Electrolytic Cell PDF

Electrochemistry electrochemical electrolytic cell cell Ø electrochemical changes in chemical systems; Ø chemical reactions that produce electricity and the changes associated with the passage of electrical current through matter. Ø Redox reactio...

Electrochemistry electrochemical electrolytic cell cell Ø electrochemical changes in chemical systems; Ø chemical reactions that produce electricity and the changes associated with the passage of electrical current through matter. Ø Redox reactions (transfer of electrons) occur in these systems. Ø Many metals may be purified or electroplated using electrochemical methods; Ø Devices such as automobiles, smartphones, electronic tablets, watches, pacemakers, and many others use batteries for power; Ø Batteries use chemical reactions that produce electricity spontaneously and that can be converted into useful work; Ø All electrochemical systems involve the transfer of electrons in a reacting system; Ø In many systems, the reactions occur in a region known as the cell, where the transfer of electrons occur at electrodes. ⟵ Current flow Key Electron flow Battery ⟶ Electrolyte Typical electrolytic cell  Terminologies in electrolytic cell: 1. Electrolytic cell: Ø apparatus for carrying out electrolysis; Ø Cell/device that produces chemical reactions by passage of direct current; Ø System of electrodes and electrolytes in which electric current is used to drive non-spontaneous redox reactions; Ø Reaction system in which redox reaction occurs through the use of electric current. 2. Electrolysis: Ø Process in which electrical energy is used to cause non-spontaneous chemical reaction to occur; Ø Decomposition of an ionic system/electrolyte by the passage of direct electrical current; Ø Process whereby electric current is used to cause redox reaction to occur. 3. Electrodes (anode and cathode) Ø Metal stripes/carbon plates (graphite) through which electric current enters or leaves electrolyte. Ø Can be active or inert; Ø Inert electrode does not take part in the electrolytic reaction and its size is not affected; Ø Example: platinum, graphite Ø Active electrode takes part in the electrolytic process and its size changes. All other metals (Cu, Zn, Pb, Sn) 4. anode: Ø Positive electrode through which current enters or electron leaves electrolyte; Ø Oxidation (electron loss)occurs at the anode. 5. cathode: Ø Negative electrode by which current leaves or electron enters electrolyte; Ø Reduction (electron gain)takes place at the cathode. 6. electrolyte: Ø Compound that conducts electricity in molten or aqueous state; Ø undergoes decomposition in the electrolytic process. Types of electrolyte: Strong electrolyte: Ø undergoes complete ionization/dissociation in aqueous/molten state; Ø Good conductors of electricity. Ø Examples: CuSO4, ZnSO4, HCl, H2SO4, HNO3, NaCl Weak electrolyte: Ø Slightly/partially ionize in aqueous/molten state; Ø Poor conductors of electricity; H2O, NH3, CH3COOH, CH3NH2 Non-electrolyte: Ø Does not ionize in aqueous/molten state; Ø Non-conductors of electricity; Ø Examples: sugar, benzene, alcohol.  Ionic theory of electrolysis (movement of ions) Ø When electrolyte is melted/dissolved in water, some, if not all, of species dissociate into freely moving ions. Ø Anions move to anode (+); Ø Cations move to cathode (-); Ø Number of electrical charges carried by ion equals to valency of the corresponding atom/group; Ø Na+ (Na has valency of 1); Ø Cu2+(Cu has valency of 2); Ø SO&' % (SO4 has valency of -2). Mechanism of electrolysis: Ø Involves decomposition of electrolyte and discharge of ions at electrodes; Ø When electrolyte is aqueous, it is dissociated with water; Ø CuSO&' %()*) &. Ø CuSO4(aq) ⟶ SO&' %()*) + Cu ()*). ' Ø H2O(l) ⟶ H()*) + OH()*) Ø When electrolyte is fused/molten, electrolyte dissociated alone: Ø Fused CaCl2 Ø CaCl2 ⟶ Ca2+ + 2Cl – Electrochemical series; Ø Serial arrangement of metallic elements/ions according to specified conditions Ø Order shows tendency of one metal to reduce ions of any other metal below it in the series. Cations: (to cathode) K+, Na+, Ca2+, Mg2+, Al3+, Zn2+, Fe2+, Sn2+, Pb2+, H+, Cu2+, Hg2+, Ag+, Au+ increasing preference for discharge(decreasing electropositivity) Anions: (to anode) F –, SO&' , NO' , Cl –, Br–, I –, OH – % 1 increasing preference for discharge(decreasing electronegativity)  Factors that determine discharge of ions: 1. Concentration of ions; 2. Nature of electrodes; 3. Position of ions in electrochemical series; 4. Size and duration of current passed. 