Redox Equilibria - Edexcel PDF

14. Redox Equilibria Electron flow Electrochemical cells A cell has two half–cells. The two half cells have to be connected with a salt bridge. Simple half cells will consist of a metal (acts an electrode) and a solution of a compound containing that metal (eg Cu and CuSO4). These two half cells wil...

14. Redox Equilibria Electron flow Electrochemical cells A cell has two half–cells. The two half cells have to be connected with a salt bridge. Simple half cells will consist of a metal (acts an electrode) and a solution of a compound containing that metal (eg Cu and CuSO4). These two half cells will produce a small voltage if connected into a circuit. (i.e. become a Battery or cell). Salt bridge copper electrode Zinc electrode 1M copper sulphate solution 1M zinc sulphate solution Why does a voltage form? In the cell pictured above When connected together the zinc half-cell has more of a tendency to oxidise to the Zn2+ ion and release electrons than the copper half-cell. (Zn Zn2+ + 2e-) More electrons will therefore build up on the zinc electrode than the copper electrode. A potential difference is created between the two electrodes. The zinc strip is the negative terminal and the copper strip is the positive terminal. This potential difference is measured with a high resistance voltmeter, and is given the symbol E. The E for the above cell is E= +1.1V. Why use a high resistance voltmeter? The voltmeter needs to be of very high resistance to stop the current from flowing in the circuit. In this state it is possible to measure the maximum possible potential difference (E). The reactions will not be occurring because the very high resistance voltmeter stops the current from flowing. Salt bridge The salt bridge is used to connect up the circuit. The free moving ions conduct the charge. A salt bridge is usually made from a piece of filter paper (or material) soaked in a salt solution, usually potassium nitrate. The salt should be unreactive with the electrodes and electrode solutions.. E.g. potassium chloride would not be suitable for copper systems as chloride ions can form complexes with copper ions. A wire is not used because the metal wire would set up its own electrode system with the solutions and wires do not allow the flow of ions. What happens if current is allowed to flow? If the voltmeter is removed and replaced with a bulb or if the circuit is short circuited, a current flows. The reactions will then occur separately at each electrode. The voltage will fall to zero as the reactants are used up. The most positive electrode will always undergo reduction. Cu2+ (aq) + 2e-  Cu(s) (positive as electrons are used up) The most negative electrode will always undergo oxidation. Zn(s)  Zn2+ (aq) + 2e- (negative as electrons are given off) N Goalby chemrevise.org 1 Cell Diagrams Electrochemical cells can be represented by a cell diagram: Zn(s) | Zn2+ (aq) | | Cu2+ (aq) | Cu (s) E= +1.1V Most oxidised form is put next to the double line The solid vertical line represents the boundary between phases e.g. solid (electrode) and solution (electrolyte) The double line represents the salt bridge between the two half cells the voltage produced is indicated the more positive half cell is written on the right if possible (but this is not essential) Systems that do not include metals. If a system does not include a metal that can act as an electrode, then a platinum electrode must be used and included in the cell diagram. It provides a conducting surface for electron transfer A platinum electrode is used because it is unreactive and can conduct electricity. e.g. for Fe2+ (aq)  Fe3+ (aq) + e- there is no solid conducting surface, a Pt electrode must be used. If the system contains several species e.g. MnO4- + 8H+ + 5e- Mn2+ + 4H2O then in the cell diagram the balancing numbers, H+ ions and H2O can be left out. The cell diagram is drawn as: | | Fe3+ (aq), Fe2+ (aq) |Pt | |MnO4- , Mn2+ |Pt or if on left hand side Still