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IB CHEMISTRY Topic 9 Redox processes Higher level 9.1 Oxidation and reduction OBJECTIVES • Oxidation and reduction can be considered in terms of oxygen gain/hydrogen loss, electron transfer or change in oxidation number. • An oxidizing agent is reduced and a reducing agent is oxidized. • Variable oxidation numbers exist for transition metals and for most main-group nonmetals. • The activity series ranks metals according to the ease with which they undergo oxidation. • The Winkler Method can be used to measure biochemical oxygen demand (BOD), used as a measure of the degree of pollution in a water sample. • Deduction of the oxidation states of an atom in an ion or a compound. • Deduction of the name of a transition metal compound from a given formula, applying oxidation numbers represented by Roman numerals. • Identification of the species oxidized and reduced and the oxidizing and reducing agents, in redox reactions. • Deduction of redox reactions using half-equations in acidic or neutral solutions. • Deduction of the feasibility of a redox reaction from the activity series or reaction data. • Solution of a range of redox titration problems. • Application of the Winkler Method to calculate BOD. 1. REDOX CALCULATIONS Oxidation GAIN OF OXYGEN 2Mg + O2 ——> 2MgO magnesium has been oxidised as it has gained oxygen REMOVAL (LOSS) OF HYDROGEN C2H5OH ——> CH3CHO + H2 ethanol has been oxidised as it has ‘lost’ hydrogen Reduction GAIN OF HYDROGEN C2H4 + H2 ——> C2H6 ethene has been reduced as it has gained hydrogen REMOVAL (LOSS) OF OXYGEN CuO + H2 ——> Cu + H2O copper(II) oxide has been reduced as it has ‘lost’ oxygen However as chemistry became more sophisticated, it was realised that another definition was required Oxidation/Reduction in terms of electrons Oxidation and reduction are not only defined as changes in O and H ... OXIDATION Removal (loss) of electrons ‘OIL’ species will get less negative or more positive REDUCTION Gain of electrons ‘RIG’ species will become more negative or less positive REDOX When reduction and oxidation take place Oxidizing agent = substance that causes oxidation to occur. The oxidizing agent is reduced. Reducing agent = substance that causes reduction to occur The reducing agent is oxidized. Which one is the oxidising/reducing agent? Mg + 2H+ Mg2+ + H2 Oxidised Reducing agent Reduced Oxidising agent Cr2O72- + 14H+ + 6Fe2+ 2Cr3+ + 6Fe3+ + 7H2O Reduced Oxidising agent Oxidised Reducing agent 2MnO4¯ + 5C2O42- + 16H+ 2Mn2+ + 10CO2 + 8H2O Reduced Oxidising agent Oxidised Reducing agent 2S2O32- + I2 S4O62- + 2I¯ Oxidised Reducing agent Reduced Oxidising agent REDOX agents Oxidising agents • Oxygen • Chlorine and other halogens • Iron (III) ion • Hydrogen peroxide (H2O2) • Manganate (VII) ion (Permanganate ion MnO4-) • Dichromate (VI) ion (Cr2O72-) Reducing agents • Hydrogen • Carbon and carbon monoxide • Metals • Iron (II) ion • Iodide ion • Thosulfate ion (S2O32-) Oxidation states Used to... tell if oxidation or reduction has taken place work out what has been oxidised and/or reduced construct half equations and balance redox equations ATOMS AND SIMPLE IONS The number of electrons which must be added or removed to become neutral atoms Na in Na = 0 neutral already ... no