1. Concentration of ions: Ø Ion is discharged more readily when its concentration is high and if only the two competing ions are closely positioned in the electrochemical series; Ø Effect of concentration becomes less important as the position of competing ions becomes further apart in the series; Ø OH – ions are discharged in preference to Cl – ions when dilute NaCl/HCl is electrolyzed using platinum electrode; Ø When concentration of NaCl/HCl have higher concentration of Cl – ions, they are discharged at the anode before OH – ions; Ø closeness in the electrochemical series; Ø For cations, H+ ions are discharged instead of Na+ ions at the cathode; Ø Far apart in the series. 2. Nature of electrodes; Ø Active electrodes take part in the electrolytic process; Ø Inactive ones do not; Ø strong affinity for certain ions by some electrodes may influence ionic discharge; Ø Electrolysis of NaCl(aq) using Pt electrodes, H+ ions are preferentially discharged at the cathode; Ø When Hg cathode is used, the Hg will tend to associate with the Na+ ions to form Na amalgam (Na/Hg); Ø Discharge of Na+ ions require less energy than H+ ions; Ø When electrode is similar to the ions in the electrolyte, electrode enters solution; Ø Example, when Cu anode is used in electrolysis of CuSO4, SO&' % or OH- will not be discharged; Ø Cu atoms from anode goes into solution as Cu2+; Ø Cu atoms give up electrons readily than SO&' - % or OH. 3. Position of ions in electrochemical series; Ø When electrolyte contains more than one ion with the same type of charge (positive/negative), competition for which ion will be discharged arises; Ø Position of ion in the series determines this; Ø Less electropositive cations easily gain electrons than more electropositive ones (they remain in solution as positive ions); Ø K+, Na+, Ca2+ are never discharged unless there is no competition for H+ or other cations in solution; Ø These ions are only discharged when their molten salts are electrolyzed; Ø Less electronegative anions are discharged because they lose electrons readily than more electronegative ones; ' Ø SO&' % and NO1 are never discharged from solution due to discharge of OH-  Illustrations on ions discharge: 4. Size and duration of current passed: Ø This is proportional to the amount of ions discharged at the electrodes. Ø This has to do with quantitative aspect of electrolysis and hence, Faradays laws of electrolysis.  Illustrations on ions discharge: 1. Electrolysis of dilute NaCl using Platinum/graphite electrodes Decomposition of electrolyte: NaCl ⟶ CI – + Na+ H2O ⟶ OH – + H+ Migration of ions: CI – and OH – move to anode Na+ and H+ move to cathode Discharge of ions OH – are preferentially discharged to CI – Reason: Ø Its position in the electrochemical series or OH – ions are less electronegative than Cl – ions or OH – ions can readily oxidize/loss electrons. H+ are discharged in preference to Na+ ions Reason: H+ ions are less electropositive or H+ ions can easily undergo reduction or gain electron. At anode or anodic reaction: Ø OH – ions are oxidized; Ø 1 volume of O2 is evolved; Ø 4OH – ⟶ O2 + 2H2O + 4e – At cathode/cathodic reaction: H+ ions are reduced; 2 volumes of H2 is produced; 2H+ + 2e – ⟶ H2 or 4H+ + 4e – ⟶ 2H2 Overall reaction: 4H+ + 4OH – ⟶ 2H2 + O2 + 2H2O 4H2O ⟶ 2H2 + O2 + 2H2O 2H2O ⟶ 2H2 + O2 ü Resulting solution is neutral; ü Solution has no effect on indicators; ü Process is environmentally friendly. 