dilute sulphuric acid of density 1.28-1.3 g/𝒄𝒎𝟑 The hydrometer is an apparatus used to measure the density of liquids. 4 DR/Mohamed Arafa Fi Eljaib 01008925892 If the acid density decreases to lower than 1.2 g/𝒄𝒎𝟑 , this mean that the battery needs recharging. The anode and cathode plates are separated from each other by a number of insulating sheets. At Anode: (oxidation) 𝑝𝑏(𝑠) + 𝑠𝑜4−2 (𝑎𝑞.) → 𝑝𝑏𝑠𝑜4 (𝑠) + 2𝑒 − At Cathode: (reduction) + 𝑝𝑏𝑜2 (𝑠) + 4𝐻(𝑎𝑞.) + 𝑆𝑜4−2 (𝑎𝑞.) + 2𝑒 − → 𝑃𝑏𝑠𝑜4 (𝑠) + 2𝐻2 𝑜(𝑙) Overall Reaction: + 𝑝𝑏(𝑠) + 𝑃𝑏𝑜2 (𝑠) + 4𝐻(𝑎𝑞.) + 2𝑆𝑜4−2 (𝑎𝑞.) → 2𝑝𝑏𝑠𝑜4 (𝑠) + 2𝐻2 𝑜(𝑙) During discharging this battery leads to: 1) Dilution of 𝐻2 SO4 density decrease or acid consumed as a result of the water produced. 2) Crystals of 𝑃𝑏𝑠𝑜4 deposit on the two electrodes. Recharging the cell: By connecting the battery to an external source of electricity (direct electric current) of slightly higher potential, where the oxidation-reduction reactions are reversed leading to: Changing of lead sulphate (𝑃𝑏𝑠𝑜4 ) to lead (Pb) and lead dioxide (𝑃𝑏𝑜2 ). Increasing the concentration of sulphuric acid. 2) Lithium lon Battery Ecell = 3 Volt Advantages: Dry and rechargeable Uses: Mobile phones. Laptop computers. -Anode: (-ve) Made from lithium graphite ( 𝐿𝐼𝐶6 ) - Cathode: (+ve) Made from lithium cobalt oxide ( 𝐿𝑖𝐶𝑜𝑂2 ) -The isolator: Made from a very thin plastic layer to separate the two electrodes, but it allows ions to pass through it. - Electrolyte:Anhydrous electrolytic solution of lithium phosphorous hexa fluoride (𝐿𝐼𝑃𝐹6 ) in which the three layers are dipped. 5  DR/Mohamed Arafa Fi Eljaib 01008925892 + At Anode: (oxidation)𝐿𝑖𝑐6 (𝑠) → 𝑐6 (𝑠) + 𝐿𝑖(𝑎𝑞.) + 𝑒− + At Cathode: (reduction) 𝐶𝑜𝑂2 (𝑠) + 𝐿𝑖(𝑎𝑞.) + 𝑒 − → 𝐿𝑖𝐶𝑜𝑂2 (𝑠) Total Reaction: 𝐿𝑖𝐶6 (𝑠) + 𝐶𝑜𝑂2 (𝑠) → 𝐶6 (𝑠) + 𝐿𝑖𝐶𝑜𝑂2 (𝑠) Points of Mercury Cell Fuel Cell Lead-acid Lithium ion Typeof cell Primary Primary secondary secondary Ecell 1.35 volt 1.23 volt 2 volt 3 volt Anode Zinc (Zn) Hollow spongy Lithium container in lead (pb) graphite which 𝐻2 gas (𝐿𝐼𝐶6 ) is supplied Cathode Mercury(ii) Hollow Lead plates Lithium oxide (Hgo) container in packed cobalt which 𝑂2 gas with lead oxide(𝐿𝑖𝐶𝑜𝑂2 ) is supplied oxide (𝑝𝑏𝑂2 ) Electrolyte Potassium Potassium Dilute Anhydrous hydroxide(KOH) hydroxide(KO sulphuric lithium H) acid phosphorous (𝐻2 𝑆𝑂4 ) hexa fluoride (𝐿𝑖𝑝𝑓6 ) Cell reaction Zn + HgO 2𝐻2 + 𝑂2 𝑝𝑏 + 𝑝𝑏𝑜2 𝐿𝑖𝐶6 + 𝐶𝑜𝑂2 → ZnO + Hg → 2𝐻2 𝑂 + 4𝐻+ → 𝐶6 + 2𝑆𝑜4−2 + 𝐿𝑖𝐶𝑜𝑂2 → 2𝑝𝑏𝑆𝑂4 + 2𝐻2 𝑜 The Corrosion of Metals corrosion of metals happens due to the formation a galvanic cell, in which the anode is the