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Addition of oxygen /electronegative element to a substance or removal of hydrogen / electropositive element from a substance.

The highest oxidation number (O.N) of N is +5 since it has five electrons in the valence shell (3s² 2p³ ) and its minimum O.N. is -3 since it can be accept three more electrons to acquire the nearest inert gas (Ne) configuration.

The highest O.N. of S is +6 since it has six electrons in the valence shell (3s² 2p⁴) and its minimum O.N. is -2 since it needs two more electrons to acquire the nearest inert gas (Ar) configuration.

Likewise the maximum O.N. of Cl is +7 since it has seven electrons in the valance shell (3s² 2p⁵) and its minimum O.N. is -1 since it needs only one more electron to acquire the nearest (Ar) Noble gas configuration.

The oxidation number is expressed by putting a Roman numeral representing the oxidation number in parenthesis after the symbol of the metal in the molecular formula. Thus aurous chloride and auric chloride are written as Au(I)Cl and Au(III)Cl₃. Similarly, stannous chloride and stannic chloride are written as Sn(II)Cl₂ and Sn(IV)Cl₄. 

Oxidation: An increase in the oxidation number 

Reduction: A decrease in the oxidation number

Calculation of oxidation number: 

• O. S. of all the elements in their elemental form (in standard state) is taken as zero O. S. of elements in Cl₂, F₂, O₂, P₄, O₃, Fe(s), H₂, N₂, C(graphite) is zero.
• Common O. S. of elements of group one (1st) is one. Common O. S. of elements of group two (2nd) is two.
• For ions composed of only one atom, the oxidation number is equal to the charge on the ion.
• The oxidation number of oxygen in most compounds is –2. While in peroxides (e.g., H₂O₂, Na₂O₂), each oxygen atom is assigned an oxidation number of –1, in super oxides (e.g., KO₂,RbO₂) each oxygen atom is assigned an oxidation number of –(½).
• In oxygen difluoride (OF₂) and dioxygen difluoride (O₂F₂), the oxygen is assigned an oxidation number of +2 and +1,respectively.
• The oxidation number of hydrogen is +1 but in metal hydride its oxidation no. is –1.
• In all its compounds, fluorine has an oxidation number of –1.
• The algebraic sum of the oxidation number of all the atoms in a compound must be zero.
• In polyatomic ion, the algebraic sum of all the oxidation numbers of atoms of the ion must equal the charge on the ion. 

Heavy water is used as moderate and coolant in nuclear reactor.

(i) Nitric Acid HNO₃: 

Here Oxidation number of N is HNO₃ = +5; But Maximum oxidation number of N = +5; Minimum oxidation number of N = -3 Since the oxidation number of N in HNO₃ is maximum (+5), 

therefore it can only decrease by accepting electrons. Hence HNO₃ acts only as an oxidizing agent. 

(ii) Nitrous Acid HNO₂: 

Here Oxidation number of N in HNO₂ = +3 Maximum oxidation number of N = +5; Minimum oxidation number of N = -3 Thus, the oxidation number of N can either increase by losing electrons or can decrease by accepting electrons. 

Therefore, HNO₂ acts both as an oxidizing as well as a reducing agent.

A reagent which can increase the oxidation number of an element in a given substance. These reagents are called as oxidants also. 

A reagent which lowers the oxidation number of an element in a given substance. These reagents are also called as reductants.

(i) By Galvanization: Coating of a less reactive metal with a more reactive metal e.g. coating of iron surface with Zn to prevent rusting of iron. 

(ii) By greasing /oiling (to keep away the object from the contact of air & moisture.) 

(iii) By painting (to keep away the object from the contact of air & moisture.)

x be ON to be find K₄MO₂Cl₈ = 4(+1) + (x)2 + 8(-1) = 0 

4 + 2x – 8 = 0 

=> 2x – 4 = 0 

=> x = +2

Na₂PdCl₂ = 2(+1) + x + 4(-1) = 0 

=> 2 + x - 4 = 0 

=> x = +2

Na₂S₂O₃ : 2(+1) + 2(oxidation no. of S) + 3(-2) = 0 

2(oxidation no. of S) = 4

P₂O₅ : 2 x (oxidation no. of P) + 5(-2) = 0 oxidation of P = pp = +5

(i) Let x be the oxidation no. of chromium in K₂CrO₄

O.N. of K = +1

O.N. of O = – 2

\(\therefore\) 2 x (+1) + \(\mathrm{x}\) + 4 x (– 2) = 6

or \(\mathrm{x}\) – 6 = 0 or \(\mathrm{x}\) = +6

Hence, oxidation no. of Cr in K₂CrO₄ = 6.

