Worked Solutions
Chapter 12
Question 1
Use the oxidation number rules to find the oxidation number of (a) Cl in KClO4 (b) S in
Na2SO4 (c) C in CO2 (d) H in CaH2 (e) Zn in ZnCl2 (f) Cr in K2Cr2O7 (g) C in C12H22O11
(h) P in H3PO3 (i) S in H2SO4 (j) Cu in CuO.
Answer:
(a) Let the oxidation number of Cl = x
The oxidation number of K = +1 (from rule 7)
The oxidation number of O = - 2 (from rule 6)
Therefore +1 + x + 4(- 2) = 0 (from rule 2)
Therefore x = -1 + 8 = +7
(b) Let the oxidation number of S = x
The oxidation number of Na = +1 (from rule 7)
The oxidation number of O = - 2 (from rule 6)
Therefore 2(+1) + x + 4(- 2) = 0 (from rule 2)
Therefore x = -2 + 8 = +6
(c) Let the oxidation number of C = x
The oxidation number of O = - 2 (from rule 6)
Therefore x + 2(- 2) = 0 (from rule 2)
Therefore x = +4
(d) Let the oxidation number of H = x
The oxidation number of Ca = +2 (from rule 7)
Therefore +2 + 2x = 0 (from rule 2)
Therefore 2x = -2
x = -1
(e) Let the oxidation number of Zn = x
The oxidation number of Cl = -1 (from rule 8)
Therefore + x + 2(- 1) = 0 (from rule 2)
Therefore x = +2
1
(f) Let the oxidation number of Cr = x
The oxidation number of K = +1 (from rule 7)
The oxidation number of O = - 2 (from rule 6)
Therefore 2(+1) + 2x + 7(- 2) = 0 (from rule 2)
Therefore 2x = -2 +14 = +12
x=+6
(g) Let the oxidation number of C = x
The oxidation number of H = +1 (from rule 5)
The oxidation number of O = - 2 (from rule 6)
Therefore 12(+x) + 22(+1) + 11(- 2) = 0 (from rule 2)
Therefore 12x = -22 - 22 = 0
x=0
(h) Let the oxidation number of P = x
The oxidation number of H = +1 (from rule 5)
The oxidation number of O = - 2 (from rule 6)
Therefore 3(+1) + x + 3(- 2) = 0 (from rule 2)
Therefore x = -3 + 6 = +3
(i) Let the oxidation number of S = x
The oxidation number of H = +1 (from rule 5)
The oxidation number of O = - 2 (from rule 6)
Therefore 2(+1) + x + 4(- 2) = 0 (from rule 2)
Therefore x = -2 + 8 = +6
(j) Let the oxidation number of Cu = x
The oxidation number of O = - 2 (from rule 6)
Therefore + x - 2 = 0 (from rule 2)
Therefore x = + 2
Question 2
Assign oxidation numbers to each atom in the following: (a) Al2O3(b) NaH (c) HNO3 (d)
K2CrO4 (e) Co2O (f) FeO (g) NH3 (h) S8 (i) Mg(NO3)2 (j) PO43- (k) Fe3+ (l) OF2 (m) H2O2 (n)
Na2B4O7 (o) C2O42- (p) C2H4O2 (q) KHSO4 (r) Cr2(SO4)3 (s) Fe3O4 (t) CoF63-.
2
Answer:
(a)
Let the oxidation number of Al = x
The oxidation number of O = - 2 (from rule 6)
Therefore 2x + 3(- 2) = 0 (from rule 2)
Therefore 2x = +6
Therefore x = + 3
(b)
The oxidation number of Na = +1 (from rule 7)
The oxidation number of H = -1 (from rule 5)
(c)
The oxidation number of H = +1 (from rule 5)
Let the oxidation number of N = x
The oxidation number of O = - 2 (from rule 6)
Therefore +1 + x + 3(- 2) = 0 (from rule 2)
Therefore x = - 1 + 6 =+5
(d)
The oxidation number of K = +1 (from rule 7)
Let the oxidation number of Cr = x
The oxidation number of O = - 2 (from rule 6)
Therefore 2(+1) + x + 4(- 2) = 0 (from rule 2)
Therefore x = - 2 + 8 = +6
(e)
Let the oxidation number of Co = x
The oxidation number of O = - 2 (from rule 6)
Therefore 2x - 2 = 0 (from rule 2)
Therefore x = +1
(f)
Let the oxidation number of Fe = x
The oxidation number of O = - 2 (from rule 6)
Therefore x - 2 = 0 (from rule 2)
Therefore x = +2
(g)
Let the oxidation number of N = x
The oxidation number of H = +1 (from rule 5)
Therefore x + 3(+1) = 0 (from rule 2)
Therefore x = - 3
(h)
The oxidation number of S = 0 (from rule 1)
(i)
The oxidation number of Mg = +2 (from rule 7)
Let the oxidation number of N = x
The oxidation number of O = - 2 (from rule 6)
Therefore +2 + 2x + 6(- 2) = 0 (from rule 2)
Therefore 2x = -2 + 12 = +10
Therefore x = +5
3
(j)
Let the oxidation number of P = x
The oxidation number of O = - 2 (from rule 6)
Therefore x + 4(- 2) = -3 (from rule 4)
Therefore x = +8 – 3 = +5
(k)
The oxidation number of Fe = +3 (from rule 3)
(l)
The oxidation number of O = +2 (from rule 6)
Let the oxidation number of F = x
Therefore 2x + 2 = 0(from rule 2)
Therefore x = -1
(m)
The oxidation number of H = +1 (from rule 5)
