Redox Reactions Class 11 MCQ Online Test With Answers Questions

Answer: (a) + 7
Explanation:
Cl show different oxidation state as -1 to +7 due to vacant d orbital. As oxygen is more electronegative than Cl. Oxygen size is small hence its more electronegative and show -2 oxidation states.
Here Cl₂O₇ then equation is: 2x + 7 × (-2) = 0
x = +7 hence oxidation state of Cl is +7. I think you get your answer how to find oxidation state.

Answer: (b) Formation of H₂O₂ by the oxidation of H₂O.
Explanation:
Autoxidation is any oxidation that occurs in presence of oxygen. The term is usually used to describe the degradation of organic compounds in air (as a source of oxygen). Autoxidation produces hydroperoxides and cyclic organic peroxides. These species can react further to form many products. The process is relevant to many phenomena including aging, paint, and spoilage of foods, degradation of petrochemicals, and the industrial production of chemicals. Autoxidation is important because it is a useful reaction for converting compounds to oxygenated derivatives, and also because it occurs in situations where it is not desired (as in the destructive cracking of the rubber in automobile tires or in rancidification).

Water automatically gets oxidised to hydrogen peroxide.

Answer: (c) Oxidation Potential
Explanation:
The magnitude of the electrode potential of a metal is a measure of its relative tendency to lose or gain electrons. i.e., it is a measure of the relative tendency to undergo oxidation (loss of electrons) or reduction (gain of electrons).
M → Mⁿ⁺ + ne^– (oxidation potential)
Mⁿ⁺ + ne^– → M (reduction potential)

Answer: (b) More with K₂Cr₂O₇
Explanation:
The reason due to which the amount of Fe oxidised will be more with ​K₂Cr₂O₇ is:
the change in the oxidation state (or number) or n factor is greater with KMnO₄
Also, ​K₂Cr₂O₇ is a very strong oxidising agent and holds the ability to take the electrons but ​KMnO₄ is more stronger than ​K₂Cr₂O₇.

Answer: (a) Formation of slaked lime from quick lime
Explanation:
Here, in this reaction
CaO + H₂​O →Ca(OH)₂
Oxidation number doesn’t change so its not a redox reaction.

Answer: (c) +3
Explanation:
First to find oxidation number of Nitrogen in N₂H₄
Oxidation number of H = +1
Let oxidation number of nitrogen be x
2x + 4(1) = 0
2x = -4
x = -2
Each nitrogen atom has -2 oxidation number. So taken both nitrogen atoms in account gives oxidation number -4.
Change in oxidation number of nitrogen on losing 10 mol of electrons (considering no change in oxidation number of hydrogen atoms)
-4 – (-10) = +6
Therefore, oxidation number of 2 nitrogen atoms in compound Y is +6. Hence, oxidation number of each nitrogen atom will be +3 in new compound Y.

Answer: (c) 8.1
Explanation:
The volume can be calculated :
N₁V₁ = N₂V₂
N₁ = Normality of H₂SO₄ = 0.5 × 2 = 1 N
V₁ = Volume of H₂SO₄
Molar mass of copper(II) carbonate = 123.5 g
N₂ = Normality of copper (II) carbonate = (0.5×2)/(123.5) N
V₂ = Volume of copper (II) carbonate = 1000 mL
So, after applying the formula,
1 × V₁ = (0.5×2)/(123.5)×1000
Hence, V₁ = 8.09 mL
= approx. 8.1 mL

Answer: (a) 0
Explanation:
CO (carbonyl) is a neutral ligand, hence oxidation state of Cr in Cr (CO)₆ is zero

Answer: (a) Formation of Fe(CO)₅ from Fe.
Explanation:
Oxidation number of Fe in Fe(CO)₅ is zero.
In both Fe and Fe(CO)₅, the oxidation state of iron is zero.
3Fe + 4H₂O → Fe₃O₄ + 4H₂
team
rusting
Fe → Fe₂O₃.xH₂O
(+3)
CuSO₄(aq) + Fe (s) → FeSO₄(aq) + Cu(s)
(0) (+2)

