1.0 Fundamentals. Fig Schematic diagram of an electrochemical cell.

Transcription

1 1 1.0 Fundamentals This chapter intrduces the electrchemical cell, its cmpnents, basic definitins, and the prcesses that take place during electrlysis. The difference between thermdynamics and kinetics is explained thrugh the cncepts f reductin ptential and verptentials. 1.1 The electrchemical cell Cnsider the electrchemical cell shwn belw where the electrlyte is stagnant. It cnsists f zinc and cpper electrdes immersed in slutins f ZnSO 4 and CuSO 4 respectively. The tw electrdes are cnnected thrugh a metal wire (see Fig. 1.1). ZnSO 4 Slutin CuSO 4 Slutin Fig Schematic diagram f an electrchemical cell. A prus membrane is used t prevent mixing f the electrlyte. Electrns are prduced in ne electrde, which crrespnds t the disslutin f Zn: Zn = Zn e (1.1) These electrns are transferred t the ther electrde thrugh an electrical cnductr and are cnsumed by the reactin, which crrespnds t the depsitin f Cu: Cu e = Cu (1.2) The electrde where disslutin (r xidatin because f increase in the xidatin number) f Zn takes place is called the ande, while the electrde where depsitin (r reductin because f decrease in the xidatin number) f Cu takes place is called the cathde. Reactins 1 and 2 individually are called half-cell reactins; tgether they are als called, fr bvius reasns, redx reactins. In an electrchemical cell, a minimum f tw half cell reactins must ccur, ne at the cathde and ne at the ande. Electrlyte in ande and cathde chamber in cells with diaphragms/separatrs is called anlyte and cathlyte respectively. Fr electrchemical reactins t ccur, reactants must exist in a charged r inized state. Inizatin can be achieved by either disslving metal cmpunds in slvents such as water and mlten salts. Fr example, (i) Cu and Zn are electrdepsited frm aqueus slutins f CuSO 4 and ZnSO 4 respectively and (ii) aluminium is electrdepsited frm Al 2 O 3 disslved in a mlten salt, crylite (Na 3 AlF 6 ).

2 Mass transfer f ins Let us nw lk clsely at the inic activity at the ande-electrlyte interface (AEI) and the cathde/electrlyte interface (CEI) in Fig Accumulatin f Zn 2+ takes place at the AEI because f the disslutin f Zn. Very sn the ande will be cmpletely surrunded by Zn 2+. Similarly, the cpper depsitin reactin will gradually deplete Cu 2+ at the CEI, ultimately resulting in a situatin where there is n Cu 2+ at the CEI. Cnsequently, the Zn disslutin reactin wuld stp at the ande since the newly-frmed Zn 2+ wuld be repelled by the Zn 2+ surrunding the ande. The absence f Cu 2+ at the CEI means the cpper depsitin reactin will stp. That is, after a shrt while, the halfcell reactins shuld cme t a grinding halt. Hwever, it has been bserved that the redx reactins cntinue in the galvanic cell till the cmplete disslutin f the Zn ande. Hw d we explain this paradx? Clearly frces must be acting in the electrlyte t ensure that Zn 2+ is cntinuusly transprted frm the AEI t the bulk electrlyte. Similarly, Cu 2+ must be transprted frm the bulk electrlyte t the CEI fr the cathdic half cell reactin t cntinue. A fllw-up questin therefre arises: What are the mechanisms fr the transprt f Zn 2+ and Cu 2+? There are three, namely, electrmigratin, diffusin and natural cnvectin. In electrmigratin, the catins and anins mve parallel and anti-parallel t the directin f the electric field, as shwn in Fig. 1.2a. Diffusin causes mvement f ins frm regins f higher t lwer cncentratins (see Fig. 1.2b). Natural cnvectin ccurs because f density difference between the electrlyte at the AEI (r CEI) and bulk electrlyte. Because f the higher Zn 2+ cncentratin at the AEI, the density f electrlyte at the AEI is greater than the bulk anlyte density. This causes the electrlyte near the AEI t sink; liquid rushes frm the bulk t fill the vid resulting in a cunter-clckwise circulatin f anlyte, as shwn in Fig The tpic n mass transfer will explain the different mechanisms f mass transfer and hw they affect electrlysis. Fig Electrchemical cell depicting transprt f ins by (a) electrmigratin and (b) diffusin. The arrws shw the directin f electric field and mtin f ins. Fig Natural cnvectin in an electrchemical cell. In many cases, as will be shwn later, mass transprt f ins is the rate cntrlling step in cmmercial redx reactins, be it in electrwinning f metals r in galvanic cells. This results in the cncept f the limiting current density, which is the maximum current that can flw thrugh a cell. T further speed up the transprt f ins, frced cnvectin is als used, fr example by sparging the electrlyte with air bubbles.