The oxidation number of O = - 1 (from rule 6)
(n)
The oxidation number of Na = +1 (from rule 7)
Let the oxidation number of B = x
The oxidation number of O = - 2 (from rule 6)
Therefore +2 + 4x + 7(- 2) = 0 (from rule 2)
Therefore 4x = – 2 + 14 = +12
Therefore x = +3
(o)
Let the oxidation number of C = x
The oxidation number of O = - 2 (from rule 6)
Therefore 2x + 4(- 2) = -2 (from rule 4)
Therefore 2x = – 2 + 8 = +6
Therefore x = +3
(p)
Let the oxidation number of C = x
The oxidation number of H = +1 (from rule 5)
The oxidation number of O = - 2 (from rule 6)
Therefore 2x + 4(+1) + 2(- 2) = 0 (from rule 2)
Therefore 2x = +4 - 4 = 0
Therefore x = 0
(q)
The oxidation number of K = +1 (from rule 7)
The oxidation number of H = +1 (from rule 5)
Let the oxidation number of S = x
The oxidation number of O = - 2 (from rule 6)
Therefore +1 + 1 + x + 4(- 2) = 0 (from rule 2)
Therefore x = - 2 + 8 = +6
(r)
Let the oxidation number of Cr = x
The oxidation number of S = +6 (from part (q))
The oxidation number of O = - 2 (from rule 6)
4
Therefore 2x + 3(+6) + 12(- 2) = 0 (from rule 2)
Therefore 2x = -18 + 24 = +6
Therefore x = +3
(s)
Let the oxidation number of Fe = x
The oxidation number of O = - 2 (from rule 6)
Therefore 3x + 4(- 2) = 0 (from rule 2)
Therefore 3x = +8
Therefore x = +8/3
(t)
Let the oxidation number of Co = x
The oxidation number of F = - 1 (from rule 8)
Therefore x + 6(- 1) = -3 (from rule 4)
Therefore x = - 3 + 6 = +3
Question 3
What is the systematic name of (a) CuCl2 (b) CuCl (c) Cr2O3 (d) TiCl4 (e) TiCl2 (f) V2O5
(g) NiNO3.6H2O (h) NiSO4 (i) KMnO4 (j) K2CrO4?
Answer:
(a) Let the oxidation number of Cu = x
The oxidation number of Cl = -1 (from rule 8)
Therefore x + 2(- 1) = 0 (from rule 2)
Therefore x = 2
The name of the compound is therefore copper(II) chloride.
(b) Let the oxidation number of Cu = x
The oxidation number of Cl = -1 (from rule 8)
Therefore x - 1 = 0 (from rule 2)
Therefore x = +1
The name of the compound is therefore copper(I) chloride.
(c) Let the oxidation number of Cr = x
The oxidation number of O = -2 (from rule 6)
Therefore 2x + 3(- 2) = 0 (from rule 2)
Therefore x = 3
The name of the compound is therefore chromium(III) oxide.
5
(d) Let the oxidation number of Ti = x
The oxidation number of Cl = -1 (from rule 8)
Therefore x + 4(- 1) = 0 (from rule 2)
Therefore x = 4
The name of the compound is therefore titanium(IV) chloride.
(e) Let the oxidation number of Ti = x
The oxidation number of Cl = -1 (from rule 8)
Therefore x + 2(- 1) = 0 (from rule 2)
Therefore x = 2
The name of the compound is therefore titanium(II) chloride.
(f) Let the oxidation number of V = x
The oxidation number of O = -2 (from rule 6)
Therefore 2x + 5(- 2) = 0 (from rule 2)
Therefore x = 5
The name of the compound is therefore vanadium(V) oxide.
(g) Let the oxidation number of Ni = x
The oxidation number of NO3 = -1 (from rule 4)
Therefore x - 1 = 0 (from rule 2)
Therefore x = 1
The name of the compound is therefore nickel(I) nitrate-6-water.
(h) Let the oxidation number of Ni = x
The oxidation number of SO4 = -2 (from rule 4)
Therefore x - 2 = 0 (from rule 2)
Therefore x = 2
The name of the compound is therefore nickel(II) sulfate.
6
(i) Let the oxidation number of Mn = x
The oxidation number of K = +1 (from rule 7)
The oxidation number of O = -2 (from rule 6)
Therefore +1 + x + 4(- 2) = 0 (from rule 2)
Therefore x = _7
The name of the compound is therefore potassium manganate(VII).
(j) Let the oxidation number of Cr = x
The oxidation number of K = +1 (from rule 7)
The oxidation number of O = -2 (from rule 6)
Therefore 2(+1) + x + 4(- 2) = 0 (from rule 2)
Therefore x = + 6
The name of the compound is therefore potassium chromate(VI).
Question 4
What is the formula of
(a) chromium(III) chloride (b) manganese(IV) oxide (c) iron(II) chloride
(d) nickel(II) sulfide (f) cobalt(II) chloride (g) iron(II) sulfate-7-water
(h) copper(II) nitrate-3-water (i) sodium dichromate(VI) (j) silver(I) nitrate
(k)S copper(I) sulfate?
Answer:
(a) The chromium(III) ion has a charge of +3.
The chloride ion has a charge of –1.
