1. AN INTRODUCTION TO ELECTRODEPOTENTIALS KNOCKHARDY PUBLISHING 2008 SPECIFICATIONS

2. INTRODUCTION This Powerpoint show is one of several produced to help students understand selected topics at AS and A2 level Chemistry. It is based on the requirements of the AQA and OCR specifications but is suitable for other examination boards. Individual students may use the material at home for revision purposes or it may be used for classroom teaching if an interactive white board is available. Accompanying notes on this, and the full range of AS and A2 topics, are available from the KNOCKHARDY SCIENCE WEBSITE at... www.knockhardy.org.uk/sci.htm Navigation is achieved by... either clicking on the grey arrows at the foot of each page orusing the left and right arrow keys on the keyboard ELECTRODE POTENTIALS

3. CONTENTS Types of half cells Cell potential The standard hydrogen electrode Measuring electrode potentials The electrochemical series Combining half cells Cell diagrams Uses of E° values ELECTRODE POTENTIALS

4. Before you start it would be helpful to… Recall the definitions of oxidation and reduction Be able to balance simple ionic equations Have a knowledge of simple circuitry ELECTRODE POTENTIALS

5. TYPES OF HALF CELL These are systems involving oxidation or reduction METALS IN CONTACT WITH SOLUTIONS OF THEIR IONS ReactionCu 2+ (aq) + 2e¯ Cu(s) Electrodecopper SolutionCu 2+ (aq) (1M) - 1M copper sulphate solution Potential+ 0.34V ReactionZn 2+ (aq) + 2e¯ Zn(s) Electrodezinc SolutionZn 2+ (aq) (1M) - 1M zinc sulphate solution Potential- 0.76V

6. TYPES OF HALF CELL These are systems involving oxidation or reduction GASES IN CONTACT WITH SOLUTIONS OF THEIR IONS Reaction2H + (aq) + 2e¯ H 2 (g) Electrodeplatinum SolutionH + (aq) (1M) - 1M HC l or 0.5M H 2 SO 4 Gashydrogen at 100 kPa (1atm) pressure Potential0.00V ReactionC l 2 (aq) + 2e¯ 2C l ¯(g) Electrodeplatinum SolutionC l ¯(aq) (1M) - 1M sodium chloride Gaschorine at 100 kPa (1atm) pressure Potential+ 1.36V

7. TYPES OF HALF CELL These are systems involving oxidation or reduction SOLUTIONS OF IONS IN TWO DIFFERENT OXIDATION STATES ReactionFe 3+ (aq) + e¯ Fe 2+ (aq) Electrodeplatinum SolutionFe 3+ (aq) (1M) and Fe 2+ (aq) (1M) Potential+ 0.77 V

8. TYPES OF HALF CELL These are systems involving oxidation or reduction SOLUTIONS OF IONS IN TWO DIFFERENT OXIDATION STATES ReactionFe 3+ (aq) + e¯ Fe 2+ (aq) Electrodeplatinum SolutionFe 3+ (aq) (1M) and Fe 2+ (aq) (1M) Potential+ 0.77 V SOLUTIONS OF OXIDISING AGENTS IN ACID SOLUTION ReactionMnO 4 ¯ + 8H + (aq) + 5e¯ Mn 2+ (aq) + 4H 2 O(l) Electrodeplatinum SolutionMnO 4 ¯(aq) (1M) and Mn 2+ (aq) (1M) and H + (aq) Potential+ 1.52 V

9. CELL POTENTIAL Each electrode / electrolyte combination has its own half-reaction which sets up a potential difference The value is affected by... TEMPERATURE PRESSURE OF ANY GASES SOLUTION CONCENTRATION Measurementit is impossible to measure the potential of a single electrode… BUT... you can measure the potential difference between two electrodes it is measured relative to a reference cell under standard conditions The ultimate reference is the STANDARD HYDROGEN ELECTRODE. However, as it is difficult to set up, secondary standards are used.