need to add any electrons cations Na in Na+ = +1 need to add 1 electron to make Na+ neutral anions Cl in Cl¯ = -1 need to take 1 electron away to make Cl¯ neutral Q. What are the oxidation states of the elements in the following? a) C (0) b) Fe3+ (+3) c) Fe2+ (+2) d) O2- (-2) e) He (0) f) Al3+ (+3) Rules for Determining Oxidation State 1. Free elements are assigned an oxidation state of zero. 2. The sum of the oxidation states of all that atoms in a species must be equal to the net charge on the species. 3. The alkali metals (Li, Na, K, Rb, and Cs) in compounds are always assigned an oxidation state of +1. 4. Fluorine in compounds is always assigned an oxidation state of -1. 5. The alkaline earth metals (Be, Mg, Ca, Sr, Ba, and Ra) and also Zn and Cd in compounds are always assigned an oxidation state of +2. 6. Hydrogen in compounds is assigned an oxidation state of +1, unless assigned to a metal then it is -1. Oxygen in compounds is assigned an oxidation state of -2, except for H2O2, where it is -1. Halogen in compounds is assigned an oxidation state of -1. 7. 8. Oxidation states in molecules MOLECULES The SUM of the oxidation states adds up to ZERO ELEMENTS H in H2 = 0 both are the same and must add up to Zero COMPOUNDS C in CO2 = O in CO2 = +4 -2 1 x +4 and 2 x -2 = Zero Explanation • because CO2 is a neutral molecule, the sum of the oxidation states must be zero • for this, one element must have a positive OS and the other must be negative HOW DO YOU DETERMINE WHICH IS THE POSITIVE ONE? • the more electronegative species will have the negative value • electronegativity increases across a period and decreases down a group • O is further to the right than C in the periodic table so it has the negative value Problems: A. The oxidation states of the elements other than O, H or F are SO2 O = -2 2 x -2 = - 4 overall neutral S = +4 NH3 H = +1 3 x +1 = +3 overall neutral N=-3 NO2 O = -2 2 x -2 = - 4 overall neutral N = +4 NH4+ H = +1 4 x +1 = +4 overall +1 N=-3 IF7 F = -1 7 x -1 = - 7 overall neutral I = +7 Cl2O7 O = -2 7 x -2 = -14 overall neutral Cl = +7 NO3¯ O = -2 3 x -2 = - 6 overall -1 N = +5 NO2¯ O = -2 2 x -2 = - 4 overall -1 N = +3 SO32- O = -2 3 x -2 = - 6 overall -2 S = +4 S2O32- O = -2 3 x -2 = - 6 overall -2 S = +2 S4O62- O = -2 6 x -2 = -12 overall -2 S = +2½ ! (10/4) MnO42- O = -2 4 x -2 = - 8 overall -2 Mn = +6 What is odd about the value of the oxidation state of S in S4O62- ? An oxidation state must be a whole number (+2½ is the average value) (14/2) (4/2) Naming with oxidation states THE ROLE OF OXIDATION STATE IN NAMING SPECIES To avoid ambiguity, the oxidation state is often included in the name of a species manganese(IV) oxide shows that Mn is in the +4 oxidation state in MnO2 sulphur(VI) oxide for SO3 S is in the +6 oxidation state dichromate(VI) for Cr2O72- Cr is in the +6 oxidation state phosphorus(V) chloride for PCl5 P is in the +5 oxidation state phosphorus(III) chloride for PCl3 P is in the +3 oxidation state Q. Name the following... PbO2 lead(IV) oxide SnCl2 tin(II) chloride SbCl3 antimony(III) chloride TiCl4 titanium(IV) chloride BrF5 bromine(V) fluoride Determining redox with oxidation states OXIDATION AND REDUCTION IN TERMS OF ELECTRONS REDUCTION in O.S. Species has been REDUCED Q. INCREASE in O.S. Species has