3 3 Anther phenmenn ccurs as the redx reactins cntinue. Psitive charge builds up in the anlyte due t the generatin f Zn 2+ by the andic reactin. Similarly, the cathlyte becmes prgressively negative as Cu 2+ is cnsumed by the cathdic reactin. Hwever, even thugh ins exist in bulk slutin, the slutin itself is always electrneutral. Cnsequently, tw events ccur simultaneusly t neutralize the charged electrlyte: (i) Zn 2+ will be transprted twards the cathlyte and (ii) SO 4 2- will mve twards the anlyte. The electrneutrality equatin is: z i C i = 0 (1.3) Here z i and C i are charge and cncentratin f in i respectively. z i is negative fr anins and psitive fr catins. In reality, the electrlyte is nt truly neutral. There is an electric duble layer as bth the AEI and the CEI are charged, but it is usually ignred since its dimensin (f the rder f nm) is much smaller than the extent f the bulk electrlyte. The charged electrde/electrlyte interface, hwever, has a significant effect n the kinetics and ccurrence f half-cell reactins and will be discussed separately in a fllwing chapter. As a cnsequence f in transprt due t electrneutrality cnsideratins, an interesting questin arises because f the presence f Zn 2+ in the anlyte: Will Zn c-depsit with Cu at the cathde? Again, this imprtant questin will be discussed in later chapters. The transprt f ins in the electrlyte raises anther fascinating pint. Is there a crrelatin between the rates f electrn flw between the metallic wire jining the electrde and the rate f transprt f ins? A charge balance in the cpper depsitin reactin suggests that the rate f flw f electrns in the circuit has t be equal t the rate f flw f inic charge thrugh Cu 2+. Here tw electrns/time flwing t the cathde requires the cnsumptin and hence transprt f 1 Cu 2+ /time. Since the electrns fr the depsitin f cpper cmes frm the ande, the implicit cnsequence is that the rate f the andic and cathdic reactins must be equal. T understand this phenmenn, let us assume that the rate f the andic reactin is faster than the rate f the cathdic reactin. Accrdingly, electrns are being pumped t the cathde at a higher rate than it can be cnsumed at the cathde, resulting in increasing accumulatin f electrns in the cathde. This will result in the cathde becming prgressively mre negative, which, in turn, will ppse further transprt f electrns frm the ande t the cathde. Cnsequently, the rate f the andic reactin will gradually slw dwn until its rate becmes equal t the cathdic reactin, after which the andic and cathdic reactins will prceed at the same slwer rate Faraday s law The rate f half cell reactins is given by Faraday s law, which states that a charge crrespnding t ne Faraday (96500 C) results in the electrlytic prductin (r cnsumptin) f ne gram equivalent f species in a half cell reactin. This law is easy t implement in the case f a half cell reactin such as Cu e = Cu. Here, C prduces 1 gram equivalent f Cu (atmic weight f Cu/valency), that is, 65/2 g f Cu. Hwever, as a rule it is much easier t see Faraday s law in terms f reactin stichimetry since the cncept f gram equivalent can be cnfusing t nn-chemists. Fr example, hw d we determine the mlar rate f (i) xygen r H + prduced r (ii) H 2 O cnsumed by the half cell reactin: 2H 2 O = O 2 + 4H + + 4e. By fllwing reactin stichimetry, we can simply say: Charge crrespnding t 4 mles f electrns (4F = 4 x = 3.86 x 10 5 C) prduces ne mle f xygen x 10 5 C prduces 1 mle f O x 10 5 C prduces 4 mles f H x 10 5 C cnsumes 2 mles f H 2 O.

4 Classificatin f electrlysis cells There are tw types f electrlysis cells. In the first type, current spntaneusly flws between the electrdes nce they are jined by a metal wire. That is, nce the electrdes are electrnically cnnected, the redx reactins ccur spntaneusly. In essence, these cells prduce electrical energy and are knwn as galvanic cells, a prminent example f which is batteries. The phenmenn f crrsin als ccurs by the frmatin f a galvanic cell between the crrding metal and its mist envirnment. The secnd type f cells requires the applicatin f a vltage fr the redx reactins t ccur. These cells are used fr the depsitin f metals, prductin f gases such as chlrine, hydrgen etc. Fr metal depsitin reactins at the cathde, these cells are further classified as electrwinning and electrdepsitin cells. Electrwinning is used fr the bulk prductin f metals frm leach liqur btained frm the hydrmetallurgical prcessing f metallic res. Electrwinning cells are used fr the prductin f a number f primary metals such as Cu, Zn, Ni, and Al. Electrdepsitin cells are used fr depsiting small quantities f metals n t a substrate fr altering their surface prperties such as increasing crrsin and abrasin resistance, r simply fr aesthetics, as in the gld plating f silver jewellery. Hereafter, fr fundamental cnsideratins, unless specifically stated, electrdepsitin cells and electrwinning cells will be used interchangeably Cell vltage Fr current t flw in the circuit, a ptential (vltage) difference must exist between the tw electrdes. In galvanic cells, the half cell reactins ccur spntaneusly and a ptential difference is autmatically set-up nce the electrdes are cnnected. In electrwinning cells an external ptential difference must be applied fr the half cell reactins t ccur. Tw questins therefre arise: What is the vltage that we can btain frm a galvanic cell given specific ande and cathde materials? What is the vltage that needs t be applied in an electrwinning cell? There are tw appraches fr determining the cell vltage. One can take all pssible cmbinatins f electrdes, make the cells and then measure the vltage between the electrdes, an unenviably tedius jb. Alternately, ne can measure the ptential f a half-cell reactin, knwn as the half cell ptential, such as thse given by equatins 1.1 and 1.2. By half cell ptential f equatin 1, we mean the ptential difference that ccurs between a Zn electrde and the slutin cntaining Zn 2+. Or, the ptential difference between the Cu electrde and the slutin cntaining Cu 2+ Cell vltage can then be determined by adding the half cell ptentials f the apprpriate andic and cathdic half-cell reactins. Hwever, befre we prceed with this apprach, there is a need t first understand the rigin f half cell ptential, its definitin, and the manner in which it is measured. 1.2 Thermdynamics This sectin discusses the rigin f the half cell ptential and intrduces the ntin and methdlgy fr determining standard and nn-standard reductin ptentials Origin f half cell ptential T understand the rigin f half cell ptential, cnsider a Cu electrde which is immersed in a slutin f CuSO 4. Yu can check that the electrde weight remains unchanged with time. This suggests that the tw pssible reactins: (i) Cu e Cu (depsitin), and (ii) Cu Cu e (disslutin) ccur at the same rate. Hwever this equilibrium state is nt achieved instantaneusly. When the Cu electrde is immersed, bth the disslutin and depsitin reactins can ccur. Nw assume that the

