CHEM1901/3 Answers to Problem Sheet 10
1.
(a)
Al(s) | Al3+(aq) || Sn2+(aq) | Sn(s)
The half reactions and potentials are:
Sn2+(aq) + 2e-  Sn(s)
Al3+(aq) + 3e-  Al(s)
E° = 0.14 V
E° = 1.68 V
The Al(s) | Al3+(aq) half cell is on the left so this is the oxidation half cell so the half cell
reaction is reversed. The Sn2+ | Sn(s) half cell is on the right so is the reduction half cell.
(i)
(ii)
Al(s)  Al3+(aq) + 3eSn2+(aq) + 2e-  Sn(s)
E° = 1.68 V
E° = 0.14 V
To balance the electrons, these must be combined with 2 × (i) and 3 × (ii):
2Al(s) + 3Sn2+(aq)  2Al3+ + 3Sn(s)
E° = {(+1.68) + (0.14)} V = 1.54 V
Al(s) is the reductant and Sn2+(aq) is the oxidant.
(b)
Pt(s) | Fe2+(aq), Fe3+(aq) || Ag+(aq) | Ag(s)
The half reactions and potentials are:
Fe3+(aq) + e-  Fe2+(aq)
Ag+(aq) + e-  Ag(s)
E° = 0.77 V
E° = 0.80 V
The Pt(s) is an inert electrode in contact with the Fe2+(aq), Fe3+(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)
(ii)
Fe2+(aq)  Fe3+(aq) + eAg+(aq) + e-  Ag(s)
E° = 0.77 V
E° = 0.80 V
These can then be combined to give the overall cell reaction:
Fe2+(aq) + Ag+(aq)  Fe3+(aq) + Ag(s)
E° = {(0.77) + (+0.80)} V = +0.03 V
Fe2+(aq) is the reductant. Ag+(aq) is the oxidant.
(c)
Pt(s) | MnO4–(aq), H+(aq), Mn2+(aq) || Sn4+(aq), Sn2+(aq) | Pt(s)
The half reactions and potentials are:
MnO4-(aq) + 8H+(aq) + 5e-  Mn2+(aq) + 4H2O(l)
Sn4+(aq) + 2e-  Sn2+(aq)
E° = 1.51 V
E° = 0.15 V
The Pt(s) is an inert electrode in contact with the MnO4-(aq), Mn2+(aq) solution. This
half is on the left so is the oxidation half cell, so the half cell reaction is reversed. The
Sn4+(aq) | Sn2+(aq) half cell is on the right so is the reduction half cell.
(i)
(ii)
Mn2+(aq) + 4H2O(l)  MnO4-(aq) + 8H+(aq) + 5eSn4+(aq) + 2e-  Sn2+(aq)
E° = 1.51 V
E° = 0.15 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:
2Mn2+(aq) + 8H2O(l) + 5Sn4+(aq)  2MnO4-(aq) + 16H+(aq) + 5Sn2+(aq)
E° = {(1.51) + (+0.15)} V = 1.36 V
Mn2+(aq) is the reductant. Sn4+(aq) is the oxidant. As E° < 0, the cell reaction is not
spontaneous as written and will occur in the opposite direction with MnO4- oxidising
Sn2+(aq).
2.
(a)
The two half cells are:
Mg(s)  Mg2+(aq) + 2e-(aq)
Pb2+(aq) + 2e-(aq)  Pb(s)
E° = 2.36 V (reversed as oxidation required)
E° = 0.13 V
The overall reaction is:
Mg(s) + Pb2+(aq)  Mg2+(aq) + Pb(s)
E° = {(2.36) + (0.13)} V = +2.23 V
As E° > 0, the reaction should occur.
(b)
The two half cells are:
Sn(s)  Sn2+(aq) + 2e2H+(aq) + 2e-  H2(g)
E° = +0.14 V (reversed as oxidation required)
E° = 0.00 V (by definition)
The overall reaction is:
Sn(s) + 2H+(aq)  Sn2+(aq) + H2(g)E° = {(+0.14) + (0.00)} V = +0.14 V
As E° > 0, the reaction should occur.
(c)
The two half cells are:
SO42-(aq) + 4H+(aq) + 2e-  SO2(g) + 2H2O(l)
E° = +0.20 V
2+
4+
Sn (aq)  Sn (aq) + 2e E° = 0.15 V (reversed as oxidation required).
.