been OXIDISED State if the changes involve oxidation (O) or reduction (R) or neither (N) Fe2+ —> Fe3+ O +2 to +3 I2 F2 C2O42H2O2 H2O2 Cr2O72Cr2O72SO42- —> —> —> —> —> —> —> —> I¯ F2O CO2 O2 H2O Cr3+ CrO42SO2 R R O O R R N R 0 to -1 0 to -1 +3 to +4 -1 to 0 -1 to -2 +6 to +3 +6 to +6 +6 to +4 Balancing redox half equations STEPS: Atom 1. Balance the atoms of the oxidation species Oxn no. 2. Write oxidation states e-1 3. Balance with electrons H+ 4. Balance charge with H+ H2O 5. Balance H with H2O Problem 1: Balance the half reaction of Iron(II) being oxidised to iron(III). Step 1 Step 2 Step 3 Fe2+ +2 Fe2+ ——> Fe3+ +3 ——> Fe3+ + e¯ Problem 2: Step 1 Step 2 Step 3 Step 4 Step 5 MnO4¯ being reduced to Mn2+ in acidic solution. MnO4¯ ———> Mn2+ +7 +2 MnO4¯ + 5e¯ ———> Mn2+ MnO4¯ + 5e¯ + 8H+ ———> Mn2+ MnO4¯ + 5e¯ + 8H+ ———> Mn2+ + 4H2O Problem 3: Step 1 Step 2 Cr2O72- being reduced to Cr3+ in acidic solution. Cr2O72- ———> Cr2O72- ———> 2 Cr at +6 Cr3+ 2Cr3+ 2 Cr at +3 Step 3 Step 4 Cr2O72- + 6e¯ Cr2O72- + 6e¯ + 14H+ Step 5 Cr2O72- + 6e¯ + 14H+ ——> 2Cr3+ ——> 2Cr3+ ——> 2Cr3+ + 7H2O Combining half equations A combination of two ionic half equations, one involving oxidation and the other reduction, produces a REDOX equation. The equations are balanced as follows... Step 1 Step 2 Step 3 Step 4 Write out the two half equations Multiply the equations so that the number of electrons in each is the same Add the two equations and cancel out the electrons on either side If necessary, cancel any other species which appear on both sides Problem 1: The reaction between manganate(VII) and iron(II) Step 1 Fe2+ ——> MnO4¯ + 5e¯ + 8H+ ——> Fe3+ + e¯ Mn2+ + 4H2O Oxidation Reduction Step 2 5Fe2+ MnO4¯ + 5e¯ + 8H+ ——> ——> 5Fe3+ + 5e¯ Mn2+ + 4H2O multiplied by 5 multiplied by 1 Step 3 MnO4¯ + 5e¯ + 8H+ + 5Fe2+ ——> Step 4 MnO4¯ + 8H+ + 5Fe2+ ——> Mn2+ + 4H2O + 5Fe3+ + 5e¯ Mn2+ + 4H2O + 5Fe3+ 2. TITRATION CALCULATIONS Titration of Iron with manganate(VII) 5Fe2+ + MnO4- + 8H+ 5Fe3+ + Mn2+ + 4H2O purple colourless Manganate(VII) ion is the oxidizing agent oxidizing iron ions. Titration of with iodine-thiosulfate reaction Iodine is usually supplied in the from of KI. Oxidation forces I- ions to I2 by oxidizing agents such as KMnO4 or K2Cr2O7. I-(aq) + oxidizing agent I2(aq) + reduced product Starch is now added to react with the I2 formed making a blue-black complex: This is titrated with thiosulfate, removing it from this complex to make the solution clear. IODOMETRY Oxidation -1 state Reaction IAnalyte 0 I2 Oxidizing agent oxidized reduced IODIMETRY Oxidation 0 state Reaction I2 Analyte -1 Reducing agent Example: Vitamin C I reduced oxidized 3. THE WINKLER METHOD Biological Oxygen Demand (BOD) Definition: BOD is the measure of the dissolved oxygen (in ppm) needed to biologically decompose the organic matter in water over 5 days, at 20°C. • Fish require about 3ppm of dissolved oxygen (0.003gdm-3) • Maximum oxygen solubility in water is 0.009gdm-3 BOD (ppm) Quality of water <1 Almost pure water 5 Doubtful purity 10 Unacceptable quality 100-400 100-10 000 Waste from untreated sewage Waste water from meat-processing plant Measurement of BOD Using the Winkler Method: 1. Water sample is saturated with O2 so known concentration is reached (9ppm) by shaking with air. 2. The sample is incubated at a fixed temperature (20°C) for 5 days to allow the organic matter present to decay by microbial action. 