5 disslutin reactin is faster than the depsitin reactin. Fr example, 6 Cu 2+ /s are depsiting as Cu and 8 Cu atm/s are disslving as Cu 2+. The net effect is that 2 mles/s f Cu frm the electrde disslves in the electrlyte as Cu 2+. Cnsequently, at the end f the 1 st secnd, there are 4 excess electrns in the metal side f the metal/electrlyte interface (MEI) and 4 excess psitive charges due t 2 excess Cu n the slutin side f the MEI. These excess charges line up in parallel, shwn in Fig. 1.4 belw, resulting in the frmatin f an electric duble layer (EDL) at the MEI. 5 Cu electrde l _ Electrlyte cntaining Cu 2+ x Fig Electric duble layer at a metal-electrlyte interface. The duble layer electric field will retard the mtin f Cu 2+ in +x directin and speed up Cu 2+ in the x directin. In the 2 nd secnd, therefre, the rate f disslutin Cu atms will decrease t, say 7 Cu 2+ /s. Simultaneusly, the electric field will hasten the transprt f Cu 2+ frm the bulk t the interface and hence increase the rate f the depsitin reactin. Since the charge n the slutin side f the duble layer will nw becme less psitive, thereby reducing the duble layer charge, which, in turn, will again speed up the disslutin reactin and slw dwn the depsitin reactin. Cmpetitin between the disslutin and depsitin reactins will cntinue until equilibrium is established, leading t the frmatin f a stable duble layer f charge. This equilibrium duble layer, which acts like a capacitr, gives rise t the half cell ptential, which is als referred t as electrde-slutin ptential. In general, sme metals have a greater inclinatin t stay in slutin as ins, that is, in their xidized state. Other metals may prefer t stay as the metal phase, that is, in their reduced state. The sign f the electric field in the duble layer depends n the natural tendency f metal t be in its inic r metallic state. Fr reactin 1, since Zn has a tendency t disslve, the EDL will have psitive charge in the electrde and negative charge in the electrlyte. Electric duble layer is a feature f all fluid-slid interfaces, irrespective f whether the slid is a cnductr, semi-cnductr, r insulatr. Fr the same fluid, say water, the magnitude f the ptential drp acrss the EDL will be highest fr a gd cnductr such as a metal. Bckris et al. prvide a deep and insightful discussin n EDLs Measurement f Half-Cell Ptential We knw frm electrical engineering that ptential is always measured relative t the earth, which is fr cnvenience, assigned zer ptential. The electrchemical analg f earth is the hydrgen evlutin reactin: H + + e = 0.5 H 2, which has been assigned a standard half cell ptential f zer. The half cell ptential f all ther reactins is measured with respect t the H 2 evlutin reactin; such a cell is shwn belw, in Fig The cell cnsists f the metal whse standard half cell ptential (fr the reactin M n+ + ne = M) is t be determined. The hydrgen evlutin reactin ccurs at the platinum electrde.