Balancing the electrons gives the overall reaction as:
SO42-(aq) + 4H+(aq) + Sn2+(aq)  SO2(g) + 2H2O(l) + Sn4+(aq)
The cell potential is E° = {(+0.20) + (0.15)} V = +0.05 V.
As E° > 0, the reaction should occur but the value is very small so an equilibrium
mixture will form.
(d)
The two half cells are:
MnO4-(aq) + 8H+(aq) + 5e-  Mn2+(aq) + 4H2O(l)
E° = 1.51 V
+
H2O2(aq)  O2(g) + 2H (aq) + 2e E° = 0.70 V (reversed as oxidation required).
.
Balancing the electrons gives the overall reaction as:
2MnO4-(aq) + 16H+(aq) + 5H2O2(aq)  2Mn2+(aq) + 8H2O(l) + 5O2(g) + 10H+(aq)
From this, the H+(aq) on the right hand side can be cancelled:
2MnO4-(aq) + 6H+(aq) + 5H2O2(aq)  2Mn2+(aq) + 8H2O(l) + 5O2(g)
3.
(a)
Keq =
The cell potential is E° = {(+1.51) + (0.70)} V = +0.81 V. As E° > 0, the reaction
should occur.
Sn2+(aq) + 2Ag+(aq)
Sn4+(aq) + 2Ag(s)
[Sn4+ (aq)]
[Sn 2+ (aq)][Ag + (aq)]2
The two half cells are:
Sn2+(aq)  Sn4+(aq) + 2eAg+(aq) + e-  Ag(s)
E = 0.15 V (reversed as oxidation required)
E = +0.80 V
Hence, E = ((-0.15) + (0.80)) V = 0.65 V.
 RT 
E = 
 ×2.303 log(K c ) , the equilibrium constant at T = 298 K is:
 nF 
K c =10nFE ° 2.303RT =10(2×96485×0.65) (2.303×8.314×298) =9.6×1021
(b)
Keq =
MnO2(s) + 4H+(aq) + 2Cl–(aq)
Mn2+(aq) + 2H2O(l) + Cl2(g)
[Mn2+ (aq)][Cl 2 (g)]
[H + (aq)]4 [Cl - (aq)]2
The two half cells are:
MnO2(s) + 4H+(aq) + 2e-  Mn2+(aq) + 2H2O(l) E = 1.23 V
2Cl-(aq)  Cl2(g) + 2eE = -1.36 V (reversed as oxidation required)
Hence, E = ((1.23) + (-1.36)) V = -0.13 V.
 RT 
E = 
 ×2.303 log(K c ) , the equilibrium constant at T = 298 K is:
 nF 
K c =10nFE ° 2.303RT =10(2×96485×-0.13) (2.303×8.314×298) = 4.0×10-5
4.
(a)
Al(s) | Al3+ (0.18 M) || Fe2+ (0.85 M) | Fe(s)
The two half cells are:
Al(s) Al3+(aq) + 3eFe2+(aq) + 2e-  Fe(s)
E° = 1.68 V (on the left so oxidation half cell)
E° = 0.44 V
Hence, E° = ((+1.68) + (-0.44)) V = 1.24 V
The overall reaction is 2Al(s) + 3Fe2+(aq)  2Al3+(aq) + 3Fe(s) which involves 6e- and
[Al 3+ (aq)]2
has Q =
.
[Fe 2+ (aq)]3
Using the Nernst equation,
Ecell = E 0 -
2.303RT
log(Q )
nF
= (1.24 V) -
2.303 × (8.314 J K -1 mol -1 ) × (298 K)
(6 × 96485 C mol -1 )
 (0.18)2 
log 
,
 (0.85)3 


= 1.25 V
(b)
Ag(s) | Ag+ (0.34 M) || Cl2(g, 0.55 atm) | Cl– (0.098 M) | Pt(s)
The two half cells are:
Ag(s) Ag+(aq) + eCl2(g) + 2e-  2Cl-(aq)
E° = 0.80 V (on the left so oxidation half cell)
E° = 1.36 V
Hence, E° = ((-0.80) + (+1.36)) V = 0.56 V
The overall reaction is 2Ag(s) + Cl2(g)  2Ag+(aq) + 2Cl-(aq) which involves 2e- and has
Q=
[Ag + (aq)]2 [Cl - (aq)]2
.
[Cl 2 (g)]
The pressure of Cl2(g) is given as 0.55 atm. Using the ideal gas law PV = nRT or
n P
(0.55 atm)
concentration = =
=
= 0.0225 M .