The formula is CrCl3
(b) The manganese(IV) ion has a charge of +4.
The oxide ion has a charge of –2.
The formula is MnO2.
(c) The iron(II) ion has a charge of +2.
The chloride ion has a charge of –1.
The formula is FeCl2.
(d) The nickel(II) ion has a charge of +2.
The sulfide ion has a charge of –2.
The formula is NiS.
7
(e) The cobalt(II) ion has a charge of +2.
The chloride ion has a charge of –1.
The formula is CoCl2
(f) The iron(II) ion has a charge of +2.
The sulfate ion has a charge of –2.
The formula is FeSO4.7H2O
(g) The copper(II) ion has a charge of +2.
The nitrate ion has a charge of –1.
The formula is Cu(NO3)2.3H2O
(h) The sodium ion has a charge of + 1
The dichromate ion has a charge of –2.
The formula is Na2Cr2O7
(i) The silver(I) ion has a charge of +1.
The nitrate ion has a charge of –1.
The formula is AgNO3.
(j) The copper(I) ion has a charge of +1.
The sulfate ion has a charge of –2.
The formula is Cu2SO4
Question 5
What is (a) oxidised (b) reduced in each of the following redox reactions?
(i) C + O2 → CO2
C + O2 → CO2
0
0
+4 -2
(a) The oxidation number of carbon increases from 0 to +4. Therefore, carbon is oxidised.
(b) The oxidation number of oxygen decreases from 0 to -2. Therefore, oxygen is reduced.
(ii) 2Na + F2 → 2NaF
2Na + F2 → 2NaF
0
0
+1 -1
(a) The oxidation number of sodium increases from 0 to +4. Therefore, sodium is oxidised.
8
(b) The oxidation number of fluorine decreases from 0 to -1. Therefore, fluorine is reduced.
(iii)
2H+ + Zn → Zn2+ + H2
2H+ + Zn → Zn2+ + H2
+1
0
+2
0
(a) The oxidation number of zinc increases from 0 to +2. Therefore, zinc is oxidised.
(b) The oxidation number of hydrogen decreases from +1 to 0. Therefore, hydrogen is
reduced.
(iv)
PCl3 + Cl2 → PCl5
+3 -1
0
+5 -1
(a) The oxidation number of phosphorus increases from +3 to +5. Therefore, phosphorus is
oxidised.
(b) The oxidation number of chlorine decreases from 0 to -1. Therefore, chlorine is reduced.
(v)
ZnS + 2O2 → ZnSO4
+2 -2
0
+2 +6-2
(a) The oxidation number of ssulfur increases from -2 to +6. Therefore, sulfur is oxidised.
(b) The oxidation number of oxygen decreases from 0 to -2. Therefore, oxygen is reduced.
(vi)
2C2H2 + 5O2 → 4CO2 + 2H2O
-1 +1
0
+4 -2
+1 -2
(a) The oxidation number of carbon increases from –1 to +4. Therefore, carbon is oxidised.
(b) The oxidation number of oxygen decreases from 0 to -2. Therefore, oxygen is reduced.
(vii)
CuO + H2 → Cu + H2O
+2 -2
0
0
+1 -2
(a) The oxidation number of hydrogen increases from 0 to +1. Therefore, hydrogen is
oxidised.
(b) The oxidation number of copper decreases from +2 to 0. Therefore, copper is reduced.
Question 6
Using oxidation numbers, determine which of the following are redox reactions. For each
redox reaction, state what is (a) oxidised (b) reduced (c) the oxidising agent (d) the reducing
agent.
9
(i) Sn + 4 HNO3 → SnO2 + 4NO2 + 2H2O
+1+5 –2
0
+2 -2
+4 -2
+1 -2
This is a redox reaction, because some oxidation numbers changed as a result of the reaction.
(a) The oxidation number of tin increases from 0 to +2. Therefore, tin is oxidised.
(b) The oxidation number of nitrogen decreases from +5 to +4. Therefore, nitrogen is reduced.
(c) Nitrogen is the oxidising agent, because it is reduced itself.
(d) Tin is the reducing agent, because it is oxidised itself.
(ii) N2O4 → 2NO2
+4 –2
+4 –2
This is not a redox reaction, because there is no change in any of the oxidation numbers
(iii) 2Fe2+ + Cl2 → 2Fe3+ + 2Cl+2
0
–1
+3
This is a redox reaction, because the oxidation numbers changed as a result of the reaction.
(a) The oxidation number of iron increases from +2 to +3. Therefore, iron is oxidised.
(b) The oxidation number of chlorine decreases from 0 to -1. Therefore, chlorine is reduced.
(c) Chlorine is the oxidising agent, because it is reduced itself.
(d) Iron is the reducing agent, because it is oxidised itself.
(iv) Cl2 + 2OH- → Cl- + ClO- + H2O
0
-2 +1
-1
+1 -2
+1 -2
This is a redox reaction, because some oxidation numbers changed as a result of the reaction.
(a) The oxidation number of chlorine increases from 0 to +1. Therefore, chlorine is oxidised.
(b) The oxidation number of chlorine decreases from 0 to -1. Therefore, chlorine is reduced.
(c) Chlorine is the oxidising agent, because it is reduced itself.
(d)Chlorine is the reducing agent, because it is oxidised itself.
