JEE Main Chemistry - Oxidation Reduction
Exam Duration: 60 Mins Total Questions : 30
In the reaction \(2Mn{ O }_{ 4 }^{ - }+5{ C }_{ 2 }{ O }_{ 4 }^{ 2- }+16{ H }^{ + }\longrightarrow 2M{ n }^{ 2+ }+8{ H }_{ 2 }O+A\) the product A is
- (a)
CO2
- (b)
C
- (c)
CO
- (d)
\(C{ O }_{ 3 }^{ 2- }\)
CO2
In oxygen difluoride the oxidation number of oxygen is
- (a)
-2
- (b)
-1
- (c)
+2
- (d)
+1
The formula of oxygen defluoride is OF2. The oxidation number of oxygen is +2
The oxidation number of hydrogen in calcium hydride is
- (a)
+1
- (b)
-1
- (c)
+2
- (d)
+2
The formula of calcium hydride is CaH2. The alkaline earth metals have oxidation number +2, hence the oxidation number of hydrogen in CaH2 is -1
The equivalent weight of KMnO4 in neutral medium is
- (a)
31.6
- (b)
158
- (c)
52.66
- (d)
26.33
\(2KMnO4+H2O\longrightarrow 2Mn{ O }_{ 2 }+2KOH+3[O]\) molecular weight of KMnO4 = 158
Equivalent weight = \(\frac { 2\times 158 }{ 6 } =52.66\)
The oxidation state of the two nitrogen in NH4 NO3 is
- (a)
+ 2 and + 3
- (b)
+ 5 and + 5
- (c)
+1.5 and 2.5
- (d)
0.5 and 1.5
The oxidation number of nitrogen in NH4+ is x-4 =1 or x=+5 and in NO-3 is
x-6=-1 or x=+5.
Which one of the following equations represents the change \(Mn{ O }_{ 4 }^{ - }\longrightarrow Mn{ O }_{ 2 }\) in basic solution?
- (a)
\(Mn{ O }_{ 4 }^{ - }+4{ H }_{ 2 }O+3e\longrightarrow Mn{ O }_{ 2 }+2{ H }_{ 2 }O+{ OH }^{ - }\)
- (b)
\(Mn{ O }_{ 4 }^{ - }+4{ OH }^{ - }\longrightarrow Mn{ O }_{ 2 }+2{ H }_{ 2 }O+{ 3e }\)
- (c)
\(Mn{ O }_{ 4 }^{ - }+4{ H }^{ + }-3e\longrightarrow Mn{ O }_{ 2 }+2{ H }_{ 2 }O\)
- (d)
\(Mn{ O }_{ 4 }^{ - }+4{ H }^{ + }\longrightarrow Mn{ O }_{ 2 }+2{ H }_{ 2 }O+3e\)
Only in this reaction OH- remains in the product
In a redox reaction, the oxidant
- (a)
shares electrons with the reductant
- (b)
gains electrons from the reductant
- (c)
transfers electrons to the reductant
- (d)
neither shares nor transfers electrons to the reductant
Oxidant or oxidising agent is an acceptor of electrons and gets itself reduced whereas reductant or reducing agent is a donor of electons and gets itself oxidised.
In which one of the following the oxidation number of sulphur is zero?
- (a)
H2SO3
- (b)
H2SO4
- (c)
H2SO5
- (d)
(CH3)2SO
In (CH3)2SO the oxidation number of sulphur can be calculated as
(4-3)x2+x-2=0 or x=0
Hydrogen peroxide can act as a
- (a)
reducing agent as well as oxidising agent
- (b)
reducing agent only
- (c)
oxidising agent only
- (d)
neither oxidising nor reducing agent
As oxidising agent:
H2O2 ⟶ H2O+[O] x4
PbS+4[0] ⟶ PbSO4
__________________
4H2O2+Pbs⟶4H2O+PbSO4
As a reducing agent:
H2O2⟶H2O+[O]
O3⟶O2+[O]
__________________
H2O2+O3⟶H2O+2O2
The oxidation numbers of oxygen in H2O and H2O2 are
- (a)
- 1 and - 1
- (b)
- 2 and - 1
- (c)
- 2 and - 2
- (d)
- 1 and - 2
By presumption oxidation number of oxygen is -2 in all compounds except in OF2 where it is +2 and in peroxides like BaO2 and H2O2 where it is -l.
