Chemistry Problem Set 6 Answers Professor: C. E. Loader. = 4.0 x 10 3 mol L JK 1 mol 1 x K.

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1 Problem Set 6 Answers Professor: C. E. Loader 1. For the cell at 25 ºC: Zn(s) Zn 2+ (4.0 x 10 mol L 1, aq) Cd 2+ (0.20 mol L 1, aq) Cd(s) (a) Write the cell reaction. (b) Calculate the cell voltage. (c) Calculate the equilibrium constant for the cell reaction (a) banode: Zn(s) Zn e E o o = v cathode: Cd e Cd(s) E r o = 0.40 v cell reaction: Zn(s) + Cd 2+ (aq) Zn 2+ (aq) + Cd(s) E o = V (b) E = E o F ln Q, Q = E = E o 2F ln Q (c) [Zn 2+ ] [Cd 2+ ] = 4.0 x 10 mol L mol L JK 1 mol 1 x K E = 0.60 ln (2.0 x 10 2 ) 2 x C mol 1 E = ( ) v = volts at equilibrium, E = 0, Q = K E o = + F ln K = 0.60 v ln K = 28.0, K = 1.49 x Explain why the E o of the half cell reaction: Cu(s) Cu e, does not equal the sum of the E o 's for Cu(s) Cu 2+ + e and Cu 2+ Cu + + e. Comparing the two paths Path a. Cu Cu e E o = 0.40 v Path b. Cu Cu + + e o E 1 = v Cu + Cu 2+ + e o E 2 = 0.15 v If G o is independent of path, then clearly the E o 's are not additive since path a. involves 2 electrons while path b. involves 1 electron. The G o 's, however, are the same for the two paths (within experimental error) Path a. G o = nfe o = (2)(96500 C mol 1 )( 0.40 V) = 65.6 kj mol 1 Path b. G o o o = nf( E + ) = (1)(96500 C mol 1 )( V) = 64.8 kj mol 1 1 E 2 Page 1 of 8 printed pages

2 The difference reflects the E o uncertainty in tables.. If the cell: Zn(s) ZnSO 4(aq, 1.00 mol L 1 ) CuSO 4(aq, mol L 1 ) Cu(s) were allowed to completely discharge at 25 ºC, what would be the final concentrations of ZnSO 4 and CuSO 4? (a) anode: Zn(s) Zn e E o = v cathode: Cu e Cu(s) E o = 0.40 v cell reaction: Zn(s) + Cu 2+ (aq) Zn 2+ + Cu(s) E o = 1.00 v F ln [Zn2+ ] [Cu 2+ ] E = O = E o at equilibrium [Zn or K = 2+ ] = exp( Eo F = 8.2 x 10 [Cu 2+ ] ) Let the starting (initial) concentrations be c Zn 2+, c Cu 2+ Then from stoichiometry of cell reaction [Cu 2+ ] = c Cu 2+ x [Zn 2+ ] = + x c Zn 2+ and let x = concentration which has reacted. c Zn 2+ + x c = 8.2 x 10 = Cu 2+ x [Zn 2+ ] [Cu 2+ ] Solving for x; see that since K is so large (so [Cu 2+ ] must be very small) that x { so that [Zn 2+ ] = , then solve for [Cu 2+ ] c Zn mol L 1 K [Cu 2+ ] = = 1.82 x 10 4 mol L 1 [Zn 2+ ] = 1.50 mol L 1 The reaction will appear to go to completion. 4. From the cell Ag(s) AgCl(s) Cl (aq) Ag + (aq) Ag(s) and appropriate E o data, calculate the solubility product of AgCl(s) at 25 ºC. Ag(s) + Cl (aq) AgCl(s) + e E o = v Ag + (aq) + e Ag(s) E o = v Ag + (aq) + Cl (aq) AgCl(s) o E cell = v E o = F ln a AgCl(s) concentrations) = [Ag + ][Cl ] F ln 1 K sp (since E = 0 at equilibrium, [Ag + ], Cl ] are equilibrium where since a AgCl(s) 1, K sp = [Ag+][Cl ] Page 2 of 8 printed pages

