Redox Reactions MCQ Quiz - Objective Question with Answer for Redox Reactions - Download Free PDF

CONCEPT:

Oxidation State

• The oxidation state (or oxidation number) of an element in a compound is the charge it would have if all bonds were ionic.
• Rules for determining oxidation states:
  • The oxidation state of an element in its standard form is 0 (e.g., O₂, H₂).
  • For monoatomic ions, the oxidation state is the same as the ion charge.
  • Oxygen usually has an oxidation state of -2, except in peroxides where it is -1 and in superoxides where it is -½.
  • Hydrogen usually has an oxidation state of +1 when bonded to non-metals and -1 when bonded to metals.
  • The sum of oxidation states in a neutral compound is 0, while in polyatomic ions it equals the ion charge.

EXPLANATION:

• In KO₂ (potassium superoxide):
  • Potassium (K) has an oxidation state of +1.
  • Oxygen in superoxide has an oxidation state of -½.
• In H₂O₂ (hydrogen peroxide):
  • Hydrogen (H) has an oxidation state of +1.
  • Oxygen in peroxide has an oxidation state of -1.
• In H₂SO₄ (sulfuric acid):
  • Hydrogen (H) has an oxidation state of +1.
  • Oxygen has an oxidation state of -2.
  • To balance the molecule:

    2(+1) + x + 4(-2) = 0
    ​x = +6

Therefore, sulfur (S) has an oxidation state of +6.

Therefore, the oxidation states of the underlined elements in KO₂, H₂O₂, and H₂SO₄ are respectively +1, -1, and +6.

CONCEPT:

Oxalic Acid Oxidation and Iodometric Titration

• The oxidation of oxalic acid by potassium permanganate (KMnO₄) in acidic medium is a redox reaction, where KMnO₄ acts as the oxidizing agent and oxalic acid is the reducing agent.
• The general reaction for the oxidation of oxalic acid is:

  C₂H₂O₄ + KMnO₄ + H⁺ → CO₂ + Mn²⁺ + K⁺

• After the oxidation, the remaining KMnO₄ is reacted with excess KI, liberating iodine (I₂) in the process. The liberated iodine is then titrated with sodium thiosulfate (Na₂S₂O₃) to determine the amount of iodine released, which helps calculate the moles of KMnO₄ that reacted with oxalic acid.
• Using the volume of Na₂S₂O₃ solution used in titration, we can calculate the percentage of purity of the oxalic acid sample.

EXPLANATION:

• Step 1: Moles of KMnO₄ reacted
  • Moles of KMnO₄ = Molarity × Volume = 0.1 M × 50 mL = 0.1 × 0.05 = 0.005 moles
• Step 2: Iodine liberated by the remaining KMnO₄
  • From the stoichiometry of the reaction between KMnO₄ and KI, 1 mole of KMnO₄ liberates 3 moles of iodine (I₂).
  • Thus, moles of iodine (I₂) liberated = 3 × 0.005 = 0.015 moles
• Step 3: Moles of Na₂S₂O₃ used in titration
  • Moles of Na₂S₂O₃ = Molarity × Volume = 0.05 M × 10 mL = 0.05 × 0.01 = 0.0005 moles
• Step 4: Calculate moles of iodine (I₂) titrated
  • According to the reaction between iodine and sodium thiosulfate, 1 mole of iodine reacts with 2 moles of Na₂S₂O₃.
  • So, moles of iodine = 0.0005 moles / 2 = 0.00025 moles of I₂ were liberated.
• Step 5: Calculate the moles of oxalic acid reacted
  • Each mole of oxalic acid reacts with 2 moles of KMnO₄.
  • Moles of oxalic acid = (0.00025 moles of I₂) × (1 mole of oxalic acid / 3 moles of KMnO₄) = 0.0000833 moles
• Step 6: Calculate the mass of oxalic acid used
  • Mass of oxalic acid = moles × molar mass = 0.0000833 moles × 90.03 g/mol = 0.0075 g
• Step 7: Calculate the percentage purity
  • Percentage purity = (pure oxalic acid mass / sample mass) × 100 = (0.0075 g / 8.25 g) × 100 = 8.6%

Therefore, the percentage of purity of oxalic acid is 8.6%.