(ii) Let \(\mathrm{x}\) be the oxidation no. of Cr in K₂Cr₂O₇

Oxidation no. of K = +1

Oxidation no. of O = – 2

\(\therefore\) 2 × (+ 1) + 2\(\mathrm{x}\) + 7 × (– 2) = 0

or  2\(\mathrm{x}\) - 12 = 0

or \(\mathrm{x}\) = +6

Hence, oxidation no. of Cr in K₂Cr₂O₇

= +6

(iii) Let \(\mathrm{x}\) be the oxidation no. of Cr in CrO₂CI₂.

Oxidation no. of CI = – 1

\(\therefore\) \(\mathrm{x}\) + 2 x (-2) + 2 x (-1) = 0

or \(\mathrm{x}\) - 6 = 0

or \(\mathrm{x}\) = +6

Hence oxidation no. Cr in CrO₂CI₂

= + 6.

(iv) Let \(\mathrm{x}\) be the oxidation no. of Cr in Cr₂(SO₄)₃.

Oxidation no. of SO₄^2– = – 2

\(\therefore\) 2\(\mathrm{x}\) + 3 × (– 2) = 0

or 2\(\mathrm{x}\) = +6

or \(\mathrm{x}\) = +3

Hence oxidation no. of Cr in Cr₂(SO₄)₃ = +3

(v) Let, \(\mathrm{x}\) be the oxidation no. of Mn in Mno₄^-

Oxidation of O = – 2

\(\therefore\) \(\mathrm{x}\) + 4 x (-2) = -1

\(\mathrm{x}\) - 8 = -1

\(\mathrm{x}\) = -1 + 8 = +7

1. All atoms in the elemental or molecular state have ‘zero’ oxidation state. 

2. Elements of group I and II in the periodic table always have ‘+1’ and ‘+2’ oxidation states respectively. 

3. Hydrogen ‘+1’ oxidation state in all its compounds except metal hydrides where it is ‘-1’ 

4. Oxygen has been assigned an oxidation number of ‘-2’ in all its compounds except peroxides and oxygen fluoride. In peroxide it is ‘-1’ and in oxygen fluoride it is ‘+2’. 

5. Halogens generally have ‘-1’ oxidation state. Except fluorine, other halogens may have positive oxidation states in their oxides and inter halogen compounds (Fluorine always has ‘-1’ state). 

6. The algebraic sum of oxidation numbers of all the elements in a compound is zero and that in an ion is equal to the net charge on the ion.

According to oxidation number, 

Oxidation: It is defined as the process that involves increase in oxidation number of the element in the given substance. 

Reduction: It is defined as the process that involves decrease in oxidation number of the element in the given substance.

No of moles of KMnO₄ present in 20ml of 0.1M

KMnOH solution = \(\frac{20}{1000}\) x 0.1 

= 2 x 10^-3

The balanced equation for the redox reactions is:

2KMnO₄ + 5H₂O₂ + 3H₂SO₄ → K₂SO₄ + 2MnSO₄ + 8H₂O + 5O₂ …(1)

From the equation, 2 moles of KMnO₄ = 5 moles of H₂O₂

2 x 10^-3 moles of KMnO₄ will react with H₂O₂ = \(\frac{5}{2}\) x 2 x 10^-3 = 5 x 10^-3 moles Molecular wt. of H₂O₂ = 34

Amount of H₂O₂ actually present 

= 34 x 5 x 10^-3 

= 0.17g

Percentage purity of in 20 ml is 

= \(\frac{0.17}{0.20}\) x 100 = 85

No of moles of KMnO₄ present in 20 mL of 0.1M

KMnO₄ solution = \( \frac{20}{1000}L\times 0.1\,M\)

= 2 x 10^-3 mol.