with more oxidised form near double line A comma separates the oxidised from the reduced species. Pt | Mn2+ , MnO4- | | If a half equation has several physical states then the solid vertical line should be used between each different state boundary. | | O2 | H2O, OH- | Pt 4e- + 2H2O (l) +O2 (g) 4OH- (aq) Cl2 (g) + 2e-  2Cl- (aq) | | Cl2 | Cl- | Pt As the phase line also separates the oxidised and reduced terms a comma is not necessary here. Measuring the electrode potential of a cell It is not possible to measure the absolute potential of a half electrode on its own. It is only possible to measure the potential difference between two electrodes. To measure it, it has to be connected to another half-cell of known potential, and the potential difference between the two half-cells measured. by convention we can assign a relative potential to each electrode by linking it to a reference electrode (hydrogen electrode), which is given a potential of zero Volts N Goalby chemrevise.org 2 The standard hydrogen electrode The potential of all electrodes are measured by comparing their potential to that of the standard hydrogen electrode. H2 gas at 100kPa The standard hydrogen electrode (SHE) is assigned the potential of 0 volts. Metal electrode e.g. Fe Salt bridge KNO3 (aq) Pt electrode The hydrogen electrode equilibrium is: H2 (g) 2H+ (aq) + 2e- 1M HCl In a cell diagram the hydrogen electrode is represented by: Pt |H2 (g) | H+ (aq) Solution containing metal ions (e.g. Fe2+) at 1 mol dm-3 concentration Components of a standard hydrogen electrode. To make the electrode a standard reference electrode some conditions apply: 1. Hydrogen gas at pressure of 100 kPa 2. Solution containing the hydrogen ion at 1.00 mol dm-3 (solution is usually 1 mol dm-3 HCl) 3. Temperature at 298K Standard conditions are needed because the position of the redox equilibrium will change with conditions. For example, in the equilibrium: Mn+(aq) + n eM(s) An increase in the concentration of Mn+ would move the equilibrium to the right, so making the potential more positive. Standard Electrode Potentials Because the equilibrium does not include a conducting metal surface a platinum wire is used which is coated in finely divided platinum. (The platinum black acts as a catalyst, because it is porous and can absorb the hydrogen gas.) Secondary standards The standard hydrogen electrode is difficult to use, so often a different standard is used which is easier to use. These other standards are themselves calibrated against the SHE. This is known as using a secondary standard - i.e. a standard electrode that has been calibrated against the primary standard. The common ones are: silver / silver chloride E = +0.22 V calomel electrode E = +0.27 V E The standard conditions are : All ion solutions at 1.00 mol dm-3 temperature 298K gases at 100 kPa pressure No current flowing When an electrode system is connected to the hydrogen electrode system, and standard conditions apply the potential difference measured is called the standard electrode potential, Standard electrode potentials are found in data books and are quoted as Li+(aq) | Li (s) E= -3.03V more oxidised form on left They may also be quoted as half equations Li+ (aq) + eLi (s) E= -3.03V but again the more oxidised form is on the left H2 gas at 100kPa Salt bridge KNO3 (aq) Pt electrode Pt electrode 1 mol dm-3 HCl Pt|H2|H+||Fe3+,Fe2+|Pt 1M Fe2+ and 1M Fe3+ Note: in the electrode system containing two solutions it is necessary to use a platinum electrode and both ion solutions must be of a 1 mol dm-3 concentration, so [Fe2+] = 1 mol dm-3 and [Fe3+] = 1 mol dm-3. N Goalby chemrevise.org 3 Calculating the EMF of a cell Mg(s) | Mg2+ (aq) | | Cu2+ (aq) | Cu (s) use the equation E= +1.1V In order to calculate the Ecell, we must use ‘standard electrode potentials’ for the half cells. Each half cell has a standard electrode potential value Mg2+ (aq)| Mg(s) E= -2.37V Cu2+ (aq) | Cu (s) E = +0.34V Ecell= Erhs - Elhs For the cell diagram