Hg2 2+ (aq) + H2(g) 2 Hg(l) + 2H + (aq)

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Magnus Perry
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1 The potential difference between two electrodes in a cell is called the electromotive force, or The EMF of a voltaic cell is called the The cell voltage of a voltaic cell will be a Note: We are used to spontaneous processes having negative values for energy terms. However, since batteries were historically measured looking at the change in energy of the surroundings rather than for the reaction, the opposite sign convention applies. Work We can now write an expression for the maximum work attainable by a voltaic cell. Let n be the number of mole electrons transferred in the overall balanced cell reaction. The maximum work for molar amounts of reactants is: LP#5. The EMF of a voltaic cell with the following reaction is V. Hg2 2+ (aq) + H2(g) 2 Hg(l) + 2H + (aq) Calculate the max work that can be done by this cell when 0.500g H2 is consumed. Step 1: Identify the number of electrons transferred during the overall reaction. Oxid ½ Rxn: Red. ½ Rxn: Step 2: Compute the maximum work per mole. Wmax = -(n)(f)(ecell) = Step 3: Determine the total energy based on the total moles of e - transferred

You can look at the oxidation half-reaction as the reverse of a corresponding reduction half-reaction.

2 Standard Cell EMF s and Standard Electrode Potentials A cell emf is a measure of the driving force of the cell reaction. A reduction potential is a measure of the in the reduction half-reaction. You can look at the oxidation half-reaction as the reverse of a corresponding reduction half-reaction. The oxidation potential for an oxidation half-reaction (the reverse reaction) is the of the reduction potential. By convention, the Table of Standard Electrode Potentials are tabulated as reduction potentials. Standard Reduction Potentials E 0 cell = standard cell potential under standard conditions, Voltage listed at standard conditions are at temperature: concentration: 19-14

3 The actual cell voltage will depend upon a) b) c) Individual potentials cannot be measured, only differences in potential. By convention, for values in the table, all others must be compared to a standard reference reaction The reference reaction is the reduction of H + (aq) ions to produce H2(g): 2H + (aq, 1M) + 2e - H2(g, 1 atm) Eºred = This is known as the: Standard electrode potentials are measured relative to this hydrogen reference. A cell will always include one oxidation reaction (at the anode) and one reduction reaction (at the cathode). When we reverse a reaction to get an oxidation half reaction, the sign of the potential must be changed. We can then sum the energy potentials the same way we have treated other energy variable (e.g. G & H) this semester. E cell = Note: The book uses the convention E cell = E cathode - E anode Where all potentials are reduction potentials! Both processes will give the same correct answer

4 Lets look at a section of an Activity Series for metals and hydrogen ion: Standard Potential, (Eºred) in Volts Reduction Half- Reaction Li + (aq) + e - Li(s) Zn 2+ (aq) + 2e - Zn(s) 0 2H + (aq) + 2e - H2(g) 0.34 Cu 2+ (aq) + 2e - Cu(s) 0.80 Ag + (aq) + e - Ag(s) The more positive the value of Eºred the greater the driving force for reduction Reactions with positive E values - want to undergo:. - are good: Reactions with negative E values - prefer to undergo: - are good: The net difference between the standard reduction potentials of the two reactions, is the excess potential that can be used to drive electrons through the cell, Eºcell Active metals are. The more active they are, the greater the oxidation potential for the metal. Since tables usually list reduction potentials, the more active they are, the more negative the reduction potential for the ion

5 When comparing Zinc with Copper in the activity series, which reaction would we expect to happen? Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) OR Zn 2+ (aq) + Cu(s) Zn(s) + Cu 2+ (aq) Zinc is higher on the activity series. Zinc is more easily oxidized than copper. We would expect the reaction to happen. Example: Consider the zinc-copper cell described earlier. Zn( s) Zn 2 ( aq) Cu The two half-reactions are: ( aq) Cu( s) Oxidation ½ Rxn: Reduction ½ Rxn: Find the oxidation potential for this half reaction. Zn(s) Zn 2+ (aq) + 2e - E o ox = Find the reduction potential for this half reaction. Cu 2+ (aq) + 2e - Cu(s) E o red = Find the overall cell potential. 2 Eº cell = The electrode potential is an intensive property whose value is independent of the amount of species in the reaction. Thus, the electrode potential for the half-reaction would be: 2Cu 2+ (aq) + 4e - 2Cu(s) E o = 19-17

