CHEM1901/3 Answers to Problem Sheet 10

Transcription

1 CHEM1901/3 Answers to Problem Sheet 10 1 (a) Al(s) Al 3+ (aq) Sn 2+ (aq) Sn(s) The half reactions and potentials are: Sn 2+ (aq) + 2e - Sn(s) Al 3+ (aq) + 3e - Al(s) E = 014 V E = 168 V The Al(s) Al 3+ (aq) half cell is on the left so this is the oxidation half cell so the half cell reaction is reversed The Sn 2+ Sn(s) half cell is on the right so is the reduction half cell (i) Al(s) Al 3+ (aq) + 3e - E = 168 V (ii) Sn 2+ (aq) + 2e - Sn(s) E = 014 V To balance the electrons, these must be combined with 2 (i) and 3 (ii): 2Al(s) + 3Sn 2+ (aq) 2Al Sn(s) E = {(+168) + (014)} V = 154 V Al(s) is the reductant and Sn 2+ (aq) is the oxidant Pt(s) Fe 2+ (aq), Fe 3+ (aq) Ag + (aq) Ag(s) The half reactions and potentials are: Fe 3+ (aq) + e - Fe 2+ (aq) Ag + (aq) + e - Ag(s) E = 077 V E = 080 V The Pt(s) is an inert electrode in contact with the Fe 2+ (aq), Fe 3+ (aq) solution This half is on the left so is the oxidation half cell, so the half cell reaction is reversed The Ag + (aq) Ag(s) half cell is on the right so is the reduction half cell (i) Fe 2+ (aq) Fe 3+ (aq) + e - E = 077 V (ii) Ag + (aq) + e - Ag(s) E = 080 V These can then be combined to give the overall cell reaction: Fe 2+ (aq) + Ag + (aq) Fe 3+ (aq) + Ag(s) E = {(077) + (+080)} V = +003 V Fe 2+ (aq) is the reductant Ag + (aq) is the oxidant (c) Pt(s) MnO 4 (aq), H + (aq), Mn 2+ (aq) Sn 4+ (aq), Sn 2+ (aq) Pt(s) The half reactions and potentials are: MnO - 4 (aq) + 8H + (aq) + 5e - Mn 2+ (aq) + 4H 2 O(l) Sn 4+ (aq) + 2e - Sn 2+ (aq) E = 151 V E = 015 V

2 The Pt(s) is an inert electrode in contact with the MnO 4 - (aq), Mn 2+ (aq) solution This half is on the left so is the oxidation half cell, so the half cell reaction is reversed The Sn 4+ (aq) Sn 2+ (aq) half cell is on the right so is the reduction half cell (i) Mn 2+ (aq) + 4H 2 O(l) MnO - 4 (aq) + 8H + (aq) + 5e - E = 151 V (ii) Sn 4+ (aq) + 2e - Sn 2+ (aq) E = 015 V To balance the electrons, these must be combined with 2 (i) and 5 (ii): These can then be combined to give the overall cell reaction: 2Mn 2+ (aq) + 8H 2 O(l) + 5Sn 4+ (aq) 2MnO 4 - (aq) + 16H + (aq) + 5Sn 2+ (aq) E = {(151) + (+015)} V = 136 V Mn 2+ (aq) is the reductant Sn 4+ (aq) is the oxidant As E < 0, the cell reaction is not spontaneous as written and will occur in the opposite direction with MnO 4 - oxidising Sn 2+ (aq) 2 (a) Mg(s) Mg 2+ (aq) + 2e - (aq) Pb 2+ (aq) + 2e - (aq) Pb(s) E = 236 V (reversed as oxidation required) E = 013 V The overall reaction is: Mg(s) + Pb 2+ (aq) Mg 2+ (aq) + Pb(s) E = {(236) + (013)} V = +223 V As E > 0, the reaction should occur Sn(s) Sn 2+ (aq) + 2e - 2H + (aq) + 2e - H 2 (g) E = +014 V (reversed as oxidation required) E = 000 V (by definition) The overall reaction is: Sn(s) + 2H + (aq) Sn 2+ (aq) + H 2 (g) E = {(+014) + (000)} V = +014 V As E > 0, the reaction should occur (c) SO 2-4 (aq) + 4H + (aq) + 2e - SO 2 (g) + 2H 2 O(l) E = +020 V Sn 2+ (aq) Sn 4+ (aq) + 2e - E = 015 V (reversed as oxidation required) Balancing the electrons gives the overall reaction as: SO 4 2- (aq) + 4H + (aq) + Sn 2+ (aq) SO 2 (g) + 2H 2 O(l) + Sn 4+ (aq) The cell potential is E = {(+020) + (015)} V = +005 V As E > 0, the reaction should occur but the value is very small so an equilibrium