above Ecell = 0.34 - -2.37 = + 2.71 V Using electrode potentials The most useful application of electrode potentials is to show the direction of spontaneous change for redox reactions The easiest way to use electrode potentials is as follows: The more negative half cell will always oxidise (go backwards) For any two half equations Mg2+ (aq) + 2e-  Mg(s) E= -2.37V Cu2+ (aq) + 2e-  Cu (s) E = +0.34V The reaction would be Mg + Cu2+  Cu + Mg 2+ The more positive half cell will always reduce (go forwards) The most negative electrode will oxidise and go from right to left The half equation is therefore Zn(s)  Zn2+ (aq) +2eElectrons are given off (lost) and travel to positive electrode The more positive electrode will reduce and go from left to right Fe2+ (aq) +2e-  Fe(s) Electrons arrive at this electrode and are absorbed (gained) If we want to work out the Ecell that corresponds to this spontaneous change then use Ecell = Ered – Eox A spontaneous change will always have a positive Ecell Zn2+(aq) + 2e-  Zn(s) E= - 0.76V Fe2+(aq) + 2e-  Fe(s) E= -0.44V To get the full equation of the reaction add the two half reactions together, cancelling out the electrons. Zn + Fe2+  Fe + Zn2+ Using series of standard electrode potentials Most strong reducing agents found here oxidation As more +ve increasing tendency for species on left to reduce, and act as oxidising agents Most strong oxidising agents found here Li+ + e-  Li Mn2+ + 2e-  Mn -3.03V -1.19V 2H+ + 2e-  H2 0V Ag+ + e-  Ag F2 + 2e-  2F- +0.8V +2.87 reduction As more -ve increasing tendency for species on right to oxidise, and act as reducing agents If we want to work out the Ecell from two standard electrode potentials then use Ecell = Ered – Eox The most powerful reducing agents will be found at the most negative end of the series on the right (ie the one with the lower oxidation number) The most powerful oxidising agents will be found at the most positive end of the series on the left (ie the one with the higher oxidation number) N Goalby chemrevise.org 4 Example 1 Use electrode data to explain why fluorine reacts with water. Write an equation for the reaction that occurs. oxidise First apply idea that more positive Eo will reduce (go forward) and more negative Eo will oxidise (go backwards) O2(g) + 4H+(aq) + 4e– → 2H2O(I) Eo+1.23V F2(g) + 2e– → 2F–(aq) Eo +2.87V reduce Explanation to write As Eo F2/F- > Eo O2/H2O, and Ecell is a positive value of +1.64V, F2 will oxidise H2O to O2 work out Ecell and quote it as part of your answer Ecell = Ered - Eox = 2.87-1.23 =1.64V Equation 2F2(g) + 2H2O(I) → Remember to cancel out electrons in full equation 4F–(aq) + O2(g) + 4H+(aq) Example 2 Use data from the table to explain why chlorine should undergo a redox reaction with water. Write an equation for this reaction. Cl2(aq) + 2e– → 2Cl–(aq) Eo+1.36V + – 2HOCl(aq) + 2H (aq) + 2e → Cl2(aq) + 2H2O(I) Eo+1.64V H2O2(aq) + 2H+(aq) + 2e– → 2H2O(I) Eo +1.77V + – O2(g) + 4H (aq) + 4e → 2H2O(I) Eo +1.23V reduce First select relevant half equations by considering the Eo values and applying the idea that more positive Eo will reduce (go forward) and more negative Eo will oxidise (go backwards). Cl2(aq) + 2e– → 2Cl– (aq) Eo+1.36V + – O2(g) + 4H (aq) + 4e → 2H2O(I) Eo +1.23V oxidise Explanation to write As Eo Cl2/Cl- > Eo O2/H2O, Cl2 will oxidise H2O to O2 Equation 2Cl2(g) + 2H2O(I) → 4Cl–(aq) + O2(g) + 4H+(aq) Example 3 Suggest what reactions occur, if any, when hydrogen gas is bubbled into a solution containing a mixture of iron(II) and iron(III) ions. Explain your answer. First select relevant half equations by considering the Eo values and applying the idea that more positive Eo will reduce (go forward) and more negative Eo will oxidise (go backwards) Fe3+ (aq) + e– → Fe2+ (aq) 2H+(aq) + 2e– → H2(g) Fe2+ (aq) + 2e– → Fe(s) Eo +0.77V Eo 0.00V Eo–0.44V reduce Fe3+ (aq) + e– → Fe2+ (aq) 2H+(aq) + 2e– → H2(g) Eo +0.77V Eo 0.00V oxidise Explanation to write Fe3+ will be reduced to Fe2+ by H2 oxidising to H+ because