6 6 Fig Cnfiguratin f an electrchemical cell fr measuring the standard half cell ptential fr the reactin M n+ + ne = M. The standard half-cell reductin ptential (E ) is defined when (i) activity f H + and M n+ = 1, (ii) p H2 = 1 atm, and (iii) temperature = 25 C. Please remember that the half cell reductin ptential is an equilibrium quantity when the metal is in equilibrium with its in in slutin. A snapsht f the electrchemical series, which lists the reductin ptential f different half-cell reactins, is given in Table 1.1. Table 1.1. A snapsht f the electrchemical series shwing the standard reductin ptentials Electrde reactin (M n+ /M) E (V) Li + + e = Li Ba e = Ba Mg e = Mg Zn e = Zn H + + e = 0.5 H Cu e = Cu 0.34 Fe 3+ + e = Fe Ag + + e = Ag 0.81 O 2 + 4H + + 4e = 2H 2 O 1.34 The meaning f the sign f E can be btained frm the relatinship: G = -nfe. Clearly M n+ /M reactins that have a negative E will tend t have the metal, M in its inic (disslved) state. Similarly M n+ /M reactins that have a psitive E will tend t depsit M frm a slutin cntaining M n+. Unless therwise stated the symbl E will represent reductin ptential. Reductin ptential may als, at places, be explicitly stated as E (M n+ /M). During the measurement f reductin ptentials, there is n current flwing thrugh the cell since the vltmeter has a very high resistance. That is, n net half cell reactin is taking place at the electrdes. Reductin ptentials, standard r nn-standard, therefre represent a state f equilibrium and hence are thermdynamic quantities. It shuld als be nted that it is nt necessary that M n+ /M reactins with negative (psitive) reductin ptential will always have the metal in its inic state. Fr example, if we set up a galvanic cell cnsisting f Ba and Zn electrdes immersed in BaSO 4 and ZnSO 4 respectively, then Ba will be the ande and Zn will be the cathde; the andic and cathdic reactins being Ba = Ba e and Zn e = Zn respectively. Hence the reductin ptential is a reflectin f the relative tendency f a metal M 1 t stay in a metallic r inic state with respect t anther metal M 2. Anther imprtant pint is that a metal M can exist as different ins in slutins f different ph and each metal-metal in pair will have a different reductin ptential. Fr example, in acidic slutins, Zn 2+ is present, while ZnO 2 2- exists in strngly basic slutin. The standard reductin ptential f the

7 half cell reactin ZnO H 2 O + 2e = Zn + 4OH - is V cmpared t vlts fr the Zn e = Zn reactin Reference electrdes (RE) It is nt cnvenient t use a standard hydrgen electrde (SHE) fr experiments. We need cmpact and prtable reference electrdes, examples f which are the standard calmel electrde (SCE) and Ag/AgCl electrde. The half cell ptential f each f these REs is defined with respect t SHE. Fr example, E (SCE) = V with respect t SHE The USP f reference electrdes is that their ptential des nt change with current, an imprtant characteristic that will be clear in the chapter n electrchemical experiments Calculatin f E cell Nw that we knw the half cell reductin ptentials, hw d we calculate the ptential f an electrchemical cell (r, the redx ptential), E cell. There is mre than ne way f calculating E cell. We will fllw the simpler methd, which des nt require memrizing the frmula fr E cell. The fllwing steps shuld be fllwed: Write the andic and cathdic reactins as they ccur, that is, xidatin reactin at the ande and reductin reactin at the cathde. Write dwn the "actual" standard half cell ptentials f the andic and cathdic reactin, defined as E a and E c respectively. Let us assume metal M 1 is xidized at the ande and metal M 2 is reduced at the cathde. Then E a = -E (M n+ 1 / M 1 ) and E c = E (M n+ 2 / M 2 ). E cell = E a + E c T illustrate the methd fr calculating E cell, cnsider the fllwing examples f a Cu/Zn galvanic cell and electrwinning f cpper Cu/Zn galvanic cell Ande: Zn = Zn e E a = -E (Zn 2+ /Zn) = 0.76 V Cathde: Cu e = Cu E c = E (Cu 2+ /Cu) = 0.34 V 7 Cell reactin: Zn + Cu 2+ = Zn 2+ + Cu E cell = E a + E c = 1.10 V Electrwinning f Cu Ande: 2H 2 O = 4H + + O 2 + 4e Cathde: Cu e = Cu E a = -E (O 2 /H 2 O) = V E c = E (Cu 2+ /Cu) = 0.34 V Cell reactin: 2H 2 O + 2Cu 2+ = 4H + + O 2 + 2Cu E cell = E a + E c = V One can als represent E cell purely in terms f reductin ptential. Fr the verall reactin Zn + Cu 2+ = Zn 2+ + Cu, E cell = E (Cu 2+ /Cu) - E (Zn 2+ /Zn). That is, E cell = reductin ptential f cathdic reactin reductin ptential f andic reactin. It is nt imprtant which cnventin yu chse, but yu must be cnsistent. One majr prblem fr students in electrchemistry is messing up the sign (- r +) f ptentials, as will be evident later. It is recmmended that the methd suggested in the beginning f this sectin may be fllwed.

