Electrochemistry

Electrochemistry Half-reactions Electrodes Cells Reduction Potentials The Electrochemical Series Cell Potentials and Thermodynamic Functions

Some Terms • Galvanic or voltaic cell • Electrolytic cell • Oxidation(what happens to the reducing agent) • Reduction(what happens to the oxidizing agent) • Half Reactions • Redox Couple • Anode • Cathode • Standard Electrode Potential

Exercise E6.1 • Identify the species that have undergone oxidation and reduction in the reaction CuS(s) + O2(g) ! Cu(s) + SO2(g) • Solution: The Cu(II) on the left is reduced from the +2 to the 0 oxidation state. The S2- on the left has been oxidized from the -2 to the +4 oxidation state. The O on the left has been reduced from the 0 to the -2 oxidation state.

Exercise E6.2 • Express the formation of H2O from H2 and O2 in acid solution as the difference of two reduction half-reactions. • Solution: The overall reaction is 2 H2(g) + O2(g)!2 H2O(l) • Oxygen is reduced according to O2!2 H2O and in acid solution this is balanced by adding H+ and e!to getO2(g) + 4 H+(aq) + 4e!!2 H2O(l) • The reaction to be subtracted from this is 4 H+(aq) + 4 e!!2 H2(g)

Exercise E6.3 • Express the oxidation of NADH (nicotinamide adenine dinucleotide, which participates in the chain of oxidations that constitutes respiration) to NAD+by oxygen, when the latter is reduced to H2O2 in aqueous solution, as the difference of two reduction half-reactions. • Solution: O2 is the oxidizing agent; therefore O2 is reduced: O2 + 2 H+ + 2 e!! H2O2 • The reverse oxidation reaction to be subtracted from this is then NAD+ + H+ + 2 e!!NADH

Reactions at electrodes • Left: Galvanic cell. Electrons are deposited on the anode (so it is neg) and collected from the cathode (so it is positive) • Right: Electro-lytic cell. Elec-trons are forced out of the anode (positive) and into the cathode (negative)

Types of Electrodes • (a) metal/metal ion electrode • (b) metal/ insoluble salt electrode • (c) gas electrode • (d) redox electrode

Exercise E6.4 • Write the half-reaction and the reaction quotient for a chlorine-gas electrode. • Solution: Written as a reduction, the half-reaction at the chlorine electrode is ½Cl2(g) + e!! Cl!(aq) • And the reaction quotient then becomes Q = a(Cl!) / a(Cl2 )½

Types of Electrode (continued)

Exercise E6.5 • Write the half-reaction and the reaction quotient for the calomel electrode, Hg(l)|Hg2Cl2(s)|Cl!(aq), in which mercury(I) chloride (calomel) is reduced to mercury metal in the presence of chloride ions. This electrode is a component of instruments used to measure pH, as explained later. • Solution: The reduction of Hg2Cl2 to Hg is given by Hg2Cl2(s) + 2 e!!2 Hg(l) + 2 Cl!(aq) • Notice that everything except the chloride ion is a pure solid or liquid. Therefore only the chloride appears in the reaction quotient. Q = a(Cl!)2

Varieties of Cell • The two basic types are concentration cells and chemical cells. • Concentration cells are either electrolyte concentration cells, where the electrode compartments are identical except for the concentrations of the electrolytes, or electrode concentration cells, in which the electrodes themselves have different concentrations, such as amalgams or gas electrodes at different pressures. • Most cells are chemical cells.

Two Versions of the Daniell Cell

Constructing a Daniell Cell

Two Practical Cells • At left is a pri-mary cell (used once only). • At right is a secon-dary cell (may be re-charged)

Cell Notation • In the version of the Daniell cell with the porous pot there is a liquid junction. This is denoted as Zn(s)|ZnSO4(aq):CuSO4(aq)|Cu(s) • When the liquid junction potential has been essentially eliminated by use of a salt bridge the Daniell cell is denoted Zn(s)|ZnSO4(aq)||CuSO4(aq)|Cu(s) • Other punctuation in cell notations includes a comma to separate two species present in the same phase.

Exercise E6.6 • Give the notation for a cell in which the oxidation of NADH by oxygen could by studied. • Solution: Refer to Exercise E6.3, where the half-reactions are O2 + 2 H+ + 2 e!!H2O2 and NAD+ + H+ + 2 e!!NADH • Putting the oxidation half-reaction into conven-tional cell notation, as written on the left, it becomes Pt|NADH(aq), NAD+(aq), H+(aq) • The reduction half-reaction is written Pt|O2(g)|H+(aq),H2O2(aq) • Putting these together, Pt|NADH(aq), NAD+(aq), H+(aq)|| H2O2(aq), H+(aq)|O2(g) | Pt