corroded metal (more active) The anode is iron. Thecathode is carbon (or other impurities). The electrolyte is water that contains many dissolved ions. +2 At Anode: (oxidation) 2𝐹𝑒(𝑠) → 2𝐹𝑒(𝑎𝑞.) + 4𝑒 − − At Cathode: (reduction) 2𝐻2 𝑂(𝐿) + 𝑂2 (𝑔) + 4𝑒 − → 4𝑂𝐻(𝑎𝑞.) Oxygen (of air) is reduced to a hydroxide group (OH-) +2 − 2𝐹𝑒(𝑎𝑞.) + 4𝑂𝐻(𝑎𝑞.) → 2𝐹𝑒(𝑂𝐻)2 (𝑠) Iron (II) hyd. is oxidized by oxygen (dissolved in water) to iron (llI) hydroxide. 2𝐹𝑒(𝑂𝐻)2 (𝑠) + 1⁄2 𝑂2 (𝑔) + 𝐻2 𝑂(𝑙) → 2𝐹𝑒(𝑂𝐻)3 (𝑠) 6  DR/Mohamed Arafa Fi Eljaib 01008925892 Factors that Cause Corrosion: Factors specific to the metal itself:The contact of metals with each other: E.g. The contact between aluminum and copper, aluminum is corroded firstly.the contact between iron and copper, iron is corroded firstly. The external factors (factors specific to the surroundings) Water, oxygen and salts are the external factors that cause metal corrosion. Protection of Iron against Corrosion 1. Painting iron with an organic material as oil, varnish or primer, but this method is not effective on the long run. 2. Covering iron with metals that resist corrosion. Cathodic Protection Anodic Protection Cover iron with less active element Cover iron with more active element (tin) (Sn) if the tin layer is scratched, Galvanizing: cover iron with zinc iron will corrode even easier than iron Sacrifying Electrode: like Mg alone. Ex:iron pipes that are buried in moist soil. Ex: food cans Electrolytic Cells "Systems in which electrical energy (provided by an external source) is changed to chemical energy in the form of a non-spontaneous oxidn- reducn reaction." The anode is +ve at which oxidation takes place. The cathode is -ve at which reduction takes place. Electrolysis"It is the chemical decomposition of a sub, due to passing an electric current in the electrolyte." When an external battery is operated, +ve ions (cations) travel to the Cathode (-ve electrode) to be neutralized by gaining electrons (Reduction), -ve ions (anions) travel to the Anode (+ve electrode) to be neutralized by losing electrons (Oxidation). Example: Electrolysis of copper (II) chloride +2 − 𝐶𝑢𝐶𝑙2 (𝑠) → 𝐶𝑢 (𝑎𝑞.) + 2𝐶𝑙(𝑎𝑞.) &𝐶𝑢+2 (𝑎𝑞.) → 𝐶𝑎𝑡ℎ𝑜𝑑𝑒(𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛) +2 0 𝐶𝑢(𝑎𝑞.) + 2𝑒 − → 𝐶𝑢(𝑠) − 2𝐶𝑙(𝑎𝑞.) → Anode (Oxidation) − 2𝐶𝑙(𝑎𝑞.) → 𝐶𝑙2 (𝑔) + 2𝑒 − 7 DR/Mohamed Arafa Fi Eljaib 01008925892 Electrolysis of aqueous solution To get product at cathode To get at anode 1-reduction pot. Of H2O more than 1-oxidation pot. Of H2O more than reduction pot. Of cation of groups oxidation pot. Of 1A,2A,3A 