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Class 11 Chemistry Chapter 8 Redox Reactions MCQs

Questions
1 Identify disproportionation reaction
A CH4 + 2O2→ CO2 + 2H2O
B CH4 + 4Cl2→ CCl4 + 4HCl
C 2F2 + 2OH-→ 2F- + OF2 + H2O
D 2NO2 + 2OH–→ NO2- + NO3- + H2O

Answer:2NO2 + 2OH–→ NO2- + NO3- + H2O
2 The largest oxidation number exhibited by an element depends on its outer electronic configuration. With which of the following outer electronic configurations the element will exhibit the largest oxidation number?
A 3d1 4s2
B 3d3 4s2
C 3d5 4s1
D 3d5 4s2

Answer:3d5 4s2
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3 Which of the following arrangements represent increasing oxidation number of the central atom?
A CrO2- , CIO-3, CrO2-4 , MnO-4
B CIO-3, CrO2-4 , MnO-4 , CrO-2
C CrO2+4 , MnO-4 , CrO-2 , CIO3-
D CrO24-, MnO4- , CrO2- , CIO3-

Answer:CrO2- , CIO-3, CrO2-4 , MnO-4
4 In which of the following compounds, an element exhibits two different oxidation states.
A NH2OH7
B NH4NO3
C N2H4
D N3H

Answer:NH4NO3
5 The oxidation number of an element in a compound is evaluated on the basis of certain rules. Which of the following rules is not correct in this respect?
A The oxidation number of hydrogen is always +1.
B The algebraic sum of all the oxidation numbers in a compound is zero.
C An element in the free or the uncombined state bears oxidation number zero.
D In all its compounds, the oxidation number of fluorine is – 1

Answer:The oxidation number of hydrogen is always +1.
6 Thiosulphate reacts differently with iodine and bromine in the reactions given below:

2S2O32- + I2 → S4O62- + 2I–

S2O32-+ 2Br2+ 5H2O → 2SO42- + 2Br–+ 10 H+

Which of the following statements justifies the above dual behaviour of thiosulphate?

A Bromine is a stronger oxidant than iodine.
B Bromine is a weaker oxidant than iodine.
C Thiosulphate undergoes oxidation by bromine and reduction by iodine in these reactions.
D Bromine undergoes oxidation and iodine undergoes a reduction in these reactions.

Answer:Bromine is a stronger oxidant than iodine.
7 Using the standard electrode potential, find out the pair between which redox reaction is not feasible.

E°Values: Fe3+/ Fe2+ = +0.77; I2/I- = +0.54;

Cu2+/ Cu = -0.34; Ag+ /Ag = + 0.80 V

A Fe3+ and I-
B Ag+ and Cu
C Fe3+ and Cu
D Ag and Fe3+

Answer:Ag and Fe3+
8 Eᶱvalues of some redox couples is given below. Based on these values

choose the correct option.

Eᶱ values : Br2/Br–= + 1.90; Ag+/Ag(s) = + 0.80

Cu2+/Cu(s) = + 0.34; I2(s)/I– = + 0.54

A Cu will reduce Br–
B Cu will reduce Ag
C Cu will reduce I–
D Cu will reduce Br2

Answer:Cu will reduce Br2
9 The more positive the value of Eᶱ, the greater is the tendency of the species to get reduced. Using the standard electrode potential of redox couples given below to find out which of the following is the strongest oxidising agent. Eᶱ Values : Fe3+/Fe2+ = + 0.77; I2(s)/I– = + 0.54;

Cu2+/Cu = + 0.34; Ag+/Ag = + 0.80V

A Fe3+
B I2(s)
C Cu2+
D Ag

Answer:Ag
10 Which of the following is not an example of redox reaction?
A CuO + H2→ Cu + H2O
B Fe2O3 + 3CO → 2Fe + 3CO2
C 2K + F2→ 2KF
D BaCl2 + H2SO4→ BaSO4 + 2HCl

Answer:BaCl2 + H2SO4→ BaSO4 + 2HCl
11 he oxidation number of an element in a compound is evaluated on the basis of certain rules. Which of the following rules is not correct in this respect?
A The oxidation number of hydrogen is always +1.
B The algebraic sum of all the oxidation numbers in a compound is zero.
C An element in the free or the uncombined state bears oxidation number zero.
D In all its compounds, the oxidation number of fluorine is -1.

Answer:The oxidation number of hydrogen is always +1.
12 In which of the following compounds, an element exhibits two different oxidation states.
A NH2OH
B NH4NO3
C N2H4
D N3H

Answer:NH4NO3
13 The largest oxidation number exhibited by an element depends on its outer electronic configuration. With which of the following outer electronic configurations the element will exhibit largest oxidation number?
A 3d14s2
B 3d2 4s2
C 3d54s1
D 3d54s2

Answer:3d54s2
14 Identify disproportionation reaction
A CH4 + 202 → C02 + 2H20
B CH4 + 4C12 → CC14 + 4HCl
C 2F2 + 20H–→2F– + OF2 + H20
D 2N02 + 20H– → N02 + NO–3 + H20

Answer:2N02 + 20H– → N02 + NO–3 + H20
15 The oxidation number of an element in a compound is evaluated on the basis of certain rules. Which of the following rules is not correct in this respect?
A The oxidation number of hydrogen is always +1.
B The algebraic sum of all the oxidation numbers in a compound is zero.
C An element in the free or the uncombined state bears oxidation number zero.
D In all its compounds, the oxidation number of fluorine is – 1.

Answer:The oxidation number of hydrogen is always +1.
16 In which of the following compounds, an element exhibits two different oxidation states.
A NH2OH
B NH4NO3
C N2H4
D N3H

Answer:N2H4
17 The more positive the value of E⊖ , the greater is the tendency of the species to get reduced. Using the standard electrode potential of redox couples given below find out which of the following is the strongest oxidising agent.

Redox Reactions

A Fe3+
B I2 (s)
C Cu2+
D Ag+

Answer:Ag+
18 E⊖ values of some redox couples are given below. On the basis of these values choose the correct option.

Redox Reactions

A Cu will reduce Br–
B Cu will reduce Ag
C Cu will reduce I–
D Cu will reduce Br2

Answer:Cu will reduce Br2
19 Using the standard electrode potential, find out the pair between which redox reaction is not feasible.

Redox Reactions

A Fe3+ and I–
B Ag+ and Cu
C Fe3+
D Ag and Fe3+

Answer:Ag and Fe3+
20 Identify disproportionation reaction
A CH4 + 2O2 → CO2 + 2H2O
B CH4 + 4Cl2→ CCl4 + 4HCl
C 2F2 + 2OH– → 2F – + OF2 + H2O
D 2NO2 + 2OH– ⎯→ NO2– + NO3– + H2O

Answer:2NO2 + 2OH– ⎯→ NO2– + NO3– + H2O

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