Presentation on theme: "Redox reagents, equations, titrations, and electrolysis."— Presentation transcript:
1 Redox reagents, equations, titrations, and electrolysis. Redox ReactionsRedox reagents, equations, titrations, and electrolysis.
2 Index Redox Reactions Electrochemical Series Writing Redox Equations Redox TitrationsElectrolysis
3 Redox EquationsRedox reactions include reactions which involve the lossor gain of electrons.The reactant giving away (donating) electrons is calledthe reducing agent (which is oxidised)The reactant taking (accepting) electrons is calledthe oxidising agent (which is reduced)Both oxidation and reduction happen simultaneously,however each is considered separately usingion-electron equations.O.I.L. R.I.G.Oxidation is loss, reduction is gain of electrons
4 e.g. e.g. Al Al3+ + 3e- ½O2 + 2e- O2- Note that, in general, ½Cl2 + e- Cl-e.g.Mg Mg2+ + 2e-Al Al e-Metals on the LHS of the Periodic Table ionise byelectron loss and are called reducing agentsNon-metals on the RHS of the Periodic Table ioniseby electron gain and are called oxidising agents
5 Cells and Redox A metal lower A metal higher in the series IonbridgeA metal higherin the seriesA metal lowerin the seriesIons of metal higher in ECSIons of metal lower in ECSMetal atoms will be oxidised.Metal atoms are the reducing agent.Metal ions in solution will be reduced,Metal ions are the oxidising agent.e.g. Mg Mg e-E.g. Cu e- CuOverall redox equationMg (s) + Cu 2+ (aq) Mg 2+ (aq) + Cu (s)
6 Cells and Redoxmagnesium(s) + silver nitrate(aq) magnesium nitrate(aq) + silver(s).The reducing agent in this reaction is the Mg as it willdonate electrons to the silver ions .The oxidising agent is the Ag+ ions as they accept electronsfrom the MgOxidation:Mg(s) Mg2+(aq) e-Half equationsor ion-equationsReduction:2Ag+(aq) + 2e- 2Ag(s)Mg (s) + 2Ag+ (aq) Mg2+ (aq) + 2Ag (s)Redox equation, electrons cancel out
7 Redox and the Electrochemical Series Eo/V Oxidising agents-3.02v-2.71v-2.37v-0.13v0.00v+0.34v+0.80vLi+(aq) + e Li(s)Na+(aq) + e Na(s)Mg2+(aq) + 2e Mg(s)Pb2+(aq) + 2e Pb(s)2H+(aq) +2e H2(g)Cu2+(aq) + 2e Cu(s)Ag+(aq) + e Ag(s)Increasing powerful reducing agent(write the reaction backwards)Mg2+(aq) + 2e Mg(s)Hydrogen referenceIncreasing powerful oxidising agent(write the reaction as it appears)Ag+(aq) + e Ag(s)Considering the two ion-equations,Mg 2+ (aq) Mg (s) + 2e- and Ag + (aq) + e- Ag ,Mg, being higher up the electrochemical series, would act as the reducing agent. (i.e. the ion-electron equation would be written backwards).While Ag would be written as it appears in the electrochemical series.Mg(s) + 2Ag+(aq) Mg2+(aq) + 2Ag(s)
8 Na(s) + H2O(l) NaOH(aq) + ½H2(g) Writing REDOX equationsConsider the reaction between sodium and water:Na(s) + H2O(l) NaOH(aq) + ½H2(g)Consider how the ions are formed in this reactionNa(s) Na+(aq) + e-H2O(l) + e- OH-(aq) + ½H2(g)Na(s) Na+(aq)+ e-H2O(l) + e- OH-(aq) + ½H2(g)and, we could say that a watermolecule must be acceptingthe electronA sodium atom losesan electron
9 OIL RIG These are called ion-electron equations Na(s) Na+(aq) + e-OILRIGH2O(l) + e- OH-(aq) + ½H2(g)These are called ion-electron equations(or ionic half equations).