𝐹 − , 𝑆𝑂3−2 , 𝑆𝑂4−2 , 𝑁𝑂3− , 𝑆2 𝑂3−2 Na+,K+,Mg+2,Ca+2,Al+3 The reduction of H2O occur at The oxidation of H2O occur at cathode and produced H2 anode and produced O2 2- reduction pot. Of H2O less than 2-Oxidation pot.of H2O less than reduction pot. Of cation oxid pot. +2 + +3 Cu ,Ag ,Au ,Cd ,Zn +2 +2 𝐶𝑙 − , 𝐵𝑟 − , 𝐼 − The reduction occurs for this ions The Oxidation occur to this ions Faraday's 1 Law: (1832) "The mass of material (solid or gas) formed or consumed at any electrode is directly proportional to the quantity of electricity that passes in the electrolyte (solution or melt)." If different quantities of electricity are passed in the same electrolyte, the masses of substances precipitated or evolved at any of the two electrodes will be directly proportional to the quantity of electricity. Experiment to prove Faraday's 1st law: Faraday's 2d Law: (1833) "The masses of different materials formed or consumed by the same amount of electricity that passes in different electrolytes (connected in series) are directly proportional to their equivalent masses. Mass of element 1 Equivalent mass of element 1 = Mass of element 2 Equivalent mass of element 2 8  DR/Mohamed Arafa Fi Eljaib 01008925892 Rules Used for Solving Problems 1) Q = I x T (Coulomb) (Ampere) (Second) the gram atomic mass 2)The gram equivalent mass = number of charges of the ion (Z) (Valency) 3) 1F= 96500C 4) 1 F will ppt. or dissolve or evolve the equivalent weight of any element in grams. 𝒒𝒖𝒂𝒏𝒕𝒊𝒕𝒚 𝒐𝒇 𝒆𝒍𝒆𝒄𝒕𝒓𝒊𝒄𝒊𝒕𝒚 5) Mass of sub ppt or evolved = x Equivalent weight 𝟗𝟔𝟓𝟎𝟎 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑖𝑚𝑡𝑒𝑛𝑠𝑖𝑡𝑦×𝑡𝑖𝑚𝑒 𝑎𝑡𝑜𝑚𝑖𝑐 𝑚𝑎𝑠𝑠 6) Mass (gram) = × 96500 𝑣𝑎𝑙𝑒𝑛𝑐𝑦 𝒄𝒖𝒓𝒓𝒆𝒏𝒕 𝒊𝒎𝒕𝒆𝒏𝒔𝒊𝒕𝒚×𝒕𝒊𝒎𝒆 7) QF = 𝟗𝟔𝟓𝟎𝟎 8)when the same quantity of electricity pass (series connection) 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝐴 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝐵 = 𝑒𝑞𝑢𝑖𝑣𝑒𝑙𝑒𝑛𝑡 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐴 𝑒𝑞𝑢𝑖𝑣𝑒𝑙𝑒𝑛𝑡 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐵 9)A rule used in some special cases: In case of one mole& One mole= One gram atom= One gm/atom Q (in Faraday) = Z F = valency ×F 𝑙𝑖𝑏𝑒𝑟𝑎𝑡𝑒𝑑 10)1 F×no.of moles of electrons → 1 mol 9  DR/Mohamed Arafa Fi Eljaib 01008925892 1) What is the weight (mass) of gold and chlorine produced by passing 10000 Coulomb of electricity through a solution of gold(IIl) chloride, if the following reactions take place at the electrodes: [Au=196.98, CI=35.45] +3 0 𝐴𝑈(𝑎𝑞.) + 3𝑒 − → 𝐴𝑈(𝑠) 3𝐶𝐿−(𝑎𝑞.) → 3/2𝐶𝐿2 (𝑔) + 3𝑒 − Solution Mass of Au =? , Mass of Cl =? Q=10000 