(v) NaOH + HCl → NaCl + H2O
+1 –2 +1
+1 –1
+1 -1
+1 -2
This is not a redox reaction, because there is no change in any of the oxidation numbers
(vi) MgCl2 + 2AgBr → MgBr2 + 2AgCl
+1 –1
+2 -1
+2 -1
+1 -1
This is not a redox reaction, because there is no change in any of the oxidation numbers
(vii) As2O3 + 2I2 + 2H2O → As2O5 + 4I- + 4H+
+3
-2
0
+1 –2
+5 -2
–1
+1
This is a redox reaction, because some oxidation numbers changed as a result of the reaction.
10
(a) The oxidation number of arsenic increases from +3 to +5. Therefore, arsenic is oxidised.
(b) The oxidation number of iodine decreases from 0 to -1. Therefore, iodine is reduced.
(c) Iodine is the oxidising agent, because it is reduced itself.
(d) Arsenic is the reducing agent, because it is oxidised itself.
Question 7
Using oxidation numbers, balance each of the following equations:
MnO4- + Fe2+ + H+ → Mn2+ + Fe3+ + H2O
(a)
Answer:
1
MnO4- + Fe2+ + H+ → Mn2+ + Fe3+ + H2O
+7
2
-2
+2
+1
Mn + 5e- → Mn
+7
Fe - 1e →
Reduced
Fe
+2
Oxidised
+3
Mn + 5e- → Mn …….. (1)
+7
+2
C - 2e →
-
–2
4
+1 –2
+3
+2
-
3
+2
C ……..
(2)
0
Multiply half equation (1) by 1, and multiply half equation (2) by 5
Mn + 5e- → Mn
- 5e- → 5Fe
5Fe
-----------------------------Mn + 5Fe → Mn + 5Fe
5
MnO4- + 5Fe
→ Mn2+ + 5Fe
6
MnO4- + 5Fe2+ + H+ → Mn2+ + 5Fe3+ + H2O
Balancing by inspection:
MnO4- + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O
Cr2O72- + CH3OH + H+
(b)
→ Cr3+ + HCHO + H2O
Answer:
1
Cr2O72- + CH3OH + H+ → Cr3+ + HCHO + H2O
+6
2
-2
-2 +1 –2 +1
Cr + 3e- → Cr
+6
+3
C - 2e- →
C
-2
+1
+3
+1 0 +1 -2
+1 –2
Reduced
Oxidised
0
11
3
Cr2 + 6e- → 2Cr …... (1)
+6
0
C - 2e →
-
C …….. (2)
–2
4
0
Multiply half equation (1) by 1, and multiply half equation (2) by 3
Cr2 + 6e- → 2Cr
3C - 6e- → 3C
-----------------------------Cr2 + 3C → 2Cr + 3C
5
Cr2O72- + 3CH3OH → 2Cr3+ + 3HCHO
6
Cr2O72- + 3CH3OH + H+ → 2Cr3+ + 3HCHO + H2O
Balancing by inspection:
Cr2O72- + 3CH3OH + 8H+ → 2Cr3+ + 3HCHO + 7H2O
(c)
ClO3- + C6H12O6 → Cl- + CO2 + H2O
Answer:
1
ClO3- + C6H12O6 → Cl- + CO2 + H2O
+5
2
0 +1 –2
-2
Cl + 6e → Cl
+5
-1
C - 4e- →
C
0
3
-1
-
+4 -2
+1 –2
Reduced
Oxidised
+4
Cl + 6e → Cl …….. (1)
-
+5
-1
C6 - 24e- → 6C …….. (2)
0
4
+4
Multiply half equation (1) by 4, and multiply half equation (2) by 1
4Cl + 24e- → 4Cl
C6
- 24e- → 6C
-----------------------------4Cl + C6 → 4Cl + 6C
5
4ClO3- + C6H12O6 → 4Cl- + 6CO2 + H2O
6
4ClO3- + C6H12O6 → 4Cl- + 6CO2 + H2O
Balancing by inspection:
4ClO3- + C6H12O6 → 4Cl- + 6CO2 + 6H2O
12
(d)
NO3- + C → NO2 + CO2 + O2-
Answer:
1
NO3- + C → NO2 + CO2 + O2+5
2
-2
0
–2
Reduced
+4
C - 4e →
-
0
C
Oxidised
+4
N + 1e- → N ……….. (1)
+5
+4
C - 4e →
C …….. (2)
0
+4
-
4
+4 -2
N + 1e- → N
+5
3
+4 -2
Multiply half equation (1) by 24 and multiply half equation (2) by 1
4N + 4e- → 4N
C - 2e- →
C
-----------------------------4N + C → 4N + C
5
4NO3- + C → 4NO2 + CO2 + O2-
6
4NO3- + C → 4NO2 + CO2 + O2Balancing by inspection:
4NO3- + C → 4NO2 + CO2 + 2O2-
(e)
Cr2O72- + S + H+
→ Cr2O3 + SO2 + OH-
Answer:
1
Cr2O72- + S + H+
+6
2
-2
0
+1
Cr + 3e → Cr
-
+6
+3
S - 4e- →
S
0
3
4
→ Cr2O3 + SO2 + OH+3
-2
+4 -2
–2 +1
Reduced
Oxidised
+4
Cr2 + 6e → Cr 2 …….. (1)
-
+6