V RT (0.08206 L atm K -1 mol -1 ) × (298) K
Using the Nernst equation,
Ecell = E 0 -
2.303RT
log(Q )
nF
= (0.56 V) = 0.57 V
2.303 × (8.314 J K -1 mol -1 ) × (298 K)
(2 × 96485 C mol -1 )
 (0.34) 2 (0.098)2 
log 
,


(0.0225)


5.
The relevant half cell reactions and reduction potentials are:
Cu2+(aq) + 2e-  Cu(s)
Fe2+(aq) + 2e-  Fe(s)
E° = 0.34 V
E° = 0.44 V
Sn4+(aq) + 2e-  Sn2+(aq)
Ag+(aq) + e-  Ag(s)
E° = 0.15 V
E° = 0.80 V
Zn2+(aq) + 2e-  Zn(s)
Fe2+(aq) + 2e-  Fe(s)
E° = 0.76 V
E° = 0.44 V
In each case, the half reaction with the lowest electrode potential is reversed.
(a)
For the first cell:
(i)
The Fe(s) | Fe2+(aq) cell is reversed giving the overall cell reaction:
Fe(s) + Cu2+(aq)  Fe2+(aq) + Cu(s)
Cu(s) | Cu2+(aq) half cell.
Fe(s) | Fe2+(aq) half cell.
(ii)
Reduction occurs at the cathode:
Oxidation occurs at the anode:
(iii)
Electrons flow from the anode to the cathode: from the Fe(s)  Cu(s) electrode.
(iv)
E° = ((0.34) + (+0.44)) V = +0.78 V
(b)
For the second cell:
(i)
The Pt | Sn4+(aq), Sn2+(aq) cell is reversed giving the overall cell reaction:
Sn2+(aq) + 2Ag+(aq)  Sn4+(aq) + Ag(s)
Ag(s) | Ag+(aq) half cell.
Pt | Sn4+(aq), Sn2+(aq) half cell.
(ii)
Reduction occurs at the cathode:
Oxidation occurs at the anode:
(iii)
Electrons flow from the anode to the cathode: from the Pt(s)  Ag(s) electrode.
(iv)
E° = ((0.80) + (0.15)) V = +0.65 V
(c)
For the third cell:
(i)
The Zn | Zn2+(aq) cell is reversed giving the overall cell reaction:
Zn(s) + Fe2+(aq)  Zn2+(aq) + Fe(s)
Fe(s) | Fe2+(aq) half cell.
Zn | Zn2+(aq) half cell.
(ii)
Reduction occurs at the cathode:
Oxidation occurs at the anode:
(iii)
Electrons flow from the anode to the cathode: from the Zn(s)  Fe(s) electrode.
(iv)
E° = ((0.44) + (0.76)) V = +0.32 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 K, Ecell is:
RT
Ecell = E 0 -2.303×
log(Q )
nF
= (0.32 V) – 2.303 ×
(8.314 J K -1mol -1 )×(298 K)
0.1
log(
) = 0.26 V
(2×96485 C mol -1 )
1.0×10-3
6.
Reduction takes places at the cathode:
Mg2+(l) + 2e-  Mg(s)
Oxidation takes places at the anode:
Cl-(l)  ½ Cl2 + e-
7.
The molar mass of Cl2(g) is (2 × 35.45) g mol-1 = 70.9 g mol-1 so 1000 kg corresponds to
1000 × 103 g
= 14100 mol.
70.9 g mol
The half cell for chlorine is Cl2(g) + 2e-  2Cl-(aq) so the number of moles of electrons
required is (2 × 14100) mol = 28200 mol. The charge on this amount of electrons is
28200F.
The time required to deliver this charge with a current of 3 × 104 A is therefore:
t=
Q
28200F
=
= 90700 s = 25.2 hours .
I
(3×104 A)
The volume occupied by 14100 mol of Cl2 can be obtained using the ideal gas equation
PV = nRT:
V=
8.
(a)
nRT (14100 mol) × (0.08206 L atm-1 K -1 mol-1 ) × (298 K)
=
= 345000 L
P
(1.00 atm)
Dipole-dipole and dispersion forces are acting between the molecules in these
compounds.
(b)
9.
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 CH2F2.
However, dispersion forces increase with the size of the electron cloud and, hence,
with the atomic number. These are therefore largest in CH2Br2. The dispersion
forces must be more important than the dipole-dipole interactions in these
compounds so the boiling points increase in the order CH2F2 < CH2Cl2 < CH2Br2.
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.