C, At. Mass of Au = 196.98, Valency of Au = 3, At. Mass of Cl = 35.45, Valency of Cl = 1 𝒒𝒖𝒂𝒏𝒕𝒊𝒕𝒚 𝒐𝒇 𝒆𝒍𝒆𝒄𝒕𝒓𝒊𝒄𝒊𝒕𝒚 𝑎𝑡𝑜𝑚𝑖𝑐 𝑚𝑎𝑠𝑠 Mass of Au = × = 6.8 grams 𝟗𝟔𝟓𝟎𝟎 𝑣𝑎𝑙𝑒𝑛𝑐𝑦 10000 35.5 Mass of Cl = × = 3.67 grams 96500 1 2-What is the quantity of electricity in coulombs is needed to separate 5.6 grams of iron from iron (III) chloride solution. [ Fe= 55.86] Cathode reaction is 𝐹𝑒 +3 + 3𝑒 − → 𝐹𝑒 0 SolutionQ=? , Mass of Fe = 5.6 g, At. Mass of Fe =55.86, Valency = 3 𝒒𝒖𝒂𝒏𝒕𝒊𝒕𝒚 𝒐𝒇 𝒆𝒍𝒆𝒄𝒕𝒓𝒊𝒄𝒊𝒕𝒚 𝑎𝑡𝑜𝑚𝑖𝑐 𝑚𝑎𝑠𝑠 Mass of Fe = × 𝟗𝟔𝟓𝟎𝟎 𝑣𝑎𝑙𝑒𝑛𝑐𝑦 5.6×96500×3 Q= = 29022.5 Coulomb 55.86 2) Aluminum metal is produced from electrolysis of alumina (𝐴𝐿2 𝑂3 ) Calculate the time needed to separate 454 grams of Al metal, knowing that the strength of the current used is 11.2 Ampere. [Al= 27] SolutionTime =? , Mass of Al = 454 g, At. Mass of Al = 27, Valency of Al = 3, Current strength 11.2 Amp I (amp.) x T (sec.) Atomic Mass Mass of Al = × 96500 Valency 11.2 x T 27 454 = × 96500 3 10 DR/Mohamed Arafa Fi Eljaib 01008925892 T = 434632 Seconds 4) How many Faradays are needed to separate 3 grams of iron from an iron(lIl) sulphate solution. [Fe= 56] Solution Q =? Faraday,Mass of Fe = 3 g, At. Mass of Fe =56, Valency= 3 Atomic Mass Mass of Fe = Q (Faraday) X Valency 56 3 =Qx 3 Q = 0.16 Faraday 5)when a quantity of electricity equals 4F is passed in CuSO4 Solution, the number of moles of deposited copper will be Solution +2 0 𝐶𝑈(𝑎𝑞.) + 2𝑒 − → 𝐶𝑈(𝑠) 2F →1mol 4F →xmol No. of moles= 2 mol 6)what is the maximum mass of aluminum can be found on the cathode of the electrolytic cell of aluminum oxide melt, if 5 mol of electrons are passed in it? Solution 𝐴𝑙2 𝑂3 → 2𝐴𝑙 +3 + 3𝑂−2 2𝐴𝑙 +3 + 6𝑒 → ́ 2𝐴𝑙 6F→2 mol 5F→x mol No. of moles= 1.66 mol Mass of Al = no.of moles × molar mass Mass of Al= 1.66 × 27 = 45 g 8) How many Faradays are needed to precipitate one gram atom of +2 0 copper, according to the reactio 𝐶𝑈(𝑎𝑞.) + 2𝑒 − → 𝐶𝑈(𝑠) Solution Q =? Faraday, Valency of Cu = 2, One gram atom One mole, No of moles = 1 Q=ZF Q = 2 Faraday 11 DR/Mohamed Arafa Fi Eljaib 01008925892 Applications of Electrolysis 1) Electroplating: "It is a process of formation of a thin layer of a certain metal on the surface of another metal to give it a shiny appearance or protect it from corrosion. 1) Electroplating a spoon by a Layer of Silver 1-Anode: Pure Silver Rod. ‫اللي هطلي بيه‬ 2-Cathode: spoon. ‫اللي هطليه‬ 3- Electrolyte: A solution containing silver ions. E.g. Silver nitrate𝐴𝑔𝑁𝑂3. At Anode: The silver anode is oxidized forming silver ions in the 0 + solution. 