0
S - 4e- →
S ………. (2)
0
+4
Multiply half equation (1) by 2, and multiply half equation (2) by 3
2Cr2 + 12e- → 2Cr2
3S - 12e- → 3S
13
-----------------------------2Cr2 + 3S → 2Cr 2 + 3S
5
2Cr2O72- + 3S + H+
→ 2Cr2O3 + 3SO2 + OH-
6
2Cr2O72- + 3S + H+
→ 2Cr2O3 + 3SO2 + OH-
Balancing by inspection:
→ 2Cr2O3 + 3SO2 + 2OH-
2Cr2O72- + 3S + 2H+
(f)
→ Mn2+ + CH3COOH + H2O
MnO4- + C2H5OH + H+
Answer:
1
+7
2
-2
-2 +1 -2 +1
+2
Mn + 5e → Mn
0 +3 0 –2 –2 +1
+1 –2
Reduced
+2
C - 2e- →
-2
C
Oxidised
0
Mn + 5e → Mn
-
+7
........ (1)
+2
- 4e- →
C2
–2
4
+1
-
+7
3
→ Mn2+ + CH3COOH + H2O
MnO4- + C2H5OH + H+
C2
........ (2)
0
Multiply half equation (1) by 4, and multiply half equation (2) by 5
4Mn + 20e- → 4Mn
- 20e- → C2
5C2
-----------------------------4Mn + 5C2 → 4Mn + 5C2
5
4MnO4- + 5C2H5OH + H+
→ 4Mn2+ + 5CH3COOH + H2O
6
4MnO4- + 5C2H5OH + H+
→ 4Mn2+ + 5CH3COOH + H2O
Balancing by inspection:
4MnO4- + 5C2H5OH + 12H+
→ 4Mn2+ + 5CH3COOH + 11H2O
Question 8
Balance each of the following equations, and name the substances acting as oxidising agent
and as reducing agent in each case:
(a)
Cr2O72- + C2H5OH + H+
→ Cr3+ + CH3COOH + H2O
Answer:
1
Cr2O72- + C2H5OH + H+
+6
-2
-2 +1 –2 +1
+1
→ Cr3+ + CH3COOH + H2O
+3
0 +1 0 -2 –2 +1
+1 –2
14
2
Cr + 3e- → Cr
+6
+3
C - 2e →
-
-2
3
C
Oxidised
0
Cr2 + 6e- → 2Cr
+6
........ (1)
0
- 4e →
C2
-
C2
–2
4
Reduced
........ (2)
0
Multiply half equation (1) by 2, and multiply half equation (2) by 3
2Cr2 + 12e- → 4Cr
3C2
- 12e- → 3C2
-----------------------------2Cr2 + 3C2 → 4Cr + 3C2
5
2Cr2O72- + 3C2H5OH + H+
→ 4Cr3+ + 3CH3COOH + H2O
6
2Cr2O72- + 3C2H5OH + H+
→ 4Cr3+ + 3CH3COOH + H2O
Balancing by inspection:
2Cr2O72- + 3C2H5OH + 16H+
→ 4Cr3+ + 3CH3COOH + 11H2O
Oxidising agent: Chromium, because it is reduced
Reducing agent: Carbon, because it is oxidised
(b)
→ Mn2+ + O2 + H2O
MnO4- + H2O2 + H+
Answer:
1
+7
2
-2
+1 -1
+1
O - 1e →
-1
+1 –2
Reduced
O
Oxidised
0
Mn + 5e- → Mn
+7
O2
0
+2
-
........ (1)
+2
- 2e →
-
–1
4
+2
Mn + 5e- → Mn
+7
3
→ Mn2+ + O2 + H2O
MnO4- + H2O2 + H+
O2
........ (2)
0
Multiply half equation (1) by 2, and multiply half equation (2) by 5
2Mn + 10e- → 2Mn
5O2
- 10e- →
5O2
-----------------------------2Mn + 5O2 → 2Mn + 5O2
15
5
2MnO4- + 5H2O2 + H+ → 2Mn2+ + 5O2 + H2O
6
2MnO4- + 5H2O2 + H+ → 2Mn2+ + 5O2 + H2O
Balancing by inspection:
2MnO4- + 5H2O2 + 6H+ → 2Mn2+ + 5O2 + 8H2O
Oxidising agent: Manganese, because it is reduced
Reducing agent: Oxygen, because it is oxidised
Question 9
22 cm3 of a potassium manganate(VII) solution required 25 cm3 of a
0.01 M ammonium iron(II) sulfate solution for complete reaction.
Calculate the concentration in mol l-1 of the potassium manganate(VII) solution.
Answer:
MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
V1 x M1 x n2 = V2 x M2 x n1
22 x M1 x 5 = 25 x 0.01 x 1
M1 = 25 x 0.01 x 1 / (22 x 5)
= 0.0023 mol l-1
Concentration of potassium manganate(VII) solution = 0.0023 mol l-1
Question 10
24 cm3 of a potassium manganate(VII) solution required 25 cm3 of a
0.0108 M ammonium iron(II) sulfate solution for complete reaction.
Calculate the concentration in (a) mol l-1 (b) g l-1 of the potassium manganate(VII) solution.