8 8 Please nte that E cell is nt a reductin r xidatin ptential. It represents the ptential fr the verall reactin the way it is written. Fr example, E cell = V represents the verall reactin 2H2O + 2Cu 2+ = 4H + + O 2 + 2Cu. Fr the reactin, 4H + + O 2 + 2Cu = 2H2O + 2Cu 2+, E cell = V Reactin multiplicatin and E T get the cell reactin, we have t equate the number f electrns invlved in bth the andic and cathdic reactins, which, in turn, requires multiplying either ne r bth half cell reactins with different integers. Fr example, in the electrwinning f cpper, the cathdic reactin has t be multiplied by 2. This raises the questin: D the fllwing reactins have the same E 0? 1. Cu e = Cu 2. 2Cu e = 2Cu Using the relatinship, G = -nfe, yu can shw that E 0 fr reactins 1 and 2 are same. In reactins 1 and 2 n is 2 and 4 respectively. In essence, multiplying a half cell reactin, say, by 2, dubles the value f G, but leaves E unchanged Significance f sign f E cell What is the significance f the sign f E cell? Again frm the relatinship, G cell = -nfe cell, a psitive E cell implies that the half-cell reactins will ccur spntaneusly if the electrdes are cnnected with a metallic wire, much like a battery. That is, the cell acts as a surce f electrical energy. Similarly, a negative E cell implies that energy has t be applied fr the half-cell reactins t ccur, as in electrwinning r electrdepsitin cells. Anther pint f cnfusin is n, the number f electrns in a cell reactin. It is easy t define n fr a half cell reactin, but slightly cnfusing fr the verall (cell) reactin because the electrns are absent in the latter. Fr a cell reactin, n refers t the number f electrns ne gets after multiplying the half cell reactins with integers t ensure that the number f electrns are same in bth the andic and cathdic reactin. Fr example, in the cpper electrwinning reactin, n = 2 fr the cathdic reactin, but n = 4 fr the cell reactin Half-cell/cell ptential under nn-standard cnditins E fr a half cell reactin crrespnds t standard cnditins, that is (i) a n+ = 1 M, (ii) T = 298 K, and (iii) fr reactins invlving gases, the partial pressure f the gas shuld be 1 atm. Hw d we determine the reductin ptential, E, under nn-standard cnditins, that is when inic activities r temperature r gas partial pressures are different frm thse under standard cnditins? Here we use the Nernst equatin. The Nernst equatin can be written either fr calculating nn-standard (i) half-cell reductin ptentials r (ii) cell reductin ptentials, E cell 1. The Nernst equatin fr the Cu e = Cu reactin is given in equatin Anther frequently used term is Decmpsitin Ptential, the ptential abve which the current rises appreciably. The magnitude f decmpsitin ptential is greater than E cell fr electrwinning cells.

9 9 E Cu2+ Cu = E Cu2+ Cu RT ln a Cu 2F a Cu 2+ = 0.34 RT ln a Cu (1.4) 2F a Cu 2+ where, a 2+ is the activity f Cu 2+ in the electrlyte., T is the temperature (in K) and F = C. Cu a Cu = 1 fr pure cpper. O 2 + 4H + + 4e = 2H 2 O E O 2 H2 O = E O 2 H2 O RT ln 1 2F P O 2 a H + 4 = 1.23 RT ln 1 2F P O 2 a H + 4 (1.5) 2H 2 O + 2Cu 2+ = 4H + + O 2 + 2Cu E cell = E cell RT 4 ln P O2 a H + 4F 2 (1.6) Please nte that activity and cncentratin are als represented by curly and square brackets respectively. Fr example, 2 2 a == + 2+ Cu { Cu } = + γ ± Cu {Cu 2+ } = activity f Cu 2+ [Cu 2+ ] = cncentratin f Zn 2+ Since activity f an in cannt be measured, define γ ±, the mean inic activity f the salt, in this case CuSO 4, in the electrlyte. Unit f cncentratin is mlality (mles/kg slvent), but usually, with little errr, mlarity (mles/litre) can be used as a unit f cncentratin. a Cu Determinatin f γ ± γ ± be determined by using the Debye-Huckel equatin: lgγ ± = A z + z I 0.5 (1.7) A = fr H 2 O at 25 C, z + and z - are the valence f the catin and anin respectively that cnstitute the salt. Equatin 1.8 hlds gd fr I 10-3 M At higher cncentratins use the extended Debye Huckel Law Limitatins f E/E I = Inic strength = 0.5 C i z i 2 (1.8) lgγ ± = = A z +z I I 0.5 (1.9) E/E has been measured under equilibrium cnditins, i.e., when there is n net current flwing thrugh the cell. This implies, fr example, that there is n net metal disslutin taking place at the

