Fundamentals of Electrochemistry CHEM*7234 CHEM 720 Assignment #1 ANSWERS. [ Ag + Ê Á =

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1 Fundamentals f Electrchemistry CHEM*7234 CHEM 720 Assignment #1 ANSWERS 1. (a Ande Reactin Cd Æ Cd 2 2e Cathde Reactin Ag e Æ Ag E = V E = V Overall Reactin: Cd 2Ag Æ Cd 2 2Ag One apprach is t first determine the standardcell ptential E cell = E cathde E ande = (0.403 = V Then we apply the Nernst equatin t the verall reactin. E cell = E cell = V E ande = E ande Ê ln RT nf ln Ê = V [ ] Cd2 [ Ag ] 2 a Cd a Cd 2 = Ê ln ( An equivalent prcedure is t determine ptentials fr the ande and cathde separately, and then cmbine them t find the cell ptential. E cathde = E cathde = V RT Ê nf ln a Ag a Ag = lnê = lnê E cell = E cathde E ande = ( = V

2 (b As we have nly been given the standard cell ptential at this temperature, we nly have the ptin f applying the Nernst Equatin t the cmplete cell reactin and btain E cell = E cell = V Cd 2 [ ] RT Ê 2F ln [ Ag ] 2 = ( ( 323 2( ln Ê ( By fllwing the lead f the given paper, yu shuld have been able t develp the ideas fr this specific situatin. Cu ( s1 Ag ( s AgBr ( s KBr, H 2 O Hg ( l Cu ( s2 This is the specific cell. The cpper n each end is meant t represent the cpper wires inside the ptential measuring instrument. This is, f curse, the place where the ptential difference is actually measured. Frm an electrchemical pint f view, the cell cnsists f a silversilver brmide reference electrde at the ande and a mercury electrde at the cathde. An aqueus slutin f KBr is the electrlyte. We start with the Gibbs adsrptin istherm fr the metalslutin interface. dg = charged species  G i dm i  G i dm i i uncharged species This indicates that changes in the surface energy arise frm changes in the chemical ptential f the varius charged(using electrchemical ptential and uncharged species (using chemical ptential, accrding t their Gibbs excess. We nw write ut this expressin fr the specific instance f this cell. We need t remember that we must als treat the electrns in the electrde explicitly t. dg = G Hg dm Hg G e dm e G K dm K G Br dm Br G H2 O dm H 2 O i Nw, by arguments f equilibrium we can develp a relatinship between the electrchemical ptential f electrns in the cpper wire and the electrchemical ptential f ins at the interface. Here is the develpment fr Hg ins. ( = me Hg m e ( Hg m Hg = m Hg m e = mhg m e m Hg = m Hg m e

3 3 The electrchemical ptential f electrns in tw metals are the same when the tw metals are in cntact with each ther and the system is at equilibrium. ( = me Hg m e Here is the relatinship between the chemical ptential f an uncharged species and the electrchemical ptential f its in and electrn arising frm the inizatin. Cu ( s2 Hg m Hg = m Hg m e m Hg = m Hg m e ( = mhg m e With these expressins we can rewrite the first tw terms in the main equatin. ( Cu ( G Hg dm Hg G e dm s2 È Cu ( e = GHg d m Hg m s2 e = G Hg dm Hg G Hg dm e = G Hg dm Hg Ê G e G Hg ( ( The charge density n the electrde is related t the difference in Gibbs excess at the interface (F relates culmbs and mles s M = FÊ G Hg G e We substitute int the abve expressin and we bserve that the chemical ptential f elemental mercury is fixed and cnstant. Hence, its differential is 0. Therefre dm Hg = 0 We nw lk at the next tw terms in the riginal expressin. First we nte a fundamental relatinship between a mlecule and its ins. Í Î Cu ( s2 Ge dm e ( ( dm Cu s2 e G Hg dm Hg G e dm e ( \ s M F = G e G Hg G e dm e s = M F dm Cu ( s2 e ( m KBr = m K m Br \ m K = m KBr m Br