Exercise E6.7 • Write the chemical equation for the cell in Exercise E6.6. • Solution: We can again refer to Exercise E6.3, where the half-reactions are O2 + 2 H+ + 2 e!!H2O2 and NAD+ + H+ + 2 e!!NADH • Subtracting the oxidation half-reaction from the reduction half-reaction gives NADH(aq) + O2(g) + H+(aq)!NAD+ (aq) + . H2O2 (aq)

The Cell Potential • Since w = DG =!work output, and since electrical work output = (charge) x (vol-tage) = nFE , • DG =!nFE • A spontaneous rxn has a neg.DG and a pos. E • E is intensive

The Nernst Equation • SubstitutingDG = !nFE intoDG=DGo+ RT ln Qgives • !nFE =!nFEo+ RT ln Q or • E = Eo!RT/nFln Q • At 25oC, RT/F = 0.02569 v = 25.69 mV • A practical form of the Nernst equation is • E = Eo!(25.69 mV/n) ln Q • At equilibrium, E = 0 and Q = K, so ln K =nFEo/RT =nEo/(25.69 mV)

Concentration Cells • A concentration cell derives its potential from the difference in concentration between the right and left sides. • M|M+(aq, L)||M+(aq, R)|M • The cell reaction is M+(aq, R) !M+(aq, L) • Using the Nernst equation, E = Eo - (RT/nF) ln Q • But Eo = 0 ! (Do you see why?) • ln Q = aL/aR • So for a conc. cell, E = - (RT/nF) ln (aL/aR)

Standard Electrode Potentials • Eocell can be found from DrGo using the equationDrGo = -nFEo • (or in general,DrG = -nFE) • But Eocell can also be found from values of Eo for the two electrodes involved. • Standard electrode potentials are given in Table 6.1, p. 216. • Since it is impossible to measure the potential of one electrode alone, these are all relative to H. • Using Table 6.1,Eocell = EoR - EoL

E and Spontaneity • Eo > 0 goes with K > 1, which indicates a spontaneous reactionfrom reactants in their standard state to products in their standard state. • However, the direction of a reaction can sometimes be reversed by judicious manipulation of the concentrations of product and reactant species. (That is, by altering Q.) • Any given reaction proceeds left to right when E (not Eo) > 0

Exercise E6.8 • Is the equilibrium constant for the displacement of copper by zinc greater or smaller than 1? • Solution: For this cell reaction, R (red’n): Cu2+ + 2e!! Cu EoR = +0.34 v L (ox’n): Zn2+ + 2e!! Zn EoL = !0.76 v • Eocell = EoR !EoL = +0.34 v !(!0.76 v) = +1.10 v • Since Eo is positive, the reaction is spontaneous, and K>1.

Exercise E6.9 • Calculate the equilibrium constant for the rxn Sn2+(g) + Pb(s)!Sn(s) + Pb2+(s) at 25oC • Solution: For this cell reaction, R (red’n): Sn2+ + 2e!! Sn EoR = !0.14 V L (ox’n): Pb2+ + 2e!! Pb EoL = !0.13 V • Eocell = EoR !EoL = !0.14 V !(!0.13 V) = !0.01 V • ln K =nEo/(25.69 mV) = 2 (!10 mV)/(25.69 mV) = !0.78 • K = 0.46

Exercise E6.10 • What is the equilibrium constant for the reduction of riboflavin with rubredoxin in the reaction rib(ox) + rub(red)! rib(red) + rub(ox) given that at pH = 7 the reduction potential for rub-redoxin is !0.06 V and that for riboflavin is !0.21 V • Solution: For this cell reaction, R (red’n): rib(ox) + 2 H+ + 2e!! rib(red)+ H2O L (ox’n): rub(ox) + 2e!! rib(red) • Eocell = EoR !EoL = !0.21 V !(!0.06 V) = !0.15 V • ln K = nEo/(25.69 mV) = 2 (!150 mV)/(25.69 mV) = !11.68 • K = 8.5 x 10-6 , making reactants strongly favored

The Hydrogen Electrode and pH • The potential of a hydrogen electrode is directly proportional to the pH of the solution. Consider the calomel-hydrogen cell Hg(l)| Hg2Cl2(s)| Cl!(aq)|| H+(aq)|H2(g)|Pt , for which the cell reaction is Hg2Cl2(s) + H2(g)! 2 Hg(l) + 2 Cl!(aq) + 2 H+(aq) • If the H2(g) is at standard pressure and the chloride ion activity is constant and incorporated into Eo†, the Nernst equation becomes E = Eo† - (RT/F) ln a(H+) = Eo†+ (RT ln 10/F) xpH = Eo†+ (59.15 mV) xpH • So the pH can be determined from the cell potential.