10 ande r metal depsitin taking place at the cathde. In real-life situatins, we wuld like t perate electrlytic cells at finite currents because prductivity is prprtinal t the current flwing thrugh the cell. 10 In general, the applied vltage V cell E cell Electrwinning cells: V Cell > E Cell Cu: E cell = 0.89V ; V cell 2V Al: E cell = 1.2V; V cell = 4.5 V Galvanic cells: V cell < E cell V cell (Zn-C battery) < 1.5 V In electrwinning cells, a vltage higher than E cell has t be applied, that is mre energy than what is dictated by thermdynamics has t be supplied, fr example in depsiting cpper. Similarly, we will be drawing lwer vltages than E cell in a battery. Hence it appears that E cell is nt relevant in real-life applicatins. This is nt true because E cell acts as a benchmark, that is, it can be taken as a measure f the extent f "nn-ideal" perfrmance, r vltage efficiency, f an electrlytic cell. Vltage efficiency is related t the energy efficiency f a cell and will be discussed in a later chapter. Vltage efficiency is defined as: Electrwinning cell: Vltage efficiency = E cell V cell x 100 Galvanic cell: Vltage efficiency = V cell E cell x Why is V cell different frm E cell? V cell is different frm E cell because the flw f current leads t the phenmenn f "plarizatin" fr each half-cell reactin. Plarizatin manifests itself as an "verptential," that is, an additinal ptential drp ver and abve the half-cell reductin ptential. There are three types f plarizatin: a) Activatin plarizatin represents the resistance t half-cell reactin ccurring at an electrde and is akin t the cncept f activatin energy in the case f pure chemical reactins. That is, there is an activatin energy assciated with half cell reactins at bth the ande and the cathde. Hwever, the presence f an electric duble layer adjacent t an electrde makes an electrchemical reactin different frm a chemical reactin. This electric field can either ppse r aid the mtin f the in twards r away frm the electrde depending n whether the in is an anin r a catin. Fr example, cnsider the half-cell reactin Cu e = Cu at the cathde. Activatin plarizatin has tw cmpnents here, namely, (i) the energy invlved in transprting Cu 2+ acrss the EDL and (ii) activatin energy fr the cpper reductin reactin t ccur. In this case, the electric field in the EDL pints twards the cathde and hence it aids the transprt f Cu 2+ twards the cathde. T visualize the effect f EDL assume that the Cu electrde in Fig. 1.4 c is the cathde. Thus activatin verptential at the cathde ( η a ) represents the extra energy required fr the half-cell reactin t ccur at the cathde. Similarly, ne can define the activatin verptential at the ande, η. a a One can clearly see the adverse impact f activatin plarizatin; it increases the vltage r, in ther wrds, energy cnsumptin fr a given value f current. Frtunately, as will be shwn later, this cncept can als be used t eliminate unwanted reactins, fr example the evlutin f H 2 during the electrwinning f Zn.

11 11 b) Cncentratin plarizatin ccurs when mass transfer f reacting/prduct ins cannt keep pace with rate f electrn transfer, that is, current. Fr example, cnsider the cathde during the electrwinning f cpper. Let us say current is flwing at a cnstant rate f 10 e - /s. Since electrns cannt accumulate at the cathde, 5 Cu 2+ /s are being cntinuusly supplied frm the bulk electrlyte t the cathde. The transprt f Cu 2+ depends n the mass transprt mechanism, namely, diffusin, cnvectin, and migratin. Nw let us say, we increase the current t 12 e - /s. Hwever, since flw cnditins in the electrlyte are unchanged, 5 Cu 2+ /s are still cming t the cathde. Since we are perating under cnstant current cnditins, and there can be n accumulatin f electrns, the ptential ges up at the cathde, which, in turn, increases the transprt f Cu 2+ t 6/s by enhancing the effect f migratin and diffusin 2. Cncentratin verptential is assciated, like activatin verptential, with bth the andic ( η ) and cathdic ( c η c ) reactins. Cncentratin plarizatin has adverse and beneficial effects, much like activatin verptential. In additin, as will be shwn later, it affects the quality f the depsit. Mrever, the cncept f cncentratin verptential is als used t define the maximum permissible current flw in an electrlytic cell. c) Resistance plarizatin is due the passage f current thrugh all the resistive cmpnents f the cell, namely the electrde, cnnecting wire, and the electrlyte. It manifests as the IR drp in the cell, the predminant cmpnent f which is due t the electrlyte resistance, R e. 1.4 Vltage balance Electrwinning cell The sign f the activatin and cncentratin verptentials at the electrdes are given belw: η a a, η c a psitive η a c, η c c negative Hwever, since all verptentials have the effect f increasing the applied ptential, we take the abslute values f activatin and cncentratin verptentials at the ande. Bth η a & η c depend n current density, i = I/A (A/m 2 ), where I is the cell current and A is the submerged electrde area. Fr a η a, η a c c, A refers t the ande area. Similarly, fr η a, η c c, A refers t the cathde area. V cell = E cell + η a a + η a c + η c a + η c c + IR e (1.10) Equatin 1.10 simply states that fr electrwinning cells, the applied vltage increases with current. Thus verptentials can als be viewed as a surce f vltage drp, much like electrlyte resistance. In a sense, ne can visualize electrchemical cells as an electrical circuit with several resistances in series; they d nt matter as lng as there is n current flw thrugh the cell. Hwever, during electrlysis, these resistances result in additinal vltage drps. Industrial electrchemical cells can be perated under tw mdes: (i) cnstant current r galvanstatic and (ii) cnstant vltage r ptentistatic. Overptentials wuld result in higher V cell in the a c 2 The increase in electric field with cell vltage is bvius as the magnitude f the electric field is prprtinal t ptential gradient, which increases with increasing V cell. Hwever, the effect f increasing cell vltage n diffusin is nt bvius and will be discussed in the chapter n cncentratin verptential.