3 mixture will form (d) MnO - 4 (aq) + 8H + (aq) + 5e - Mn 2+ (aq) + 4H 2 O(l) E = 151 V H 2 O 2 (aq) O 2 (g) + 2H + (aq) + 2e - E = 070 V (reversed as oxidation required) Balancing the electrons gives the overall reaction as: 2MnO 4 - (aq) + 16H + (aq) + 5H 2 O 2 (aq) 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) 2Mn 2+ (aq) + 8H 2 O(l) + 5O 2 (g) The cell potential is E = {(+151) + (070)} V = +081 V As E > 0, the reaction should occur 3 (a) Sn 2+ (aq) + 2Ag + (aq) Sn 4+ (aq) + 2Ag(s) K eq = 4+ [Sn (aq)] [Sn (aq)][ag (aq)] Sn 2+ (aq) Sn 4+ (aq) + 2e - Ag + (aq) + e - Ag(s) E = 015 V (reversed as oxidation required) E = +080 V Hence, E = ((-015) + (080)) V = 065 V RT E = 2303log( Kc ), the equilibrium constant at T = 298 K is: nf K 2303 ( ) ( ) 21 c =10 nfe RT =10 =96 10 MnO 2 (s) + 4H + (aq) + 2Cl (aq) Mn 2+ (aq) + 2H 2 O(l) + Cl 2 (g) K eq = [Mn 2+ (aq)][cl 2(g)] [H (aq)] [Cl (aq)] MnO 2 (s) + 4H + (aq) + 2e - Mn 2+ (aq) + 2H 2 O(l) E = 123 V 2Cl - (aq) Cl 2 (g) + 2e - E = 36 V (reversed as oxidation required) Hence, E = ((123) + (36)) V = -013 V RT E = 2303log( Kc ), the equilibrium constant at T = 298 K is: nf

4 K 2303 ( ) ( ) -5 c =10 nfe RT =10 = (a) Al(s) Al 3+ (018 M) Fe 2+ (085 M) Fe(s) Al(s) Al 3+ (aq) + 3e - E = 168 V (on the left so oxidation half cell) Fe 2+ (aq) + 2e - Fe(s) E = 044 V Hence, E = ((+168) + (-044)) V = 124 V The overall reaction is 2Al(s) + 3Fe 2+ (aq) 2Al 3+ (aq) + 3Fe(s) which involves 6e - and has Q = 3+ 2 [Al (aq)] 2+ 3 [Fe (aq)] Using the Nernst equation, RT Ecell = E - log( Q) nf 2303 (8314 J K mol ) (298 K) (018) = (124 V) - log ( C mol ) (085) = 125 V 2 3, Ag(s) Ag + (034 M) Cl 2 (g, 055 atm) Cl (0098 M) Pt(s) Ag(s) Ag + (aq) + e - E = 080 V (on the left so oxidation half cell) Cl 2 (g) + 2e - 2Cl - (aq) E = 136 V Hence, E = ((-080) + (+136)) V = 056 V The overall reaction is 2Ag(s) + Cl 2 (g) 2Ag + (aq) + 2Cl - (aq) which involves 2e - and has Q = [Cl 2(g)] [Ag (aq)] [Cl (aq)] The pressure of Cl 2 (g) is given as 055 atm Using the ideal gas law PV = nrt or n P (055 atm) concentration = = = = M V RT ( L atm K mol ) (298) K Using the Nernst equation, RT Ecell = E - log( Q) nf (8314 J K mol ) (298 K) (034) (0098) = (056 V) - log, ( C mol ) (00225) = 057 V

5 5 The relevant half cell reactions and reduction potentials are: Cu 2+ (aq) + 2e - Cu(s) Fe 2+ (aq) + 2e - Fe(s) Sn 4+ (aq) + 2e - Sn 2+ (aq) Ag + (aq) + e - Ag(s) Zn 2+ (aq) + 2e - Zn(s) Fe 2+ (aq) + 2e - Fe(s) E = 034 V E = 044 V E = 015 V E = 080 V E = 076 V E = 044 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) 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) Cu(s) electrode E = ((034) + (+044)) V = +078 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) Sn 4+ (aq) + Ag(s) (ii) Reduction occurs at the cathode: Ag(s) Ag + (aq) half cell Oxidation occurs at the anode: Pt Sn 4+ (aq), Sn 2+ (aq) half cell (iii) (iv) (c) (i) Electrons flow from the anode to the cathode: from the Pt(s) Ag(s) electrode E = ((080) + (015)) V = +065 V For the third cell: The Zn Zn 2+ (aq) cell is reversed giving the overall cell reaction: Zn(s) + Fe 2+ (aq) Zn 2+ (aq) + Fe(s) (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) Fe(s) electrode E = ((044) + (076)) V = +032 V

6 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 K, E cell is: = 0 RT Ecell E log( Q) nf (8314 J K mol ) (298 K) 01 = (032 V) 2303 log( ) = 026 V ( C mol ) Reduction takes places at the cathode: Mg 2+ (l) + 2e - Mg(s) Oxidation takes places at the anode: Cl - (l) ½ Cl 2 + e - 7 The molar mass of Cl 2 (g) is (2 3545) g mol = 709 g mol so 1000 kg corresponds to g = mol 709 g mol The half cell for chlorine is Cl 2 (g) + 2e - 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 = 252 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 K mol ) (298 K) = = L P (100 atm) 8 (a) Dipole-dipole and dispersion forces are acting between the molecules in these compounds

7 The electronegativity of the halogens decreases in the order F > Cl > Br so the polarity of the bonds decreases in the order C-F > C-Cl > C-Br, as shown by the dipole moments The dipole-dipole interactions are therefore largest in CH 2 F 2 However, dispersion forces increase with the size of the electron cloud and, hence, with the atomic number These are therefore largest in CH 2 Br 2 The dispersion forces must be more important than the dipole-dipole interactions in these compounds so the boiling points increase in the order CH 2 F 2 < CH 2 Cl 2 < CH 2 Br 2 9 A and C presence of OH groups will lead to H-bonding interactions B delocalized π electron density will lead to dispersion interactions D charged group with N-H bonds will lead to interaction with ions and polar groups