Explore topic-wise InterviewSolutions in .

Correct Option is: (B) oxidation

Oxidation involves loss of electrons and Reduction involves gain of electrons .

Correct option is: (B) oxidation

HCl is oxidized to Cl₂.

Oxidising agent is a substance which can gain electrons easily. F₂ is the best oxidizing agent.

Correct Option is: (A) gain of electrons

Reduction involves gain of electrons , removal of oxygen , addition of Hydrogen , and decrease in oxidation number.

Correct option is: (A) gain of electrons

Methane is oxidized owing to the addition of oxygen to it.

Reduction is a process in which gain of electrons take place.

Aluminum is oxidized because oxygen is added to it Ferrous ferric oxide (Fe₃O₄) is reduced because oxygen has been removed from it.

Oxidation is a process in which loss of electrons takes place.

i. Oxidation is an increase in the oxidation number of an element in a given substance.

e.g. Fe_(s)  → \(Fe_{(aq)}^{2+}\)

ii. Reduction is a decrease in the oxidation number of an element in a given substance.

e.g. \(Cu_{(aq)}^{2+}\) → Cu_(s)

Oxidation: Loss of electron(s) by any species is called oxidation.

Reduction : Gain of electron(s) by any species is called reduction

Increase in Oxidation Number is Oxidation and decrease in Oxidation Number is called reduction. 

i. PF₆^-

Oxidation number of F = -1

\(PF_6^-\) is an ionic species.

∴ Sum of the oxidation numbers of all atoms = – 1 

∴ (Oxidation number of P) + 6 × (Oxidation number of F) = – 1 

∴ Oxidation number of P + 6 × (-1) = -1 

∴ Oxidation number of P – 6 = – 1

Oxidation number of P in \(PF_6^-\) = +5

ii. NaIO₃

Oxidation number of Na = +1 

Oxidation number of O = -2 

NaIO₃ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0 (Oxidation number of Na) + (Oxidation number of I) + 3 × (Oxidation number of O) = 0

(+1) + (Oxidation number of I) + 3 × (-2) = 0

Oxidation number of I + 1 – 6 = 0

Oxidation number of I in NaIO₃ = +5

iii. NaHCO₃

Oxidation number of Na = +1 

Oxidation number of H = +1 

Oxidation number of O = -2

NaHCO₃ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0 

∴ (Oxidation number of Na) + (Oxidation number of H) + (Oxidation number of C) + 3 × (Oxidation number of O) = 0

∴ (+1) + (+1) + (Oxidation number of C) + 3 × (-2) = 0

∴ Oxidation number of C + 2 – 6 = 0

∴ Oxidation number of C in NaHCO₃ = +4

iv. ClF₃

Oxidation number of F = -1

ClF₃ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0 

∴ (Oxidation number of Cl) + 3 × (Oxidation number of F) = 0 

∴ Oxidation number of Cl + 3 × (-1) = 0 

∴ Oxidation number of Cl in ClF₃ = +3

v. SbF₆^-

Oxidation number of F = -1

\(SbF_6^-\) is an ionic species.

∴ Sum of the oxidation numbers of all atoms = – 1

∴ (Oxidation number of Sb) + 6 × (Oxidation number of F) = – 1

∴ Oxidation number of Sb + 6 × (-1) = -1

∴ Oxidation number of Sb in \(SbF_6^-\) = +5

vi. NaBH₄

Oxidation number of Na =+1 

Oxidation number of H = -1 (for Hydride) 

NaBH₄ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0 

∴ (Oxidation number of Na) + (Oxidation number of B) + 4 × (Oxidation number of H) = 0 

∴ (+1) + (Oxidation number of B) + 4 × (-1) = 0 

∴ Oxidation number of B + 1 – 4 = 0

∴ Oxidation number of B in NaBH₄ = +3

vii. H₂PtCl₆

Oxidation number of H = +1 

Oxidation number of Cl = -1 

H₂PtCl₆ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0 

∴ 2 × (Oxidation number of H) + (Oxidation number of Pt) + 6 × (Oxidation number of Cl) = 0