10. The ultimate reference is the STANDARD HYDROGEN ELECTRODE. However as it is difficult to set up secondary standards are used. THE STANDARD HYDROGEN ELECTRODE The standard hydrogen electrode is assigned an E° value of 0.00V. HYDROGEN GAS AT 100 kPa PRESSURE PLATINUM ELECTRODE SOLUTION OF 1M H + (aq) e.g. 1M HC l or 0.5M H 2 SO 4 298K (25°C)

11. 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. conditionstemperature 298K solution conc. 1 Molar (1 mol dm -3 ) with respect to ions gases 100 kPa pressure salt bridge filled with saturated potassium chloride solution it enables the circuit to be completed MEASUREMENT OF E ° VALUES SALT BRIDGE ZINC ZINC SULPHATE (1M) HYDROGEN (100 kPa) PLATINUM ELECTRODE HYDROCHLORIC ACID (1M)

12. E° / V F 2 (g) + 2e¯2F¯(aq) +2.87 MnO 4 ¯(aq) + 8H + (aq) + 5e¯Mn 2+ (aq) + 4H 2 O(l) +1.52 C l 2 (g) + 2e¯2C l ¯(aq) +1.36 Cr 2 O 7 2- (aq) + I4H + (aq) + 6e¯2Cr 3+ (aq) + 7H 2 O(l) +1.33 Br 2 (l) + 2e¯2Br¯(aq) +1.07 Ag + (aq) + e¯Ag(s) +0.80 Fe 3+ (aq) + e¯Fe 2+ (aq) +0.77 O 2 (g) + 2H + (aq) + 2e¯H 2 O 2 (aq) +0.68 I 2 (s) + 2e¯2I¯(aq) +0.54 Cu + (aq) + e¯Cu(s) +0.52 Cu 2+ (aq) + 2e¯Cu(s) +0.34 Cu 2+ (aq) + e¯Cu + (aq) +0.15 2H + (aq) + 2e¯H 2 (g) 0.00 Sn 2+ (aq) + 2e¯Sn(s) -0.14 Fe 2+ (aq) + 2e¯Fe(s) -0.44 Zn 2+ (aq) + 2e¯Zn(s) -0.76 LayoutIf 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 oxidise (reverses) one with a lower value THE ELECTROCHEMICAL SERIES REACTION MORE LIKELY TO WORK SPECIES ON LEFT ARE MORE POWERFUL OXIDATION AGENTS

13. E° / V F 2 (g) + 2e¯2F¯(aq) +2.87 MnO 4 ¯(aq) + 8H + (aq) + 5e¯Mn 2+ (aq) + 4H 2 O(l) +1.52 C l 2 (g) + 2e¯2C l ¯(aq) +1.36 Cr 2 O 7 2- (aq) + I4H + (aq) + 6e¯2Cr 3+ (aq) + 7H 2 O(l) +1.33 Br 2 (l) + 2e¯2Br¯(aq) +1.07 Ag + (aq) + e¯Ag(s) +0.80 Fe 3+ (aq) + e¯Fe 2+ (aq) +0.77 O 2 (g) + 2H + (aq) + 2e¯H 2 O 2 (aq) +0.68 I 2 (s) + 2e¯2I¯(aq) +0.54 Cu + (aq) + e¯Cu(s) +0.52 Cu 2+ (aq) + 2e¯Cu(s) +0.34 Cu 2+ (aq) + e¯Cu + (aq) +0.15 2H + (aq) + 2e¯H 2 (g) 0.00 Sn 2+ (aq) + 2e¯Sn(s) -0.14 Fe 2+ (aq) + 2e¯Fe(s) -0.44 Zn 2+ (aq) + 2e¯Zn(s) -0.76 Application Chlorine is a more powerful oxidising agent - it has a higher E° Chlorine will get its electrons by reversing the iodine equation C l 2 (g) + 2e¯ ——> 2C l ¯(aq) and 2I¯(aq) ——> I 2 (s) + 2e¯ Overall equation is C l 2 (g) + 2I¯(aq) ——> I 2 (s) + 2C l ¯(aq) THE ELECTROCHEMICAL SERIES AN EQUATION WITH A HIGHER E° VALUE WILL REVERSE AN EQUATION WITH A LOWER VALUE

