13.11 ANS
(a) Ni(s) | Ni2+(aq)(0.20 M) || Ag+(aq)(0.5 M) | Ag(s).
(b) Cu(s)|Cu2+(aq)(0.20 M) || PtCl62-(aq)(0.10 M), PtCl42-(aq)(0.10M), Cl-(aq) (0.40 M) |
Pt(s).
(c) Pb(s) | PbSO4(s) | SO42-(aq)(0.30M) || Cl-(aq)(0.20 M) | AgCl(s) |Ag(s).
(d) Zn(s) | Zn2+(aq)(0.125 M) || H+(aq)(0.010 M) | H2(g)(1.0 atm) | Pt(s).
c) Fe(s) + Sn2+(aq) + S2-(aq) → FeS(s) + Sn(s) (2 electron process)
(a)
Fe2+(aq) + 2e- Æ Fe(s) E0 = -0.44 V
Hg2+(aq) + 2e- Æ Hg(l)
E0 = 0.855 V
Hg2+(aq) + Fe(s) Æ Hg(l) + Fe2+(aq)
E0cell = 0.855 – (-0.440) = 1.295 V = 1.3 V.
(b)
Fe3+(aq) + e- Æ Fe2+(aq) E0 = 0.771 V
MnO4-(aq) + 8 H+(aq) + 5e- Æ Mn2+(aq) + 4 H2O(l)
E0 = 1.507 V.
MnO4-(aq) + 8 H+(aq) + 5Fe2+(aq) Æ Mn2+(aq) + 4 H2O(l) + 5Fe3+(aq)
E0cell = 1.507 – (0.771) = 0.736 V
(c)
Cl2(s) + 2e- Æ 2Cl-(aq)
E0 = 1.360 V
Au+(aq) + e- Æ Au(s)
E0 = 1.68 V
2Au+(aq) + 2Cl-(aq) Æ 2Au(s) + Cl2(s)
E0cell = 1.68 – 1.360 = 0.320 V.
ANS:
For a gralvanic cell with a copper metal in the copper sulfate solution and a silver
metal in the silver nitrate solution,
Cu(s) | Cu2+(aq)(0.60 M) || Ag+(aq)(0.45M) | Ag(s)
Ag+(aq) + e- Æ Ag(s)
E0 = 0.7994 V
Cu2+(aq) + 2e- Æ Cu(s)
E0 = 0.337 V
Cu(s) + 2Ag+(aq) Æ Cu2+(aq) + 2Ag(s)
Ecell = E0cell – (0.0592/2)log([Cu2+]/[Ag+]2)
= (0.7994 – 0.337)- (0.0296)log{(0.60/(0.45)2}
= 0.462 -0.014 = 0.448 V
ANS
(a)
Ga(aq)3+ + 3e- Æ Ga(s)
E0 = -0.53 V.
Ag(aq)+ + e- Æ Ag(s)
E0 = 0.7994 V.
3Ag(aq)+ + Ga(s) Æ 3Ag(s) + Ga(aq)3+
E0cell = 0.7994 – (-0.53) = 2.9282 V = 1.3 V.
ΔG0 = -nFEo = -3×96485 JV-1mol-1 x 1.3 V = -385 kJ mol-1.
(b)
Zn2+(aq) + 2e- Æ Zn(s)
Cr3+(aq) +3e- Æ Cr(s)
E0 = -0.763 V
E0 = -0.74 V.
2Cr3+(aq) + 3Zn(s) Æ 2Cr(s) + 3Zn2+(aq)
E0cell Æ -0.74 -(-0.763) = 0.023 V
ΔG0 = -nFEo = -6×96485 JV-1mol-1 x 0.023 V = -13 kJ mol-1.
(c)
FeS(s) + 2e- Æ Fe(s) + S2-(aq) E0 = -1.01V
Sn2+(aq) + 2e- Æ Sn(s)
E0 = -0.14 V
Sn2+(aq) + Fe(s) + S2-(aq) Æ FeS(s) + Sn(s)
E0cell = -0.14 – (-1.01) = 0.87 V
ΔG0 = -nFEo = -2×96485 JV-1mol-1 x 0.87 V = -168 kJ mol-1.
-0.236 V
Cu2+ + 2e- Æ Cu(s)
E0 = 0.337 V,
ΔGf0 (Cu2+)= 65.49 kJ mol-1 is for the reaction Cu(s) Æ Cu2+ + 2e∆Gf0 (Ni2+) = -45.6 kJ mol-1 is for the reaction Ni(s) Æ Ni2+ + 2eNi2+ + 2e- Æ Ni(s) E0 = ?
∆Gf0 (Ni2+) = nFE0 = 2x(96485) x E0
-45.6 x 103 = 2x(96485)x(Eo)
E0 = -0.236 V.
9.11 x 1036
H2O2(l) + 2H+ + 2e- Æ 2 H2O
O2(g) + 4H+ + 4e- Æ 2H2O
E0 = +1.776 V
E0 = +1.229
2 H2O2(l) Æ O2(g) + 2H2O
E0cell = 1.776 – 1.229 = 0.547 V
E0cell = RTln(K)/nF = 0.0592 logK/4
K = 10(0.547 x 4/ 0.0592) = 9.11 x 1036.
The reduction potential of “anode” for Zn is Zn2+(aq) + 2e- Æ Zn(s)
and, that for Fe is
Fe2+(aq) + 2e- Æ Fe(s) E0 = -0.44 V
E0cell = E0cathode -E0anode
E0cathode -E0anode(Zn) > E0cathode -E0anode(Fe)
The cell potential will drop.
E0 = -0.763 V
Charge = current x time.
Q=Ixt
Pd2+(aq) + 2e- Æ Pd(s) E0 = 0.987 V.
(0.0032 g) x (1 mol Pd/106 42 g) x (2 mol e-/1 mol Pd) x (96485 C / 1 mol e-) = 5.8 C
t = 5.8 C/(1.5 C s-1) = 3.9 sec.
(a)
Volume of Au coated on the substrate is (2.62 cm2) x (5.00 x 10-4 mm) x (0.1 cm/1mm)
= 1.31 x 10-4 cm3.
The mass of Au is (1.31 x 10-4 cm3) x (19.32 g cm-3) = 2.53 x 10-3 g = 2.53 mg.
(b)
Au(aq)+ + e- Æ Au(s)
(2.53x10-3 g)x(5000)x(1 mol Au/ 196.97 g)x(96485 C/1 mol e-) = 6196 C
t = 6196 C/(15.0 C s-1) = 413 sec.
(a) At the anode, the only possible oxidation half-reaction is
2H2O Æ O2(g) + 4H+ + 4eE0 = -1.229 V
and, O2 will be formed at anode.
At the cathode, one has to compare two possible reduction half-reactions
(i) 2H+ + 2e- Æ H2(g)
E0 = 0 V
(ii) SO42-(aq) + 4H+ + 2e- Æ SO2(g) + 2H2O
E0 = +0.20 V
For concentrated H2SO4 solution, SO2(g) will be formed at the cathode. However, for
diluted H2SO4 solution, H2(g) will be formed instead.
(b) 2H2O(l) Æ 2H2(g) + O2(g) 4 electrons transfer process
Q = I x t = (5.00 C s-1) x (30.0 min) x (60 s/ 1min) = 9.00 x 103 C.
(9.00 x 103 C ) x (1 mol e-/96485 C) x (1 mol O2/ 4 mol e-)= 0.0233 mol O2.
The mass of O2(g) produced from the water is (0.0233 mol ) x (32.0 g/mol) = 0.746 g.