Electrode potentials

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. The Standard Hydrogen Electrode The potential of all electrodes are measured by comparing their potential to that of the standard hydrogen electrode. The standard hydrogen electrode (SHE) is assigned the potential of 0 volts. The hydrogen electrode equilibrium is: H2 (g) 2H+ (aq) + 2eBecause the equilibrium does not include a conducting metal surface a platinum wire is used which is coated in finely divided platinum. (The platinum black is used because it is porous and can absorb the hydrogen gas.) In a cell diagram the hydrogen electrode is represented by: Pt |H2 (g) | H+ (aq) Components of a standard hydrogen electrode. To make the electrode a standard reference electrode some conditions apply: 1. Hydrogen gas at pressure of 100kPa 2. Solution containing the hydrogen ion at 1 M (solution is usually 1M HCl) 3. Temperature at 298K 4. Platinum Electrode 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 Standard conditions are needed because the position of the redox equilibrium will change with conditions. For example, in the equilibrium: Mn+ (aq) + n e- M(s) An increase in the concentration of Mn+ would move the equilibrium to the right, so making the potential more positive. Standard Electrode Potentials The standard conditions are : •All ion solutions at 1M •temperature 298K •gases at 100kPa 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, E 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) + e- Li (s) E= -3.03V but again the more oxidised form is on the left Pt electrode Pt electrode 1M HCl 1M FeSO4 and 0.5 M Fe2 (SO4 )3 Salt bridge KNO3 (aq) Pt|H2 |H+ ||Fe3+,Fe2+|Pt H2 gas at 100kPa Solution containing metal ions (e.g. Fe2+) at 1 mol dm-3 concentration Metal electrode e.g. Fe H2 gas at 100kPa 1M HCl Pt electrode Salt bridge KNO3 (aq) N Goalby chemrevise.org Note: in the electrode system containing two solutions it is necessary to use a platinum electrode and both ion solutions must be of a 1M concentration so [Fe2+] = 1M and [Fe3+ ] = 1M .

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 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 because 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. 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. 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)

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. The cell diagram is drawn as: | | Fe3+ (aq), Fe2+ (aq) |Pt Still with more oxidised form near double line A comma separates the oxidised from the reduced species. 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. | |MnO4 – , Mn2+ |Pt or if on left hand side Pt | Mn2+ , MnO4 – | | If a half equation has several physical states then the solid vertical line should be used between each different state boundary. 4e- + 2H2O (l) +O2 (g) 4OH- (aq) | | O2 | H2O, OH- | Pt 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. Calculating the EMF of a cell Mg(s) | Mg2+ (aq) | | Cu2+ (aq) | Cu (s) 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 For the cell diagram above Ecell = 0.34 – -2.37 = + 2.71 V use the equation Ecell= Erhs – Elhs

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: For any two half equations The more negative half cell will always oxidise (go backwards) Mg2+ (aq) + 2e-  Mg(s) E= -2.37V Cu2+ (aq) + 2e-  Cu (s) E = +0.34V The more positive half cell will always reduce (go forwards) The reaction would be Mg + Cu2+  Cu + Mg 2+ 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 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) To get the full equation of the reaction add the two half reactions together, cancelling out the electrons. Zn + Fe2+  Fe + Zn2+ The most negative electrode will oxidise and go from right to left The half equation is therefore Zn(s)  Zn2+ (aq) +2e- Electrons are given off (lost) and travel to positive electrode. Using series of standard electrode potentials Li+ + e-  Li -3.03V Mn2+ + 2e-  Mn -1.19V 2H+ + 2e-  H2 0V Ag+ + e-  Ag +0.8V F2 + 2e-  2F- +2.87 As more +ve increasing tendency for species on left to reduce, and act as oxidising agents As more -ve increasing tendency for species on right to oxidise, and act as reducing agents oxidation reduction Most strong reducing agents found here Most strong oxidising agents found here 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). If we want to work out the Ecell from two standard electrode potentials then use Ecell = Ered – Eox. Use electrode data to explain why fluorine reacts with water. Write an equation for the reaction that occurs. 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 Use data from the table to explain why chlorine should undergo a redox reaction with water. Write an equation for this reaction. Use the half-equations to explain in terms of oxidation states what happens to hydrogen peroxide when it is reduced. H2O2 (aq) + 2H+ (aq) + 2e– → 2H2O(I) Eo+1.77V O2 (g) + 2H+ (aq) + 2e– → H2O2 (aq) Eo +0.68V Example 1 First apply idea that more positive Eo will reduce (go forward) and more negative Eo will oxidise (go backwards) reduce oxidise Explanation to write As Eo F2 /F- > Eo O2 /H2O, F2 will oxidise H2O to O2 Equation 2F2 (g) + 2H2O(I) → 4F– (aq) + O2 (g) + 4H+ (aq) Can also work out Ecell and quote it as part of your answer Ecell = Ered – Eox = 2.87-1.23 =1.64V Remember to cancel out electrons in full equation Conventional Cell diagram for above example Pt|H2O|O2 || F2 |F-|Pt Example 2 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 reduce 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) Fe3+ (aq) + e– → Fe2+ (aq) Eo +0.77V 2H+ (aq) + 2e– → H2 (g) Eo 0.00V Fe2+ (aq) + 2e– → Fe(s) Eo–0.44V 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) Eo +0.77V 2H+ (aq) + 2e– → H2 (g) Eo 0.00V oxidise reduce Explanation to write Fe3+ will be reduced to Fe2+ by H2 oxidising to H+ because Eo Fe3+ /Fe2+ > Eo H+ /H2 Equation 2Fe3+ (aq) + H2 (g) → 2Fe2+ (aq) + 2H+ (aq) Example 3 reduce oxidise Explanation to write As Eo H2O2 /H2O > Eo O2 /H2O2 , H2O2 disproportionates from -1 oxidation state to 0 in O2 and -2 in H2O 2H2O2 (aq) → 2H2O(I) + O2 (g)

Effect of conditions on Cell voltage Ecell The effects of changing conditions on E cell can be made by applying le Chatelier’s principle. 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. 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. 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.

