Electrochemistry. Redox reactions. Half Reactions. Nernst Equation Ion selective electrodes

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1 Electrochemistry Nernst Equation Ion selective electrodes Redox reactions oxidation - loss of electrons M n+ M n+1 + e - M is oxidized - reducing agent reduction - gain of electrons N n+ + e - N n-1 N is reduced - oxidizing agent Half Reactions Fe 3+ + e - Fe 2+ Zn e Zn if mix these which will donate, which will accept electrons? can t measure equilibrium concentrations - only represent this reaction

2 -ve, ANODE +ve CATHODE Galvanic Cell Cu 2+ + Zn Cu + Zn 2+ Half reactions are: Cu e - Cu Zn - 2e - Zn 2+ transfer of electrons - electric current - can be measured half-cell with greater driving force makes other cell accept electrons Spontaneous reaction copper more easily reduced electrons flow spontaneously from zinc half cell to copper half cell Zn - 2 e- Zn 2+ (zinc dissolves) Cu 2+ +2e - Cu (copper bar gains weight) Zn electrode -ve (loses electrons) ANODE Cu electrode +ve (gains electrons) CATHODE

3 Salt Bridge Salt bridge maintains electrical neutrality by transport of ions Cu deposits leaves excess SO neutralized by K + from KCl bridge Zn dissolves to give excess Zn 2+ in solution, neutralized by Cl - from salt bridge Also SO 4 2- and Zn 2+ could migrate into bridge - does not matter which Driving Force Driving force of half-cell can t be measured - only by comparison to other half cells All potentials quoted against hydrogen half cell - assigned zero potential 2H + + 2e- H 2 if half cell causes H 2 cell to accept electrons - assigned -ve potential (Zn ) if H 2 call causes half cell to accept electrons - assigned +ve potential (Cu +0.86) Oxidizing/Reducing agents Strong oxidizing agents e.g. permanganate +ve potential Strong reducing agents, e.g. zinc -ve potential Potentials shown in next slide are for gases at 1 atm pressure, and 1M for solutions relative to hydrogen electrode

4 Which way is spontaneous? Fe 3+ + e - Fe V Zn e - Zn V 1. Spontaneous reaction has +ve potential 2. Subtract one reaction potential from other to make difference +ve i.e (-0.763) = V 3. potential of Zn has to be subtracted to make final number positive - Zn goes in reverse: Zn Zn e - Overall Reaction 2Fe e - 2 Fe 2+ E 0 1 = Zn Zn e - E 0 2 = Fe 3+ + Zn Zn Fe 2+ E 1 - E 2 = V

5 The Nernst Equation if species not in standard state, E 0 changes depending on concentrations (half reactions) E = E 0! RT [red]b ln nf [ox] a " or [products] % $ ' # [reactants]& where a ox + n e - ( b red (complete reactions) a, b coefficients in balanced equation at 25 0 C : E = E 0! ln [red]b n [ox] a Example using cell notation CuCu + (10!5 M) Sn 4 + (10!1 M)Sn 2 + (10!4 M) Pt CONVENTION : reaction proceeds from Left to Right Anode on LHS Cathode on RHS Oxidation Reduction " loss of electrons " gain of electrons E cell = E right! E left CuCu + (10!5 M) Sn 4 + (10!1 M)Sn 2 + (10!4 M) Pt write each half reaction with its potential Cu + + e - Cu Sn e - Sn 2+ balanced equation is: Sn Cu Sn Cu Volts Volts Nernst eqn to calculate each potential E L = 0.521! ln [Cu] 2 [Cu + ] 2 could use 1 s here - gives the same answer = 0.521! ln [1] = volts 2 [10!5 Potential less than E0 2 ]

6 CuCu + (10!5 M) Sn 4 + (10!1 M)Sn 2 + (10!4 M) Pt anode oxidation E R = 0.154! ln [Sn 2 + ] 2 [Sn 4 + ] 2 cathode reduction = 0.154! ln [10!4 ] = volts 2 [10!1 Potential > E ] 0 E R! E L = E cathode! E anode = 0.243! = volts opposite to expected using E 0 Cu + + e - Cu volts Sn e - Sn volts for +ve potential must be made -ve tin must go in reverse, Sn 2+ Sn e - or: 2Cu + + Sn 2+ 2Cu + Sn 4+ Subtract to give +ve number CuCu + (10!5 M) Sn 4 + (10!1 M)Sn 2 + (10!4 M) Pt Nernst equation changes predicted direction of the reaction E R is the actual tin reaction = volts E L is the actual Cu reaction = volts with concentrations in cell notation, reaction reversed cf predicted by subtraction of E 0 s Cu reaction goes in reverse Sn Cu Sn Cu + Subtract to give +ve number (0.226 is made -ve) Measurement of Potential assumed electrons actually flow during measurement undesirable to have current flow reduction or oxidation - changes concentration potentiometer principle is used