Answer:
(a) MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
V1 x M1 x n2 = V2 x M2 x n1
24 x M1 x 5 = 25 x 0.0108 x 1
M1 = 25 x 0.0108 x 1 / (24 x 5)
= 0.00225 mol l-1
Concentration of potassium manganate(VII) solution = 0.00225 mol l-1
(b) Molar mass of KMnO4 = 158
Concentration of potassium manganate(VII) solution in g l-1 = 158 x 0.00225 = 0.36 g l-1
16
Question 11
In an experiment to determine the mass of iron in an iron tablet, five iron tablets of total mass
1.8 g were dissolved in dilute sulfuric acid, and the solution made up to 250 cm3 with
deionised water. 25 cm3 of this solution required 16.5 cm3 of a 0.005 M potassium
manganate(VII) solution for complete reaction. Calculate (a) the mass of iron in one iron
tablet (b) the percentage of iron in the tablet.
Answer:
(a) MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
V1 x M1 x n2 = V2 x M2 x n1
16.5 x 0.005 x 5 = 25 x M2 x 1
M2 = 16.5 x 0.005 x 5 / (25 x 1)
= 0.0165 mol l-1
Concentration of iron(II) solution = 0.0165 mol l-1
Volume of iron(II) solution in total = 250 cm3
Moles of iron in this volume = 0.0165 / 4 = 0.004125
Mass of iron in this volume = 0.004125 x 56 = 0.231 g
Mass of iron in each tablet = 0.231 / 5 = 0.0462 g
Mass of each tablet = 1.8 / 5 = 0.36 g
(b) Percentage of iron in each tablet = 0.0462 x 100 / 0.36%
= 12.83%
Question 12
What volume of 0.02 M potassium manganate solution is required to react fully with 25cm3
of 0.04 M ammonium iron(II) sulfate solution?
Answer:
MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
V1 x M1 x n2 = V2 x M2 x n1
V1 x. 0.02 x 5 = 25 x 0.04 x 1
V1 = 25 x 0.04 x 1 / (0.02 x 5)
= 10 cm3
17
Question 13
25 cm3 of a 0.01 M iodine solution required 30 cm3 of a sodium thiosulfate solution for
complete reaction. Calculate the concentration in mol l-1 of the sodium thiosulfate solution.
Answer:
I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62- (aq)
V1 x M1 x n2 = V2 x M2 x n1
25 x 0.01 x 2 = 30 x M2 x 1
M2 = 25 x 0.01 x 2 / 30
= 0.0167 M
Concentration of sodium thiosulfate solution = 0.017 mol l-1
Question 14
25 cm3 of an iodine solution required 24.5 cm3 of a 0.01 M sodium thiosulfate solution for
complete reaction. Calculate the concentration in mol l-1 of the iodine solution.
Answer:
I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62- (aq)
V1 x M1 x n2 = V2 x M2 x n1
25 x M1 x 2 = 24.5 x 0.01 x 1
M1 = 24.5 x 0.01 x 1 / 50
= 0.0049 M
Concentration of iodine solution = 0.0049 mol l-1
Question 15
25 cm3 of bleach was diluted to 250 cm3 with deionised water. It was found that the
iodine solution produced by the reaction of 25 cm3 of the diluted bleach with acidified
potassium iodide solution required 18.5 cm3 of a 0.1 M sodium thiosulfate solution
for complete reaction. Calculate (a) the concentration in mol l-1 of the iodine solution
(b) the concentration of hypochlorite in the bleach (c) the percentage (w/v) of
hypochlorite in the bleach.
Answer:
(a)
I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62- (aq)
V1 x M1 x n2 = V2 x M2 x n1
25 x M1 x 2 = 18.5 x 0.1 x 1
M1 = 18.5 x 0.1 x 1 / 50
= 0.037 M
Concentration of iodine solution = 0.037 M
18
(b) Concentration of hypochlorite in diluted bleach solution
= 0.037 M
Concentration of hypochlorite in undiluted bleach solution
= 0.037 x 10 M
= 0.37 M
(c) Concentration of hypochlorite in g l-1
= 74.5 x 0.37 g l-1
= 27.57 g l-1
Percentage (w/v) of hypochlorite in bleach
= 27.57 x 100 / 1000
= 2.76%
Question 16
What is the oxidation number of (a) sulfur in
(i) SO2
Let the oxidation number of S = x
The oxidation number of O = - 2 (from rule 6)
Therefore x + 2(- 2) = 0 (from rule 2)
Therefore x = +4
(ii) SO3
Let the oxidation number of S = x
The oxidation number of O = - 2 (from rule 6)
Therefore x + 3(- 2) = 0 (from rule 2)
Therefore x = +6
(iii) Na2SO3
Let the oxidation number of S = x
The oxidation number of Na = +1 (from rule 7)
The oxidation number of O = - 2 (from rule 6)
Therefore 2(+1) + x + 3(- 2) = 0 (from rule 2)
Therefore x = -2 + 6 = +4
(iv) H2SO4
Let the oxidation number of S = x
The oxidation number of H = +1 (from rule 5)
The oxidation number of O = - 2 (from rule 6)
Therefore 2(+1) + x + 4(- 2) = 0 (from rule 2)
19
Therefore x = -2 + 8 = +6
What is the oxidation number of (b) nitrogen in
(i) NO2
Let the oxidation number of N = x
The oxidation number of O = - 2 (from rule 6)
Therefore x + 2(- 2) = 0 (from rule 2)
Therefore x = +4
(ii) HNO3
Let the oxidation number of N = x
The oxidation number of H = +1 (from rule 5)
The oxidation number of O = - 2 (from rule 6)
Therefore +1 + x + 3(- 2) = 0 (from rule 2)
Therefore x = -1 + 6 = +5
(i)
NO
Let the oxidation number of N = x
The oxidation number of O = - 2 (from rule 6)
Therefore x +(- 2) = 0 (from rule 2)
Therefore x = +2
(iv) NH3
Let the oxidation number of N = x
The oxidation number of H = +1 (from rule 5)
Therefore x + 3(+1) = 0 (from rule 2)
Therefore x = -3
Question 17
(b) What is the oxidation number of
(i) Cr in K2Cr2O7
Let the oxidation number of Cr = x
The oxidation number of K = +1 (from rule 7)
The oxidation number of O = - 2 (from rule 6)
Therefore 2(+1) + 2x + 7(- 2) = 0 (from rule 2)
Therefore 2x = -2 + 14 = +12
x = +6
20
(ii) Cl in KClO4
Let the oxidation number of Cl = x
The oxidation number of K = +1 (from rule 7)
The oxidation number of O = - 2 (from rule 6)
Therefore +1 + x + 4(- 2) = 0 (from rule 2)
Therefore x = -1 + 8 = +7
Question 18
Using oxidation numbers, state which of the following may be regarded as redox reactions.