2. Electrolysis of brine (conc. NaCl) Decomposition: NaCl ⟶ Na+ + Cl— H2O ⟶ H+ + OH — Migration: Cl— and OH — to anode Na+ and H+ to cathode Discharge: Cl— discharge at anode Reason: high conc. of Cl— ions and Cl— and OH — ions are closely positioned in the series. H+ discharge at cathode Reason: H+ and Na+ are positioned far apart in the series; conc. does not count. Anode: Cl— are oxidized Cl2 evolves 2 Cl— ⟶ Cl2 + 2e— Cathode: H+ are reduced H2 is liberated 2H+ + 2e— ⟶ H2 Overall equation: 2 Cl— + 2H+ ⟶ Cl2 + H2 Na+ and OH — ions remain in solution Solution is basic pH of solution is greater than 7; Solution changes red litmus blue; Phenolphthalein (pink); Methyl orange (orange). 2. Electrolysis of molten NaCl using graphite electrode NB: if Na(s) is deposited on Hg electrode, it forms amalgam (sodium amalgam) NaCl ⟶ Na+ + Cl— Na+ to cathode Cl— to anode Na+ discharge at cathode Cl— discharge at anode Anode: Cl— are oxidized Cl2 evolves 2Cl— ⟶ Cl2 + 2e— Cathode: Na+ reduce Na deposits Na+ + e— ⟶ Na Overall equation: 2Cl— + 2Na+ ⟶ Cl2 + 2Na 3. Electrolysis of dil. CuSO4 using Pt CuSO4⟶ Cu2+ + SO&' % H2O ⟶ H+ + OH — Cu2+ and H+ to cathode OH — and SO&' % to anode Cu2+ discharge at cathode OH — discharge at anode Anodic reaction: OH — oxidize 1 volume of O2 liberated 4OH — ⟶ O2 + 2H2O + 4e— Cathodic reaction: Cu2+ reduce Cu deposits Cu2+ + 2e— ⟶ Cu Overall reaction: 4OH — + 2Cu2+ ⟶ O2 + 2H2O + 2Cu Ø Blue solution fades; Ø H+ and SO&'% are left in solution; Ø Solution becomes acidic; Ø pH of solution is less than 7; Ø Electrolysis of acid (H2SO4) occurs; Ø Hydrogen gas at cathode; Ø Oxygen at anode. 4. Electrolysis of dil. CuSO4 using Cu electrode (active): CuSO4 ⟶ Cu2+ + SO&' % H2O ⟶ H+ + OH — Cu2+ and H+ to cathode OH — and SO&' % to anode NB: electrons can be supplied through the anode to the external circuit by three ways: 1. SO&' % ions in the solution could deposit as salt (sulphonation of anode); 2. OH — ions in solution could be discharged as gaseous O2 and H2O; 3. Metallic ions on surface of anode might dissolve in solution as ions, leaving electrons behind(Cu ⟶ Cu2+ + 2e—) Conversion of copper atoms (at anode) to ions is favoured as it requires least energy; SO"# ! and OH — are not discharged. Anode: Cu electrode loses electrons(oxidizes)/Cu electrode goes into solution; Cu ⟶ Cu2+ + 2e— Cathode: Cu2+ from anode accept electrons (get reduced) at cathode and are deposited; Cu2+ + 2e— ⟶ Cu No change in composition of electrolyte; Cu is merely transferred from anode to cathode during electrolysis; ' −2𝑒 ' &. +2𝑒 Cu(s) ⟶ Cu()*) ⟶ Cu(s) anode electrolyte cathode Ø Blue solution persists; Ø Size of anode electrode decreases; Ø Size of cathode electrode increases; Ø pH of solution remains the same. 5. Electrolysis of dil. H2SO4/acidified H2O/H2O between graphite electrode; 6. Electrolysis of dil. CuSO4 using carbon electrode; 7. Electrolysis of molten lead(II)bromide, CuBr2, CuBr, NaCl using Pt electrode; 8. Electrolysis of dil. SnCl2 using Sn electrode 9. Concentrated solution of sodium chloride is electrolyzed using carbon electrode. Predict the products at the cathode and anode and pH of the final solution. 10. Electrolysis of NaOH(aq) and CaCl2(aq) between Pt electrodes.  Applications of electrolysis: 1. Electroplating (of metals); 2. Extraction of metals (Na, K, Mg, Ca, Al, Zn); 3. Purification of metals (Cu, Hg, Ag, Au); 4. Preparation of compounds/elements (H2, Cl2, O2, NaOH, NaClO3) Electroplating/electrodeposition: Ø Process of coating an inferior metal with a superior one  copper plating a key Ø The metallic object to be plated/coated is set as the cathode (-) electrode (Key); Ø Strip of the plating metal/metal to be used for the plating is set as the anode (+) (Copper); Ø Salt of the plating metal is the electrolyte (CuSO4); Ø As the current is passed through the cell, the plating metal dissolves at the anode; Ø The ions produced migrate to the cathode; Ø The ions are discharged and deposited as a layer on the object at the cathode; Anodic reaction: Cu(s) ® Cu&. ()*) + 2e – Cathodic reaction: Cu&. ()*) + 2e ® Cu(s) –  silver plating of spoon Ø Cathode is the spoon Ø Anode is silver rod Ø Electrolyte is a solution of silver salt (AgNO3) equations for the process: Ag(s) (anode) + e – ® Ag % ("#) (in solution) + e ® Ag(s) (cathode) – Or Anodic reaction: Ag(s) ® Ag. ()*) + e – Cathodic reaction: Ag. ()*) + e ® Ag(s) – Precautions: Ø The metal to be plated must be polished ( free of grease, scale, rust and dirt); Ø If this is not done, the deposit may not be well adhering to the base metal and it is likely to peel off.  