7 potentiometer battery electrodes variable resistor galvanometer measures current cell with electrodes fraction of a standard voltage from battery varied until no current flows voltage required to stop flow matches potential being measured ph meter is a potentiometer - measures voltage w/o current flow electrodes half cells that do not involve pure metal in reaction: conducting electrode is usually inert Pt to conduct electrons potentiometric measurements: choose a suitable electrode whose potential depends on specie being measured e.g. a) Ag Ag + Cu Cu ++ Zn Zn ++ M M n+ b) Pt redox couple such as Cr 2 O 7 2! + 14H + + 6e! " #" $" 2Cr H 2 O E 0 = 1.33 Volts electrodes (cont.) Potential must be measured relative to a reference electrode E cell = E cathode - E anode (E right - E left ) hydrogen is standard ref electrode - but difficult to use need another reference electrode needs to have constant potential not affected by ions in solution

8 Saturated calomel electrode (SCE) Hg Hg 2 Cl 2 KCl sat E = 0.242V vs SHE Hg 2 Cl 2 + 2e! " 2Hg + 2Cl! mercurous chloride E = E 0! lg[cl! ] 1 potential constant with small current flow why? If the electrode accepts electrons: Hg +! Hg mercurous mercury solid Hg 2 Cl 2 dissolves to resaturate the solution If the electrode produces electrons: Hg! Hg + mercury mercurous but solution is saturated with Hg 2 Cl 2 so merurous ion precipitates as Hg 2 Cl 2 because [Cl - ] is high - small changes in [Cl - ] do not affect potential significantly Cl - depresses solubility of Hg 2 Cl 2 by common ion effect to maintain constant ionic strength and constant potential

9 practical device electrode constructed to dip directly into analytical solution salt bride replaced by fiber - acts like salt bridge small [Cl - ] leaks into solution, but not usually important silver/silver chloride ref electrode Ag AgCl Cl! volts at 25 C, w.r.t. SHE silver chloride immersed in saturated KCl saturated with AgCl As long as Cl- doesn't take part in reaction can be used as a reference electrode. Titration MnO 4 - with Fe 2+ deter mn of Fe in soln - titrate w/ std permanganate MnO 4! + 5Fe H +! Mn Fe H 2 O Fe must be in Fe 2+ state - reduce w/ stannous (see next) add known increments KMnO 4, measure potential of Pt electrode vs SCE as titration proceeds. plot of potential vs mls of titrant potential determined by Nernst eq n at different conc ns

reduction Fe 3+ to Fe 2+ 2Fe 3+ + SnCl 2! 4 + 2Cl!! 2Fe 2+ 2! + SnCl 6 then must destroy tin II with mercury II calomel SnCl 2-4 + 2HgCl 2! 4! SnCl 2! 6 + Hg 2 Cl 2 (s) + 4Cl!

10 reduction Fe 3+ to Fe 2+ 2Fe 3+ + SnCl 2! 4 + 2Cl!! 2Fe 2+ 2! + SnCl 6 then must destroy tin II with mercury II calomel SnCl HgCl 2! 4! SnCl 2! 6 + Hg 2 Cl 2 (s) + 4Cl! enough tin II added to complete reduction of iron III but if too much excess Sn II, Hg metal will form, not calomel SnCl HgCl 2! 4! SnCl 2! 6 + Hg(l) + 2Cl! will react with MnO 4 - and interfere with permanganate titration Jones reductor Amalgam of Zn and Hg in a column (zinc shot) Zn + Hg 2+! Zn 2+ + Hg pass iron Fe 3+ through column to reduce it to Fe 2+ 1M H 2 SO 4 as the solvent Zn is a powerful reducing agent will reduce almost anything Zn e!! Zn(s) E 0 =!0.764V Harris, 6 edn p358, fig Redox titration calculations MnO 4! + 5Fe H +! Mn Fe H 2 O - After adding aliquot of MnO 4 - reaction comes to eq m potentials of both half reactions are equal Calculate potential of reaction with half reaction for iron... [C] of both species known - (each mmole of MnO 4 will oxidize 5 mmole Fe 2+ ) Fe 3+ + e!! Fe 2+ E = 0.771! lg [Fe2+ ] [Fe 3+ ]

11 add drop titrant - know amount Fe 2+ converted to Fe 3+ - " (1 mmole MnO 4! 5 mmole Fe 2+ ) known ratio # $ calculate E from Nernst E = E 0 ( lg Fe2+ 1 Fe 3+ at equivalence point Fe 2+ Fe 3+ MnO 4 ( + 5Fe H +! Mn Fe H 2 O 1 5 x + x " 1 5 C ( 1 # $ 5 x % & ' ( C ( x) C is [Fe 3+ ] - know this because all Fe 2+ converted to Fe 3+ x is negligible compared to C, in terms of [] but not in potential eq m will affect potential - can solve for x % & ' by equating two Nernst equations - obtains equilibrium constant must be equal and opposite at equilibrium E = 0.771! lg [Fe2+ ] 5 [Fe 3+ ] i.e.(1.51! 0.771) = = [Mn 2+ ] = 1.51! lg 5 5 [MnO! 4 ][H + ] 8 [Mn 2+ ] lg [MnO! 4 ][H + ]! [Mn 2+ ][Fe 3+ ] 5 log [MnO! 4 ][H + ] 8 [Fe 2+ ] 5 lg [Fe2+ ] 5 [Fe 3+ ] = lg K eqm lg K eqm = 62.6, K eqm = 5 " # 1 substitute x, (C-x) and 5 C! 1 $ % 5 x & ' ( into eq m constant exp n to calc x use either half reaction to calculate potential using Nernst after equivalence point have Mn 2+ formed and excess MnO 4 - E = E 0! lg [Mn2+ ] calculate 1 [MnO! 4 ] from volume We want a difference in potential of 0.2 V in E and E 1 for a sharp endpoint break Note: in advanced calculations, activity must be taken into account rather than just [] a = f [C] f is the activity coefficient, and depends on charge on the ion, which affects ionic strength of solution

