The Gibbs Free Energy and Cell Voltage

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1 The Gibbs Free Energy and Cell Vltage When an amunt f charge, Q, mves thrugh a ptential difference, E w = - Q E b/c wrk dne by the system E > 0 fr galvanic (vltaic) cells Recall, G = H TS = E + PV TS Fr cnstant T and P: G = E + P V - T S (usual case in electrchemical cells) But E = q + w Ttal wrk = P-V wrk + electrical wrk (w elect ) = - P V + w elect E = q + w = q + w elect - P V G = q + w elect - P V + P V - T S = = q + w elect - T S If the cnditin f reversibility is impsed upn the galvanic cells, q = q rev = T S Therefre, G = q rev + w elect - T S = T S + w elect - T S = w elect And w elect = - Q E If n mles f e (nf culmbs f charge) pass thrugh the external circuit f the galvanic cell when it is perated reversibly and if E is the reversible vltage, then G rxn = - Q E = - nf E (why G rxn? we ll prve it in a bit!) Electrical wrk is prduced when G rxn < 0 Standard States and Cell Vltages G rxn = - n F E, E (standard cell vltage): is the cell vltage in a galvanic cell in which reactants and prducts are in their standard states. (Gases at 1 bar, slutes in 1 M, metals in their pure stable states and at a specified temperature) A 6.0 Vlt battery delivers a steady current f 1.25 A fr a perid f 1.50 hurs. Calculate the ttal charge in culmbs that passes thrugh the circuit. Q = It = (1.25 C s -1 )(1.5 hurs)(3600 s hur -1 ) = 6750 C w elect = - Q E = - (6750 C)(6 V) = x 10 4 J (wrk dne n the battery) - ( x 10 4 J) = x 10 4 J wrk dne by the battery. 1

2 Half-Cell Ptentials It is a very lng and tedius jb t tabulate all the cnceivable galvanic cells We can avid the previus jb by tabulating the half-cell reductin ptentials, E These E values express the intrinsic tendency f a reductin half-reactin t ccur when the reactants and prducts are at standard states. E = E cathde E ande Hw E s are measured? Line ntatin f cell: S.H.E. Cd 2+ (aq A=1) Cd(s) Oxidatin ccurs in the SHE half-cell and this electrde is the ande: H 2 (g 1 bar) 2H + (aq,1m) + 2e 2H + (aq,1m) + 2e H 2 (g,1 bar) E (H 2 / H +, xid. = E (H + / H 2, red V Cnsider Cu 2+ (aq,1 M) Cu (s) half-cell cnnected t SHE. We measure Then we assign E fr Cu 2+ (aq,1 M) + 2e Cu (s) E = 0.34 V Nw cnsider Zn 2+ (aq,1 M) Zn (s) half-cell cnnected t SHE. We measure E = V. Then we assign E = V fr Zn 2+ (aq,1 M) + 2e Zn (s) Nw we build the cell: Zn 2+ (aq,1 M) Zn (s) Cu 2+ (aq,1 M) Cu (s) fr which E = E cathde E ande = 0.34 ( 0.76) = 1.10V (>0) - Cell is galvanic Half-cell ptentials are intensive prperties, namely independent f the amunt f the reacting species. 2

3 1. All values are relative t SHE ( = reference electrde) 2. Half-reactins are written as reductins (nly reactants are xidizing agents and nly prducts are the reducing agents) 3. The mre psitive the E the mre readily the reactin ccurs 4. Half-reactins are shwn with b/c each can ccur as reductin r xidatin 5. The half-cell that is listed higher at the table acts as the cathde Zn (s) + Cu 2+ (aq) Zn 2+ (aq) + Cu (s) strnger strnger weaker weaker reducing xidizing xidizing reducing agent agent agent agent Will Ag + xidize Zn (s) r will Zn 2+ xidize Ag (s) A vltaic cell huses the reactin between aqueus brmine and zinc metal: Br 2+ (aq) + Zn (s) Zn 2+ (aq) + 2Br - (aq) E cell = V Calculate Ebr min e given E zinc = V. The cell is vltaic E cell = V > 0 (reactin spntaneus) as written. Zn is being xidized and therefre it is the ande. E = E E E = E + E = ( 0.76) = 1.07 V = cell cathde ande cathde cell ande Ebr min e 3