3. Excess Mn2+ (aq) is added with OH- to use up remaining O2 and convert to MnO2 & H2O 2Mn2+(aq) + 4OH-(aq) + O2(g) 2MnO2(s) + 2H2O(l) 4. I- in H+(aq) is added to form I2 & water & reform Mn2+ (aq) MnO2(s) + 2I-(aq) + 4H+(aq) Mn2+(aq) + I2(aq) + 2H2O(l) All the solid precipitate must be dissolved. 5. Finally the I2 is titrated with thiosulfate (S2O32-) to release the iodide ions (a starch indicator could be used which will form a blue/black starch I2 complex that turns clear as thiosulphate turns I2 to I-). I2(aq) + 2S2O32-(aq) S4O62-(aq) +2I-(aq) 6. Use reaction stoichiometry to determine the number of moles of O2 left after 5 days. Following the red arrows, for every mole of S2O32- used, ¼ mole of O2 is used up. Then covert to mg O2 per dm3 to get DO in ppm. 7. Convert this to BOD (ppm) 9ppm less DO 8. BOD higher than 9ppm are determined by first diluting the original sample before incubation and then multiplying by the dilution factor. 9.2 Electrochemical cells OBJECTIVES Voltaic (Galvanic) cells: • Voltaic cells convert energy from spontaneous, exothermic chemical processes to electrical energy. • Oxidation occurs at the anode (negative electrode) and reduction occurs at the cathode (positive electrode) in a voltaic cell. Electrolytic cells: • Electrolytic cells convert electrical energy to chemical energy, by bringing about non-spontaneous processes. • Oxidation occurs at the anode (positive electrode) and reduction occurs at the cathode (negative electrode) in an electrolytic cell. • Construction and annotation of both types of electrochemical cells. • Explanation of how a redox reaction is used to produce electricity in a voltaic cell and how current is conducted in an electrolytic cell. • Distinction between electron and ion flow in both electrochemical cells. • Performance of laboratory experiments involving a typical voltaic cell using two metal/metal-ion half-cells. • Deduction of the products of the electrolysis of a molten salt. 4. VOLTAIC CELLS Comparison of Electrochemical Cells Cations are produced at the cathode Galvanic/voltaic Electrolytic Cations go to the cathode produces electrical current anode (-) cathode (+) salt bridge two electrodes conductive medium reduction at cathode oxidation at anode DG < 0 need power source anode (+) cathode (-) DG > 0 Voltaic cell 5. ELECTROLYTIC CELLS Electrolytic cell Testing electrolytes for conductivity Graphite and platinum electrodes are called inert electrodes as they don’t take part in the reaction Inert electrolysis - Predictions 1. If the metal is high in the reactivity series you will get hydrogen 2. If the below is low in the reactivity series you will get the metal 3. If the halide solution is concentrated you will get the halogen (chlorine, bromine, iodine). With other common negative ions you will get oxygen. Electrolysis of a molten salt (NaCl) cathode half-cell (-) REDUCTION Na+ + e- Na - anode half-cell (+) OXIDATION 2Cl- Cl2 + 2eoverall cell reaction 2Na+ + 2Cl- 2Na + Cl2 + battery Cl2 (g) escapes e- Na (l) Cl- Na+ Cl- (-) electrode halfcell Na+ + e- Na NaCl (l) Na+ Cl- (+) Na+ electrode halfcell 2Cl- Cl2 + 2e- OBJECTIVES • A voltaic cell generates an