In each of the redox reactions, state what is being oxidised and what is being reduced.
Ca + F2 → CaF2
(a)
Answer:
Ca + F2 → CaF2
0
0
+2 -1
This is a redox reaction, because the oxidation numbers changed as a result of the reaction.
The oxidation number of calcium increases from 0 to +2. Therefore, calcium is oxidised.
The oxidation number of fluorine decreases from 0 to -1. Therefore, fluorine is reduced.
NH3 + H+ → NH4+
(b)
Answer:
NH3 + H+ → NH4+
-3
+1
+1
-3 +1
This is not a redox reaction, because there is no change in any of the oxidation numbers.
Ca2+ + CO32- → CaCO3
(c)
Answer:
Ca2+ + CO32- → CaCO3
+2
+4 -2
+2 +4 -2
This is not a redox reaction, because there is no change in any of the oxidation numbers.
2S2O32- + I2 → S4O62- + 2I-
(d)
Answer:
2S2O32- + I2 → S4O62- + 2I+2
-2
0
+2.5 -2
-1
This is a redox reaction, because some of the oxidation numbers changed as a result of the
reaction.
21
The oxidation number of sulfur increases from +2 to +2.5. Therefore, sulfur is oxidised.
The oxidation number of iodine decreases from 0 to -1. Therefore, iodine is reduced.
2MnO4- + 6H+ + 5SO32- → 2Mn2+ + 3H2O + 5SO42-
(e)
Answer:
2MnO4- + 6H+ + 5SO32+7
-2
+1
→
2Mn2+ + 3H2O + 5SO42-
+4 –2
+4 –2
+2
+6 -2
This is a redox reaction, because some of the oxidation numbers changed as a result of the
reaction.
The oxidation number of sulfur increases from +4 to +6. Therefore, sulfur is oxidised.
The oxidation number of manganese decreases from +7 to +2. Therefore, manganese is
reduced.
(f)
Fe2+ + H+ + HNO3 → Fe3+ + H2O + NO2
Answer:
Fe2+ + H+ + HNO3 → Fe3+ + H2O + NO2
+2
+1 +5 –2
+1
+3
+1 –2
+4 -2
This is a redox reaction, because some of the oxidation numbers changed as a result of the
reaction.
The oxidation number of iron increases from +2 to +3. Therefore, iron is oxidised.
The oxidation number of nitrogen decreases from +5 to +4. Therefore, manganese is reduced.
Question 21
Use oxidation numbers to determine whether or not each of the following reactions is a redox
reaction, and if so, state the species reduced.
2CaCO3 + 2SO2 + O2 → 2CaSO4 + 2CO2
Answer:
2CaCO3 + 2SO2 + O2 → 2CaSO4 + 2CO2
+2 +4 -2
+4 -2
0
+2 +6 –2
+4 -2
This is a redox reaction, because some of the oxidation numbers changed as a result of the
reaction.
The oxidation number of oxygen decreases from 0 to -2. Therefore, oxygen is reduced.
4NH3 + 5O2 → 4NO + 6H2O
Answer:
4NH3 + 5O2 → 4NO + 6H2O
-3 +1
0
+2 -2
+1 –2
22
This is a redox reaction, because some of the oxidation numbers changed as a result of the
reaction.
The oxidation number of oxygen decreases from 0 to -2. Therefore, oxygen is reduced.
S2O32- + 2H+ → SO2 + S + H2O
Answer:
S2O32- + 2H+ → SO2 + S + H2O
+2 -2
+1
+6 -2
+1 –2
0
This is a redox reaction, because some of the oxidation numbers changed as a result of the
reaction.
The oxidation number of sulfur decreases from +2 to 0. Therefore, sulfur is reduced.