Answer: (b) Two
Explanation:
In alkaline medium the reduction of KMnO₄ with KI will takes place as
2 KMnO₄ + H₂O → 2 KOH + 2 MnO₂
KI + 3[O] → KIO₃
Hence the overall reaction is
KI + 2KMnO₄ + H₂O → KIO₃ + 2 KOH + 2 MnO₂
So, one mole of KI will reduced two moles of KMnO₄

Answer: (c) CrO^2-₄ → Cr₂O₂^-7
Explanation:
In VO²⁺ →V₂​O₃, V is reduced from +4 to +3 oxidation state.
In Na → Na²⁺, Na is oxidised from to +1 oxidation state.
In CrO₄^-2​ → Cr₂O₇^-2, Cr remains in same oxidation state +6.
In Zn⁺² → Zn, Zn is reduced from +2 to 0 oxidation state.

Answer: (d) 50 mL of 0.1 M C₂H₂O₄
Explanation:
2MnO₄^– + 5C₂O₄^2- + 16H+ → 2Mn²⁺ + 10CO₂ + 8H₂O
Therefore, 2 moles of MNO₄^– equivalent to 5 moles of C₂O₄^2-
20 mL of 0.1 M KMnO₄ = 2 moles of KMnO₄
Also, 50 mL of 0.1 M C₂H₂O₄ equivalent to 5 mol of C₂O₄^2-
Therefore, these are equivalent.

Answer: (c) +3
Explanation:
First to find oxidation number of Nitrogen in N₂H₄
Oxidation number of H = +1
Let oxidation number of nitrogen be x
2x + 4(1) = 0
2x = -4
x = -2
Each nitrogen atom has -2 oxidation number. So taken both nitrogen atoms in account gives oxidation number -4.
Change in oxidation number of nitrogen on losing 10 mol of electrons (considering no change in oxidation number of hydrogen atoms)
-4 – (-10) = +6
Therefore, oxidation number of 2 nitrogen atoms in compound Y is +6. Hence, oxidation number of each nitrogen atom will be +3 in new compound Y.

Answer: (c) Equal to Number of moles of electrons Lost or gained by one mole of reductant or oxidant.
Explanation:
For redox reaction it is considered as change in their oxidation number or change in their reduction number in both side of a chemical reaction.

Answer: (d) KMnO₄.
Explanation:
The electronic configuration of Mn is:
Mn(25) = [Ar]3d⁵ 4s², 4p⁰
In excited state, it can lose its all 7 electrons.
Hence, maximum oxidation sate exhibited by Mn is +7 which is in KMnO₄.

Answer: (a) Increase in oxidation number
Explanation:
Oxidation process Involves:-
Addition of O₂ or electronegative element
Removal of H/electropositive element
Loss of electrons
Increase in oxidation number

Answer: (b) Magnesium
Explanation:
Most of the metals such as Al, Cu, Fe etc. reacts with dilute acids to produce hydrogen gas but magnesium is an exception. Magnesium being an active metal liberates dihydrogen gas as it is allowed to react with dilute HCl. Thus all the given metals react with dilute acids.
Mg + 2HCl → MgCl₂ + H₂

Answer: (d) 10
Explanation:
Oxidation state of Ba in general = +2 and of O = −2
Applying formula, Sum of total oxidation state of all atoms = Overall charge on the compound.
Let oxidation state of Xe in BaXeO₆ be x.
2 + x + 6(−2) = 0,
x = 10
But oxidation state 10 is not possible for Xe. In this case the oxidation state of Xe is equal to maximum possible oxidation state for Xe = +8.

Answer: (d) Oxidation of Cu atoms.
Explanation:
When copper turnings are added to silver nitrate solution, the solution becomes brown in color after sometime because copper is more reactive than silver so it displaces silver from silver nitrate solution and form copper nitrate solution.

Answer: (d) D
Explanation:
Reduction potential of D is minimum i.e. −0.402 V. Thus oxidation potential of D is maximum i.e. to +0.402 V. D can oxidise itself and reduce other.
The aqueous solution A will be present in its ionic form and can be reduced by D as its reduction potential is higher than D.
Thus D can replace A from its Aqueous solution.