Explore topic-wise InterviewSolutions in .

Here 

2e^– + PbO₂ → [Pb(OH)₃]^– 

2H₂O + 2e^–PbO₂² → [Pb(OH)₃] + OH^– …(i) 

2OH + Cl^– → ClO^– + 2e^– + H₂O …(ii)

Add (i) & (ii) 

PbO₂ + OH^– + Cl^– + H₂O → [Pb(OH)₃]^– + ClO^–

Combination of an electrode and the solution in which it is dipped is called a half – cell.

Step I: H₂ + O₂ → H₂O 

Step II: 2H₂ + O₂ → 2H₂O

(i) (a) In this reaction, nitric oxide is formed by combination of the elemental substances, nitrogen and oxygen. So, it is an example of combination redox reaction.

(b) It involves breaking down of lead nitrate into three components. So, it is an example of decomposition redox reaction.

(c) In this reaction, hydrogen of water has been displaced by hydride ion into dihydrogen gas. So, it is an example of displacement redox reaction.

(d) This reaction involves disproportionation of NO₂ (+4 state) in to NO₂^– (+3 state) and NO₃^– (+5 state). So, it is an example of disproportionation redox reaction.

(ii) Pb₃O₄ is actually a stoichiometric mixture of 2 mol of PbO and 1 mole of PbO₂. In PbO₂, lead is present in +4 oxidation state, whereas the stable oxidation state of lead in PbO is +2. PbO₂ thus can act as an oxidant and so, it can oxidise Cl^- ion of HCl into chlorine. As we know that PbO is a basic oxide. So, the reaction.

Pb₃O₄ + 8HCI → 3PbCI₂ + CI₂ + 4H₂O can be splitted into two reactions-

2PbO + 4HCI → 2PbCI₂ + 2H₂O (Acid-base reaction)

\(\overset{+4}{Pb}O_2+4H\overset{-1}Cl\rightarrow\overset{+2}{Pb}Cl_2+\overset{o}{Cl_2}+2H_2O\) (Redox Reaction)

Since, HNO₃ itself is an oxidizing agent, so, it is unlikely that the reaction may occur between PbO₂ and HNO₃.

However, the acid-base reaction occurs between PbO and HNO₃ as :

2PbO + 4HNO₃ → 2Pb(NO₃)₂ + 2H₂O + 2H₂O 

It is the passive nature of PbO₂ against HNO₃, that makes the reaction different from the one that follows with HCl.

a. Cl₂ (E⁰ = 1.36 V) and Br₂ (E⁰ = 1.09 V)

Cl₂ has a larger positive value of E⁰ than Br₂. 

Thus,

Cl₂ is a stronger oxidizing agent than Br₂.

b. MnO₄^Θ(E⁰ = 1.51 V) and Cr₂O₇^2Θ (E⁰ = 1.33 V)

MnO₄^Θ has larger positive value of E⁰ than Cr₂O₇^2Θ. 

Thus,

MnO₄^Θ is stronger oxidizing agent than Cr₂O₇^2Θ.

a. Li (E⁰ = – 3.05 V) and Mg(E⁰ = – 2.36 V)

Li has a larger negative value of E⁰ than Mg. 

Thus, 

Li is a stronger reducing agent than Mg.

b. Zn(E⁰ = – 0.76 V) and Fe(E⁰ = – 0.44 V)

Zn has a larger negative value of E⁰ than Fe. 

Thus,

Zn is a stronger reducing agent than Fe.

A redox couple is defined as having together oxidized and reduced forms of a substance taking part in an oxidation and reduction half – reaction.

In this method two half equations are balanced separately and then added together to give balanced equation. Following steps are involved.

• Step 1: Write unbalanced equation for the redox reaction, assign oxidation number to all the atoms in the reactants and products. Divide the equation into two half equations. One half equation involves increase in oxidation number and another involves decrease in oxidation number (Write two half equations separately).
• Step 2: Balance the atoms except O and H in each half equation. Balance oxygen atom by adding H₂O to the side with less O atoms.
• Step 3: Balance H atoms by adding H⁺ ions to the side having less H atoms.
• Step 4: Balance the charges by adding appropriate number of electrons to the right side of oxidation half equation and to the left of reduction half equation.
• Step 5: Multiply half equation by suitable factors to equalize the number of electrons in two half equations. Add two half equations and cancel the number of electrons on both sides of equation.
• Step 6: If the reaction occurs in basic medium then add OH ions, equal to number of H⁺ ions on both sides of equation. The H⁺ and OH^– ions on same side of equation combine to give H₂O molecules.
• Check that the equation is balanced in both, the atoms and the charges.

1. Oxidation number method 

2. Half reaction method or ion electrode method

It is defined as having together the oxidized and reduced forms of a substance taking part in an oxidation or reduction half reaction.

(D) Zn + 2AgCN → 2Ag + Zn(CN)₂

1. Respiration 

2. Rusting 

3. Combustion of fuel

A titration by which the strength of an oxidant or reductant is determined by titrating it with standard solution of reductant or oxidant using a redox sensitive indicator is called redox titration.

Redox is an abbreviation used for the terms ‘oxidation and reduction’.

1. Take some hydrogen peroxide solution in a test tube. Show a glowing incense stick into the test tube. 

2. No .

Now add some manganese dioxide (MnO₂) into the test tube. Again show the glowing incense stick. 

Observation: Glowing incense stick flares up now.

It indicates that when manganese dioxide is added, the rate of reaction increases and oxygen is formed faster. Filter the solution using a filter paper when the reaction is completed.

The substance remaining in the filter paper is manganese dioxide itself When examined carefully it becomes clear that there is no change in its amount or property. The presence of manganese dioxide has increased the rate of the reaction. Manganese dioxide acts as a catalyst in this reaction.