6 LP#6. Consider a cell constructed of the following two half-reactions. What would the overall reaction be that would create a voltaic cell and what would the voltage of that cell be at standard conditions? 2 o Cd ( aq ) 2e Cd ( s); E 0.40 V Ag ( aq) 1e Ag( s); E o 0.80 V Which reaction should be reversed? Half reactions and associated voltages therefore are: Cd(s) Cd 2+ (aq) + 2e - ; E = Ag + (aq) + 1e - Ag(s) ; E = 0.80V We must double the silver half-reaction so that the electrons cancel. Cd(s) Cd 2+ (aq) + 2e - ; E = 0.40 V 2Ag + (aq) + 2e - 2Ag(s) ; E = Now we can add the two half-reactions. The corresponding cell notation would be: Spontaneity of Redox Reactions Eºcell = Eºred + Eºoxid Eº will be positive for the case where the reaction is: Eº will be zero for a redox reaction that is: Eº will be negative for the case where the reaction is: LP#7. Can copper be dissolved (oxidized) by acid (i.e., H + )? Really asking if the following reaction is spontaneous: Cu(s) + 2H + (aq) Cu 2+ (aq) + H2(g) This reaction can be broken down into two half-reactions: Oxidation: Cu(s) Cu 2+ (aq) + 2e - E 0 ox = Reduction: 2H + (aq) + 2e - H2(g) E 0 red = 19-18

7 The standard cell potential for this reaction is: Eº cell = Since this value is, this redox reaction is and. What if we only had the activity series without any numerical voltages? (Remember the most active at the top is most easily Since H is higher than Cu, it is more easily oxidized. Cu cannot replace it as the oxidized species. Li Zn H Cu Ag EMF, Free Energy Changes, and Equilibrium Free energy change, Gº, Kc, and Eºcell, all measure spontaneity of a reaction. What is the relationship between these variables? Gº and Eºcell Previously, we saw that G is the free energy available which equals the maximum useful work of a reaction. Remember, for a voltaic cell, work = -nfecell, so when reactants are in their standard states The Gibbs Free Energy (G) can be related to the EMF of the cell. Where n= number of moles of e- transferred F = Faraday s constant = 96,485 C/mole e - C = Coulombs Our typical unit of energy, Joules can be related to the cell EMF: Since 1 V = 1J/C 19-19

8 º and Kc The measurement of cell EMF s gives you yet another way of calculating equilibrium constants. Combining the previous equation, G o = -nfe o cell, with the equation G o = -RT lnk, (from our previous chapter) we get Or, rearranging, we get LP#8. The standard EMF for the following cell is 1.10 V. Zn( s) Zn 2 ( aq) Cu ( aq) Cu( s) Calculate the equilibrium constant Kc at 25 o C for the reaction: Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) Note that n=. Substituting into the equation relating E o cell and K gives 2 Solving for log Kc, you find: logk= Now take the antilog of both sides: K c = Effect of Concentration on Cell EMF The EMF of a voltaic cell is determined by a) the identity of the redox reaction and b) the concentrations of the reactants and products. The EMF of the cell will fall as the reactants are used up and products increase in concentration 19-20

9 The Nernst Equation This is an equation that related the EMF of a redox reaction on the concentration of reactants and products. Developed by Walther Hermann Nernst ( ), a German chemist. E cell o RT Ecell lnq nf At equilibrium concentrations of reactants and products, the EMF =. Electrons flow spontaneously in a redox reaction because the system is attempting to achieve equilibrium. When equilibrium is achieved, net electron flow is zero. 0 E o cell RT nf ln K E cell 0 RT ln K at equilibrium nf A common form of the equation does away with the natural log and puts the equation in the form of log10: E cell o 2.303RT Ecell logq Nernst Equation nf At 25 C and using base 10 logs this becomes: E = E log Q n The Nernst Equation. Allows us to determine cell potentials at 19-21

10 LP#9. A voltaic cell that utilizes the oxidation of zinc by copper ion is set up with an initial concentration of 5.0M copper ion and 0.050M zinc ion. Calculate the cell potential at 20C. Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) E 0 cell = 1.10V n= Determination of ph A ph electrode works by relating cell EMF (i.e., potential) to concentration. As the concentration of H + in the solution changes, the amount of voltage that can be measured changes. We can then calibrate the voltage to actual ph values

CONCENTRATION CELLS Since cell potentials cepend not only on the half reactions but on the concentrations, it is possible to create a cell where the two half reactions are the same, but only th e

11 CONCENTRATION CELLS Since cell potentials cepend not only on the half reactions but on the concentrations, it is possible to create a cell where the two half reactions are the same, but only th e concentrations are different. The standard cell potential would be zero: Oxidation: Cu(s) Cu 2+ (aq) + 2e - E 0 ox = Reduction: Cu 2+ (aq) + 2e - Cu(s) E 0 red = Using the Nernst Equation the actual cell potential is: 19-23

12 LP#10. Consider a cell with the shorthand notation: Al(s) Al 3+ (aq) (5.0M) Cu 2+ (aq)(0.020m) Cu(s) A) What is the cell voltage at 20C given: Al 3+ (aq) + 3e Al(s) E = 1.66V Cu 2+ (aq) + 2e Cu(s) E = V Label each part: a) anode & cathode: b) signs of electrodes; c) oxidation and reduction cell; d) electron flow; e) ion flows in beakers; f) ion flows in salt bridge, g) cathode and anode processes, h) salt bridge, i) solutions in beakers; j) electrode materials

Review: Balancing Redox Reactions. Review: Balancing Redox Reactions

Review: Balancing Redox Reactions Determine which species is oxidized and which species is reduced Oxidation corresponds to an increase in the oxidation number of an element Reduction corresponds to a