4 A standard (Pt MnO 4, H +, Mn 2+ ) half-cell cnsists f an inert electrde in cntact with a slutin cntaining MnO 4 (aq), H + (aq) and Mn 2+ (aq) ins in standard states. Such a half-reactin is assembled and cnnected t a standard Zn 2+ (aq,1 M) Zn (s) half-cell/ a) Calculate E f the cell at 25 C. In Appendix E we find: (Rxn 1): MnO 4 (aq) + 8H + (aq) + 5e Mn 2+ (aq) + 4H 2 O (l) E = 1.507V (Rxn 2): Zn 2+ (aq) + 2e Zn (s) E = V Rxn 1 is the cathde because it lies abve E (Zn 2+ /Zn), S E = E E = ( 0.762) = V cathde ande b) Write a balanced equatin fr the verall reactin We multiply (Rxn 1) by 2 and (Rxn 2) by 5: 2MnO 4 (aq) + 16H + (aq) + 5Zn (s) 2Mn 2+ (aq) + 5Zn 2+ (aq) + 8H 2 O (l) Number f e invlved: 10 Nw check: G rxn = [2 G f (Mn 2+ (aq)) +5 G f (Zn 2+ (aq))+ 8 G f (H 2 O (l)) ] - [2 G f (MnO 4 (aq)) 16 G f (H + (aq)) +5 G f (Zn (s)) ] = kj E = G n F e rxn = x10 Jml 1 = 2.27JC 1 (10)(96,485Cml ) = 2.27V Very gd match!!! Disprprtinatin A single chemical species is bth xidized and reduced This species must be able t give up electrns and accept electrns and in additin the half-reactin in which it is reduced must lie higher in the table than the half-reactin in which it is xidized If this is the case then it drives the secnd reactin t g in reverse and disprprtinatin ccurs spntaneusly 4

5 (1): Cu + (aq) + e Cu (s) E = V (2): Cu 2+ (aq) + e Cu + (aq) E = 0.153V (1) is reductin and drives (2) as xidatin 2 Cu + (aq) Cu 2+ (aq) + Cu (s) E cell = = V (>0) Which means E rxn > 0 and G rxn < 0 spntaneus. Yes it disprprtinates. Decide whether Fe 2+ (aq) in its standard state at 25 C is stable w/r/t disprprtinatin. In Appendix E we find: (1): Fe 3+ (aq) + e Fe 2+ (aq) E = 0.771V (2): Fe 2+ (aq) + 2e Fe (s) E = V In rder fr Fe 2+ (aq) t disprprtinate reactin (1) wuld have t be driven backwards (xidatin): 3Fe 2+ Fe (s) + 2Fe 3+ (aq) And E = (+0.771) = V (< 0) and G rxn > 0 (nn-spntaneus) and therefre stable against disprprtinatin. Oxygen as an xidizing agent O 2 (g) + 4H + (aq) + 4e 2H 2 O (l) E = 1.229V gd xidizing agent! O 3 (g) + 2H + (aq) + 2e O 2 (g) + H 2 O (l) E = 2.067V even better xidizing agent! H 2 O 2 (aq) + 2H + (aq) + 2e 2H 2 O (l) E = 1.776V (bleach + germicider) O 2 (g) +2H + (aq) + 2e H 2 O 2 (aq) E = 0.695V H 2 O 2 as a reducing agent can reduce nly chemical species with reductin ptentials greater than V. Relative Reactivities f Metals Metals that can displace hydrgen frm acid Fe (s) + 2H + (aq) H 2 (g) + Fe 2+ (aq) Why des this rxn g? Fe (s) Fe 2+ (aq) + 2e E = V and 2H + (aq) + 2e H 2 (g) E = 0.0 (SHE) E cell = 0.00 ( ) = V (> 0) rxn will g! 5

6 Metals that cannt displace hydrgen frm acid 2Ag (s) + 2H + (aq) 2Ag + (aq) + H 2 (g) E = 0.80V 2 x [Ag (s) Ag + (aq) + e ] and 2H + (aq) + 2e H 2 (g) E = 0.0 (SHE) E cell = 0.00 (0.80) = V (< 0) rxn will nt g! Metals that can displace hydrgen frm water 2H 2 O (l) + 2e H 2 (g) + 2OH (aq) E = V Hwever, [OH ] = 10-7 nn-standard Metals active enugh t reduce H 2 O (l) lie belw ( 0.42 V) 2Na (s) + 2H 2 O (l) 2Na + (aq) + H 2 (g) + 2OH (aq) 2 x [Na (s) Na + (aq) + e ] E = V 2H 2 O (l) + 2e H 2 (g) + 2OH (aq) E = -0.42V, E cell = (-2.71) = 2.29 V Yes, Na can! Metals that displace ther metals in slutes Zn (s) + Fe 2+ (aq) Zn 2+ (aq) + Fe (s) Zn (s) Zn 2+ (aq) + 2e E = V Fe 2+ (aq) + 2e Fe (s) E = V E cell = (- 0.76) = 0.32 V Cncentratin Effects and the Nernst Equatin G = G + RT ln Q G = nfe G = nfe G = G + RT ln Q nfe = nfe + RT ln Q Divide bth sides nf = nfe nfe RT + ln Q nf nf nf RT E E Q nf E : standard reductin ptential R: J K -1 ml -1 = V C K -1 ml -1 F: Faraday s cnstant = 96,485 C ml -1 T: abslute temperature (K) n: number f electrns in the half-reactin Q: reactin qutient = ln Nernst Equatin (expresses the net driving frce fr a reactin) 6