14. Why?The standard hydrogen electrode (SHE) is difficult to set up it is easier to choose a more convenient secondary standard the secondary standard has been calibrated against the SHE Calomel the calomel electrode contains Hg 2 C l 2 it has a standard electrode potential of +0.27V is used as the LH electrode to determine the potential of an unknown to get the value of the other cell ADD 0.27V to the measured cell potential SECONDARY STANDARDS

15. CELLS electrochemical cells contain two electrodes each electrode / electrolyte combination has its own half-reaction the electrons produced by one half reaction are available for the other oxidation occurs at the anode reduction occurs at the cathode. ANODEZn(s) ——> Zn 2+ (aq) + 2e¯OXIDATION CATHODECu 2+ (aq) + 2e¯ ——> Cu(s)REDUCTION The resulting cell has a potential difference (voltage) called the cell potential which depends on the difference between the two potentials It is affected by... current temperature pressure of any gases solution concentrations ELECTROCHEMICAL CELLS

16. zinc is more reactive - it dissolves to give ions Zn(s) ——> Zn 2+ (aq) + 2e¯ the electrons produced go round the external circuit to the copper electrode electrons are picked up by copper ions Cu 2+ (aq) + 2e¯ ——> Cu(s) As a result, copper is deposited overall reaction Zn(s) + Cu 2+ (aq) ——> Zn 2+ (aq) + Cu(s) A TYPICAL COMBINATION OF HALF CELLS ZINC SULPHATE (1M) COPPER SULPHATE (1M) E° = - 0.76V E° = + 0.34V 1.10V + _

17. These give a diagrammatic representation of what is happening in a cell. Place the cell with the more positive E° value on the RHS of the diagram. Cu 2+ (aq) + 2e¯ Cu(s)E° = + 0.34Vput on the RHS Zn 2+ (aq) + 2e¯ Zn(s)E° = - 0.76Vput on the LHS CELL DIAGRAMS Zn Zn 2+ Cu 2+ Cu

18. These give a diagrammatic representation of what is happening in a cell. Place the cell with the more positive E° value on the RHS of the diagram. Cu 2+ (aq) + 2e¯ Cu(s)E° = + 0.34Vput on the RHS Zn 2+ (aq) + 2e¯ Zn(s)E° = - 0.76Vput on the LHS ZINC IS IN CONTACT THE SOLUTIONS A SOLUTION OF WITH A SOLUTION ARE JOINED VIA A COPPER IONS IN OF ZINC IONS SALT BRIDGE TO CONTACT WITH COPPER CELL DIAGRAMS Zn Zn 2+ Cu 2+ Cu + _

19. These give a diagrammatic representation of what is happening in a cell. Place the cell with the more positive E° value on the RHS of the diagram. Cu 2+ (aq) + 2e¯ Cu(s)E° = + 0.34Vput on the RHS Zn 2+ (aq) + 2e¯ Zn(s)E° = - 0.76Vput on the LHS Draw as shown…the cell reaction goes from left to right the zinc metal dissolves Zn(s) ——> Zn 2+ (aq) + 2e¯ OXIDATION copper is deposited Cu 2+ (aq) + 2e¯ ——> Cu(s) REDUCTION oxidation takes place at the anode reduction at the cathode CELL DIAGRAMS Zn Zn 2+ Cu 2+ Cu + _

20. CELL DIAGRAMS These give a diagrammatic representation of what is happening in a cell. Place the cell with the more positive E° value on the RHS of the diagram. Cu 2+ (aq) + 2e¯ Cu(s)E° = + 0.34Vput on the RHS Zn 2+ (aq) + 2e¯ Zn(s)E° = - 0.76Vput on the LHS Draw as shown… the electrons go round the external circuit from left to right electrons are released when zinc turns into zinc ions the electrons produced go round the external circuit to the copper electrons are picked up by copper ions and copper is deposited Zn Zn 2+ Cu 2+ Cu V + _

21. CELL DIAGRAMS These give a diagrammatic representation of what is happening in a cell. Place the cell with the more positive E° value on the RHS of the diagram. Cu 2+ (aq) + 2e¯ Cu(s)E° = + 0.34Vput on the RHS Zn 2+ (aq) + 2e¯ Zn(s)E° = - 0.76Vput on the LHS Draw as shown…the cell voltage is E°(RHS) - E°(LHS) - it must be positive cell voltage = +0.34V - (-0.76V) = +1.10V Zn Zn 2+ Cu 2+ Cu V + _

