Nerst Equatio Skills to develop Eplai ad distiguish the cell potetial ad stadard cell potetial. Calculate cell potetials from kow coditios (Nerst Equatio). Calculate the equilibrium costat from cell potetials. Nerst Equatio Electrochemistry deals with cell potetial as well as eergy of chemical reactios. The eergy of a chemical system drives the charges to move, ad the drivig force give rise to the cell potetial of a system called galvaic cell. The eergy aspect is also related to the chemical equilibrium. All these relatioships are tied together i the cocept of Nearst equatio. Walther H. Nerst (1864-1941) received the Nobel prize i 1920 "i recogitio of his work i thermochemistry". His cotributio to chemical thermodyamics led to the well kow equatio correlatig chemical eergy ad the electric potetial of a galvaic cell or battery. Electric Work ad Gibb's Free Eergy Eergy takes may forms: mechaical work (potetial ad kietic eergy), heat, radiatio (photos), chemical eergy, uclear eergy (mass), ad electric eergy. A summary is give regardig the evaluatio of electric eergy, as this is related to electrochemistry. Electric Work Eergy drives all chages icludig chemical reactios. I a redo reactio, the eergy released i a reactio due to movemet of charged particles give rise to a potetial differece. The maimum potetial differec is called the electromotive force, (EMF), E ad the maimum electric work W is the product of charge q i Coulomb (C), ad the potetial E i Volt (= J / C) or EMF. W J = q E C J/C (uits) Note that the EMF E is determied by the ature of the reactats ad electrolytes, ot by the size of the cell or amouts of material i it. The amout of reactats is proportioal to the charge ad available eergy of the galvaic cell. Gibb's Free Eergy The Gibb's free eergy G is the egative value of maimum electric work, G = - W = - q E A redo reactio equatio represets defiite amouts of reactats i the formatio of also defiite amouts of products. The umber () of electros i such a reactio equatio, is related to the amout of charge trasferred whe the reactio is completed. Sice each mole of electro has a charge of 96485 C (kow as the Faraday's costat, F), q = F ad, G = - F E At stadard coditios, G = - F E
The Geeral Nerst Equatio The geeral Nerst equatio correlates the Gibb's Free Eergy G ad the EMF of a chemical system kow as the galvaic cell. For the reactio a A + b B = c C + d D ad [C] c [D] d Q = --------- [A] a [B] b It has bee show that G = G + R T l Q ad G = - F E. Therefore - F E = - F E + R T l Q where R, T, Q ad F are the gas costat (8.314 J mol -1 K -1 ), temperature (i K), reactio quotiet, ad Faraday costat (96485 C) respectively. Thus, we have R T [C] c [D] d E = E - ----- l --------- F [A] a [B] b This is kow as the Nerst equatio. The equatio allows us to calculate the cell potetial of ay galvaic cell for ay cocetratios. Some eamples are give i the et sectio to illustrate its applicatio. It is iterestig to ote the relatioship betwee equilibrium ad the Gibb's free eergy at this poit. Whe a system is at equilibrium, E = 0, ad Q eq = K. Therefore, we have, R T [C] c [D] d E = ----- l ---------, F [A] a [B] b (for equilibrium cocetratios) Thus, the equilibrium costat ad E are related. The Nerst Equatio at 298 K At ay specific temperature, the Nerst equatio derived above ca be reduced ito a simple form. For eample, at the stadard coditio of 298 K (25 ) ad log(10) coversio, the Nerst equatio becomes 0.0592 V [C] c [D] d E = E - --------- log --------- [A] a [B] b See last page for how this umber is derived!
For the cell Z Z 2+ H + H 2 Pt we have a et chemical reactio of Z(s) + 2 H + = Z 2+ + H 2 (g) ad the stadard cell potetial E = 0.763. If the cocetratios of the ios are ot 1.0 M, ad the H 2 pressure is ot 1.0 atm, the the cell potetial E may be calculated usig the Nerst equatio: 0.0592 V P(H 2 ) [Z 2+ ] E = E - ------- log ------------ [H + ] 2 with = 2 i this case, because the reactio ivolves 2 electros. The umerical value is 0.0592 oly whe T = 298 K. This costat is temperature depedet. Note that the reactivity of the solid Z is take as 1. If the H 2 pressure is 1 atm, the term P(H 2 ) may also be omitted. The epressio for the argumet of the log fuctio follows the same rules as those for the epressio of equilibrium costats ad reactio quotiets. Ideed, the argumet for the log fuctio is the epressio for the equilibrium costat K, or reactio quotiet Q. Whe a cell is at equilibrium, E = 0.00 ad the epressio becomes a equilibrium costat K, which bears the followig relatioship: E log K = -------- 0.0592 where E is the differece of stadard potetials of the half cells ivolved. A battery cotaiig ay voltage is ot at equilibrium. The Nerst equatio also idicates that you ca build a battery simply by usig the same material for both cells, but by usig differet cocetratios. Cells of this type are called cocetratio cells. Eample 1 Calculate the EMF of the cell Z(s) Z 2+ (0.024 M) Z 2+ (2.4 M) Z(s) Solutio Z 2+ (2.4 M) + 2 e = Z Reductio Z = Z 2+ (0.024 M) + 2 e Oidatio -------------------------------------------- Z 2+ (2.4 M) = Z 2+ (0.024 M), E = 0.00 - - Net reactio
Usig the Nerst equatio: 0.0592 (0.024) E = 0.00 - ------- log -------- 2 (2.4) = (-0.296)(-2.0) = 0.0592 V Discussio Uderstadably, the Z 2+ ios try to move from the cocetrated half cell to a dilute solutio. That drivig force gives rise to 0.0592 V. From here, you ca also calculate the eergy of dilutio. If you write the equatio i the reverse directio, Z 2+ (0.024 M) = Z 2+ (2.4 M), its voltage will be -0.0592 V. At equilibrium cocetratios i the two half cells will have to be equal, i which case the voltage will be zero. Eample 2 Show that the voltage of a electric cell is uaffected by multiplyig the reactio equatio by a positive umber. Solutio Assume that you have the cell Mg Mg 2+ Ag + Ag ad the reactio is: Mg + 2 Ag + = Mg 2+ + 2 Ag Usig the Nerst equatio 0.0592 [Mg 2+ ] E = E - ------ log -------- 2 [Ag + ] 2 If you multiply the equatio of reactio by 2, you will have 2 Mg + 4 Ag + = 2 Mg 2+ + 4 Ag Note that there are 4 electros ivolved i this equatio, ad = 4 i the Nerst equatio: 0.0592 [Mg 2+ ] 2 E = E - ------ log -------- 4 [Ag + ] 4 which ca be simplified as 0.0592 [Mg 2+ ] E = E - ------ log -------- 2 [Ag + ] 2 Thus, the cell potetial E is ot affected.