12 Rules for redox titrations equilibrium constant must be high so that x is small - reaction well to right (difference in E 0 of about 0.2 V should do it) measure potential for observation of the end point, or use an indicator such as MnO 4 Reagents for redox titrations Oxidizing agents Potassium Permanganate - KMnO 4 MnO! 4 + 5Fe H +! Mn Fe H 2 O, E 0 = 1.51V purple solution - self indicating Potassium Dichromate - primary standard Cr 2 O 2! H + + 6e!! 2Cr H 2 O E 0 = 1.33 Volts Ceric ion Ce 4+ + e!! Ce 3+ E 0 = V reducing agents Fe 2+ stable for short periods Fe 3+ + e!! Fe 2+ E 0 = V 2! Thiosulpate S 2 O 3 not oxidized by air (rare) S 2 O 2! 6 + 2e! 2!! 2S 2 O 3 Note: you should study the iron/cerium system Fe 2+ + Ce 4 +! Fe 3+ + Ce 4 +

13 Ion Selective Electrodes The glass electrode - for ph measurement - specific for H + ions. potential difference develops across thin glass membrane w/ sol ns of diff. ph on either side potential measured by placing ref. electrodes on each side of membrane on ref. electrode is incorporated in the glass electrode (Ag/AgCl) and the other is usually an SCE placed in sol n whose ph is to be measured. Ag AgCl HCl (H + internal) glass membrane H + (unknown) SCE The potential of this cell is given by: E cell = k RT F lga + Hunknown k is a constant - contains: potentials of the two reference electrodes potential due to H+ inside the glass membrane asymmetry potential - due to non-perfect behavior of glass membrane potential not same when ph is same on both sides of membrane changes if physical condition of membrane changes

14 Since ph = - lg a H + E cell = k! 2.303RT ph, i.e. ph = E! k cell F 2.303RT F k is determined using a buffer of known ph i.e. k = E cellstd! 2.303RT F lg ph std then ph unknown = ph std + E cell unknown! E cellstd 2.303RT F acid/alkaline error membrane glass made of Na 2 O and SiO 2 glass surface -SiO - Na + + H +! SiO! H + + Na + K for this equilibrium is large - gives silicic acid potential due to ion exchange between H + in sol n & (Na + ) ions in hydrated glass layer at solution/membrane boundary acid sol ns, H + conc n high - glass electrode responds solely to H+ (xpt very high [H + ] acid error) basic (>ph 9) H + activity small - glass electrode begins to respond to other monovalent cations e.g. Na - alkaline error Alkaline error of glass electrode

15 Acid error of glass electrode activity, a, different from [H+] electrode behaves like there are less protons available then actually added Other ISE s e.g. LiO-Si instead of NaO-Si found that different membrane compositions can enhance alkaline error electrode can be made to be more specific for Na, K, Li, etc. construction similar to H+ responding glass electrode internal solution usually chloride salt of cation of interest Solid State Membrane electrodes solid state fluoride electrode membrane single LaF 3 crystal + small qty of Eu II creates disorders in crystal, lattice defects defects correct size for F - ion F - in lattice - mobile lattice acts as semi-permeable membrane for F - alone construction similar to glass electrode Ag/AgCl internal ref. electrode

16 Fluoride Electrode selectivity ratio potential of ISE for an ion on its own: 0 ' E electrode = E M! 2.303RT lga nf M n+ 0 ' where E M depends on internal ref. electrode, filling solution, and construction of membrane E M 0 ' is a constant determined by measuring solution of known concentration if soln contains mixture of cations may respond to other cations eg. mixture of Na and K Nernst eqn has additive term for K if determining Na 0 ' E NaK = E Na! 2.303RT F lg(a Na + + k NaK a K + ) E NaK is measured potential k NaK is the selectivity ratio for potassium over sodium

17 selectivity ratio is the fraction of the sodium potential that is due to potassium 0 ' K NaK and E Na determined by use of known solutions of Na and K, and by solving simultaneous equations for K NaK and know how to calculate selectivity ratios and potential of ISE s 0 ' E Na

Chapter 18 Electrochemistry. Electrochemical Cells

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