10 Na(s) + H2O(l) NaOH(aq) + ½H2(g) Na(s) Na+(aq) + e-Electrons cancel!H2O(l) + e- OH-(aq) + ½H2(g)Reduction and oxidation occur simultaneously. Adding the two equations together gives us the overall equation for a reaction.Na(s) + H2O(l) NaOH(aq) + ½H2(g)
11 Balancing Redox equations Most redox reaction you will come across will occur inneutral or acidic conditions.1. Make sure there are the same number of atoms of eachelement being oxidised or reduce on each side of thehalf equation.2. If there are any oxygen atoms present, balance them byadding water molecules to the other side of the half-equation.3. If there are any hydrogen atoms present, balance themby adding hydrogen ions on the other side of thehalf-equation.4. Make sure the half-reactions have the same overallcharge on each side by adding electrons.For basic solutions H atoms are balanced using H2O and then the samenumber of OH- ions to the opposite side to balance the oxygen atoms
12 SO2(g) + 2H2O(l) SO42-(aq) + 4H+(aq) 1. Write down what you know….sulphur dioxide is oxidised to sulphate ionsSO2(g) SO42-(aq)2. Balance the oxygen atoms by adding water2H2O(l)SO2(g) SO42-(aq)3. Balance the hydrogen atoms by adding hydrogen ionsSO2(g) + 2H2O(l) SO42-(aq) +4H+(aq)4. Balance the charges by adding electronsSO2(g) + 2H2O(l) SO42-(aq) + 4H+(aq) +2e-charge is zero4 - and 4 + equals zero
13 Redox TitrationsTitration is a technique for measuring the concentrationof a solution. A solution of known concentration is used towork out the unknown concentration of another solution.Redox titrations involve solutions of reducing andoxidising agents.At equivalence-point of a redox titration precisely enoughelectrons have been removed to oxidise all of the reducingagent.
14 RedoxTitration 1 2 3 4 What to do: Carefully fill the burette with potassium permanganate .1Carefully pipette exactly20 ml of iron (II) sulphateinto the conical flask.Then add 20 ml 1 mol l-1 H2SO42Add the permanganate until apermanent purple colour appearsin the conical flask.3A rough titration is done first to give a roughequivalence-point (end-point), then repeatedmore accurately to give concordant results.4
15 Redox Titrations5 Fe2+ (aq) + 8H+ (aq) + MnO4- (aq) 5 Fe3+ (aq) + Mn2+ (aq) + 4H2O(l)purplecolourlessUse a standard solution of potassium permanganate to findout the unknown concentration of an iron (II) sulphate solutionx = [MnO4- (aq) ]y = [Fe 2+ (aq) ]n y = 5n x = 1V y x C yV x x C x=n yn xOrC y=V x x C xx n yV y x n x
16 Redox Titrations, Vitamin C I2 (aq) + 2e- 2I - (aq)C6H8O6 C6H6O H+ (aq) e-reductionoxidationBlue/Black (in the presence of starch)I2 (aq) + C6H8O6colourlessC6H6O H I- (aq)Iodine, those concentration is known (in the burette)acts as an oxidising agent.Vitamin C, the unknown concentration (in the conical flask)is a reducing agent.Starch is added to show when the end-point is reached.V x x C xn x=V y x C yn y
17 Electrolysis Faraday was the first person to measure the amount of electrical charge needed to deposit a certain amountof substance at an electrode.Amount ofelectricalcharge(electrons)electrolysisMass of substancedepositedElectrical charge is the amount of electrons
18 Electrolysis Q I x t Quantity of charge = current x time Q = I x t Current is the flow of an electrical chargeThe amount or quantity of charge (Q) is measured in Coulombs (C)Quantity of charge = current x timeQ = I x t96,500 coulombs is called 1 Faraday (F).The number of coulombs required to deposit 1 mole of atoms or molecules of an element is 96,500 x n. (F x n) n being either 1,2,3 or 4.The multiplying factor n, can be equated to the number of electrons associated with the production of one atom or molecule of the element.
20 Electrolysis and Hydrogen 2H+ (aq) + H2 (g)To produce 1 mole of H2, 2 moles of electrons are needed.So to produce 1 mole of H2 , x 2 C of charge is needed.It is possible to confirm that x 2 C of charge areneeded to produce 1 mole of H2 gas by electrolysing. Thevolume of hydrogen gas collected at the cathode is measuredand converted to moles using the gases molar volume.(The molar volume = 24 litres).So knowing the volume of gas collected, you can work out thenumber of moles of gas collected.
21 Gases and Electrolysis The mass or volume of an element discharged by electrolysis can be calculated from the quantity of electricity used and vice-versa.Example: A solution of HCl is electrolysed. What current is neededto produce 2.4 litres of H2 gas in 16min 5 sec? Molar volume = 24 l mol-1Since 2H e- 1 mole of gas requires 2 moles of electrons.i.e x 2 C of charge is needed to produce 1 mole of gasSince 2.4 litres is 0.1 mole of gas, so ( x 2 ) x 0.1 C is neededQ = I x tSo I = Q/t( x 2 ) x 0.1/(16 * 60) + 5Ans: 20 A