3 ln K sp = FEo = (1)(96500 Cl mol 1 ) (.5766 v) = J K 1 mol 1 x K K sp = 1.78 x (mol L 1 ) 2 5. Given that the cell below has a voltage of volts at 25 ºC, calculate the unknown [Cl ]. Pt Cl 2 (g, 1.00 atm) Cl (aq, c = mol L 1 ) Cl (x mol L 1 ) Cl 2 (1.0 atm Pt) anode: 2 Cl (0.010) Cl 2 (g, 1 atm) + 2 e cathode: Cl 2 (1.0 atm) + 2 e 2 Cl (a) 2 Cl (.010) + Cl 2 (1.0 atm) Cl 2 (1 atm) + 2 Cl o (x) E cell = 0 E = E o F ln Q ln Q = FE = (2)(96500 C mol 1 ) ( v) = J K 1 mol K Q = 2.05 ButQ = 2.05 = (1 atm) x 2 (1. atm) (.010 mol L 1 ) 2 x = 5.47 x 10 2 mol L 1 6. The following reduction potentials (25 ºC) are available for iodine containing species present in iodine titrations: I 2(s) + 2 e - î 2 I - (aq) E = V I 2(aq) + 2 e - î 2 I - (aq) E = V I (aq) + 2 e - î I - (aq) E = V (a) Calculate the equilibrium constant for the reactions (i) I 2(s) + I - (aq) î I (aq) I 2(s) + 2 e - 2 I - (aq); E = V I - (aq) I (aq) + 2 e - ; E = V I 2(s) + I - (aq) I (aq); E = V so K = 0 cell (ii) I 2(aq) + I - (aq) î I (aq) Page of 8 printed pages

4 I 2(aq) + 2 e - 2 I - (aq); E = V I - (aq) I (aq) + 2 e - ; E = V I 2(aq) + I - (aq) I (aq); E = V o E cell = F = lnk K = 6.9 x x x ln (b) Calculate the solubility in g L -1 of I 2(s) in water I 2(s) + 2 e - 2 I - (aq); K cell E = V 2 I - (aq) I 2(aq) + 2 e - ; E = V I 2(s) I 2(aq); E = V o E cell = F = lnk 8.14 x x ln K K = 1.4 x 10 - = [I 2(aq)] [I 2 (aq)] o so [I = 1.4 x 10 - mol L -1 2 (aq)] or, in g/l = 1.4 x 10 - mol L -1 x 25.8 g mol -1 = 0.55 g/l cell Note how insoluble iodine is - barely enough to colour the water. 7. Determine the EMF at 298 K for Cd(s) Cd 2+ (0.020 mol L -1 ),KCN (0.090 mol L -1 ) KCl(0.015 mol L -1 ) AgCl(s) Ag(s) Given Cd CN î Cd(CN) 4 K ASS = 1. x This question involves a complex ion equilibrium. When the solutions in the left-hand electrode are mixed complexation of the cadmium ion will take place according to the above equation. Half Cell reactions Cd(s) t Cd 2+ (aq) + 2 e - E o = 0.40 V 2 e AgCl(s) t 2 Ag(s) + 2 Cl - (aq) E o = 0.40 V Cd(s) + 2 AgCl(s) t Cd 2+ (aq) + 2 Ag(s) + 2 Cl - (aq) E o cell = V Sine the solutions are fairly dilute we will assume that the activities are approximately equal to the molar concentrations. The complexation of the cadmium ion with the cyanide ion will lower the concentration of free cadmium ion in the solution and therefore affect the E o (the equilibria are connected through the Page 4 of 8 printed pages