Eo Fe3+ /Fe2+ > Eo H+/H2 and Ecell is a positive value of +0.77V Equation 2Fe3+ (aq) + H2(g) → 2Fe2+ (aq) + 2H+(aq) Example 4 Disproportionation Use the half-equations to explain in terms of oxidation states what happens to hydrogen peroxide when it is reduced. Explanation to write As Eo H2O2/H2O > Eo O2/H2O2 and Ecell is a positive value of +1.09V , H2O2 disproportionates from -1 oxidation state to 0 in O2 and -2 in H2O reduce H2O2(aq) + 2H+(aq) + 2e– → 2H2O(I) Eo+1.77V O2(g) + 2H+(aq) + 2e– → H2O2(aq) Eo +0.68V oxidise Equation 2H2O2(aq) → O2 + 2H2O(I) N Goalby chemrevise.org 5 Disproportionation of copper(I) iodide Copper(I) iodide when reacting with sulphuric acid will disproportionate to Cu2+ and Cu metal 2Cu+  Cu + Cu2+ Cu+(aq) + e−  Cu(s) Eo = +0.52 V Cu2+(aq) + e−  Cu+(aq) Eo = +0.15 V So Eo cell = 0.52 − 0.15 = +0.37 V As Eo Cu+/Cu > Eo Cu2+/Cu+ and Ecell has a positive value of +0.37V , Cu+ disproportionates from +1 oxidation state to 0 in Cu and +2 in Cu2+ Ecell is directly proportional to the total entropy change and to lnK (where K is equilibrium constant)for a reaction A positive Ecell will lead to a positive total entropy change Effect of conditions on Cell voltage Ecell The effects of changing conditions on E cell can be made by applying le Chatelier’s principle. If current is allowed to flow, the cell reaction will occur and the Ecell will fall to zero as the reaction proceeds and the reactant concentrations drop. Effect of concentration on Ecell Looking at cell reactions is a straight forward application of le Chatelier. So increasing concentration of ‘reactants’ would increase Ecell and decreasing them would cause Ecell to decrease. Ecell is a measure of how far from equilibrium the cell reaction lies. The more positive the Ecell the more likely the reaction is to occur. Zn2+(aq) + 2e-  Zn(s) E= - 0.76V Fe2+(aq) + 2e-  Fe(s) E= -0.44V Zn + Fe2+  Fe + Zn2+ E= +0.32 Increasing the concentration of Fe2+ and decreasing the concentration of Zn2+ would cause Ecell to increase. Effect of temperature on Ecell Most cells are exothermic in the spontaneous direction so applying Le Chatelier to a temperature rise to these would result in a decrease in Ecell because the equilibrium reactions would shift backwards. If the Ecell positive it indicates a reaction might occur. There is still a possibility, however, that the reaction will not occur or will occur so slowly that effectively it does not happen. If the reaction has a high activation energy the reaction will not occur. Also if the reaction is carried out at non-standard conditions the Ecell using standard conditions may deviate. N Goalby chemrevise.org 6 Cells You should be able to work out Ecell for given half reactions. Electrochemical cells can be used as a commercial source of electrical energy Cells can be non-rechargeable (irreversible), rechargeable and fuel cells. You do not need to learn the details of most of these cells. Relevant cell information will be given. You should be able to convert between standard electrode potential half cells, full cell reactions and cell diagrams and be able to calculate potentials from given data. Cells are non-rechargeable when the reactions that occur with in them are non-reversible. Example primary non rechargeable cells Zn2+(aq) + 2e-  Zn(s) Dry Cell 2MnO2(s) + 2NH4+(aq) E = - 0.76 V +2 e- More negative half equation will oxidise → Mn2O3(s) + 2NH3(aq) + H2O(l) 2MnO2 + 2NH4++ Zn → Mn2O3 + 2NH3 + H2O + Zn2+ Overall reaction E = 0.75 V Ecell =+1.51V Example primary Lithium –manganese dioxide cell- non rechargeable Li+aq +e-  Li (s) E = - 3.04 V More negative half equation will oxidise Li+aq + MnO2 (s) +e-  LiMnO2(s) E = - 0.13 V (Mn will reduce changing oxidation state from +4 to +3) Li + MnO2 → LiMnO2 Overall reaction Ecell = E red- Eox = -0.13 - - 3.04 = 2.91 V Conventional cell diagram Li (s)| Li+aq | | Li+aq | MnO2 (s) , LiMnO2(s) | Pt (s) Example secondary rechargeable cells PbSO4 + 2e-  Pb + SO42- Lead acid Cell E= -0.356V