The reductant
- (a)
adds electrons
- (b)
gets reduced
- (c)
loses electrons
- (d)
NONE OF THE ABOVE
Reductant or reducing agent is donor of electrons.
In a redox reaction
- (a)
oxidant gets reduced
- (b)
oxidant loses electrons
- (c)
reductant gets reduced
- (d)
reductant gains electrons
Oxidant gets reduced.
In the reaction \({ I }_{ 2 }+2{ S }_{ 2 }{ O }_{ 3 }^{ 2- }\longrightarrow { 2I }^{ - }+A,\) the product A is
- (a)
SO32-
- (b)
S2O42-
- (c)
S4O62-
- (d)
SO42-
I2+2Na2S2O3 ⟶2NaI+Na2S4O6
Sodium tetrathionate
Copper is precipitated from the solution of CuSO4 by adding Zn dust. This is because
- (a)
Cu2+ is oxidised
- (b)
Cu2+ is reduced
- (c)
Zn2+ is reduced
- (d)
SO42+ is oxidised
Cu2+ is changed to Cu0 by gaining two electrons Cu2++2e ⟶ Cu0
When copper turnings are added to AgNO3 solution, the solution becomes blue because
- (a)
Cu is oxidised to Cu2+
- (b)
Cu2+ is reduced to Cu+
- (c)
Cu2+ is reduced to Cu
- (d)
Cu is oxidised to Cu+
Cu is oxidised to Cu2+ which is blue in colour.
In the reaction \({ Sn }^{ 2+ }+2{ Hg }^{ 2+ }\longrightarrow { Sn }^{ 4+ }+{ Hg }_{ 2 }^{ 2+ }\) the oxidant is
- (a)
\({ Sn }^{ 2+ }\)
- (b)
\({ Hg }_{ 2 }^{ 2+ }\)
- (c)
\({ Sn }^{ 4+ }\)
- (d)
\({ Hg }_{ 2 }^{ 2+ }\)
Oxidant is that which is reduced itself. Here \({ Hg }^{ 2+ }\)is reduced to \({ Hg }_{ 2 }^{ 2+ }\)
In the reaction \({ 2I }^{ - }+{ Cl }_{ 2 }\longrightarrow { I }_{ 2 }+2C{ l }^{ - }\) the reductant is
- (a)
I-
- (b)
Cl2
- (c)
I2-
- (d)
Cl-
Reductant is that which is oxidised itself. Here 2I- is oxidised to I2
In which one of the following reactions H2O2 acts as an oxidising agent?
- (a)
\({ 2I }^{ - }+2{ H }^{ + }+{ H }_{ 2 }{ O }_{ 2 }\longrightarrow { I }_{ 2 }+2{ H }_{ 2 }{ O }\)
- (b)
\({ Ag }_{ 2 }O+{ H }_{ 2 }{ O }_{ 2 }{ \longrightarrow 2Ag+{ H }_{ 2 }{ O } }+{ O }_{ 2 }\)
- (c)
\(2Mn{ O }_{ 4 }^{ - }+6{ H }^{ + }+5{ H }_{ 2 }{ { O }_{ 2 } }{ \longrightarrow 2M{ n }^{ 2+ }+{ 8H }_{ 2 }{ O } }+{ 5O }_{ 2 }\)
- (d)
\({ O }_{ 2 }+2{ H }_{ 2 }{ O }_{ 2 }{ \longrightarrow { 2O }_{ 2 }+{ 2H }_{ 2 }{ O } }\)
When a substance acts as an oxidising agent there is a decrease in oxidation number. In H2O2 oxidation number of oxygen is +2 and in water it is +1.
The oxidation number of sulphur in Na2S4O6 is
- (a)
2.5
- (b)
2 and 3 (two S have + 2 and the other two have + 3)
- (c)
2 and 4 (two S have + 2 and the other two have + 4)
- (d)
5 and 0 (two S have + 2 and the other two have 0)
In Na2S4O6 the oxidation state of sulphur is 2+4x-2x6=0 or 4x=10 or x=2.5.