7 Let us evaluate RT nf at 25 C (298.15K) 1 1 RT (. 8314JK ml )( K ) = = 1 nf n( 96, 485Cml ) n And the Nernst Equatin at K becmes, E = E ln Q n Recall that ln x = 2.303lg10 x and E = E lg 10 Q n Recall, at equilibrium G = 0 G = RTln K nfe = RTln K RT And E = ln K nf Let s see in depth! RT E = E ln Q nf When Q < 1 and thus [reactant] > [prduct], ln Q < 0, s E > E cell When Q = 1 and thus [reactant] = [prduct], ln Q = 0, s E = E cell When Q >1 and thus [reactant] < [prduct], ln Q > 0, s E < E cell The signs f G and E cell determine the directin f the reactin at standard cnditins. 7

8 Suppse that the Zn Zn 2+ (aq) MnO 4 (aq) Mn 2+ (aq) Pt cell is perated at ph 2.00 with [MnO 4 ]= 0.12 M, [Mn 2+ ] = M and [Zn 2+ ] = M. Calculate the cell vltage, E, at 25 C. The reactin f this cell is: 2MnO 4 (aq) + 16H + (aq) + 5Zn (s) 2Mn 2+ (aq) + 5Zn 2+ (aq) + 8H 2 O (l), E = 2.27V [ Mn ] [ Zn ] E = E ln Q = 2.27 ln [ MnO ] [ H ] (0.0010) (0.015) = 2.27 ln (0.12) (0.010) = 2.14V An electrchemical cell is set up at 25 C. One half-cell cnsists f a zinc ande immersed in a 1.0 M slutin f Zn(NO 3 ) 2. The secnd half-cell cnsists f a platinum cathde that has gaseus hydrgen bubbling ver it at a pressure f 1.00 atm in a slutin f unknwn hydrgen-in cncentratin. The bserved cell vltage is V. (a) Calculate the reactin qutient, Q Ande: Zn (s) Zn 2+ (aq) + 2e E = V Cathde: 2H + (aq) + 2e H 2 (g) E = 0.00 V (SHE) E = 0.00 (-0.762) = V The cell is nt at the standard state E = E Q = e RT nf 2 ln Q lnq = ( E E) = ( ) = nf RT x10 (b) Calculate the hydrgen-in cncentratin in the secnd half-cell Q = 2+ [ Zn ] P (1.00)(1.00) = = 5.8x10 [ H ] = 1.3x [ H ] [ H ] H 5 M Electrlytic Cells In galvanic (vltaic) cells electrns are generated at the ande (-) and they are cnsumed st the cathde (+) In an electrlytic cell, electrns cme frm an external pwer surce, which supplies t the cathde and remves them frm the ande. Cathde (-), Ande (+) 8

9 The tin-cpper reactin as the basis f a vltaic and an electrlytic cell Cncentratin Cells 9

10 In these cells the half-reactins are the same but the cncentratins are different. A cncentratin cell based n the Cu/Cu 2+ half-reactin The easiest is ph. We cnstruct a cell with cathde t be SHE and the unknwn has the same apparatus dipping int an unknwn [H + ] slutin. Oxidatin: H 2 (g 1 atm) 2H + (aq, unknwn) + 2e Reductin: 2H + (aq, 1M) + 2e H 2 (g 1 atm) Overall: 2H + (aq, 1M) 2H + (aq, unknwn) [ H ] umknwn Ecell = Ecell lg = 0.00 lg[ H ] = lg[ H [ H ] s tan dard 2 By measuring the cell ptential we can get [H + ] and cnsequently ph + ] ph Meter 10

11 A cncentratin cell cnsists f tw Ag/Ag + half-cells. In half-cell A, electrde A dips int M AgNO 3 ; in half-cell B, electrde B dips int 4.0 x 10-4 M AgNO 3. What is the cell ptential at K? Which electrde has a psitive charge? Ag + (aq, M) [half-cell A] Ag + (aq 4.0 x 10-4 M) [half-cell B] [ Ag ] B x10 Ecell = Ecell lg = 0.00 lg = V + 1 [ Ag ] A Reductin ccurs at the cathde electrde A, s electrde A has a psitive charge. 11