Importance of electroplating: Ø abrasion and wear resistant; Ø corrosion prevention; Ø Lubricity; Ø aesthetic qualities; Ø building thickness on undersized parts/form objects by electrolysis.  Purification of metals: Ø one of the most convenient and important method of refining and giving a metal high purity; Ø applicable to many metals such as Cu, Ag, Pb, Au, Ni, Sn, Zn, Al, etc. Ø the blocks of impure metal form the anode Ø thin sheets of pure metal form the cathode. Ø A solution of soluble salt of the metal is taken as an electrolyte; Ø On passing an electric current through the solution pure metal dissolves from the anode and deposits on cathode; Ø more metal ions undergo reduction and pure metal is deposited at the cathode. Ø insoluble impurities either dissolve in the electrolyte or fall at the bottom and collect as anode mud. Ø example, in refining copper, impurities like Fe and Zn dissolve in the electrolyte; Ø Au, Ag and Pt are left behind as anode mud. purification of copper. Ø thick block of impure copper is made anode; Ø thin pure copper is made cathode. Ø Copper(II) sulphate solution is used as electrolyte. Ø On passing electric current, following reactions take place: Ø Cu2+ ions (from copper sulphate solution) go to the cathode (negative electrode), where they are reduced to copper Ø gets deposited on the cathode. Cu2+ + 2𝑒 ' ® Cu Ø Copper (impure anode) forms copper ions, and these go into solution of electrolyte. Cu ® Cu2+ + 2𝑒 ' Ø the net result is transfer of pure copper from anode to the cathode; Ø Impurities like zinc, iron, etc., go into solution; Ø noble impurities like silver, gold, etc., are left behind as anode mud; Ø Copper is refined to 99.98% pure copper by electrolytic refining.  Faraday’s laws of electrolysis (looks at the quantitative aspect of electrolysis) Law 1 Ø Mass of an element discharged at the electrode is directly proportional to the quantity of electricity passed; Ø m a Q; but Q = It. Law 2 Ø When the same quantity of electricity is passed through different electrolytes, the relative number of moles of elements discharged are inversely proportional to the charges on the ions of the element; : Ø n a , where z is the charge on the element; ; Or Ø When the same quantity of electricity flows through different electrolytes the mass of each of the element liberated at an electrode is directly proportional to the chemical equivalence of the respective element; ?@ CDEC/EG?@HC or EG?@HC quantity of electricity (q) is measured in coulomb Ø 1 coulomb is the quantity of electricity passed when a current (I) of 1A flows through a conductor for 1 second, Q = It  Faraday (F) Ø charge on one mole of an electron/singly charged ion; Ø 96500 Cmol – 1 Illustrations Ø Na+ + e – ® Na; 1F is needed to deposit 1mol of Na metal. Ø Mg2+ + 2e – ® Mg; 2F is needed to deposit 1 mol of Mg. Ø 1 mol of a given substance contains its formula mass/the mass of its formula unit. This means that for Mg, 2F is required to deposit 24 g of Mg. Factors affecting the amount of products deposited at the electrode: Ø the magnitude of the steady current passed Ø the time of flow of the steady current Ø the ionic charge of the liberated element  SOLUTION TO EXAMINATION STYLE QUESTIONS 1. Calculate the mass of Cu deposited at the cathode when CuSO4 solution is electrolyzed between Cu electrodes by a current of 10 A for 1 hour. [Cu = 64; F = 96500 C mol – 1] Solution:  CuSO4 ® Cu2+ + SO&' % Cu2+ + 2e – ® Cu 2 x 96500 C deposits 64 g of Cu K% L :M L KM L KM 10 x 60 x 60 deposits = 11.94 g & L NKOMM zF C deposits M of Cu 𝐌𝐈𝐭 It deposits =m 𝐳𝐅 𝐦𝐳𝐅 𝐌𝐈𝐭 𝐦 𝐈𝐭 mzF = MIt ⟹ = ⟹ 𝐌 = 𝐌𝐳𝐅 𝐌𝐳𝐅 𝐳𝐅 Therefore, always remember: 𝐦 𝐈𝐭 𝐕 𝐍 n= = = = = CV 𝐌 𝐳𝐅 𝐕𝐦 𝐋 2. A piece of jewellery is to be gold plated by electrolysis using Au(NO3)3 solution. Draw and label a diagram to show how this could be done. If a current of 0.40 A is passed through the solution for 30.6 min, calculate the mass of