Cells Electrochemical cells can be used as a commercial source of electrical energy Cells can be non-rechargeable (irreversible), rechargeable and fuel cells. You should be able to work out Ecell for given half reactions. Example primary non rechargeable cells Dry Cell Zn2+(aq) + 2e-  Zn(s) E = – 0.76 V 2MnO2 (s) + 2NH4+ (aq) + 2 e- → Mn2O3 (s) + 2NH3 (aq) + H2O(l) E = 0.75 V More negative half equation will oxidise Overall reaction 2MnO2 + 2NH4++ Zn → Mn2O3 + 2NH3 + H2O + Zn2+ Ecell =+1.51V Example secondary rechargeable cells PbSO4 + 2e-  Pb + SO4 2- PbO2 + SO4 2- + 4H+ + 2e-  PbSO4 + 2H2O E= -0.356V E= +1.685 V Lead acid Cell Overall reaction PbO2 +Pb + 2SO4 2- + 4H+  2 PbSO4 + 2H2O Ecell= +2.04V The forward reaction occurs on discharge giving out charge. Charging causes the reaction to reverse Reversible cells only work if the product stays attached to the electrode and does not disperse 6 Example primary Lithium –manganese dioxide cell- non rechargeable Li (s)| Li+ aq | | Li+ aq | MnO2 (s) , LiMnO2(s) | Pt (s) Li+ aq +e-  Li (s) Li+ aq + MnO2 (s) +e-  LiMnO2(s) Ecell =+2.91V E = – 3.04 V E = – 0.13 V (Mn will reduce changing oxidation state from +4 to +3) Overall reaction Li + MnO2 → LiMnO2 More negative half equation will oxidise 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. Conventional cell diagram Ecell = E red- Eox = -0.13 – – 3.04 = 2.91 V. 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 The overall reaction would be reversed in the recharging state 2Ni(OH)2 + Cd(OH)2  2NiO(OH) + Cd + 2H2O Li | Li+ || Li+ , CoO2 | LiCoO2 | Pt Example secondary Lithium ion cells are used to power cameras and mobile phones. Li+ + CoO2 + e-  Li+ [CoO2 ] – E=+0.6V Li+ + e-  Li E=-3.0V Overall discharge Li + CoO2  LiCoO2 E=3.6V reaction (Co will reduce changing oxidation state from 4 to 3 Conventional cell diagram 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. The overall reaction would be reversed in the recharging state

Fuel cells 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 4e- + 2H2O +O2  4OH- E=+0.4V Overall reaction 2H2 + O2  2H2O E=1.23V 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 E cell falls. A higher pressure can help counteract this. 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 H2 from fuel Fuel cells will maintain a constant voltage over time because 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. Alkaline Conditions You should learn the details of the lithium cell and the hydrogen fuel cell in alkaline conditions. 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 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 C2H5OH + 3O2 → 2CO2 + 3H2O Equation that occurs at ethanol electrode C2H5OH + 3H2O → 2CO2 + 12H+ + 12e- Overall equation. 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;

A-level Chemistry exemplar for required practical No. 8 Measuring the EMF of an electrochemical cell Student sheet Requirements You are provided with the following:  pieces of copper and zinc foil (about 2 cm × 5 cm)  propanone  2.0 mol dm–3 NaCl solution  1.0 mol dm–3 CuSO4 solution  1.0 mol dm–3 ZnSO4 solution  emery paper or fine grade sandpaper  two 100 cm3 beakers  plastic or glass U-tube  cotton wool soaked in sodium chloride solution  voltmeter (digital or high impedance)  two electrical leads with connectors for the voltmeter at one end and crocodile clips at the other end  samples of metals. Suggested method for setting up a standard cell a) Clean a piece of copper and a piece of zinc using emery paper or fine grade sandpaper. b) Degrease the metal using some cotton wool and propanone. c) Place the copper into a 100 cm3 beaker with about 50 cm3 of 1 mol dm–3 CuSO4 solution. d) Place the zinc into a 100 cm3 beaker with about 50 cm3 of 1 mol dm–3 ZnSO4 solution. e) Lightly plug one end of the plastic tube with cotton wool and fill the tube with the solution of 2 mol dm–3 NaCl provided. f) Plug the free end of the tube with cotton wool which has been soaked in sodium chloride. Join the two beakers with the inverted U-tube so that the plugged ends are in the separate beakers. g) Connect the Cu(s)|Cu2+(aq) and Zn(s)|Zn2+(aq) half-cells by connecting the metals (using the crocodile clips and leads provided) provided to the voltmeter and read off the voltage.Suggested method for measuring comparative electrode potentials of different metals a) Clean a piece of copper using emery paper or fine grade sandpaper. b) Connect the positive terminal of the voltmeter to the copper using a crocodile clip and one of the leads. c) Cut a piece of filter paper to about the same area as the copper, moisten the filter paper with the sodium chloride solution and place on top of the copper. d) Connect the second lead to the voltmeter and use the crocodile clip on the other end of the lead to grip a piece of another metal. e) Hold the metal against the filter paper and note the voltage reading and sign. f) Repeat steps (d) and (e) with different metals and record your results in a table. g) Write the conventional representation for each of the cells that you have constructed h) Suggest how you could construct the cell with the largest EMF from the metals provided.