CONCEPT:

Iodometric Titration

• Iodometric titration is an indirect method of determining the concentration of iodine (I₂) by estimating it through the titration of iodine released from KI solution with sodium thiosulfate (Na₂S₂O₃) solution.
• The general reaction between iodine and sodium thiosulfate is:

  I₂ + 2Na₂S₂O₃ → 2NaI + Na₂S₄O₆

• In this titration, the iodine liberated from KI solution in the presence of an oxidant is titrated with Na₂S₂O₃ to calculate the amount of iodine.
• For potassium dichromate (K₂Cr₂O₇), the iodine is liberated in the reaction:

  Cr₂O₇^2– + 14H⁺ + 6I^– → 2Cr³⁺ + 3I₂ + 7H₂O

EXPLANATION:

• Given:
  • Volume of K₂Cr₂O₇ solution = 100 mL = 0.1 L
  • Volume of Na₂S₂O₃ solution = 50 mL = 0.05 L
  • Concentration of Na₂S₂O₃ solution = 0.1 N
• Step 1: Moles of Na₂S₂O₃ used:

  Moles of Na₂S₂O₃ = Normality × Volume = 0.1 N × 0.05 L = 0.005 moles

• Step 2: Using the stoichiometry of the reaction:
  • From the reaction, we know that 1 mole of K₂Cr₂O₇ liberates 3 moles of iodine (I₂) which reacts with 1 mole of Na₂S₂O₃.
  • The moles of iodine released = 0.005 moles of Na₂S₂O₃ × (1 mole I₂ / 1 mole Na₂S₂O₃) = 0.005 moles of I₂.
• Step 3: Moles of I₂ liberated by K₂Cr₂O₇:
  • From the stoichiometry of the reaction, 1 mole of K₂Cr₂O₇ releases 3 moles of iodine (I₂).
  • Therefore, the moles of K₂Cr₂O₇ = 0.005 moles / 3 = 0.00167 moles.
• Step 4: Finding the concentration of K₂Cr₂O₇ solution:
  • Concentration (C) = moles / volume = 0.00167 moles / 0.1 L = 0.0167 M

Therefore, the volume of x is 0.0167 M.

Concept:

The redox reaction given is:
3H₃AsO₃(aq) + BrO⁻₃(aq) → Br⁻(aq) + 3H₃AsO₄(aq)

Explanation:To identify the elements undergoing reduction and oxidation, we need to determine the oxidation states of the elements involved in the reaction.

•  Oxidation states of As in H₃AsO₃ and H₃AsO₄:
  • In H₃AsO₃ (Arsenious acid): As has an oxidation state of +3.
  • In H₃AsO₄ (Arsenic acid): As has an oxidation state of +5.
  • So, As is oxidized from +3 to +5.
• . Oxidation states of Br in BrO⁻₃ and Br⁻:
  •  In BrO⁻₃ (Bromate ion): Br has an oxidation state of +5.
  •  In Br⁻ (Bromide ion): Br has an oxidation state of −1.
  • So, Br is reduced from +5 to −1.

Conclusion:

Therefore, the element undergoing oxidation is Arsenic (As) and the element undergoing reduction is Bromine (Br).

CONCEPT:

Reduction Process

• Reduction is a chemical reaction that involves the gain of electrons by a molecule, atom, or ion.
• It is one half of a redox (reduction-oxidation) reaction, where the other half involves oxidation, or the loss of electrons.
• Reduction leads to a decrease in the oxidation state of the species being reduced.

EXPLANATION:

• Given the following statements about the reduction process:
  1. Gain of electrons
  2. Addition of oxygen
  3. Loss of electrons
  4. Increase in oxidation number
• The correct answer is option 1 (Gain of electrons) because:
  • Reduction involves the gain of electrons, which results in a decrease in the oxidation state of the species being reduced.
  • Option 2 (Addition of oxygen) is incorrect because it pertains to oxidation, not reduction.
  • Option 3 (Loss of electrons) is incorrect because it describes oxidation, not reduction.
  • Option 4 (Increase in oxidation number) is incorrect because reduction results in a decrease in oxidation number.

Therefore, the correct answer is Gain of electrons.

Option 4 is correct.

Key Points

• Electrons are lost in an oxidation reaction. 
• Oxidation simply means a gain of oxygen and an oxidizing agent is a substance that oxidizes something.
• The reduction is the gain of electrons. A reducing agent reduces something in a reaction.
• As when one molecule gets reduced, another molecule gets oxidized. Most of the time, Oxidation and Reduction take place simultaneously and such a reaction is called a redox reaction.
• Redox reaction example- In a reaction between hydrogen and fluorine, hydrogen is oxidized and fluorine is reduced.

The correct answer is displacement.Key Points

• The reaction Fe + CuSO₄ -> FeSO₄ + Cu involves the displacement of copper from copper sulfate by iron.
• This is because iron is more reactive than copper and can displace it from its compound.
• This is an example of a displacement reaction, which is a type of redox (reduction-oxidation) reaction where there is a transfer of electrons between the reactants.