electromotive force (EMF) resulting in the movement of electrons from the anode (negative electrode) to the cathode (positive electrode) via the external circuit. The EMF is termed the cell potential (E⁰). • The standard hydrogen electrode (SHE) consists of an inert platinum electrode in contact with 1 mol dm-3 hydrogen ion and hydrogen gas at 100 kPa and 298 K. The standard electrode potential (E⁰) is the potential (voltage) of the reduction half-equation under standard conditions measured relative to the SHE. Solute concentration is 1 mol dm-3 or 100 kPa for gases. E⁰ of the SHE is 0 V. • When aqueous solutions are electrolysed, water can be oxidized to oxygen at the anode and reduced to hydrogen at the cathode. • ∆G⁰ = -nFE⁰. When E⁰ is positive, ∆G⁰ is negative indicative of a spontaneous process. When E⁰ is negative, ∆G⁰ is positive indicative of a non-spontaneous process. When E⁰ is 0, then ∆G⁰ is 0. • Current, duration of electrolysis and charge on the ion affect the amount of product formed at the electrodes during electrolysis. • Electroplating involves the electrolytic coating of an object with a metallic thin layer. • Calculation of cell potentials using standard electrode potentials. • Prediction of whether a reaction is spontaneous or not using E⁰ values. • Determination of standard free-energy changes (∆G⁰) using standard electrode potentials. • Explanation of the products formed during the electrolysis of aqueous solutions. • Perform lab experiments that could include single replacement reactions in aqueous solutions. • Determination of the relative amounts of products formed during electrolytic processes. • Explanation of the process of electroplating. Higher level 19.1 Electrochemical cells Higher level 6. CELL POTENTIALS A voltaic cell produces a potential difference known as the electromotive force (EMF). The cell potential or electrode potential (E) is measured by comparing it to a standard which is the standard hydrogen electrode (SHE). Higher level Electromotive force (EMF) STANDARD ELECTRODE POTENTIALS (Eº) are the potential values determined by comparison with the STANDARD HYDROGEN ELECTRODE which is the potential created by 1 mole of hydrogen ions at 100kPa H2 at 298K, which has the Eº value 0.00V. 298K (25°C) Solution of 1M H+(aq) e.g. 1M HCl or 0.5M H2SO4 Hydrogen gas at 100kPa pressure Platinum electrode Glass tube with holes to allow gas to escape The standard hydrogen electrode is assigned an E° value of 0.00V. Equilibrium reached: 2H+(aq) + 2e- ⇌ H2(g) Higher level The standard hydrogen electrode (SHE) SALT BRIDGE HYDROGEN (100kPa) ZINC ZINC SULPHATE (1M) PLATINUM ELECTRODE HYDROCHLORIC ACID (1M) In the diagram the standard hydrogen electrode is shown coupled up to a zinc half cell. The voltmeter reading gives the standard electrode potential of the zinc cell. conditions salt bridge temperature 298K solution conc. 1 Molar (1 mol dm-3) with respect to ions gases 100kPa filled with saturated potassium chloride solution it enables the circuit to be completed Higher level Measurement of E⁰ values Higher level The electrochemical series E° / V F2(g) + 2e¯ 2F¯(aq) +2.87 Mn2+(aq) + 4H2O(l) +1.52 2Cl¯(aq) +1.36 2Cr3+(aq) + 7H2O(l) +1.33 2Br¯(aq) +1.07 Ag+(aq) + e¯ Ag(s) +0.80 Fe3+(aq) + e¯ Fe2+(aq) +0.77 H2O2(aq) +0.68 