∴ 2 × (+1) + (Oxidation number of Pt) + 6 × (-1) = 0

(Oxidation number of Pt) + 2 – 6 = 0

∴ Oxidation number of Pt in H₂PtCl₆ = +4

viii. H₅P₃O₁₀

Oxidation number of H = +1 

Oxidation number of O = -2

H₅P₃O₁₀ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0

∴ 5 × (Oxidation number of H) + 3 × (Oxidation number of P) +10 × (Oxidation number of O) = 0

∴ 5 × (+1) + 3 × (Oxidation number of P) + 10 × (-2) = 0

∴ 3 × (Oxidation number of P) + 5 – 20 = 0

Oxidation number of P = +\(\frac{15}{3}\)

∴ Oxidation number of P in H₅P₃O₁₀ = +5

ix. V₂O₇^4-

Oxidation number of O = -2

\(V_2O_7^{4-}\) is an ionic species.

∴ Sum of the oxidation numbers of all atoms = – 4

∴ 2 × (Oxidation number of V) + 7 × (Oxidation number of O) = – 4

∴ 2 × (Oxidation number of V) + 7 × (-2) = – 4

∴ 2 × (Oxidation number of V) = – 4 + 14

∴ Oxidation number of V = \(+\frac{10}{2}\)

∴ Oxidation number of V in \(V_2O_7^{4-}\) = +5

x.  CuSO₄

Oxidation number of Cu = +2 

Oxidation number of O = -2

CuSO₄ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0

∴ (Oxidation number of Cu) + (Oxidation number of S) + 4 × (Oxidation number of O) = 0

∴ (+2) + Oxidation number of S + 4 × (-2) = 0

∴ Oxidation number of S + 2 – 8 = 0

∴ Oxidation number of S in CuSO₄ = +6

xi. BiO₃^-

Oxidation number of O = -2

\(BiO_3^-\) is an ionic species.

∴ Sum of the oxidation numbers of all atoms = – 1 

∴ (Oxidation number of Bi) + 3 × (Oxidation number of O) = – 1 

∴ Oxidation number of Bi + 3 × (-2) = – 1 

∴ Oxidation number of Bi = – 1 + 6

∴ Oxidation number of Bi in \(BiO_3^-\) = +5

xii. CH₃OH

Oxidation number of H = +1 

Oxidation number of O = -2

CH₃OH is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0

∴ (Oxidation number of C) + 4 × (Oxidation number of H) + (Oxidation number of O) = 0

∴ (Oxidation number of C) + 4 × (+1) + (-2) = 0

∴ Oxidation number of C + 2 = 0

∴ Oxidation number of C in CH₃OH = -2

As we know that a metal having low value of E^Θ can displace the metal having high value of E^Θ. The standard reduction potential of Cu is less than Ag. So, it is not safe to stir 1 M AgNO₃ solution with copper spoon. Because, copper can displace silver from 1 M AgNO₃ solution.

(i) VO₂⁺

Let, oxidation number of V = x

\(\therefore\) x + 2(-2) = 1

x - 4 = 1

x = 1 + 4 = +5

(ii) UO₂²⁺

Let, oxidation number of U = x

\(\therefore\) x + 2 (– 2) = 2

x – 4 = 2

x = 2 + 4 = +6

(iii) Ba₂XeO₆

Let, oxidation number Xe = x

\(\therefore\) 2(2) + x + 6(– 2) = 0

4 + x – 12 = 0

x – 8 = 0

x = +8

(iv) K₄P₂O₇

Let, oxidation number of P = x

4(1) + 2(x) + 7 (– 2) = 0

4 + 2x – 14 = 0

2x – 10 = 0

2x = 10

X = \(\frac{10}{2}\) = +5

(v) K₂S

Let, oxidation number of S = x

\(\therefore\) 2(1) + x = 0

x = – 2

(i) H₂O₂

Oxidation number of O = -1 (for peroxide)

H₂O₂ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0 

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

∴ 2 × (Oxidation number of H) + 2 × (-1) = 0

∴ Oxidation number of H = \(+\frac{2}{2}\)

∴ Oxidation number of H in H₂O₂ = +1

(ii) C₄H₄O₆^2-

Oxidation number of H = +1 

Oxidation number of O = -2

C₄H₄O₆^2- is an ionic species.