Additional Information

• Neutralisation reactions involve the combination of an acid and a base to form a salt and water. 
• Reversible reactions are those that can proceed in both the forward and reverse directions.
  • The given reaction is not reversible as it proceeds in only one direction.
• Iron is a chemical element with the symbol Fe and atomic number 26.
  • ​It is a metal and is found in many minerals, including hematite and magnetite.
  • Iron is used in the production of steel, which is an important material in construction and manufacturing.

Concept:

Redox Reaction:

CuO + H2 → Cu + H2O

• Oxidation and reduction are opposite in nature and they always take place simultaneously.
• In a chemical reaction, when one substance is reduced, the other is oxidized.
• In the reaction, CuO + H2 → Cu + H2O Cu is reduced whereas Hydrogen is oxidized.
• The reactions in which oxidation and reduction take place simultaneously are called Redox- reactions.

Oxidation reaction:

• Oxidation Reaction refers to a reaction in which either the addition of Oxygen takes place or the removal of Hydrogen takes place.
• It can also be said as the process of loss of one or more electrons by atoms or ions.
• The oxidation number increases during the oxidation process.
• Example is:

Mg + O2 = Mg2O

Reduction Reaction:

• A reduction reaction refers to a reaction in which either the addition of Hydrogen takes place or the removal of oxygen takes place.

• In the process of reduction, a chemical species also gains electrons.
• The oxidation number of elements decreases during reduction.
• It is the reverse of the oxidation reaction.
• Example is:

H2 + F2 → HF

​Explanation:

Let us study the species and the oxidation number of the central elements:

BaCI2 + Na2SO4→ BaSO4 + 2NaCl

• There is no change of oxidation numbers, and thus it is not a redox reaction. It is an example of a Double decomposition reaction.

 For the reaction: 2KCIO3 → 2KCI + 3O2

• Here, Cl is getting from oxidation state +5 to -1, i.e., the oxidation number decreases and it is reduced.
• Oxygen is getting oxidised.

This is an example of a redox reaction as both oxidation and reduction take place.

2KBr + CI2 → 2 KCI + Br2

• The species Br and Cl are oxidized and reduced respectively. Hence it is an example of a redox reaction.

I2 + 2S2O3-2 →  2I- +S4O6-2

• In the above reaction, iodine is getting reduced and Sulphur is getting oxidised. Thus, it is an example of a redox reaction.

Hence, BaCI2 + Na2SO4→ BaSO4 + 2NaCl is not a redox reaction.

Additional Information

Decomposition Reaction-

• It is a reaction in which a single component breaks down into multiple products.
• Certain changes in energy in the environment have to be made like heat, light, or electricity breaking bonds of the compound. for example of the decomposition of calcium carbonate gives out CaO (Quick Lime) which is a major component of cement. CaCO3(s)→CaO(s)+CO2(g) Here, the compound Calcium carbonate when heated breaks down into Calcium Oxide and Carbon Dioxide.

Important Points

• There are two oxidation states of Sulphur in  S4O6-2, Zero and +5.​

The correct answer is  OF2.

Key Points

• The oxidation number of Oxygen in OF2 is +2.

Important Points

• The oxidation number of Oxygen in H₂O₂ is -1.
• The oxidation number of Oxygen in NA₂O₂ is -1.
• The oxidation number of Oxygen in H₂O is -2.

 correct option is 4.

 Concept:

• The exchange of electrons between an electron donor (which oxidises) and an electron acceptor occurs in oxidation-reduction or redox processes (that becomes reduced).

Disproportionation reactions,

• Also known as dismutation reactions, 
• It is also known as a redox reaction in which the same element's atoms are both reduced and oxidised from one oxidation state (OS) to two different oxidation states.

Calculations:

   2NO₂ + 2OH^- → NO₂^- +H₂O + NO₃^-

 Step 1: Oxidation state of NO2  (Reactant)

•   x+(-4)=0           
•   x+(-4)=-1       
•   x =+4

Step 2 : Oxidation state of NO2 ^- (Product)

•  x+(-4) =-1
•  x= -1 + 4           
•  x=+3  

Step 2: Oxidation state of NO3 - (Product)

•  x+(-6)=-1
•  x=-1+6
•  x=+5

From the above reaction, we can observe that the same element NO₂ is undergoing both oxidation and reduction at the same time.

so, we can conclude that it is an example of a Disproportion redox reaction.

The correct answer is  Rust and iron have the same composition.

