CHEM1101 Answers to Problem Sheet 10

Transcription

1 CHEM1101 Answers to Problem Sheet (a) 2Al(s) + 3Sn 2+ (aq) 2Al 3+ (aq) + 3Sn(s); Al(s) à Al 3+ (aq) + 3e - E = V (reverse of reduction potential) Sn 2+ (aq) + 2e - à Sn(s) E = V (reduction potential) E = V Fe 2+ (aq) + Ag + (aq) Fe 3+ (aq) + Ag(s) Fe 2+ (aq) à Fe 3+ (aq) + e - E = V (reverse of reduction potential) Ag + (aq) + e - à Ag(s) E = V (reduction potential) E = V 2. (a) 2MnO 4 - (aq) + 16H + (aq) + 5Sn 2+ (aq) 2Mn 2+ (aq) + 8H 2 O(l) + 5Sn 4+ (aq) MnO 4 - (aq) is the oxidising agent (and is itself reduced to to Mn 2+ (aq). Sn 2+ (aq) is the reducing agent (and is itself oxidised to Sn 4+ (aq)). (c) Sn 2+ (aq) à Sn 4+ (aq) + 2e - E = V (reverse of reduction potential) MnO - 4 (aq) + 8H + (aq) + 5e - E = V (reduction potential) à Mn 2+ (aq) + 4H 2 O(l) E = V 3. (a) Mg(s) à Mg 2+ (aq) + 2e - (aq) Pb 2+ (aq) + 2e - (aq) à Pb(s) E = V (reversed as oxidation required) E = 0.13 V The overall reaction is: Mg(s) + Pb 2+ (aq) à Mg 2+ (aq) + Pb(s) E = ((+2.36) + ( 0.13)) V = V As E > 0, the reaction should occur. Sn(s) à Sn 2+ (aq) + 2e - 2H + (aq) + 2e - à H 2 (g) E = V (reversed as oxidation required) E = 0.00 V (by definition) The overall reaction is: Sn(s) + 2H + (aq) à Sn 2+ (aq) + H 2 (g) E = ((+0.14) + (0.00)) V = V As E > 0, the reaction should occur. (c) SO 2-4 (aq) + 4H + (aq) + 2e - à SO 2 (g) + 2H 2 O(l) E = V Sn 2+ (aq) à Sn 4+ (aq) + 2e - E = 0.15 V (reversed as oxidation required).

2 . Balancing the electrons gives the overall reaction as: SO 4 2- (aq) + 4H + (aq) + Sn 2+ (aq) à SO 2 (g) + 2H 2 O(l) + Sn 4+ (aq) The cell potential is E = ((+0.20) + ( 0.15)) V = V. As E > 0, the reaction should occur but the value is very small so an equilibrium mixture will form. (d) MnO - 4 (aq) + 8H + (aq) + 5e - à Mn 2+ (aq) + 4H 2 O(l) E = 1.51 V H 2 O 2 (aq) à O 2 (g) + 2H + (aq) + 2e - E = 0.70 V (reversed as oxidation required). Balancing the electrons gives the overall reaction as: 2MnO 4 - (aq) + 16H + (aq) + 5H 2 O 2 (aq) à 2Mn 2+ (aq) + 8H 2 O(l) + 5O 2 (g) + 10H + (aq) From this, the H + (aq) on the right hand side can be cancelled: 2MnO 4 - (aq) + 6H + (aq) + 5H 2 O 2 (aq) à 2Mn 2+ (aq) + 8H 2 O(l) + 5O 2 (g) The cell potential is E = ((+1.51) + ( 0.70)) V = V. As E > 0, the reaction should occur. 4. (a) The strongest oxidising agent has the most positive reducing potential: Ag + (aq). The strongest reducing agent has the smallest (least positive) reducing potential. Pb(s) is the strongest reducing agent. The highest voltage will be delivered by the cell combining the strongest oxidising and reducing agents: Ag + (aq) + e - à Ag(s) Pb(s) à Pb 2+ (aq) + 2e - E = V E = V (reversed as oxidation required) Reduction occurs in the Ag(s) Ag + (s) half cell. Reduction occurs at the cathode. Oxidation occurs in the Pb(s) Pb 2+ (aq) half cell. Oxidation occurs at the anode. Electrons flow from the anode to the cathode. (c) 2Ag + (aq) + Pb(s) à Ag(s) + Pb 2+ (aq). The standard cell potential, E = ((+0.80) + (+0.13)) V = 0.93 V. RT (d) Using, E = 2.303log( c ), the equilibrium constant at T = 298 is: ( )( ) 31 c =10 E RT =10 =

3 2+ [Pb (aq)] (e) The reaction quotient, Q =. + 2 [Ag (aq)] If the anode (Pb 2+ (aq)) concentration is reduced by a factor of 1000 from the standard concentration, then [Pb 2+ (aq)] = M and [Ag + (aq)] = 1 M. The cell potential is now: RT (2.303) (8.314 J mol (298 ) = (0.93 V) - log( ) 2 ( C mol ) 1 = 1.02 V If the cathode (Ag + (aq)) concentration is reduced by a factor of 1000 from the standard concentration, then [Ag + (aq)] = M and [Pb 2+ (aq)] = 1 M. The cell potential is now: RT (8.314 J mol ) (298 ) 1 = (0.93 V) - log( ) 2 ( C mol ) = 0.75 V 5. (a) Sn 2+ (aq) + 2Ag + (aq) Sn 4+ (aq) + 2Ag(s) eq = [Sn [Sn 4+ (aq)] (aq)][ag (aq)] Sn 2+ (aq) à Sn 4+ (aq) + 2e - Ag + (aq) + e - à Ag(s) E = 0.15 V (reversed as oxidation required) E = V Hence, E = ((-0.15) + (0.80)) V = 0.65 V. RT E = 2.303log( c ), the equilibrium constant at T = 298 is: ( )( ) 21 c =10 E RT =10 = MnO 2 (s) + 4H + (aq) + 2Cl (aq) Mn 2+ (aq) + 2H 2 O(l) + Cl 2 (g) eq = [Mn 2+ (aq)][cl2 (g)] [H (aq)] [Cl (aq)]