∴ Sum of the oxidation numbers of all atoms = – 2

∴ 4 × (Oxidation number of C) + 4 × (Oxidation number of H) + 6 × (Oxidation number of O) = – 2

∴ 4 × (Oxidationnumber of C) + 4 × (+1) +6 × (-2) = -2 

∴ 4 × (Oxidation number of C) + 4 – 12 = -2 

∴ 4 × (Oxidation number of C) = – 2 + 8

∴ Oxidation number of C = \(+\frac{6}{4}\) 

∴ Oxidation number of C in C₄H₄O₆^2- = +1.5

(iii) H₂AsO₄^-

Oxidation number of H = +1 

Oxidation number of O = -2

H₂AsO₄^- is an ionic species.

∴ Sum of the oxidation numbers of all atoms = – 1

∴ 2 × (Oxidation number of H) + (Oxidation number of As) + 4 × (Oxidation number of O) = -1

∴ 2 × (+1) + Oxidation number of As + 4 × (-2) = – 1

∴ Oxidation number of As + 2 – 8 = – 1

∴ Oxidation number of As = – 1 + 6

∴ Oxidation number of As in H₂AsO₄^- = +5

(iv) Mn(OH)₃

Oxidation number of O = -2 

Oxidation number of H = +1 

Mn(OH)₃ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0 

∴ (Oxidation number of Mn) + 3 × (Oxidation number of O) + 3 × (Oxidation number of H) = 0 

∴ Oxidation number of Mn + 3 × (-2) + 3 × (+1) = 0 

∴ Oxidation number of Mn – 6 + 3 = 0 ∴ Oxidation number of Mn in Mn(OH)₃ = +3

(v) I₃^-

I₃^- is an ionic species.

∴ Sum of the oxidation numbers of all atoms = – 1 

∴ 3 × Oxidation number of I = – 1

∴ Oxidation number of I in I₃^- = \(-\frac{1}{3}\)

(vi) C₂H₅OH

Oxidation number of O = -2 

Oxidation number of H = +1

C₂H₅OH is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0

∴ 2 × (Oxidation number of C) + 6 × (Oxidation number of H) + (Oxidation number of O) = 0

∴ 2 × (Oxidationnumberof C) + 6 × (+1) + (-2) = 0 

∴ 2 × (Oxidation number of C) = – 4

∴ Oxidation number of C = \(-\frac{4}{2}\)

∴ Oxidation number of C in C₂H₅OH = -2

(vii) Na₂CO₃

Oxidation number of Na = +1 

Oxidation number of O = -2

Na₂CO₃ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0

∴ 2 × (Oxidation number of Na) + (Oxidation number of C) + 3 × (Oxidation number of O) = 0

∴ 2 × (+1) + (Oxidation number of C) + 3 × (-2) = 0 ∴ Oxidation number of C + 2 – 6 = 0 

∴ Oxidation number of C in Na₂CO₃ = +4

(viii) IO₄^-

Oxidation number of O = -2

IO₄^- is an ionic species.

∴ Sum of the oxidation numbers of all atoms = – 1 

∴ (Oxidation number of I) + 4 × (Oxidation number of O) = – 1 

∴ Oxidation number of I + 4 × (-2) = – 1 

∴ Oxidation number of I = -1 +8

∴ Oxidation number of I in IO₄^- = +7

(ix) VO₄^3-

Oxidation number of O = -2

VO₄^3- is an ionic species.