gold deposited on the jewellery. [Au = 197; F = 96500 C] (ans = 0.50 g) Solution 3. Aring is to be gold plated by electrolysis using Au(NO3)3 solution. (a)Write the reaction at the cathode; (b)Determine the time required to deposit 0.50 g gold on the ring when a current of 0.8 A is passed through the electrolyte. [Au = 197; F = 96500 C] (ans = 15.3 min) 4. (a) State three factors that influence the selective discharge of ions during electrolysis. (b) A steady current of 2 A is passed through a magnesium chloride solution for 3 hours. Determine the mass of magnesium discharged at the cathode electrode. If the initial mass of the cathode electrode is 0.25 g, what is the total mass at the cathode? (ans = 2.69, 2.94) (c) Give two practical application/uses of electrolysis. 5. A current of 2.0 A deposited 2.5 g of recently discovered metal in 24 minutes. (i) Calculate the mass of the metal deposited by 1 Faraday. (ii) Determine the atomic mass of the metal given that it is a trivalent. (iii) What is the molar mass of the metal? [1 F = 96500 C] (ans:84 g, 251 u, 251 g mol – 1) 6. A current of 1.5 A passed for 4 hours through a molten metal salt deposited 13.3 g of the metal. (i) Calculate the mass deposited by 1 Faraday of electricity; (ii) What is the oxidation state of the metal in the salt? [metal = 118.7; 1F = 96500 C mol – 1] (ans = 59.4, +2) 7. The electrolysis of molten magnesium chloride is carried out with a current of 8.00 x 10 – 4 A. what masses of magnesium and chlorine are produced in exactly 1 hour? Calculate the volume of gas evolved at the anode at s.t.p. [Mg = 24; Cl = 35.5; 1 F = 96500 C, molar gas volume at s.t.p. = 22.4 dm3 mol – 1] (ans = 3.58 x 10 – 4 g; 1.06 x 10 – 3 g; 3.34 x 10 – 4 dm3) 8. Calculate the volume of oxygen at s.t.p. when a current of 2.5 A is passed through acidified water for 1.5 min. [H = 1; O = 16; Vm = 22400 cm3; 1 F = 96500 C mol – 1] (13.1 cm3 or 0.0131 dm3) 9. What quantity of electricity will liberate 0.25 mol of oxygen molecules during the electrolysis of water? [1F = 96500 C, ans: 96500 C] 10. (a) State Faraday’s first law of electrolysis. (b) Explain briefly how electrolysis of dilute CuSO4 affects the pH of electrolytes using each of the following electrodes: (a) carbon; (b) copper. (c) Mention two industrial applications of electrolysis. 11. Describe, using a diagram, the essential components of an electrolytic cell. The diagram should include the source of electric current and conductor, positive and negative electrodes and the electrolyte.  12. A current of 1.26 A is passed through an electrolytic cell containing a dilute H2SO4 solution for 7.44 hours. Write (a) half – cell reaction equation; (b) overall reaction equation. Calculate the volume of the gas liberated at the (a) anode; (b) cathode at s.t.p. [1F = 96500; molar volume of gas = 22.4 dm3 mol – 1] (ans: 1.96 dm3, 3.92 dm3) 13. State three reasons why electroplating is important. 14. During the purification of a sample of gold, a solution containing gold ions was electrolyzed using gold electrodes. A current of 0.20 A was passed for 60 minutes and at the end of the experiment the cathode gained a mass of 0.492 g. (a) Calculate the, (i) quantity of electricity passed; (ii) mass of gold deposited by 1faraday of electricity; (iii) number of faradays required to deposit 1.0 mol of gold. (b) From the experiment above, deduce the charge on the gold ion. [Au = 197, 1Faraday = 96500 C mol – 1] 15. A current of 0.5 A was passed through acidified water for 1.75 minutes at 298 K and 6.2 x 104 Nm – 2 using platinum electrodes. (i) Write balanced chemical equation for the reaction at the (a) anode; (b) cathode. (ii) Calculate the volume of the gas evolved at the anode at s.t.p. [Vm = 22.4 dm3, 1F = 96500 C, s.p. is 1.01 x 105Nm – 2, s.t. is 273 K] 16. Reaction stops suddenly when a piece of copper metal is placed in a solution of silver trioxonitrate(V). Explain briefly this observation. 17. State the species responsible for the conduction of electricity in the following substances: (i) Cu metal; (ii) Molten copper(II) tetraoxosulphate(VI); (iii) Aqueous calcium chloride. end