Exercise E6.11 • Calculate the biological standard potential of the half-reaction O2(g) + 4 H+(aq) + 4 e!! H2O(l) at 25oC given its value of +1.23 V under thermo-dynamic standard conditions. • Solution: Just as biochemical values of DGoNcan be calculated from the thermodynamic DGo values, it is also possible to calculate biochemical EoN values from Eo values. • The relationship is EoN = Eo!(RT /nF) (ln 10)(n x pH) = Eo !(0.05915 V /n) (7n) • n = the number of electrons; n = the number of H+ ions • EoN = 1.23 V !(0.05915 V / 4)(28)= 1.23 V - 0.41 V = 0.82 V

Exercise E6.12 • What range should a voltmeter have to display changes of pH from 1 to 14 at 25oC if it is arranged to give a reading of 0 when pH = 7 ? • Solution: Each pH unit changes the cell potential by (RT ln 10/F) = 0.05916 V. An increase in pH makes the measured potential more negative and a decrease in pH makes the measured potential more positive. • To go from 7 to 14 changes the value of E by !7 x 0.05916 V = !0.414 V • To go from 7 to 1 changes the value of E by 6 x 0.05916 V = + 0.355 V • Therefore, a range of 0.77 V would be needed.

The Electrochemical Series • A species with a low standard reduction potential has a thermodynamic tendency to reduce a species with a high standard reduction potential. • More briefly, low reduces high. • Equivalently, high oxidizes low. • This is the basis for the activity series of metals. • Other couples can also be fitted into the activity series.

Activity Series of Metals potassium sodium calcium magnesium aluminum zinc chromium iron nickel tin lead copper silver platinum gold React violently with cold water React slowly with cold water React very slowly with steam but quite reactive in acid increasing reactivity React moderately with high levels of acid < HYDROGEN comes here Unreactive in acid

Exercise E6.13 • Does acidified dichromate (Cr2O72-) have a thermodynamic tendency to oxidize mercury to mercury(I)? • Solutions: The relevant Eo’s are +1.33 V for Cr2O72-|Cr3+ and +0.79 V for Hg22+|Hg • Since high oxidizes low, the answer is • Yes, Cr2O72- will oxidize Hg to Hg22+

Exercise E6.14 • Can lead displace (a) iron(II) ions, (b) copper(II) ions from solutions at 298 K? • Solutions: The relevant Eo’s are: • -0.13 V for Pb2+|Pb • -0.44 V for Fe2+|Fe • +0.34 V for Cu2+|Cu • (a) Lead is NOT lower than iron(II), so the answer is NO. • (b) Lead IS lower than copper(II), so the answer is YES

Thermodynamic Functions and E’s • Because of the relationship • DrG = - nFE • DrGo = - nFEo • It is possible to obtain the thermodynamic value of the standard reaction Gibbs energy by measuring cell potentials.

Exercise E6.15 • Estimate the standard reaction Gibbs energy of Ag+(aq) + ½ H2(g)!H+(aq) + Ag(s) given that the standard potential of the cell H2| H+||Ag+|Ag is Eo = +0.7996 V • Solution:DrGo = - nFEo • DrGo = - (1)(96.485 kC)(+0.7996 V) = -77.15 kJ

Finding Eo via DGo • Eo’s for two half-reactions can be combined directly as long as the number of electrons is the same in each half-reaction and the electrons cancel out when the half-reactions are combined. • Eo’s cannot be directly combined for half-reactions in which the electrons do not cancel out. • For instance, Eo for Cu2+|Cu+ cannot be found by directly combining Eo’s for Cu2+|Cu and Cu+|Cu . • In these cases, Eo’s may be converted to DGo’s for and the DGo’s for then directly combined.

Other Thermodynamic Values - DrHo • The van’t Hoff equation (Eq. 3.25) may be modified to give DrHo if Eo is measured at two different temperatures. • Substitute -DrGo/RT for ln K • Substitute -nFEo for DrGo • or do it in one step by replacing nFEo /RT for ln K

Other Thermodynamic Values - DrSo • With expressions for both DrGo and DrHo the entropy can also be obtained. • DrSo is given by the relation • Eo(TN) - Eo(T) DrSo = nF x . . TN - T

Exercise E6.16 • Predict the standard potential of the Harned cellPt|H2(g)|HCl(aq)|AgCl(s)|Ag(s)at 303 K from tables of thermodynamic data at 298 K. • Solution: The cell reaction is AgCl(s) + ½ H2(g)!H+(aq) + Cl!(aq) + Ag(s) • From tables of thermodynamic data, DrGo=(!131.23 + 109.79) kJ = -21.44 kJ and DrSo=0 + 56.5 + 42.55 !96.2 !½(130.684) J/K = -62.5 J/K; at 298 K, Eo =DrGo/nF = 0.2222 v. • Eo(303) = Eo(303) + DrSo (TN- T)/nF = 0.2222 v + (-62.5 J/K)(303-298)K/(96,485 C) = +0.2190 v

Electrochemistry