Key Points

• A reddish-brown deposit called rust, forms over a piece of iron when it is exposed to moist air for some time. Rust is hydrated iron (III) oxide (Fe2O3.xH2O).
• When iron metal is exposed to air for a long time, it oxidises and forms a reddish-brown colour iron oxide on its surface.
• The iron rusting formula is 4Fe + 3O2 + 6H2O → 4Fe(OH)3.

Additional Information

• The slow conversion of iron into its hydrated oxide, in the presence of moisture and air is called rusting, whereas the hydrated oxide of iron is called rust.
• When some grease or oil is applied to the surface of an iron object, then air and moisture cannot come in contact with it and hence rusting is prevented.
• The most common method of preventing the rusting of iron (or corrosion of iron) is to coat its surface with paint.
• Galvanization is the process of coating a metal with a zinc layer for protection. It is a widely used technique for protecting iron from rusting.

Concept:- 

• Rust is basically iron oxide.
• It is of reddish-brown color.
• During rusting of iron, the red-brown powder is formed.
• During rusting, iron and oxygen react in the presence of water or air moisture.
• Hydrous iron oxide and iron oxide hydroxide are present in the rust.
• Generally, refined iron goes through corrosion.
• Rust is an iron oxide, usually red oxide formed by the redox reaction of iron and oxygen in the presence of water or air moisture. Rust consists of hydrated iron oxides Fe2O.
• An entire iron can be converted into rust with time.
• During rusting in iron, Red-brown powder is coated on the iron.

Explanation: -

The chemical reaction is :

There is an anodic dissolution or oxidation of iron going into aqueous (water) solution:

2Fe → 2Fe2+  + 4e-

Cathodic reduction of oxygen that is dissolved into the water also occurs:

O2  + 2H2O + 4e- → 4OH- 

The iron ion and the hydroxide ion react to form iron hydroxide: 

2Fe2++ 4OH- → 2Fe(OH)2

The iron oxide reacts with oxygen to yield red rust, Fe2O3.H2O.

Conclusion: -

As the Iron oxidation state is changing from 0 to +3 hence, it is an oxidation reaction.

So, the correct option is (3).

Additional Information

• Many other metals are corroded but rusting of these metals is not called rust.
• The most familiar form of rust is the reddish coating that forms flakes on iron and steel (Fe2O3), but rust also comes in other colors, including yellow, brown, orange, and green. The different colors reflect various chemical compositions of rust.
• Galvanization is used to prevent the rusting of iron.
• In galvanization:-
  •  a layer of zinc is coated on iron.
• Special weathering steel alloys can be used, in which rusting takes place slowly.

The correct option is 3, i.e I and III only. 

•  The oxidation number of Oxygen in most compounds is -2.
• When Oxygen is bonded to Fluorine in compounds like Oxygen difluoride(OF₂) and dioxygen diflouride(O₂F₂), the oxygen is assigned an oxidation number of +2 and +1 respectively. Hence, statement II is incorrect.
• In peroxide, an Oxygen atom is assigned an oxidation number of -1 and in Superoxide, each oxygen atom is assigned an oxidation number of -1/2.
• Oxidation number denotes the oxidation state of an element in a compound.

Copper doesn’t replace iron from its solution.

• This is because Copper is less reactive than iron.
• A more reactive metal displaces the less reactive metal from its solution.
• In the given reaction, Iron (Fe) displaces Copper (Cu) by giving up two electrons and gets oxidized and forms a new compound called ferrous sulfate.
• Copper gets reduced.
• Iron is a reducing agent, meaning that it loses 2 electrons.
• CuSO₄ is an oxidizing agent, taking 2 electrons.
• The final solution turns green on exposure to moist air.
• The reaction is a redox reaction as it involves both loss and gain of electrons by different elements.

The correct answer is Option 2. 

Key Points Oxidation is the loss of hydrogen. Reduction is the gain of hydrogen.

• Reduction Reaction is always accompanied by an oxidation reaction where the reactant loses one or more electron.
• Reduction-oxidation reactions are often called redox equations.
• A reduction reaction is only one half of a redox reaction. The other half is the oxidation reaction.​

Additional Information

•  Oxidation Reaction refers to a reaction in which either the addition of Oxygen takes place or the removal of Hydrogen takes place.
• Rusting is an example of an oxidation reaction.
  • The iron reacts with water and oxygen to form hydrated iron oxide, which we see as rust.
    iron + water + oxygen → hydrated iron oxide
• Decomposition is the process of breakdown of the complex organic matter into a simpler inorganic matter like carbon dioxide, water, and nutrients.  The fungi, bacteria, and flagellates initiate the process of decomposition and are known as decomposers.