CHEM1101 Answers to Problem Sheet 10
1.
(a)
2Al(s) + 3Sn2+(aq) → 2Al3+(aq) + 3Sn(s);
Al(s) à Al3+(aq) + 3eSn2+(aq) + 2e- à Sn(s)
(b)
(reverse of reduction potential)
(reduction potential)
Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s)
Fe2+(aq) à Fe3+(aq) + eAg+(aq) + e- à Ag(s)
2.
E° = +1.68 V
E° = -0.14 V
E° = +1.53 V
E° = -0.77 V
E° = +0.80 V
E° = +0.03 V
(reverse of reduction potential)
(reduction potential)
(a)
2MnO4-(aq) + 16H+(aq) + 5Sn2+(aq) → 2Mn2+(aq) + 8H2O(l) + 5Sn4+(aq)
(b)
MnO4-(aq) is the oxidising agent (and is itself reduced to to Mn2+(aq).
Sn2+(aq) is the reducing agent (and is itself oxidised to Sn4+(aq)).
(c)
Sn2+(aq) à Sn4+(aq) + 2eMnO4-(aq) + 8H+(aq) + 5eà Mn2+(aq) + 4H2O(l)
E° = -0.15 V
E° = +1.51 V
(reverse of reduction potential)
(reduction potential)
E° = +1.36 V
3.
(a)
The two half cells are:
Mg(s) à Mg2+(aq) + 2e-(aq)
Pb2+(aq) + 2e-(aq) à Pb(s)
E° = +2.36 V (reversed as oxidation required)
E° = −0.13 V
The overall reaction is:
Mg(s) + Pb2+(aq) à 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) à Sn2+(aq) + 2e2H+(aq) + 2e- à H2(g)
E° = +0.14 V (reversed as oxidation required)
E° = 0.00 V (by definition)
The overall reaction is:
Sn(s) + 2H+(aq) à 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- à SO2(g) + 2H2O(l)
E° = +0.20 V
2+
4+
Sn (aq) à 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) à 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- à Mn2+(aq) + 4H2O(l)
E° = 1.51 V
+
H2O2(aq) à 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) à 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) à 2Mn2+(aq) + 8H2O(l) + 5O2(g)
The cell potential is E° = ((+1.51) + (−0.70)) V = +0.81 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.
(b)
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 Pb2+(aq) + 2e-
E° = +0.80 V
E° = +0.13 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) | Pb2+(aq) half cell. Oxidation occurs at the anode.
Electrons flow from the anode to the cathode.
(c)
2Ag+(aq) + Pb(s) à Ag(s) + Pb2+(aq).
The standard cell potential, E° = ((+0.80) + (+0.13)) V = 0.93 V.
(d)
⎛ RT ⎞
Using, E° = ⎜
⎟ ×2.303log(Kc ) , the equilibrium constant at T = 298 K is:
⎝ nF ⎠
Kc =10nFE ° 2.303RT =10(2×96485×0.93) (2.303×8.314×298) = 2.8×1031
(e)
The reaction quotient, Q =
[Pb 2 + (aq)]
+
2
.
[Ag (aq)]
If the anode (Pb (aq)) concentration is reduced by a factor of 1000 from the
standard concentration, then [Pb2+(aq)] = 0.001 M and [Ag+(aq)] = 1 M. The cell
potential is now:
2+
Ecell = E 0 -
2.303RT
log(Q )
nF
= (0.93 V) -
(2.303) × (8.314 J K -1 mol-1 × (298 K)
-1
(2 × 96435 C mol )
log(
0.001
12
)
= 1.02 V
If the cathode (Ag+(aq)) concentration is reduced by a factor of 1000 from the
standard concentration, then [Ag+(aq)] = 0.001 M and [Pb2+(aq)] = 1 M. The cell
potential is now:
Ecell = E 0 -
2.303RT
log(Q )
nF
= (0.93 V) -
2.303 × (8.314 J K -1 mol -1 ) × (298 K)
-1
(2 × 96435 C mol )
log(
1
0.0012
)
= 0.75 V
5.
(a)
Keq =
Sn2+(aq) + 2Ag+(aq)
Sn4+(aq) + 2Ag(s)
[Sn4+ (aq)]
[Sn2+ (aq)][Ag + (aq)]2
The two half cells are:
Sn2+(aq) à Sn4+(aq) + 2eAg+(aq) + e- à 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° = ⎜
⎟ ×2.303log(Kc ) , the equilibrium constant at T = 298 K is:
⎝ nF ⎠
Kc =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)][Cl 2 (g)]
[H + (aq)]4 [Cl - (aq)]2
The two half cells are:
Mn2+(aq) + 2H2O(l) + Cl2(g)
MnO2(s) + 4H+(aq) + 2e- à Mn2+(aq) + 2H2O(l) E° = +1.23 V
2Cl-(aq) à Cl2(g) + 2eE° = -1.36 V (reversed as oxidation required)
Hence, E° = ((1.23) + (-1.36)) V = -0.13 V.
⎛ RT ⎞
E° = ⎜
⎟ ×2.303log(Kc ) , the equilibrium constant at T = 298 K is:
⎝ nF ⎠
Kc =10nFE ° 2.303RT =10(2×96485×-0.13) (2.303×8.314×298) = 4.0×10-5
6.
(a)
Al(s) | Al3+ (0.18 M) || Fe2+ (0.85 M) | Fe(s)
The two half cells are:
Al(s)à Al3+(aq) + 3eFe2+(aq) + 2e- à 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) à 2Al3+(aq) + 3Fe(s) which involves 6e- and
has Q =
[Al 3+ (aq)]2
[Fe2+ (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)à Ag+(aq) + eCl2(g) + 2e- à 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) à 2Ag+(aq) + 2Cl-(aq) which involves 2e- and has
[Ag + (aq)]2 [Cl- (aq)]2
Q=
.
[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) -
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)
⎝
⎞
⎟ ,
⎟
⎠
= 0.57 V
7.
The relevant half cell reactions and reduction potentials are:
Cu2+(aq) + 2e- à Cu(s)
Fe2+(aq) + 2e- à Fe(s)
E° = +0.34 V
E° = −0.44 V
Sn4+(aq) + 2e- à Sn2+(aq)
Ag+(aq) + e- à Ag(s)
E° = +0.15 V
E° = +0.80 V
Zn2+(aq) + 2e- à Zn(s)
Fe2+(aq) + 2e- à 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) à 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) à 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) à Sn4+(aq) + Ag(s)
Ag(s) | Ag+(aq) half cell.
Pt | Sn2+(aq), Sn4+(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) à 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) à 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) à 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.
(8.314 J K -1mol -1 )×(298 K)
0.1
log(
) = 0.26 V
-1
-3
(2×96485 C mol )
1.0×10
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- à 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:
nRT (14100 mol) × (0.08206 L atm-1 K -1 mol -1 ) × (298 K)
V=
=
= 345000 L
P
(1.00 atm)