2I¯(aq) +0.54 Cu+(aq) + e¯ Cu(s) +0.52 Cu2+(aq) + 2e¯ Cu(s) +0.34 Cu2+(aq) + e¯ 2H+(aq) + 2e¯ Cu+(aq) H2(g) +0.15 0.00 Sn2+(aq) + 2e¯ Fe2+(aq) + 2e¯ Sn(s) Fe(s) -0.14 -0.44 Zn2+ (aq) + 2e¯ Zn(s) -0.76 MnO4¯(aq) + 8H+(aq) + 5e¯ Cl2(g) + 2e¯ Cr2O72-(aq) + I4H+(aq) + 6e¯ Br2(l) + 2e¯ O2(g) + 2H+(aq) + 2e¯ I2(s) + 2e¯ Layout REACTION MORE LIKELY TO WORK SPECIES ON LEFT ARE MORE POWERFUL OXIDATION AGENTS If species are arranged in order of their standard electrode potentials you get a series that shows how good each substance is at gaining electrons. All equations are written as reduction processes ... i.e. gaining electrons A species with a higher E° value oxidises (reverses) one with a lower value Higher level The electrochemical series E° / V F2(g) + 2e¯ 2F¯(aq) +2.87 Mn2+(aq) + 4H2O(l) +1.52 2Cl¯(aq) +1.36 2Cr3+(aq) + 7H2O(l) +1.33 2Br¯(aq) +1.07 Ag+(aq) + e¯ Ag(s) +0.80 Fe3+(aq) + e¯ Fe2+(aq) +0.77 O2(g) + 2H+(aq) + 2e¯ H2O2(aq) +0.68 2I¯(aq) +0.54 Cu+(aq) + e¯ Cu(s) +0.52 Cu2+(aq) + 2e¯ Cu(s) +0.34 Cu2+(aq) + e¯ 2H+(aq) + 2e¯ Cu+(aq) H2(g) +0.15 0.00 Sn2+(aq) + 2e¯ Fe2+(aq) + 2e¯ Sn(s) Fe(s) -0.14 -0.44 Zn2+ (aq) + 2e¯ Zn(s) -0.76 MnO4¯(aq) + 8H+(aq) + 5e¯ Cl2(g) + 2e¯ Cr2O72-(aq) + I4H+(aq) + 6e¯ Br2(l) + 2e¯ I2(s) + 2e¯ AN EQUATION WITH A HIGHER E° VALUE WILL REVERSE AN EQUATION WITH A LOWER VALUE Application Chlorine is a more powerful oxidising agent - it has a higher E° Chlorine will get its electrons by reversing the iodine equation Cl2(g) + 2e¯ ——> Overall equation is 2Cl¯(aq) Cl2(g) + and 2I¯(aq) 2I¯(aq) ——> I2(s) + 2e¯ ——> I2(s) + 2Cl¯(aq) Will be reduced, gain electrons Will be oxidized, lose electrons • E° value can be used to predict the feasibility of redox and cell reactions. • In theory ANY REDOX REACTION WITH A POSITIVE E° VALUE WILL WORK. • In practice, it proceeds if the E° value of the reaction is greater than + 0.40V. • An equation with a more positive E° value reverse a less positive one. Higher level Predicting with E⁰ values Write out the equations Cu2+(aq) + 2e¯ Cu(s) ; E° = +0.34V Sn2+(aq) + 2e¯ Sn(s) ; E° = -0.14V the half reaction with the more positive E° value is more likely to work it gets the electrons by reversing the half reaction with the lower E° value Cu2+(aq) ——> Cu(s) therefore Sn(s) the overall reaction is Cu2+(aq) + Sn(s) ——> and Sn2+(aq) ——> Sn2+(aq) + Cu(s) the cell voltage is the difference in E° values... (+0.34) - (-0.14) = + 0.48V If the value is positive the reaction will be spontaneous Higher level Problem 1: What happens if an Sn(s) / Sn2+(aq) and a Cu(s) / Cu2+(aq) cell are connected? 7. GIBBS Higher level ΔG⁰ = -nFE⁰ n = number of moles F = Faraday constant, charge per mole, 96 500 C/mol For non-standard conditions (Option C) we use the Nernst Equation (in data booklet): E = E⁰ – (RT/nF)lnQ Higher level Electrode potential (E⁰) and free energy change (ΔG) Sn4+(aq) + 2e- ⇌ Sn2+(aq) Fe3+(aq) + e- ⇌ Fe2+(aq) E⁰ = +0.015V E⁰ = +0.77V E⁰ = E⁰cathode + E⁰anode = 0.77 - 0.015 = + 0.62V ΔG⁰ = -nFE⁰ = -(2mol e- x 96 500 C/mol x 0.62V) = -119660 V C (J = VC) = -1.2 x 102 kJ Higher level Problem 1: Calculate Gibbs free energy for the following voltaic cell at standard conditions: Higher level 8. PRODUCTS OF ELECTROLYSIS 1. If the metal is high in the reactivity series you will get hydrogen 2. If