12 galvanstatic mde and lwer cell current in the ptentistatic mde, cnclusins that be verified frm the vltage balance equatin stated abve Galvanic Cell Equatin 1.11 shws the vltage balance in a galvanic cell, fr example a battery used in trch r a cell phne. V cell = E cell η a a η a c η c a η c c IR e (1.11) Here the vltage written n the battery is E Cell. When current is drawn frm the battery t light a trch r run the cell phne, the vltage between the ande and the cathde (V cell ) will always be lwer than E cell because f plarizatin effects. 1.5 Purbaix (Eh ph) diagram The Purbaix diagram is a plt f reductin ptential (E r Eh) versus ph. It is cnstructed by pltting the Nernst equatin fr half cell reactin pertaining t ne metal species. Fr the zinc-water system shwn in Fig. 1.6, the fllwing half-cell reactins are used t cnstruct the Purbaix diagram 3 : (i) Zn e = Zn, (ii) Zn(OH) 2 + 2H + + 2e = Zn + 2H 2 O, (iii) HZnO H + + 2e = Zn + 2H 2 O (iv) ZnO H + + 2e = Zn + 2H 2 O, (v) Zn(OH) 2 + 2H + = Zn H 2 O, (vi) Zn(OH) 2 = HZnO H +, (vii) HZnO 2 - = ZnO H + The O 2 and H 2 evlutin half-cell reactins are superimpsed n the Purbaix diagram. 12 O 2 + 4H + + 4e = 2H 2O 2H + + 2e = H 2 Zn Fig Purbaix diagram fr the Zn-H 2 O system 3. Hrizntal lines crrespnd t half-cell reactins that nly invlve reductin f metal species, as in reactin (i). Slanted lines crrespnd t half-cell reactins that invlve bth reductin f metal species and H +, as in reactins (ii) t (iv). Vertical lines represent chemical reactins, as in reactin (v) t (vii). A Purbaix diagram tells yu the pssible half-cell reactins as a functin f ph. At ph = 3 and 15, the thermdynamically feasible half cell reactins are Zn e = Zn and ZnO H + + 2e = Zn + 2H 2 O respectively. 3

13 Example prblems In all wrked example, unless it is specifically mentined, the ptentials are with respect t the standard hydrgen electrde (SHE). Example 1-1 Cnsider the electrchemical cell: Zn Zn 2+ Cu 2+ Cu. The zinc and cpper slutins are bth prduced by disslving the apprpriate sulfates. The cncentratin f zinc sulfate in the left half-cell is ml/l and the cncentratin f cpper sulfate in the right half-cell is ml/l. Each cell cntains ml/l H 2 SO 4. Calculate the electrchemical ptential f this cell at 298 K using the Debye-Huckel thery. Given E (Zn 2+ /Zn) = V and E (Cu 2+ /Cu) = 0.34 V. The cell reactin is: Zn + Cu 2+ = Zn 2+ + Cu with E cell = 1.10 V. Assume 1 dm 3 (1 litre) f bth anlyte and cathlyte. Square and curly brackets dente cncentratin and activity respectively. Given: Anlyte: [Zn 2+ ] = M, [H + ] = 0.01, [SO 4 2- ] = 0.01 M Cathlyte: [Cu 2+ ] = M, [H + ] = 0.01, [SO 4 2- ] = M E cell = E cell lg Zn2+ {Cu 2+ } {Zn 2+ } = γ ±,ZnSO4 [Zn 2+ ] {Cu 2+ } = γ ±,CuSO4 [Cu 2+ ] (e1-1.1) (e1-1.2) (e1-1.3) Anlyte 2 I = 0.5 i=1 C i z 2 i = 0.5[Zn 2+ ](+2) [H + ](+1) [SO 2 4 ]( 2) 2 = M (e1-1.4) Using extended Debye-Huckel law, that is, lg γ ±,ZnSO4 = A z +z I I 0.5 = = 0.32 (e1-1.5) γ ±,ZnSO4 = 0.48 (e1-1.6) Cathlyte 2 I = 0.5 i=1 C i z 2 i = 0.5[C 2+ ](+2) [H + ](+1) [SO 2 4 ]( 2) 2 = M (e1-1.7) lg γ ±,CuSO4 = A z +z I I 0.5 = = 0.27 (e1-1.8) γ ±,CuSO4 = 0.54 (e1-1.9) Frm e and e1-1.3, {Zn 2+ } = = M, {Cu 2+ } = = M (e1-1.10)