In which of the following oxidation number of chlorine is+ 5?
- (a)
Cl2O7
- (b)
\({ ClO }_{ 3 }^{ - }\)
- (c)
\({ ClO }^{ - }\)
- (d)
\({ ClO }_{ 4 }^{ - }\)
(b) x-6=-1
or x=+5
(a) 2x-14=0
x=+7
(c) x-2=-1
x=+1
(d) x-8=-1
x=+7
The oxidation number of carbon in C12H22O11 is
- (a)
12
- (b)
22
- (c)
11
- (d)
Zero
12C + 22 - 22 = 0
12C = 0 or C = 0
The oxidation number of C in CH4 , CH3 Cl, CH2Cl2 , CHCl3 are CCl4 is respectively
- (a)
0, 2, -2, 4, 4
- (b)
2, 4, 0, -2, -4
- (c)
-4, -2, 0, 2, 4
- (d)
4, 2, 0, -2, -4
For the following reaction in the acidic solution \(Mn{ O }_{ 4 }^{ - }+8{ H }^{ + }+5e\longrightarrow Mn^{ 2+ }+4{ H }_{ 2 }O\) which of the following gives the oxidation numbers of the manganese on each side of the equation.
- (a)
+ 4 to + 2
- (b)
-1 to + 2
- (c)
+ 7 to + 5
- (d)
+ 7 to + 2
The brown ring complex compound is formulated as \([Fe\left( { H }_{ 2 }O \right) _{ 5 }(NO)]{ SO }_{ 4 }.\) The oxidation state of iron in this compound is
- (a)
0
- (b)
1
- (c)
2
- (d)
3
Fe + 0 X 5 + 0 - 2 = 0
Fe = + 2
The charge on coblt in \([Co\left( { CN }_{ 6 } \right) ]^{ -3 }\) is
- (a)
- 3
- (b)
- 6
- (c)
+ 3
- (d)
+ 6
Oxidation numbr of Fe in \({ K }_{ 3 }Fe({ CN })_{ 6 }\) is
- (a)
+ 1
- (b)
+ 2
- (c)
+ 3
- (d)
+ 4
3 + Fe - 6 = 0
Fe = + 3
Oxidation numbr of oxygen atom in potassium peroxide is
- (a)
0
- (b)
\(????-\frac { 1 }{ 2 } \)
- (c)
- 4
- (d)
- 2
KO2 = 0
or K - 4 = 0
K = 4
The colour of K2Cr2O7 changes from red orange to lemon orange yellow on treatment with aqueous KOH because of
- (a)
reduction of Cr (VI) to Cr (III)
- (b)
formation of chromium hydroxide
- (c)
conversion of dichromate ion to chromate
- (d)
oxidation of potassium hydroxide to potassium peroxide
\({ K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }+2KOH\Longleftrightarrow 2{ K }_{ 2 }{ Cr }{ O }_{ 4 }+{ H }_{ 2 }O\)
Which substance is serving as a reducing agent in the following reaction?\(14{ H }^{ + }+{ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }+3\quad Ni\longrightarrow 2{ Cr }^{ 3+ }+7{ H }_{ 2 }O+3{ Ni }^{ 2+ }\)
- (a)
\({ H }_{ 2 }O\)
- (b)
Ni
- (c)
\({ H }^{ + }\)
- (d)
\({ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }\)
Which of the following is a redox reaction?
- (a)
\(Nacl+KN{ O }_{ 3 }\longrightarrow Na{ NO }_{ 3 }+Kcl\)
- (b)
\(Ca{ C }_{ 2 }{ O }_{ 4 }+2HCl\longrightarrow CaC{ l }_{ 2 }+{ H }_{ 2 }{ C }_{ 2 }{ O }_{ 4 }\)
- (c)
\(Mg\left( OH \right) _{ 2 }+2N{ H }_{ 4 }Cl\longrightarrow Mg{ Cl }_{ 2 }+2N{ H }_{ 4 }OH\)
- (d)
\(Zn+2AgCN\longrightarrow 2Ag+Zn\left( CN \right) _{ 4 }\)