Important MCQs on Redox Reactions with Explanation

Ans. (a) + 7

Explanation:  Cl shows different oxidation states as -1 to +7 due to the vacant d orbital. As oxygen is more electronegative than Cl. Oxygen size is small hence it is more electronegative and shows -2 oxidation states.

Here Cl₂O₇ then equation is: 2x + 7 × (-2) = 0

x = +7 hence the oxidation state of Cl is +7. 

Ans. (b) Formation of H₂O₂ by the oxidation of H₂O.

Explanation: Autoxidation is any oxidation that occurs in the presence of oxygen. It is used to refer to the degradation of organic compounds in air (as a source of oxygen). Autoxidation produces hydroperoxides and cyclic organic peroxides. These species may react further in order to form many other products. The process is relevant to many phenomena including aging, paint, spoilage of foods, degradation of petrochemicals, and the industrial production of chemicals. 

Ans. (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. In other words, it can be described as 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)

Ans. (b) More with K₂Cr₂O₇

Explanation: The reason why the amount of Fe oxidized will be more with K₂Cr₂O₇ is the change in the oxidation state (or number) or n factor is greater with KMnO₄.  Also, K₂Cr₂O₂ is a very strong oxidizing agent and holds the ability to take electrons but KMnO₄ is stronger than K₂Cr₂O₇.

Ans. (a) Formation of slaked lime from quicklime

Explanation: Here, in this reaction

CaO + H₂O →Ca(OH)₂

The oxidation number doesn’t change so it's not a redox reaction.

Ans. (c) +3

Explanation: First to find the oxidation number of Nitrogen in N₂H₄

Oxidation number of H = +1

Let the oxidation number of nitrogen be x

2x + 4(1) = 0

2x = -4

x = -2

Each nitrogen atom has a -2 oxidation number. So taking both nitrogen atoms into account gives oxidation number -4.

The change in oxidation number of nitrogen on losing 10 mol of electrons while considering no change in oxidation number of hydrogen atoms

-4 – (-10) = +6

Therefore, the oxidation number of 2 nitrogen atoms in compound Y is +6. Hence, the oxidation number of each nitrogen atom will be +3 in the new compound Y.

Ans. (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

Ans. (a) 0

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

Ans. (a) Formation of Fe(CO)₅ from Fe.

Explanation: The oxidation number of Fe in Fe(CO)₅ is zero.

The oxidation state of iron is zero in both Fe and Fe(CO)₅.

3Fe + 4H₂O → Fe₃O₄ + 4H₂

Fe → Fe₂O₃.xH₂O

(+3)

CuSO₄(aq) + Fe (s) → FeSO₂(aq) + Cu(s)

(0) (+2)

Ans. (b) Two

Explanation: In an alkaline medium, the reduction of KMnO₄ with KI will take place as

2 KMnO₄ + H₂O → 2KOH + 2MnO₂

KI + 3[O] → KIO₃

Hence the overall reaction is

KI + 2KMnO₄ + H₂O → KIO₃ + 2KOH + 2 MnO₂

So, one mole of KI will reduce two moles of KMnO₄.

Ans. (c) Equal to the number of moles of electrons lost or gained by one mole of reductant or oxidant.

Explanation: For redox reactions, it is considered a change in their oxidation number or a change in their reduction number on both sides of a chemical reaction.

Ans. (d) KMnO₄.

Explanation: The electronic configuration of Mn is:

Mn(25) = [Ar]3d5 4s2, 4p0

It can lose all 7 electrons in an excited state.

Hence, the maximum oxidation state exhibited by Mn is +7 which is in KMnO₄.

Ans. (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.

Ans. (b) Magnesium

Explanation: Most metals such as Al, Cu, Fe, etc. react with dilute acids to produce hydrogen gas but magnesium is an exception. Magnesium, being an active metal, liberates hydrogen gas as it is allowed to react with dilute HCl. Thus, all rest of the given metals react with dilute acids.

Mg + 2HCl → MgCl₂ + H₂

Ans. (d) 10

Explanation: Oxidation state of Ba in general = +2 and of O = −2

Applying the formula, the Sum of the total oxidation state of all atoms = the overall charge on the compound.

Let the 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 the maximum possible oxidation state for Xe = +8.

Ans. (d) Oxidation of Cu atoms.

Explanation: When copper turnings are added to silver nitrate solution, the solution becomes brown in color after some time because copper is more reactive than silver so it displaces silver from silver nitrate solution and leads to the formation of copper nitrate solution.

Ans. (d) D

Explanation: The reduction potential of D is minimum i.e. −0.402 V. Thus, the oxidation potential of D is maximum which is +0.402 V. D can oxidize itself and reduce others.

The aqueous solution A will be there in its ionic form and it can be reduced by D as its reduction potential is way higher than D. 

Thus, it is concluded that D can replace A from its aqueous solution.