When solids are made into small pieces or powder, their surface area increases. As a result the number of molecules undergoing effective collision also increases. Hence the rate of reaction increases.

1. 0 to +2

2. Zn

3. +1 to 0

4. H Here HCI is the oxidizing agent and Zn is the reducing agent. „

Take equal volume of dil. HCI in two beakers. Add a small piece of marble into one and marble powder of equal mass into the other. Reaction rate is greater when marble powder is used. Rate of reaction increases when surface area increases

Mg – 2, 8, 2

Cl – 2, 8,7

1.  No

2.  O

3.  2

4.  2 MgO

5.  No

6.  2

7.  2 Mg + O₂ → 2MgO

8. Yes

As the concentration of reactants increases, the number of molecules per unit volume and the number of effective collisions increase. Consequently the rate of reaction increases.

It burns with sparkling light

Test tube 1: Reaction is faster in con. HCl 

Test tube 2: Reaction rate is slow

E°cell = (E°_cathode (SRK) - E°_anode (SRP))

(SRP means Standard Reduction Potential)

= E°_{Ni⁺/Ni²⁺} - E°_Fe²⁺/Fe 

= +0.049V - (-0.88V)

= 1.37V

Materials required for the experiment. Zn, Mg, dil. HCl arid test tubes. 

Procedure: 

Take equal volume of dil. HCl in two test tubes. Add Zn to one and Mg of same mass to the other. Hydrogen gas is produced in both the test tubes. Rate of reaction is faster in the test tube containing Mg.

Materials required: Mg, dil. HCl, Con. HCl and test tubes. 

Procedure: Take magnesium ribbons of equal mass in two test tubes. Add concentrated HCl to one test tube and dilute HCl to the other in equal volume

Both are equal

In a chemical reaction mass is neither created nor destroyed. This is the law of conservation of mass. This law was proposed by the scientist Antoine Lavoisier. That is, the total mass of the reactants will be equal to the total mass of the products.

a) Reactants 18 g, Product 18 g 

b) Reactants 36 g, Products 36 g

Zn(s)|Zn²⁺(aq)||Fe²⁺(aq)|Fe(s);

E°_cell = -0.44 - (-0.76)

 = 0.32V

Zn(s)|Zn²⁺(aq)||Cu²⁺(aq)|Cu(s)

E°_cell = 0.34 - (-0.76) = +1.1V

For maximum emf anode should have minimum reduction potential and cathode should have maximum reduction potential. Hence A²⁺/A couple should be anode and C²⁺/C couple acts as cathode.

Cell reaction A + C²⁺ → A²⁺ + C

Direction of flow of electrons is from A to C in external and from C to A internal circuit. Direction of flow of current is opposite to the direction of flow of electrons.

Correct option is: (A) +6, -2

Te has six electrons in their valence shell. It can accept two electron and form noble gas (Xe) configuration or it can released six electrons and form Noble gas (Kr) configuration.

Thus, Te has highest and lowest oxidation states are +6 and -2 respectively.

Correct option is: (A) +6, -2

Correct option is: (C) -1

Fluorine has atomic number 9 and  it is highly electronegative atom in the periodic table. It has required one electron to complete their octet. Thus it can exhibit oxidation state of -1

Correct option is: (C) -1

Correct option is: (B) -1

Let say oxidation number of Hydrogen is x 

then, 1 x (+2) + 2 x (x) = 0

2x = -2

 = -1

Correct option is: (B) -1

O.N. of S = + 6 in HSO₄^–.

A. Mercury(II) chloride :

HgCl₂

B. Thallium(I) sulphate :

Tl₂SO₄

C. Tin(IV) oxide :

SnO₂

D. Chromium(III) oxide :

Cr₂O₃

Ox. no. of P in P₄ = 0

Ox. no. of P in H₃PO₄

3(+1) + x + 4(-2) = 0 

+3 + x - 8 = 0 

x = +5

It is Oxidation.

Given: E°(Zn²⁺/Zn) = 0.76V, E°(Fe²⁺/Fe) = 0.44V and E°(Ni²⁺/Ni) = 0.25V

(i) The reduction potential of iron is more than that of zinc. Therefore, iron will be reduced. In other words Zn²⁺ will not be reduced. 

(ii) The reduction potential of Ni²⁺ is more than that of Iron. Therefore, Ni²⁺ will be reduced by iron.

Zn(s)|Zn²⁺(1M)|Pb²⁺(1M)|Pb(s)

Cr₂O₇^2- + C₂H₂O^- → C₂H₂O + Cr³⁺  ...(i)

Here Cr₂O₇^2- → 2Cr³⁺

i.e., Cr₂O₇^2- → 2Cr³⁺ + 7H₂O

6e^- + 14H⁺ + Cr₂O₇^2- → 2Cr³⁺ + 7H₂O

[H₂O + C₂H₄O^- → C₂H₄O₂ + 2H⁺ + 3e^-] x 2    ...(ii)

10H⁺ + Cr2O₇^2- + 2C₂H₄O^- → 2C₂H₄O₂ + 2Cr³⁺ + 5H₂O

I₂(s) + 10HNO₃(conc.) → 2HIO₃(Iodic acid) + 10NO₂ + 4H₂O

H₂SO₃:

2(+1) + x + 3(-2) = 0 

x = +4 

Ox. state of is S in H₂SO₃.

The above equation represents a combination reaction.

The different types of redox reactions are 

(i) Combination reactions 

(ii) Decomposition reactions 

(iii) Displacement reactions 

(iv) Disproportionation reactions.

Oxidaton number changes from +3 to + 6.

(B) Both oxidised and reduced