Ecell =+2.91V The forward reaction occurs on discharge giving out charge. Charging causes the reaction to reverse PbO2 + SO42- + 4H+ + 2e-  PbSO4 + 2H2O E= +1.685 V Overall reaction PbO2 +Pb + 2SO42- + 4H+  2 PbSO4 + 2H2O Ecell= +2.04V Reversible cells only work if the product stays attached to the electrode and does not disperse Example secondary nickel–cadmium cells are used to power electrical equipment such as drills and shavers. They are rechargeable cells. The electrode reactions are shown below. NiO(OH) + H2O + e-  Ni(OH)2 + OH– E = +0.52 V (Ni will reduce changing oxidation state from 3 to 2) Cd(OH)2 + 2e-  Cd + 2OH– E = –0.88 V (Cd will oxidise changing oxidation state from 0 to 2) Overall reaction discharge 2NiO(OH) + Cd + 2H2O  2Ni(OH)2 + Cd(OH)2 E= +1.40V Ecell = E red- Eox = +0.52 - - 0.88 = + 1.40 V Example secondary Lithium ion cells are used to power cameras and mobile phones. Li+ + CoO2 + e-  Li+[CoO2] - E=+0.6V Li+ + e-  Li (Co will reduce changing oxidation state from 4 to 3 E=-3.0V Overall discharge Li + CoO2  LiCoO2 E=3.6V reaction Conventional cell Li | Li+ || Li+ , CoO2 | LiCoO2 | Pt diagram The overall reaction would be reversed in the recharging state The reagents in the cell are absorbed onto powdered graphite that acts as a support medium. The support medium allows the ions to react in the absence of a solvent such as water. Water would not be good as a solvent as it would react with the lithium metal. N Goalby chemrevise.org 7 Fuel cell A fuel cell uses the energy from the reaction of a fuel with oxygen to create a voltage Hydrogen Fuel cell (potassium hydroxide electrolyte) 4e- + 4H2O  2H2 +4OH- E=-0.83V H2 from fuel 4e- + 2H2O +O2  4OH- E=+0.4V Overall reaction 2H2 + O2  2H2O E=1.23V 2e- + 2H+ H2 E=0V 4e- + 4H+ +O2  2H2O E=1.23V Overall 2H2 + O2  2H2O E=1.23V In acidic conditions these are the electrode potentials. The Ecell is the same as alkaline conditions as the overall equation is the same O2 from air H2O +heat Using standard conditions: The rate is too slow to produce an appreciable current. Higher temperatures are therefore used to increase rate but the reaction is exothermic so by applying le chatelier would mean the emf falls. A higher pressure can help counteract this Fuel cells will maintain a constant voltage over time as they are continuously fed with fresh O2 and H2 so maintaining constant concentration of reactants. This differs from ordinary cells where the voltage drops over time as the reactant concentrations drop Advantages of Fuel cells over conventional petrol or diesel-powered vehicles (i) less pollution and less CO2. (Pure hydrogen emits only water whilst hydrogen-rich fuels produce only small amounts of air pollutants and CO2). (ii) greater efficiency; Hydrogen is readily available by the electrolysis of water, but this is expensive. To be a green fuel the electricity needed would need to be produced from renewable resources Limitations of hydrogen fuel cells (i) expensive (ii) storing and transporting hydrogen, in terms of safety, feasibility of a pressurised liquid and a limited life cycle of a solid ‘adsorber’ or ‘absorber’ (iii) limited lifetime (requiring regular replacement and disposal) and high production costs, (iv) use of toxic chemicals in their production Hydrogen can be stored in fuel cells (i) as a liquid under pressure, (ii) adsorbed on the surface of a solid material, (iii) absorbed within a solid material; Ethanol fuel cells Ethanol fuel cells have also been developed. Compared to hydrogen fuel cells they have certain advantages including. Ethanol can be made from renewable sources in a carbon neutral way Raw materials to produce ethanol by fermentation are abundant Ethanol is less explosive and easier to store than hydrogen. New petrol stations would not be required as ethanol is a liquid fuel. Equation that occurs at oxygen electrode 4e- + 4H+ +O2  2H2O E=1.23V Equation that occurs at ethanol electrode C2H5OH + 3H2O → 2CO2 + 12H+ + 12eOverall equation C2H5OH + 3O2 → 2CO2 + 3H2O Methanol can also be used in fuel cells N Goalby chemrevise.org 8 Redox titrations Manganate redox titration The redox titration between Fe2+ with MnO4– (purple) is a very common exercise. This titration is self indicating because of the significant colour change from reactant to product The purple colour of manganate can make it difficult to see the bottom of meniscus in the burette MnO4-(aq) + 8H+ (aq) + 5Fe2+ (aq)  Mn2+ (aq) + 4H2O (l) + 5Fe3+ (aq) Purple colourless If the manganate is in the burette then the end point of the titration will be the first permanent pink colour Colourless  purple Choosing correct acid for manganate titrations. The acid is needed to supply the 8H+ ions. Some acids are not suitable as they set up alternative redox reactions and hence make the titration readings inaccurate. Only use dilute sulfuric acid for manganate titration Insufficient volumes of sulfuric acid will mean the solution is not acidic enough and MnO2 will be produced instead of Mn2+ MnO4-(aq) + 4H+(aq) + 3e-  MnO2 (s) + 2H2O The brown MnO2 will mask the colour change and lead to a greater (inaccurate) volume of manganate being used in the titration Using a weak acid like ethanoic acid would have the same effect as it cannot supply the large amount of hydrogen ions needed (8H+) It cannot be conc HCl as the Cl- ions would be oxidised to Cl2 by MnO4- as the Eo MnO4-/Mn2+ > Eo Cl2/ClMnO4-(aq) + 8H+(aq) + 5e–  Mn2+ (aq) + 4H2O(l) E+1.51V Cl2(aq) +2e–  2Cl–(aq) E +1.36V This would lead to a greater volume of manganate being used and poisonous Cl2 being produced It cannot be nitric acid as it is an oxidising agent. It oxidises Fe2+ to Fe3+ as Eo NO3-/HNO2> Eo Fe3+/Fe2+ NO3- (aq) + 3H+(aq) + 2e–  HNO2(aq) + H2O(l) Eo +0.94V Fe3+ (aq)+e–  Fe2+ (aq) Eo +0.77 V This would lead to a smaller volume of manganate being used be able to perform calculations for these titrations and for others when the reductant and its oxidation product are given. Manganate titration example A 2.41g nail made from an alloy containing iron is dissolved in 100cm3 acid. The solution formed contains Fe(II) ions. 10cm3 portions of this solution are titrated with potassium manganate (VII) solution of 0.02M. 9.80cm3 of KMnO4 were needed to react with the solution containing the iron. What is the percentage of Iron by mass in the nail? MnO4-(aq) + 8H+(aq) + 5Fe2+  Mn2+(aq) + 4H2O + 5Fe3+ Step1 : find moles of KMnO4 moles = conc x vol 0.02 x 9.8/1000 = 1.96x10-4 mol Step 2 : using balanced equation find moles Fe2+ in 10cm3 = moles of KMnO4 x 5 = 9.8x10-4 mol Step 3 : find moles Fe2+ in 100cm3 = 9.8x10-4 mol x 10 = 9.8x10-3 mol Step 4 : find mass of Fe in 9.8x10-3 mol mass= moles x RAM = 9.8x10-3 x 55.8 = 0.547g Step 5 ; find % mass %mass = 0.547/2.41 x100 = 22.6% N Goalby chemrevise.org 9 Other useful manganate titrations With hydrogen peroxide Ox H2O2  O2 + 2H+ + 2eRed MnO4-(aq) + 8H+(aq) + 5e-  Mn2+ (aq) + 4H2O Overall 2MnO4-(aq) + 6H+(aq) + 5H2O2  5O2 + 2Mn2+ (aq) + 8H2O With ethanedioate Ox C2O42-  2CO2 + 2eRed MnO4-(aq) + 8H+(aq) + 5e-  Mn2+ (aq) + 4H2O Overall 2MnO4-(aq) + 16H+(aq) + 5C2O42-(aq)  10CO2(g) + 2Mn2+(aq) + 8H2O(l) The reaction between MnO4- and C2O42- is slow to begin with (as the reaction is between two negative ions) To do as a titration the conical flask can be heated to 60o C to speed up the initial reaction. With Iron (II) ethanedioate both the Fe2+ and the C2O42- react with the MnO41MnO4- reacts with 5Fe2+ and 2 MnO4- reacts with 5C2O42MnO4-(aq) + 8H+(aq) + 5Fe2+  Mn2+ (aq) + 4H2O + 5Fe3+ 2MnO4-(aq) + 16H+(aq) + 5C2O42-  10CO2 + 2Mn2+ (aq) + 8H2O So overall 3MnO4-(aq) + 24H+(aq) + 5FeC2O4  10CO2 + 3Mn2+ (aq) + 5Fe3+ + 12H2O So overall the ratio is 3 MnO4- to 5 FeC2O4 Step1 : find moles of KMnO4 moles = conc x vol 0.0189 x 23.45/1000 = 4.43x10-4 mol Step 2 : using balanced equation find moles FeC 2O4.2H2O in 25cm3 = moles of KMnO4 x 5/3 (see above for ratio) = 7.39x10-4 mol Step 3 : find moles FeC2O4.2H2O in 250 cm3 = 7.39x10-4 mol x 10 = 7.39x10-3 mol Step 4 : find mass of FeC2O4.2H2O in 7.39x10-3 mol mass= moles x Mr = 7.39x10-3 x 179.8 = 1.33g A 1.412 g sample of impure FeC2O4.2H2O was dissolved in an excess of dilute sulfuric acid and made up to 250 cm3 of solution. 