The balanced chemical equation for the redox reaction is : 

2KMnO₄ + 5H₂O₂ + 3H₂SO₄ → 2MnSO₄ + K₂SO₄ + 8H₂O + 5O₂ .....(1)

According to above balance chemical equation -

∵ We have required 5 mole pure H₂O₂ to reduce 2 mol KMnO₄.

∴ Number of moles of pure H₂O₂ required to reduce 2 x 10^-3 mol KMnO₄ 

= \(\frac{5}{2}\) x 2 x 10^-3 mol

= 5 x 10^-3 mol of pure H₂O₂

Molecular weight of H₂O₂ = 34 g/mol

∴ Mass of pure H₂O₂ = 5 x 10^-3 mol x 34g/mol

= 0.17 g

But,

Amount of H₂O₂ actually precent = 0.2g

∴ Percentage purity of H₂O₂ = \(\frac{0.17}{0.20}\times 100\)

= 85

Hence,

The purity of H₂O₂ was 85%.

Oxidation is the process of loss of electrons. Reduction is the process of gain of electrons. The atom which loss electron is called the reducing agent and the atom which gains electron is called the oxidizing agent.

Oxidizing agent – O 

Reducing agent-C

a. Mg/Mg²⁺

Here,

Mg loses two electrons to form Mg²⁺ ion.

Mg_(s) ⟶ Mg²⁺_(aq) + 2e^-

Hence,

Mg/Mg²⁺ is an oxidized state.

b. Cu/Cu²⁺

Here,

Cu loses two electrons to form Cu²⁺ion.

Cu_(s) ⟶ Cu²⁺_(aq) + 2e^-

c. O₂/O^2- 

Here,

Each O^2- gains two electrons to form O^2- ion.

Hence,

O₂/O^2- 

d. Cl₂/Cl^- is in a reduced state.

Here,

Each Cl gains one electron to form Cl^- ion.

Cl_2(g) + 2e^- ⟶2 − Cl^-_(aq)

Hence,

Cl₂/Cl^- is in a reduced state.

(a) Oxidized atom — Mg 

Equation of oxidation — Mg → Mg²⁺ + 2e^- 

Reduced atom — F 

Equation of reduction — F + e^- → F^- 

Oxidizing agent — F 

Reducing agent — Mg 

(b) Oxidized atom — Na

Equation of oxidation — Na → Na + 1e^- 

Reduced atom — Cl

Equation of reduction — Cl + 1e^- → Cl^- 

Oxidizing agent — Na 

Reducing agent — Cl

2Na(s) + H₂(g) → 2NaH(s)

In the above reaction, the compound formed is an ionic compound, which may also be represented as Na+ H^– (s). 

This suggests that one half reaction in this process is:

2Na(s) → 2Na⁺ (g) + 2e^– (Oxidation) and the other half reaction is:

H₂(g) + 2e^– → 2H^– (g) (Reduction) So, it is evident that sodium is oxidised and hydrogen is reduced in this reaction. So, the complete reaction is a redox reaction.

3Na → 2Na⁺ +2e^– (oxidation)

Cl₂ + 2e^– > 2Cl^– (reduction) 

Na is oxidized to Na⁺ 

Cl₂ is reduced to Cl^–

Na is the reducer 

Cl₂ is the oxidizer.

A freshly cut apple is almost white but it turns reddish brown after sometime. It is due to the fact that the cut surface is exposed to air due to reaction with oxygen in the surrounding air, known as oxidation.

Cu lies below hydrogen in the activity series.

At anode, loss of electrons occurs which means oxidation. So, it is called oxidation electrode. At cathode, gain of electrons occurs which means reduction, so, it is called reduction electrode.

Since the electrode potential of Cu is higher than that of Al, therefore, Cu has ahigher tendency to get reduced and hence Cu electrode acts as a cathode.

In an electrochemical cell, electrons release at anode and move towards cathode, which make the anode negatively charged.

Correct option is: (D) All of these

Electrode potential is the tendency of an electrode to accept or to lose electrons. The potential of an electrode is the potential difference between it and the electrolyte surrounding the electrode.

The factors affecting the magnitude of electrode potential is 

i) Nature of metal or electrode.

ii) The concentration of metal ions in solution

iii) Temperature

Correct option is: (D) All of these

At 1 M concentration of Cu²⁺(aq), will electrode potential become equal to its standard electrode potential.

Greater the value of more negative standard reduction potential of the metal, greater will be its reducing power.

