Complete and balance the following equations, and identify the oxidation and reducing agents:
a) Cr2O72-(aq) + I- (aq) → 2Cr +3 + IO3- (acidic solution)
b) MnO4-(aq) + CH3OH(aq) →Mn2+(aq) + HCOOH(aq) (acidic solution)
c) I2(s) + ClO- (aq) → IO3- (aq) +Cl- (aq) (acidic solution)
d) As2O3 (s) + 2NO3-(aq) → 2H3AsO4 +N2O3 (acidic solution)
e) MnO4- + H2O +Br- → 2MnO2 + 2OH- + BrO3- (basic solution)
f) Pb(OH)42- (aq) + ClO- (aq) → PbO2(s) + Cl – (aq) (basic solution)
One Class Solution:-
Cr2O72- + I- + → Cr +3 + IO3- (acidic solution)
+6 +3
Cr2O72-→ Cr +3 REDUCTION
-1 +5
I-→ IO3- OXIDATION
Cr2O72- +14 H+ +6 e- → 2Cr +3 + 7H2O
I- + 3H2O → IO3- + 6H++ 6e-
Cr2O72- + 14 H++ I- + 3H2O → 2Cr +3 + IO3- + 6H+ + 7H2O
+6OH- → 6OH-
Cr2O72- + 8H++ I- + 3H2O → 2Cr +3 + IO3- + 7H2O
b)MnO4- + CH3OH →Mn2+ + HCOOH (acidic solution)
+7 +2
MnO4- →Mn2+ REDUCTION
-2 +2
CH3OH → HCOOH OXIDATION
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O ∕. 4
CH3OH + H2O → HCOOH + 4H+ +4 e- ∕ .5
4MnO4- + 32H+ + 20e-+ 5CH3OH + 5H2O→ 4Mn2+ + 5HCOOH + 20H+ +20 e- + 16H2O
+20OH- → 20OH-
4MnO4- + 12H+ + 5CH3OH + → 4Mn2+ + 5HCOOH + 11H2O
c)I2 + OCl- → IO3- + Cl- (acidic solution)
0 +5
I2→ IO3- OXIDATION
+1 -1
OCl- → Cl- REDUCTION
0 +5
I2 + 3H2O → 2IO3- + 6H+ + 5e- / 2
+1 -1
ClO- + H+ + 2 e-→ Cl- + H2O /5
2I2 + 6H2O + 5ClO- + 5H+ + 10 e- → 4IO3- + 12H+ + 10e- + 5Cl- + 5H2O
+5H+ → +5H+
2I2 + 6H2O + 5ClO- → 4IO3- + 7H+ + 5Cl- + 5H2O
d)As2O3 + NO3-→ H3AsO4 + N2O3 (acidic solution)
+3 +5
As2O3→2H3AsO4 OXIDATION
+5 +3
2NO3-→ N2O3 REDUCTION
As2O3 + 5H2O → 2H3AsO4 + 4H++ 4e- /4
2NO3- + 6H+ +4e- → N2O3 + 3H2O /4
As2O3 + 2NO3- + 6H+ +4e- +5H2O → 2H3AsO4 + 4H++ 4e- + N2O3 + 3H2O
+4 OH → +4OH
As2O3 + 2NO3- + 2H+ +2H2O → 2H3AsO4 + N2O3
e)MnO4- + Br- → MnO2 + BrO3- (basic solution)
+7 +4
MnO4- + 2H2O +3e- → MnO2 + 4OH- /2
-1 +5
Br- + 6OH- → BrO3-+ 3H2O + 6e- /1
2MnO4- + 4H2O +6e- + Br- + 6OH- → 2MnO2 + 8OH-+ BrO3-+ 3H2O + 6e-
+ 6 H+ → + 6 H+
2MnO4- + H2O +Br- → 2MnO2 + 2OH-+ BrO3-
f) Pb(OH)42- + ClO- → PbO2 + Cl –
+2 +4
Pb(OH)42- → PbO2 + H2O +2 e- oxidation/1
+1 -1
ClO- + H2O +2 e- → Cl – +2 OH- reduction /1
Pb(OH)42- + ClO- + H2O + 2 e- → PbO2 + H2O + Cl – + 2OH- + 2 e-
+H+ → +H+
Pb(OH)42- + ClO- → PbO2 + Cl – + 2OH-+H2O