25.0 cm3 of this solution decolourised 23.45 cm3 of a 0.0189 mol dm–3 solution of potassium manganate(VII). Calculate the percentage by mass of FeC2O4.2H2O in the original sample. Step 5 ; find % mass %mass = 1.33/1.412 x100 = 94.1% Thiosulfate redox titration The redox titration between I2 and thiosulfate S2O32- is a common exercise. 2S2O32-(aq) + I2 (aq) yellow/brown sol  2I- (aq) + S4O62-(aq) A starch indicator is added near the end point when the iodine fades a pale yellow to emphasise it. With starch added the colour change is from blue/black to colourless. colourless sol The starch should not be added until nearly all the iodine has reacted because the blue complex formed with high concentrations of iodine is insoluble and does not re-dissolve as more thiosulfate is added. N Goalby chemrevise.org 10 Uncertainty Readings and Measurements Readings the values found from a single judgement when using a piece of equipment Measurements the values taken as the difference between the judgements of two values (e.g. using a burette in a titration) The uncertainty of a reading (one judgement) is at least ±0.5 of the smallest scale reading. The uncertainty of a measurement (two judgements) is at least ±1 of the smallest scale reading. Calculating Apparatus Uncertainties Each type of apparatus has a sensitivity uncertainty To decrease the apparatus uncertainties you can either decrease the sensitivity uncertainty by using apparatus with a greater resolution (finer scale divisions ) or you can increase the size of the measurement made. balance  0.001 g volumetric flask  0.1 cm3 25 cm3 pipette  0.1 cm3 burette  0.05 cm3 Calculate the percentage uncertainty for each piece of equipment used by % uncertainty =  uncertainty x 100 Measurement made on apparatus e.g. for pipette % uncertainty = 0.05/ 25 x100 Uncertainty of a measurement using a burette. If the burette used in the titration had an uncertainty for each reading of +/– 0.05 cm3 then during a titration two readings would be taken so the uncertainty on the titre volume would be +/– 0.10 cm3. To calculate the maximum percentage apparatus uncertainty in the final result add all the individual equipment uncertainties together. Reducing uncertainties in a titration Replacing measuring cylinders with pipettes or burettes which have lower apparatus uncertainty will lower the % uncertainty If looking at a series of measurements in an investigation the experiments with the smallest readings will have the highest experimental uncertainties. To reduce the uncertainty in a burette reading it is necessary to make the titre a larger volume. This could be done by: increasing the volume and concentration of the substance in the conical flask or by decreasing the concentration of the substance in the burette. Reducing uncertainties in measuring mass Using a balance that measures to more decimal places or using a larger mass will reduce the % uncertainty in weighing a solid. Weighing sample before and after addition and then calculating difference will ensure a more accurate measurement of the mass added. Calculating the percentage difference between the actual value and the calculated value If we calculated an Mr of 203 and the real value is 214, then the calculation is as follows: Calculate difference 214-203 = 11 % = 11/214 x100 =5.41% If the %uncertainty due to the apparatus < percentage difference between the actual value and the calculated value then there is a discrepancy in the result due to other errors. If the %uncertainty due to the apparatus > percentage difference between the actual value and the calculated value then there is no discrepancy and any difference in the results can be explained by the sensitivity of the equipment. N Goalby chemrevise.org 11