14 14 Substituting e in e1-1.1, we get: E cell = lg = 1.09 V This prblem suggests that yu can E cell = E cell t a gd apprximatin in many cases, as in cncentratins d nt significantly changee cell. Example 1-2 A vltaic cell is cnstructed that uses the fllwing reactin Ni + 2Ag + = Ni Ag. a) Write the half reactins & indicate the andic r cathdic reactins. b) Calculate E cell c) State whether the reactin is spntaneus. Given: E (Ni 2+ /Ni) = -0.28V, E (Ag + /Ag) = +0.80V a) The half cell reactins are: Ande: Ni = Ni e E a = -E (Ni 2+ / Ni) = 0.28 V Cathde: Ag + + e = Ag E c = E (Ag + / Ag) = 0.80 V b) Cell reactin: Ni + Ag + = Ni 2+ + Ag E cell = E a + E c = 1.08 V c) G cell = nfe cell = = 212 kj. Hence the cell reactin is spntaneus. Example 1-3 Calculate the ptential f a silver wire, with respect t SCE, immersed in a 1.0 mm AgNO3 slutin. Ignre activity cefficients. The ptential f SCE with respect t the SHE is V. The relevant reactin is Ag + + e = Ag, E (Ag + / Ag) = 0.80 V, [Ag + ] = M. E(Ag + /Ag) SHE = lg 1 = 0.71 V [0.001] E(Ag + /Ag) SCE = E(Ag + /Ag) SHE E (SCE) = = 0.47 V T ensure that yu d nt make any mistakes in cnverting frm SHE scale t SCE scale, please d the fllwing: (i) Mark E(Ag + /Ag) = 0.71 V, E (SCE) = V, and E (H + /H 2 ) = 0 n the right side f a vertical line. This line is the SHE scale. (ii) In a SCE scale, E (SCE) =0. Hence cnvert all values frm SHE scale t SCE scale by subtracting V frm them. Write these values n the left hand side f the line. (iii) The value next t E(Ag + /Ag) n the left hand side f the line is E(Ag + /Ag) SCE Learning: Example 1-3 shws an electrchemical technique fr determining in cncentratin in slutin by measuring half-cell ptential. Example 1-4 The fllwing equatin applies fr the electrde reactin antimny-antimny (III) xide electrde: Sb 2 O 3 + 6H + + 6e = Sb(s) + 3H 2 O E (Sb 3+ /Sb) SHE = V Calculate the ph f a slutin, which gives rise t a ptential f V vs. SCE n such an electrde.

15 15 First cnvert V frm SCE scale t SHE scale. Therefre, E(Sb 3+ /Sb) SCE = E(Sb 3+ /Sb) SHE - E (SCE) r, E(Sb 3+ /Sb) SHE = = V Nw, E(Sb 3+ /Sb) SHE = E (Sb 3+ /Sb) SHE lg 1 {H + } 6 r, = ph r, ph = 10.5 Learning: Example 1-4 shws an electrchemical technique fr determining ph f a slutin by measuring cell ptential. Example 1-5 A nvel ph sensitive electrde can be cnstructed by andizing a cpper wire, which xidizes t Cu +2 in a slutin cntaining OH -, resulting in the frmatin f an insluble Cu(OH) 2 cating n the wire. A basic slutin yields a ptential f V vs. SCE with this electrde. Calculate the ph f the slutin. The relevant equilibria are: Cu e = Cu(s) E = 0.34 V; K sp [Cu(OH) 2 ] = 4.8 x Unlike example 1-4, ph des nt cme directly frm the given half-cell reactin, but frm the slubility prduct. E(Cu 2+ /Cu) SCE = E(Cu 2+ /Cu) SHE - E (SCE) E(Cu 2+ /Cu) SHE = = V E(Cu 2+ /Cu) SHE = E (Cu 2+ /Cu) SHE lg 1 Cu 2+ (e1-5.1) r, = lg{cu 2+ } {Cu 2+ } = M Nw, K sp = {Cu 2+ }{OH 2 } 2 = {Cu 2+ } {H + } 2 = 4.8 x {H + } = Hence ph = lg{h + } = 10.2 Please nte that even thugh in reality the cpper is being xidized, that is, Cu(s) = Cu e reactin is taking place, we have cnsidered the reductin reactin Cu e = Cu(s). Yu can check that even if cnsider the Cu = Cu e, e1-5.1 and hence {Cu 2+ } will remain unchanged.

16 16 Example 1-6 The slubility prduct f AgBr(s) is determined in the fllwing manner. A silver cathde is suspended in a saturated slutin f AgBr. This half-cell is cnnected by a salt bridge t a standard hydrgen electrde, which is the ande. The ptential f this galvanic cell is vlt. Calculate the K sp f AgBr. Given E (Ag + /Ag) = 0.8 V K sp (AgBr) = {Ag + }{Br } Frm the disslutin reactin, AgBr(s) = Ag + + Br -, we cnclude that, at equilibrium, {Br } = {Ag + } K sp (AgBr) = {Ag + } 2 (e1-6.1) {Ag + } can be calculated frm the ptential f the half cell reactin Ag + + e = Ag, E(Ag + /Ag)= 0.44V. E(Ag + /Ag) = E (Ag + /Ag) lg 1 Ag + r, {Ag + } = 10 6 M (e1-6.2) Substituting e1-6.2 in e1-6.1, we get, K sp (AgBr) = 10 12