4 Nw, because there must be equilibrium at the reference electrde interface, where the reactin AgBr(s e Æ Ag(s Br is ccuring, then this implies the fllwing relatinship: 4 m AgBr m e ( Cu ( s1 = mag m Br And since d m AgBr = dm Ag = 0 d m e ( Cu ( s1 = d mbr This allws us t write the fllwing: G K dm K G Br dm Br = G K d m KBr m Br = G K dm KBr G K dm Br G Br dm Br = G K dm KBr ( G Br G K dm e Since the interfacial charge density arises frm this difference in Gibbs excess, we have s M = F( G Br G K S ur riginal expressin nw takes n the frm [ ] G Br dm Br ( Cu ( s1 ( ( G dm G dm = G dm K K Br Br K KBr s M F dm Cu s1 e dg = s M F dm Cu s2 e = s M F È Í Î ( ( GK dm KBr s M ( dme dm Cu ( s1 e F dm e Cu s1 ( ( GH2 O dm H2 O G K dm KBr G H2 O dm H 2 O Since we can assume that the tw cpper wires have the same cmpsitin, their electrchemical ptential differences is just the ptential difference between the tw wires. This means dm e ( Cu ( s1 Cu ( s2 dme ( = FdE dg = s M de G K dm KBr G H2 O dm H2 O

5 5 We nw call in the GibbsDuhem equatin. We d this because the chemical ptential f the water and the KBr as linked. Yu can t change ne withut changing the ther; yu increase KBr cncentratin nly by decreasing H 2 O cncentratin. We btain X KBr dm KBr X H2 O dm H 2 O = 0 dm H2 O = X KBr X H2 O dm KBr Hence, we can rewrite the equatin in the final frm we seek. Ê dg = s M de G K X KBr G X H2 O H2 O dm KBr

6 3 (a The charge density n the metal is balanced by the charges brught t the interface by the anins and catins. The charge, Z, n each species must be crrectly accunted fr, alng with the fact that the excess electrde charge density is f electrns (negatively charged. With tw values we can calculate the third. s M (µc/cm 2 ZFG K (µc/cm 2 ZFG I (µc/cm s M = z F G K z F G I z FG I = s M FG K Using this we can calculate the Gibbs excess charge density fr the idide ins. The data is given in the table belw and the K and I Gibbs excess are graphed in the accmpanying figure. In the indicated regin f the graph, the idide in is superequivalent. If yu calculate the slpe f the curve in this regin, yu will find it t be abut This means that when there is change in the electrde charge density, there is a 27% greater change in the idide Gibbs excess. There is a greaterthanequivalent change in the idide Gibbs excess. Hence, it is superequivalent. This pints t the fact that the idide is being specifically adsrbed n the electrde at these ptentials. A similar examinatin f the ptassium Gibbs excess at psitive ptentials will shw that it is NOT specifically adsrbed. 6

7 3(b. The differential capacity is determined by finding the derivative f the E vs charge density curve at every pint. We can apprximate this derivative by taking differences. We d this and assign the capacity t the midpint f the f the tw ptential values used in each calculatin. This is quite easily dne with a spreadsheet prgram. The results fr the three sets f data are given in the table belw and graphed n the accmpanying curve. The pzc is at the minimum f the lw cncentratin differential capacity curve: abut 0.5 V. The duble layer capacitance, C DL, is related t the capacitance f the Helmhltz layer, C H, and that f the diffuse layer, C D, as the sum f reciprcals. Away frm the pzc, the diffuse layer capacity is largest and the verall capacity is cntrlled by that f the Helmhltz layer. Hwever, in dilute slutins, the diffuse layer capacitance drps cnsiderably near the pzc s that the duble layer capacitance is nw determined by it. This is the cause f the dip in the differential capacity curve and why it nly happens in dilute slutins. 1 C DL = 1 C H 1 C D E(V C(0.916 C(0.100 C( [NaF] = M [NaF] = M [NaF] = M