Chemistry NCERT Exemplar Solutions Class 11th Chapter Eight

(a) let x= oxidation number of   sulphur, and   +1 is oxidation number of Na, -2 is oxidation number of O, also we can assume total charge on compound = 0 then solving we get.

+2 + 2x - 6 = 0 

x = +2. 

(b) let x= oxidation number of   sulphur, and   +1 is   oxidation number of Na, -2 is oxidation number of O, also we can assume total charge on compound = 0 then solving we get.

+2 + x - 6 = 0

x = +4

(c) let x= oxidation number of   sulphur, and   +1 is   oxidation number of Na, -2 is oxidation number of O, also we can assume total charge on compound = 0 then solving we get.

+2 + x - 6 = 0

x = +4

(d)  let x= oxidation number of sulphur, and  +1 is   oxidation number of Na, -2 is oxidation number of O, also we can assume total charge on compound = 0 then solving we get.

+2 + x - 8=0

x = +6.

We can balance the given reaction by ion electron method as

Cl₂O₇ (g) + H₂O₂ (aq) → ClO₂^- +O₂ (acidic medium)

This is a Short answer type question as classified in NCERT Exemplar

Balancing bu ion electron method

2 * { Cl₂O₇ + 6H⁺ + 8e^- → 2ClO₂^- + 3H₂O

8  * { H₂O₂ → O₂ + 2H⁺ + 2e^- 

The balanced chemical is given as-

2Cl₂O₇ + 12H⁺ + 16e^- → 4ClO₂^- + 6H₂O

                         8H₂O₂ → 8O₂ + 16H⁺ + 16e^-

          2Cl₂O₇ + 8H₂O₂ → 4ClO₂^- + 6H₂O + 8O₂ + 4H⁺

This is a Short answer type question as classified in NCERT Exemplar

(i) 2Mn0^–₄ + 5S0₂ + 2H₂0 + H⁺?5HS0^–₄ + 2Mn²⁺
Balancing by ion-electron method:

(ii) We can balance the given reaction by oxidation number method-

Balancing by oxidation number method as to make the electron gain and loss equal as given

3N₂H₄ + 4ClO₃^- → 6NO + 4Cl^- + 6H₂O

The balanced chemical is given as-

(iii) We can balance the given reaction by ion electron method as

Cl₂O₇(g) + H₂O₂ (aq) → ClO₂^- +O₂ (acidic medium)

Balancing bu ion electron method

2 * { Cl₂O₇ + 6H⁺ + 8e^- → 2ClO₂^- + 3H₂O

8  * { H₂O₂ → O₂ + 2H⁺ + 2e^- 

The balanced chemical is given as-

2Cl₂O₇ + 12H⁺ + 16e^- → 4ClO₂^- + 6H₂O

                         8H₂O₂ → 8O₂ + 16H⁺ + 16e^-

          2Cl₂O₇ + 8H₂O₂ → 4ClO₂^- + 6H₂O + 8O₂ + 4H⁺

This is a Long answer type question as classified in NCERT Exemplar

We can Measure the electrode potential of the given species by connecting the redox couple of the given species with standard hydrogen electrode. If it is positive, the electrode of the given species acts as reductant and if it is negative, it acts as an oxidant.  Find the electrode potentials of the other given species in the same way, compare the values and determine their comparative strength as an reductant or oxidant. Example Measurement of standard electrode potential of electrode E? _Zn2+/Zn using SHE as a reference electrode.

Types of Redox Reactions

The different types of Redox Reactions are:

Decomposition Reaction

Combination Reaction

Displacement Reaction

Disproportionation Reactions