Using oxidation numbers, balance the following equation:
Cr2O72- + Cl- + H+ → Cr3+ + Cl2 + H2O
Answer:
1
Cr2O72- + Cl- + H+ → Cr3+ + Cl2 + H2O
+6 -2
2
-1
Cl - 1e →
1
Oxidised
0
+6
–1
+1 –2
Reduced
Cl
Cr2 + 6e- → 2Cr
2Cl
0
+3
-
4
+3
Cr + 3e- → Cr
+6
3
+1
........ (1)
+3
- 2e → Cl2 ........ (2)
-
0
Multiply half equation (1) by 1, and multiply half equation (2) by 3
Cr2 + 6e- → 2Cr
6Cl
- 6e- → 3Cl2
-----------------------------Cr2- + 6Cl- → 2Cr3+ + 3Cl2
5
Cr2O72- + 6Cl- → 2Cr3+ + 3Cl2
6
Cr2O72- + 6Cl- + H+ → 2Cr3+ + 3Cl2 + H2O
Balancing by inspection:
Cr2O72- + 6Cl- + 14H+ → 2Cr3+ + 3Cl2 + 7H2O
23
Question 22
4.41 g of ammonium iron (II) sulfate crystals, (NH4)2SO4.FeSO4.6H2O, was dissolved in
deionised water to which some sulfuric acid had been added. The solution was then made up
accurately to 250 cm3. A pipette was used to measure 25 cm3 of this solution into a conical
flask and a further 10 cm3 of dilute sulfuric acid were added. A burette was filled with a
solution of potassium manganate(VII) and a number of titrations were carried out. The mean
titration result was 22.5 cm3. The equation for the titration reaction is
MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
(i) Calculate the concentration of the potassium manganate (VII) solution (i) in mol l-1,
(ii) in g l-1.
Answer:
(i) 4.41 g = 4.41 / 392 = 0.01125 moles in 250 cm3
= 0.045 moles per litre
MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
V1 x M1 x n2 = V2 x M2 x n1
22.5 x M1 x 5 = 25 x 0.045 x 1
M1 = 25 x 0.045 x 1 / (22.5 x 5)
= 0.01 mol l-1
Concentration of potassium manganate(VII) solution = 0.01 mol l-1
(ii) Concentration in g / l = 0.01 x 158 = 1.58 g / l
Question 23
The following experiment was carried out to find the mass of iron in an iron tablet. Five iron
tablets of total mass 1.75 g were dissolved in dilute sulfuric acid, and the solution made up to
250 cm3 with deionised water. 25 cm3 of this solution required 18.5 cm3 of a previously
standardised 0.005 M potassium manganate(VII) solution for complete reaction. The equation
for the reaction is
MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
(f) Assuming that the tablets are exactly equal in mass, calculate the mass of iron in each
tablet.
Answer:
MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
V1 x M1 x n2 = V2 x M2 x n1
18.5 x 0.005 x 5 = 25 x M2 x 1
24
M2 = 18.5 x 0.005 x 5 / (25 x 1)
= 0.0185 mol l-1
Concentration of iron(II) solution = 0.0185 mol l-1
Volume of iron(II) solution in total = 250 cm3
Moles of iron in this volume = 0.0185 / 4 = 0.004625
Mass of iron in this volume = 0.004165 x 56 = 0.259 g
Mass of iron in each tablet = 0.259 / 5 = 0.0518 g
Question 24
In an experiment to determine the concentration of a sodium thiosulfate solution, a 25.0 cm3
portion of a 0.05 M solution of iodine (previously standardised) was pipetted into a conical
flask and the sodium thiosulfate solution was titrated against it. Three titrations were carried
out and the titration results were 28.6 cm3, 28.2 cm3 and 28.3 cm3. The equation for the
titration reaction is
I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62- (aq)
Show clearly, using oxidation numbers, where oxidation and reduction occur in this reaction.
Answer:
(a)
I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62- (aq)
0
+2 -2
-1
+2.5 –2
The oxidation number of sulfur increases from +2 to +2.5. Therefore, sulfur is oxidised.
The oxidation number of iodine decreases from 0 to -1. Therefore, iodine is reduced.
(g) Calculate the concentration of the sodium thiosulfate solution (i) in mol l-1, (ii) in g l-1.
Answer:
Average titre = 28.2 + 28.3 / 2 = 28.25 cm3
I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62- (aq)
V1 x M1 x n2 = V2 x M2 x n1
25 x 0.05 x 2 = 28.25 x M2 x 1
M2 = 25 x 0.05 x 2 / 28.25
= 0.0885 M
Concentration of sodium thiosulfate solution = 0.0885 mol l-1
= 0.0885 x 158 = 13.98 g l-1
Question 25
In an experiment to determine the concentration of sodium hypochlorite in domestic bleach,
25 cm3 of the bleach was first diluted to 250 cm3 with deionised water. It was found that the
iodine solution produced by the reaction of 25 cm3 of the diluted bleach with acidified
25
potassium iodide solution required 19.4 cm3 of a 0.16 M sodium thiosulfate solution for
complete reaction.
The equation for the reactions are:
ClO-(aq) + 2I-(aq) + 2H+(aq) → Cl-(aq) + I2(aq) + H2O(l)
I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62- (aq)
(e) Calculate the concentration of sodium hypochlorite in the household bleach in moles per
litre.
Express this concentration in terms of % w/v.
Answer:
I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62- (aq)
V1 x M1 x n2 = V2 x M2 x n1
25 x M1 x 2 = 19.4 x 0.16 x 1
M1 = 19.4 x 0.16 x 1 / 50
= 0.0621 M
Concentration of iodine solution
= 0.0621 M
Concentration of hypochlorite in diluted bleach solution
= 0.0621 M
Concentration of hypochlorite in undiluted bleach solution
= 0.0621 x 10 M
= 0.621 M
= 74.5 x 0.621 g l-1
= 46.26 g l-1
Percentage (w/v) of hypochlorite in bleach
= 46.26 x 100 / 1000
= 4.63%
26