22. USE OF E o VALUES - WILL IT WORK? E° values 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

23. USE OF E o VALUES - WILL IT WORK? What happens if an Sn(s) / Sn 2+ (aq) and a Cu(s) / Cu 2+ (aq) cell are connected? Write out the equationsCu 2+ (aq) + 2e¯ Cu(s) ; E° = +0.34V Sn 2+ (aq) + 2e¯ Sn(s) ; E° = -0.14V An equation with a more positive E° value reverse a less positive one

24. USE OF E o VALUES - WILL IT WORK? What happens if an Sn(s) / Sn 2+ (aq) and a Cu(s) / Cu 2+ (aq) cell are connected? Write out the equationsCu 2+ (aq) + 2e¯ Cu(s) ; E° = +0.34V Sn 2+ (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 therefore Cu 2+ (aq) ——> Cu(s) and Sn(s) ——> Sn 2+ (aq) An equation with a more positive E° value reverse a less positive one

25. USE OF E o VALUES - WILL IT WORK? What happens if an Sn(s) / Sn 2+ (aq) and a Cu(s) / Cu 2+ (aq) cell are connected? Write out the equationsCu 2+ (aq) + 2e¯ Cu(s) ; E° = +0.34V Sn 2+ (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 therefore Cu 2+ (aq) ——> Cu(s) and Sn(s) ——> Sn 2+ (aq) the overall reaction isCu 2+ (aq) + Sn(s) ——> Sn 2+ (aq) + Cu(s) the cell voltage is the difference in E° values... (+0.34) - (-0.14) = + 0.48V An equation with a more positive E° value reverse a less positive one

26. USE OF E o VALUES - WILL IT WORK? An equation with a more positive E° value reverse a less positive one Will this reaction be spontaneous? Sn(s) + Cu 2+ (aq) ——> Sn 2+ (aq) + Cu(s) Write out the appropriate equations Cu 2+ (aq) + 2e¯ Cu(s) ; E° = +0.34V as reductions with their E° values Sn 2+ (aq) + 2e¯ Sn(s) ; E° = - 0.14V The reaction which takes place will involve the more positive one reversing the other i.e.Cu 2+ (aq) ——> Cu(s) and Sn(s) ——> Sn 2+ (aq) The cell voltage will be the difference in E° values and will be positive... (+0.34) - (- 0.14) = + 0.48V If this is the equation you want then it will be spontaneous If it is the opposite equation (going the other way) it will not be spontaneous

27. USE OF E o VALUES - WILL IT WORK? An equation with a more positive E° value reverse a less positive one Will this reaction be spontaneous? Sn(s) + Cu 2+ (aq) ——> Sn 2+ (aq) + Cu(s) Split equation into two half equationsCu 2+ (aq) + 2e¯ ——> Cu(s) Sn(s) ——> Sn 2+ (aq) + 2e¯ Find the electrode potentialsCu 2+ (aq) + 2e¯ Cu(s); E° = +0.34V and the usual equationsSn 2+ (aq) + 2e¯ Sn(s); E° = - 0.14V Reverse one equation and its signSn(s) ——> Sn 2+ (aq) + 2e¯ ; E° = +0.14V Combine the two half equationsSn(s) + Cu 2+ (aq) ——> Sn 2+ (aq) + Cu(s) Add the two numerical values(+0.34V) + (+ 0.14V) = +0.48V If the value is positive the reaction will be spontaneous

28. REVISION CHECK What should you be able to do? Recall the different types of half cells Recall the structure of the standard hydrogen electrode Recall the methods used to calculate standard electrode electrode potentials Write balanced full and half equations representing electrochemical processes Know that a reaction can be spontaneous if it has a positive E  value Calculate if a reaction is feasible by finding its E  value CAN YOU DO ALL OF THESE? YES NO

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31. © 2009 JONATHAN HOPTON & KNOCKHARDY PUBLISHING THE END AN INTRODUCTION TO ELECTRODEPOTENTIALS