∴ Sum of the oxidation numbers of all atoms = – 3

∴ (Oxidation number of V) + 4 × (Oxidation number of O) = – 3

∴ Oxidation number of V + 4 × (-2) = – 3

∴ Oxidation number of V = -3 + 8

∴ Oxidation number of V in VO₄^3- = +5

(x) Ni₂O₃

Oxidation number of O = -2

Ni₂O₃ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0 

∴ 2 × (Oxidation number of Ni) + 3 × (Oxidation number of O) = 0 

∴ 2 × (Oxidation number of Ni) + 3 × (-2) = 0 

∴ 2 × (Oxidation number of Ni) = +6 

∴ Oxidation number of Ni = +\(\frac{6}{2}\)

∴ Oxidation number of Ni in Ni₂O₃ = +3

(xi) K₃[Fe(CN)₆]

Oxidation number of K = +1 

Oxidation number of CN group = -1

K₃[Fe(CN)₆] is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0

∴ 3 × (Oxidation number of K) + (Oxidation number of Fe) + 6 × (Oxidation number of CN group) = 0

∴ 3 × (+1) + Oxidation number of Fe + 6 × (-1) = O

∴ Oxidation number of Fe + 3 – 6 = 0

∴ Oxidation number of Fe in K₃[Fe(CN)₆] = +3

First of all the oxidation number of each metallic element in its compound is calculated. 

Then these compounds are represented as stock notation:

Representation in which oxidation number of an atom is denoted by Roman numeral in parentheses after the chemical symbol is called Stock notation. 

This name was given after the German scientist, Alfred Stock.

e.g. Au¹⁺ Cl^1- → Au(I)Cl

The Stock notation is used to specify the oxidation number of the metal.

Al acts as anode, Ni acts as cathode.

Oxidation

[Al(s) → Al³⁺(aq) + 3e^-] x 2

Reduction

[Ni²⁺(aq) + 2e^- → Ni(s)] x 3

Net ionic equation

2Al(s) + 3Ni²⁺(aq) → 2Al³⁺(aq) + 3Ni(s)

i. Cu₂S + O₂ → 2Cu + SO₂

ii. Sulphur undergoes an increase in the oxidation state from -2 (in Cu₂S) to +4 (in SO₂).

iii. Copper accepts one electron and undergoes a decrease in the oxidation state from +1(in Cu₂S) to 0 (in Cu). Oxygen accepts two electrons and undergoes a decrease in the oxidation state from 0 (in O₂) to -2 (in SO₂).

Yes, it can be fractional. For example oxidation no. of Pb in Pb₃O₄ = +\(\frac{8}{3}\)

Hg²⁺ is reduced to Hg while O^2- is oxidized to O₂.

CH₄ is oxidized to CO₂ while O₂ is reduced to H₂O.

\(2\overset{o}{K}(s) +\overset{o}{Cl}(g)\rightarrow2\overset{+1}{K}\overset{-1}{C}(s)\)

Oxidation half reaction

K → K⁺ + e^- (Loss electron)

Reduction half reaction

Cl₂ + 2e^- → 2Cl^- (Gain electron)

(i) H₂S(g) + Cl₂ → 2HCl(g) + S(s)

In this reaction, H₂S is oxidised because a more electronegative element i.e. chlorine is added to hydrogen (or a more electropositive element, hydrogen has been removed from S). Chlorine is reduced due to addition of hydrogen to it.

So, H₂S undergoes oxidation and Cl₂ undergoes reduction in this reaction.

(ii) 3Fe₃O₄(s) + 8 Al(s) → 9Fe(s) + 4Al₂O₃(s)

In this reaction aluminium is oxidation because oxygen is added to it. Ferrous ferric oxide (Fe₃O₄) is reduced because oxygen has been removed from it.

So, Al undergoes oxidation and Fe₃O₄ undergoes reduction in this reaction.

(iii) \(2\overset{o}{Na}(s)+\overset{o}{H_2}(g)\rightarrow2\overset{+1}{N}a\overset{-}{H}(s)\)

In this reaction, sodium is oxidised because oxidation number increases and H₂ is reduced because oxidation number decreases.

So, Na undergoes oxidation and H₂ undergoes reduction in this reaction.

Mg → Mg²⁺ + 2e^-