\(\therefore\) Increasing order of reducing power:

X < Z < Y

Redox reaction is reaction in which oxidation and reduction take place simultaneously, 

e.g. Zn(s)+ Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

If H⁺ or any acid appears on either side of the chemical equation, the reaction takes place in the acidic solution. If OH^- or any base, appears on either side of the chemical equation, the solution is basic. If neither H+, OH^- nor any acid or base is present in the chemical equation, the solution is neutral.

It is an apparent or the actual charge possessed by an atom of the element in the molecule.

Fe in [Fe(CN)₆]^4-

Let the oxidation number of Fe in [Fe(CN)₆]^4- bexand CN be -1 i.e..,

\(\overset{x}{Fe}\,\,\overset{-1}{(CN)_6}\)

∴ Sum of oxidation numbers of all the atoms in

[Fe(CN)₆]^4- 

⇒ x + 6(-1) = -4 

⇒ x - 6 = -4

x = +2

Thus, the oxidation number of Fe in [Fe(CN)₆]^4- 

Fe is +2.

P in H₃P₂O₇^- , let the oxidation number of P in H₃P₂O₇^- be x.

H₃P₂O₇^- = 3(+1) + 2 (x) = -1 +7(-2) 

or 2x – 11 = -1; 

x = + 5

N in NO₃^- , let the oxidation number of N in NO₃^- be x. Writing the oxidation number of each atom above its symbol.

\(\overset{x}{N}\,\,\overset{-2}{O_3}\)

x + 3(-2) = -1

x – 6 = -1, 

x = +5

S = +6 in PbSO₄

U = +5 in U₂O₇^4-

B = +3 in B₄O₇^2-

Cr = +6 in CrO₄^2-

A reaction in which an ion (or an atom) in a compound is replaced by an ion (or an atom) of another element is called displacement reaction.

e.g. X + YZ → XZ + Y

i. SO₂ is a neutral molecule.

∴ Sum of oxidation numbers of all atoms of SO₂ = 0

∴ (Oxidation number of S) + 2 × (Oxidation number of O) = 0

∴ Oxidation number of S + 2 × (- 2) = 0

∴ Oxidation number of S in SO₂ = 0 – (- 4)

∴ Oxidation number of S in SO₂ = +4

ii. \(SO_4^{2-}\) is an ionic species.

∴ Sum of the oxidation numbers of all atoms of \(SO_4^{2-}\) = – 2

∴ (Oxidation number of S) + 4 × (Oxidation number of O) = – 2

∴ Oxidation number of S in \(SO_4^{2-}\) = – 2 – 4 × (-2) = – 2 + 8

∴ Oxidation number of S in \(SO_4^{2-}\) = +6

Therefore, decreasing order of oxidation state of N is

N₂O₄ and NO₂^- > HN₃ > N₂H₄ > NH₃

(i) E^ΘFe²⁺_/Fe = – 0.44 V 

E^ΘCu²⁺_/Cu = +0.34 V

\(\because\) As we know that a metal having more negative value of E^Θ can displace the metal having less negative or positive value of E^Θ from their salt solution. So, Fe can displace Cu from CuSO₄ solution.

But, in the reaction, it is showing that Cu displaces Fe from FeSO₄ solution which is not feasible. So, the given reaction does not occur.

(ii) E^ΘZn²⁺_/zn = – 0.76 V 

E^ΘCu²⁺_/Cu = + 0.34 V 

E^Θ_Ag+Ag = + 0.80 V

Similarly, Zn has negative E^Θ and Cu has positive E^Θ . Zinc can displace copper from aqueous CuSO₄ solution. But silver has highest +ve value of E^Θ . Than Cu, so, Ag cannot displace Cu from CuSO₄ solution.

(iii) Since, Cu has lesser E^Θ than Ag. So, copper can displace Ag from AgNO₃ solution due to which the solution turns blue.

Option : a. Sn^2⊕ is undergoing oxidation

Fluorine is the strongest oxidizing agent.

Lithium (Li) is best reducing agent because it has lowest standard reduction potential, i.e., Li⁺ is most stable.

In a disproportionation reaction an element in one oxidation state is simultaneously oxidized and reduced.

F- ions cannot be converted to F₂ by chemical means because fluorine is the strongest oxidizing agent.