5 casdmium ion concentration). Since K ASS is so large it is reasonable to assume complete reaction so the concentration after mixing are Cd 2+ (aq, very << 0.20 mol L -1 ) + 4 CN - (aq, (4 x 0.20 mol L -1 ) t Cd(CN) 4 2- (aq, 0.20 mol L -1 ) [Cd(CN) K ASS = 2 = 1. x (aq)] [Cd 2+ (aq)][cn (aq)] 4 and [Cd 2+ (aq)] = so the assumption is valid. [Cd(CN) 2 4 (aq)] (1. x )[CN(aq)] = (0.020) 4 (1. x ) x (0.010) = 1.54 x mol 1 Now, E cell = E o cell V - V and E cell = 1.05 V :Ans F ln[cd2+ (aq)][cl (aq)] 2 = 7. (a) Determine the EMF of the cell Sn(s) SnSO 4(aq, M) Na 2SO 4(aq, 0.00 M) PbSO 4(s) Pb(s) Sn(s) t Sn 2+ (aq) + 2 e - 2 x ln[(1.54 x ) x (0.015) 2 ] E o = V 2 e - + PbSO 4(s) t Pb(s) + SO 4 2- (aq) E o = V Sn(s) + PbSO 4(s) t Pb(s) + SO 4 2- (aq) + Sn 2+ (aq) E o = V The concentration of the solutions in the cell are not standard but can be assumed to be close to the molality and the activity of the solutions. Using the Nernst equation E cell = E o cell - F lnq = x ln(0.010 x 0.00) = V:Ans (b) When this cell is discharged what is [Sn 2+ ]? When the cell is discharged the voltage drops to 0 and E cell = 0 and the cell is at equilibrium so Q = K eq. 0 = E o cell - F ln K eq = x ln K eq = V so K eq =.7 x 10-8 and the reaction is spontaneous in the reverse direction. Since the equilibrium constant is so small hardly any Sn 2+ ion is present at equilibrium and in this cell will be used up and therefore also using up and equivalent amount of sulfate ion K eq =.7 x 10-8 = [Sn 2+ (aq)][so 4 2- (aq)] = [Sn 2+ (aq)]( ) so [Sn 2+ (aq)] =.7x mol L 1 ) = 1.69 x 10-6 mol L -1 :Ans Note: [Sn 2+ (aq)] is << mol L -1 and the assumption is valid. 8. The EMF of the cell Pt Hg(l) Hg 2+ (aq), KCN(aq) Fe + (aq), Fe 2+ (aq) Pt is E = V when [KCN(aq)] = mol L 1, [Hg 2+ (aq)] =.4 x 10 mol L 1 ; [Fe + (aq)] = mol L 1 ; [Fe 2+ (aq)] = mol L 1. Page 5 of 8 printed pages

6 Calculate K ASS for Hg 2+ (aq) + 4 CN (aq) î Hg(CN) 4 (aq).910 Cell reaction Hg(l) t Hg 2+ (aq) + 2 e - E o = V 2 Fe + (aq) + 2 e - t 2 Fe 2+ (aq) E o = V 2 Fe + (aq) + Hg(l) t 2 Fe 2+ (aq) + Hg 2+ (aq) E o cell = V This reaction involves the formation of a complex ion Hg(CN) 4 (aq) and it is the equilibrium associated with this that controls the Hg 2+ (aq) ion concentration. K ASS is [Hg(CN) the association constant for the equilibrium so K ASS = 4 ] [Hg and the true 2+ (aq)][cn (aq)] 4 [Hg 2+ (aq)] in the cell will not be that stated (as the solutions were made) but the value dictated by this equilibrium and can be calculated from the cell potential. So the real concentrations in the left-hand cell are: E cell = E o cell - F lnq = x ln [Fe2+ (aq)] 2 [Hg 2+ (aq)] [Fe + (aq)] 2 2 x ln (0.060)2 [Hg 2+ (aq)] (0.020) V = V solving gives [Hg 2+ (aq)] = 6.07 x 10-7 mol L -1. Clearly the Hg 2+ (aq) reacts almost completely and the concentrations are as shown. Hg 2+ (aq, 5.99 x 10-7 mol L -1 ) + 4 CN (aq, (4 x 0.004) mol L -1 ) Hg(CN) 4 (aq, x 10-7 mol L -1 ) Hg 2+ (aq, 5.99 x 10-7 mol L -1 ) + 4 CN (aq, mol L -1 ) Hg(CN) 4 (aq, mol L -1 ) [Hg(CN) K ASS = 4 ] (0.004) = =.4 x [Hg :Ans 2+ (aq)][cn (aq)] 4 (6.07 x 10 7 )(0.0114) The cell has E CELL = V at 25 ºC. Ag(s) AgCl(s) Cl (aq, mol L 1 ) (Ag +, mol L 1 ) Ag(s) When mol L 1 NH is added to the mol L 1 Ag +, E CELL falls to V. (ie. One litre of solution contains mol NH and mol of Ag+). (a) Calculate K sp for AgCl(s). Taking the initial cell the cell reaction is Ag(s) + Cl - (aq) t AgCl(s) + e - e - + Ag + (aq) t Ag(s) Ag + (aq) + Cl - (aq) t AgCl(s) measured E cell = V The cell is effectively a concentration cell with the silver ion concentration on the LHS dictated by the solubility product for AgCl(s). So the above can be rewritten as Page 6 of 8 printed pages