the metal is low in the reactivity series you will get the metal 3. If the halide solution is concentrated you will get the halogen (chlorine, bromine, iodine). With other common negative ions you will get oxygen. OH- > Cl- > SO42- Higher level Inert electrolysis – General predictions Cathode Formation of the metal or formation of hydrogen Anode Formation of the non-metal, formation of oxygen or oxidation of the electrode Higher level Products of electrolysis 1. Determine which products go the cathode and which go to the anode 2. For the cathode determine which cation has the most positive electrode potential. This will be reduced (gain e-1). The higher the positive potential value the more naturally a reaction will occur (atom + e-1 ion). 3. For the anode reverse the reactions and electrode potential values before determining which is the most positive and likely to occur (ion atom + e-1). 4. Include all setting out!!! (as seen in examples below) Higher level Using standard electrode potentials to determine products of electrolysis Higher level Electrolysis of CuSO4(aq) Ions present: Cu2+, SO42-, H2O(l) From the data booklet (water must be on it’s own on one side of the equation, polyatomic ions don’t react). Cathode: Anode: Final equation: Most positive -1.23 Most positive Higher level Electrolysis of NaCl(aq) Ions present: Na+, Cl-, H2O(l) From the data booklet (water must be on it’s own on one side of the equation). Cathode: Anode: Most positive -1.23 Most positive BUT… -1.36 BUT for concentrations of >25% Cl2 is preferred. Final equation: Water can be electrolyzed with weak sulphuric acid, (or weak NaOH). Ions present: SO42-, H2O(l), H+ From the data booklet (water must be on it’s own on one side of the equation, polyatomic ions don’t react). Cathode: Anode: Final equation: Most positive -1.23 Most positive Higher level Electrolysis of water Higher level Factors affecting relative amounts of products in electrolysis Higher level Electroplating 1. The item to be coated is placed on the negative electrode so that metal cations will be reduced. 2. The anode may be a metal block, often the solution is cyanide. 3. The anode dissolves and releases cations into the solution. 4. Temperature, current, constitution and concentration of solution all controlled. Higher level Electroplating Jewellery – cheaper to coat with gold than make out of pure gold Tin cans – coating with tin protects the steal cans from rusting Car parts and tools – coating with nickel and chromium decreases friction, prevents rust, and looks shiny Galvanizing – coating with zinc prevents rusting. Used on car bodies, pipes, wire, sheeting. Higher level Electroplating applications Q = It Q = charge (measured in C – coulombs), the Faraday constant which is 96 500 C/mol I = current (measured in A – amps) t = time (measured in s – seconds) Higher level Quantitative electrolysis Q = It = 1.50A x 3.25hr x 60min/hr x 60s/min = 17550C 96500C : 1mol 17500C : Xmol ∴ X = 0.1813 moles of electrons Cu2+(aq) + 2e- Cu(s) 2mol e- : 1 mol Cu 0.1813 : X ∴ X = 0.0910 moles of Cu m(Cu) = nM = 0.0910mol x 63.55g/mol = 5.78g Cu Higher level Problem 1: Calculate the mass, in g, of copper produced when a current of 1.50 A is passed through a solution of aqueous copper(II) sulphate for 3.25 hours.