Eample 3 The stadard cell potetial de for the reactio Fe + Z 2+ = Z + Fe 2+ is -0.353 V. If a piece of iro is placed i a 1 M Z 2+ solutio, what is the equilibrium cocetratio of Fe 2+? Solutio The equilibrium costat K may be calculated usig ( E )/0.0592 K = 10 = 10-11.93 = 1.210-12 = [Fe 2+ ]/[Z 2+ ]. Sice [Z 2+ ] = 1 M, it is evidet that [Fe 2+ ] = 1.2E-12 M. Eample 4 From the stadard cell potetials, calculate the solubility product for the followig reactio: AgCl = Ag + + Cl - Solutio There are Ag + ad AgCl ivolved i the reactio, ad from the table of stadard reductio potetials, you will fid: AgCl + e = Ag + Cl -, E = 0.2223 V - - - -(1) Sice this equatio does ot cotai the species Ag +, you eed, Ag + + e = Ag, E = 0.799 V - - - - - - (2) Subtractig (2) from (1) leads to, AgCl = Ag + + Cl -... E = - 0.577 Let K sp be the solubility product, ad employ the Nerst equatio, log K sp = (-0.577) / (0.0592) = -9.75 K sp = 10-9.75 = 1.810-10 This is the value that you have bee usig i past tutorials. Now, you kow that K sp is ot always measured from its solubility. Cofidece Buildig Questios I the lead storage battery, Pb PbSO 4 H 2 SO 4 PbSO 4, PbO 2 Pb would the voltage chage if you chaged the cocetratio of H 2 SO 4? (yes/o) Aswer... Yes! Hit... The et cell reactio is Pb + PbO 2 + 2 HSO 4 - + 2 H + 2 PbSO4 + 2 H 2 O
- Pb + PbO 2 + 2 HSO 4 + 2 H + 2 PbSO4 + 2 H 2 O ad the Nerst equatio - E = E - (0.0592/2)log{1/{[HSO 4 ] 2 [H + ] 2 }}. Choose the correct Nerst equatio for the cell Z(s) Z 2+ Cu 2+ Cu(s). A. E = E - 0.0296 log([z 2+ ] / [Cu 2+ ]) B. E = E - 0.0296 log([cu 2+ ] / [Z 2+ ]) C. E = E - 0.0296 log(z / Cu) D. E = E - 0.0296 log(cu / Z) Aswer... A Hit... The cell as writte has Reductio o the Right: Cu 2+ + 2 e = Cu oidatio o the left: Z = Z 2+ + 2 e Net reactio of cell is Z (s) + Cu 2+ = Cu (s) + Z 2+ The stadard cell potetial E is 1.100 V for the cell, Z(s) Z 2+ Cu 2+ Cu(s). If [Z 2+ ] = 0.01 M, ad [Cu 2+ ] = 1.0 M, what is E or EMF? Aswer... 1.159 V Hit... A likely wrog result is 1.041 V. The term that modifies E is -(0.059/)log{[Z 2+ ]/[Cu 2+ ]} ( = 2 i this case). Uderstadably, if the cocetratio of Z 2+ is low, there is more tedecy for the reactio, Z = Z 2+ + 2 e. The logarithm of the equilibrium costat, log K, of the et cell reactio of the cell Z(s) Z 2+ Cu 2+ Cu(s)... E = 1.100 V is A. 1.100 / 0.0291 B. -1.10 / 0.0291 C. 0.0291 / 1.100 D. -0.0291 / 1.100 E. 1.100 / 0.0592 Aswer... A Hit... Use the Nerst equatio i the form 0 = 1.100-0.0296 log ([Z 2+ ] / [Cu 2+ ]) The Nerst equatio is useful for the determiatio of equilibrium costats. Uderstadig is the key. Take time to uderstad it, there is o poit i rushig. cchieh@uwaterloo.ca
R T F R: Uiversal Gas Costat (Joules/(Mol DegreesKelvi)) T: Temperature, (DegreesKelvi) F: Faraday costat (Coulomb/Mol) : Moles of Electros 1 Joule=1 Coulomb Volt Joules Mol DegreesKelvi DegreesKelvi 1 Coulombs Mol Joules Mol DegreesKelvi DegreesKelvi 1 Mol Coulombs Coulomb Volt Mol * DegreesKelvi DegreesKelvi 1 Mol Coulombs Volts ElectroMoles Add i rest of equatio 8.31447 298.15 At 25C (298.15K): 0.025693 = 96485 l (ReactioQuotiet) 0.025693 l(10) log (ReactioQuotiet) 0.05915 log (ReactioQuotiet)