7 LHS Ag(s) t Ag + (aq, mol L -1 ) + e - RHS e - + Ag + (aq, mol L -1 ) t Ag(s) Ag + (aq, mol L -1 ) t Ag + (aq, LHS mol L -1 ) measured E cell = and the solubility product expression is AgCl(s) t Ag + (aq) + Cl - (aq); K sp = [Ag + (aq)][cl - (aq)] so [Ag + (aq)] = so = 0 V - K sp = 1.8 x :Ans Alternatively, For the cell E cell = E o cell - F ln( K 1 x ln ks x [Ag + (aq)][cl (aq)] ) 1 1 x ln x = E o cell - giving E o cell = V E o = F lnk cell reaction so = 1 x ln K cell reaction K = x 10 9 cell reaction but K cell reaction = so K sp =1.8 x :Ans (b) Calculate K STAB for the reaction Ag + (aq) + 2 NH (aq) Ag(NH ) [Ag + (aq)][cl (aq)] = 1 K sp] Ksp [Cl (aq)] The concentration of Ag + (aq) after the ammonia has been added can be calculated as before = 0 V - Ksp x ln [Ag + (aq)] RHS giving [Ag + (aq)] =7.5 x 10-8 mol L -1 RHS The silver ion reacts almost completely with the ammonia so we can get the other concentrations: Ag + (aq) + 2 NH (aq) Ag(NH ) x 10-8 mol L (2 x 0.100) = 0.00 mol L mol L -1 K stab = [Ag(NH ) 2 + (aq)] [Ag + (aq)][nh ] 2 = (7.5 x 10 8 ) x (0.00) 2 = 1.5 x 10 7 :Ans (c) Calculate the solubility of AgCl(s) in mol L 1 NH. The solubility is governed by the solubility product and the stability constant. The silver ion concentration is common to both. Ag + (aq) + 2 NH (aq) Ag(NH ) 2+ (aq); K stab = 1.57 x 10 7 AgCl(s) t Ag + (aq) + Cl - (aq); K sp = 1.75 x adding gives Page 7 of 8 printed pages

8 AgCl(s) + 2 NH (aq) Ag(NH ) 2+ (aq) + Cl - (aq); K overall = (1.57 x 10 7 ) x (1.75 x ) = 2.75 x 10 - Now if x mol L -1 of AgCl dissolves we have [NH (aq)] = ( x); [Ag(NH ) 2+ (aq)] = x; [Cl - (aq)] = x. Now, K overall = [Ag(NH ) 2 + (aq)][cl (aq)] [NH (aq)] 2 = taking the square root of both sides meaningful). This gives x =.29 x 10 - mol L -1 x 2 ( x) 2 = 2.75 x 10 x ( x) = (only the +ve answer is Solubility of AgCl in mol L -1 ammonia solution =. x 10 - mol L -1 : Ans Or, if x mol L -1 of AgCl dissolves the answer to (b) shows that most of the silver ion in solution will be in the form of the complex ion (K stab = x 10 7 ) so that the [Cl - (aq)] concentration is equal to x and [Ag(NH ) 2+ (aq)] = x - [Ag + (aq)] or Ksp [Cl (aq)] approximately x since K stab is large. So the remaining [Ag + (aq)] = = [Ag(NH ) + 2 (aq)] [Ag + (aq)][nh (aq)] = 2 x x ( x) 2 Now K stab = = 1.51 x 10 7 This gives x =.29 x 10 - mol L -1 as above. 1.75x10 10 x (d) Calculate the solubility of AgCl(s) in a solution that is mol L 1 in NH and mol L 1 in KCl. Using the same overall equilibrium as in part (c), the only change is that [Cl - (aq)] = ( x). x( x) 1.75 x ( x) 2 x(0.070) 1.75 x (0.070) 2 The approximation is OK. = 1.51 x If we first assume that x<< then = 1.51 x 10 7 then x = 1.8 x 10-4 mol L -1 : Ans Page 8 of 8 printed pages