𝐴𝑔(𝑠) → 𝐴𝑔(𝑎𝑞.) + 𝑒− At Cathode: Silver ions in the solution will be reduced and deposit as + 0 silver metal on the spoon. 𝐴𝑔(𝑎𝑞.) + 𝑒 − → 𝐴𝑔(𝑠) Mass of anode decrease and Mass of cathode increase the concentration of solution doesn’t change 2) Preparation of Aluminum: The aluminum ore is called bauxite (𝐴𝐿2 𝑂3 ). To obtain pure Al, the ore is dissolved in cryolite containing some fluorospar. Cryolite (𝑁𝑎3 𝐴𝐼𝐹6 ): Acts as a solvent for bauxite. Fluorospar (𝐶𝑎𝐹2 ): Acts as a flux, that decreases the melting point of the mixture from 2045°C to 950°C. - Anode: Carbon rods (grapite). -Cathode: A steel container plated by a layer of carbon (graphite). -Electrolyte: Molten bauxite. Cell Reactions: 𝐴𝐿2 𝑂3 (𝑠) → 2𝐴𝐿+3 (𝑎𝑞.) + 3𝑂 −2 Cathode (Reduction) 2𝐴𝐿+3 − (𝑎𝑞.) + 6𝑒 → 2𝐴𝐿(𝐿) −2 Anode (Oxidation)3𝑂(𝑎𝑞.) → 3⁄2 𝑂2 (𝑔) + 6𝑒 − Overall Reaction: 2𝐴𝐿+3 −2 3 (𝑎𝑞.) + 3𝑂(𝑎𝑞.) → 2𝐴𝐿(𝐿) + ⁄2 𝑂2 (𝑔) Give Reason: The anode (carbon rods) is changed from time to time. 12  DR/Mohamed Arafa Fi Eljaib 01008925892 because the oxygen gas evolved at the anode reacts with the carbon rods changing them to carbon monoxide (CO) and carbon dioxide (𝐶02 ) gases.3⁄2 𝑂2 (𝑔) + 2𝐶(𝑠) → 𝐶𝑂(𝑔) + 𝐶𝑂2 (𝑔) 3-Purification of metals: e.g. Copper Structure of the Cell: - Anode: Impure copper rod. -Cathode: Pure copper rod. - Electrolyte: Copper (II) sulphate solution, which ionizes in water as follows -At Anode: The anode itself is oxidized 0 +2 𝐶𝑈(𝑠) → 𝐶𝑈(𝑎𝑞.) + 2𝑒 − 0 +2 𝐹𝑒(𝑠) → 𝐹𝑒(𝑎𝑞.) + 2𝑒 − 0 +2 𝑍𝑛(𝑠) → 𝑍𝑛(𝑎𝑞.) + 2𝑒 − Ag and Au Not Oxidized. Why? Because they are difficultly oxidized compared to copper as they are below copper in the E.C.S. So they fall below the anode. Note: This method is used for separation of some rich materials as Ag and Au from copper. At Cathode:Copper ions (𝐶𝑢+2 ) are only reduced and deposit on the +2 0 cathode. 𝐶𝑈(𝑎𝑞.) + 2𝑒 − → 𝐶𝑈(𝑠) +2 +2 𝑍𝑛(𝑎𝑞.) 𝐴𝑛𝑑 𝐹𝑒(𝑎𝑞.) Not reduced. Why? Because they are difficultly reduced compared to copper as they are above copper in the E.C.S. So they stay as ions in the solution. Galvanic Cell: Electrolytic Cell: 13 DR/Mohamed Arafa Fi Eljaib 01008925892 Changes chemical energy into Changes electrical energy into electrical energy. chemical energy. Contains a salt bridge. Contains an external battery. Produce a spontaneous The current is non spontaneous. current. The anode is the +ve electrode The anode is the -ve electrode (where oxidation takes place). (where oxidation takes place). The cathode is the-ve electrode The cathode is the +ve (where reduction takes place). electrode ( reduction ). The Ecell is a negative value. The Ecell is a positive value. 14