4 MnO 2 (s) + 4H + (aq) + 2e - à Mn 2+ (aq) + 2H 2 O(l) E = V 2Cl - (aq) à Cl 2 (g) + 2e - E =.36 V (reversed as oxidation required) Hence, E = ((1.23) + (.36)) V = V. RT E = 2.303log( c ), the equilibrium constant at T = 298 is: ( )( ) -5 c =10 E RT =10 = (a) Al(s) Al 3+ (0.18 M) Fe 2+ (0.85 M) Fe(s) Al(s)à Al 3+ (aq) + 3e - Fe 2+ (aq) + 2e - à Fe(s) Hence, E = ((+1.68) + (-0.44)) V = 1.24 V E = V (on the left so oxidation half cell) E = 0.44 V The overall reaction is 2Al(s) + 3Fe 2+ (aq) à 2Al 3+ (aq) + 3Fe(s) which involves 6e - and 3+ 2 [Al (aq)] has Q = [Fe (aq)] Using the Nernst equation, RT (8.314 J mol ) (298 ) (0.18) = (1.24 V) - log ( C mol ) (0.85) = 1.25 V 2 3, Ag(s) Ag + (0.34 M) Cl 2 (g, 0.55 atm) Cl (0.098 M) Pt(s) Ag(s)à Ag + (aq) + e - E = 0.80 V (on the left so oxidation half cell) Cl 2 (g) + 2e - à 2Cl - (aq) E = V Hence, E = ((-0.80) + (+1.36)) V = 0.56 V The overall reaction is 2Ag(s) + Cl 2 (g) à 2Ag + (aq) + 2Cl - (aq) which involves 2e - and has [Ag (aq)] [Cl (aq)] Q =. [Cl 2(g)] The pressure of Cl 2 (g) is given as 0.55 atm. Using the ideal gas law PV = nrt or n P (0.55 atm) concentration = = = = M. V RT ( L atm mol ) (298) Using the Nernst equation,

5 RT (8.314 J mol ) (298 ) (0.34) (0.098) = (0.56 V) - log ( C mol ) (0.0225) = 0.57 V 7. The relevant half cell reactions and reduction potentials are: 2 2, Cu 2+ (aq) + 2e - à Cu(s) Fe 2+ (aq) + 2e - à Fe(s) Sn 4+ (aq) + 2e - à Sn 2+ (aq) Ag + (aq) + e - à Ag(s) Zn 2+ (aq) + 2e - à Zn(s) Fe 2+ (aq) + 2e - à Fe(s) E = V E = 0.44 V E = V E = V E = 0.76 V E = 0.44 V In each case, the half reaction with the lowest electrode potential is reversed. (a) (i) For the first cell: The Fe(s) Fe 2+ (aq) cell is reversed giving the overall cell reaction: Fe(s) + Cu 2+ (aq) à Fe 2+ (aq) + Cu(s) (ii) Reduction occurs at the cathode: Cu(s) Cu 2+ (aq) half cell. Oxidation occurs at the anode: Fe(s) Fe 2+ (aq) half cell. (iii) (iv) (i) Electrons flow from the anode to the cathode: from the Fe(s) à Cu(s) electrode. E = ((+0.34) + (+0.44)) V = V For the second cell: The Pt Sn 4+ (aq), Sn 2+ (aq) cell is reversed giving the overall cell reaction: Sn 2+ (aq) + 2Ag + (aq) à Sn 4+ (aq) + Ag(s) (ii) Reduction occurs at the cathode: Ag(s) Ag + (aq) half cell. Oxidation occurs at the anode: Pt Sn 2+ (aq), Sn 4+ (aq) half cell. (iii) (iv) (c) (i) Electrons flow from the anode to the cathode: from the Pt(s) à Ag(s) electrode. E = ((+0.80) + ( 0.15)) V = V For the third cell: The Zn Zn 2+ (aq) cell is reversed giving the overall cell reaction: Zn(s) + Fe 2+ (aq) à Zn 2+ (aq) + Fe(s)

6 (ii) Reduction occurs at the cathode: Fe(s) Fe 2+ (aq) half cell. Oxidation occurs at the anode: Zn Zn 2+ (aq) half cell. (iii) (iv) Electrons flow from the anode to the cathode: from the Zn(s) à Fe(s) electrode. E = (( 0.44) + (+0.76)) V = V As the concentrations are not standard, the Nernst equation must be used to calculate the cell potential. The reaction involves the movement of 2 electrons. At 298, E cell is: = 0 RT E log( Q) (8.314 J mol ) (298 ) 0.1 = (0.32 V) log( ) = 0.26 V ( C mol ) The molar mass of Cl 2 (g) is ( ) g mol = 70.9 g mol so 1000 kg corresponds to g = mol g mol The half cell for chlorine is Cl 2 (g) + 2e - à 2Cl - (aq) so the number of moles of electrons required is ( ) mol = mol. The charge on this amount of electrons is 28200F. The time required to deliver this charge with a current of A is therefore: t = Q I 28200F = = s = 25.2 hours. ( A) The volume occupied by mol of Cl 2 can be obtained using the ideal gas equation PV = nrt: V = nrt (14100 mol) ( L atm mol ) (298 ) = = L P (1.00 atm)