Solutions-Class-11-Science-Chemistry-Chapter-6-Redox Reactions-Maharashtra Board

(b) -1 in Cl^a and +1 in Cl^b

(d) BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl

(b) A₃ (BC₄)₂

(c) 1, 6, 2, 6

(a) Sn²⁺ is undergoing oxidation

(d) +4

(d) Mn(IV)O₂

(c) \(-\frac{1}{2}\)

(a) F

(a) Reduction

H₂SO4 is a neutral molecule.

Sum of oxidation number of all atoms of H₂SO₄ = 0

∴ 2(oxidation number of H) + oxidation number of S + 4(oxidation number of O) = 0

∴ 2(+1) + x + 4(-2) = 0

∴ x = + 6

∴ Oxidation number of S = +6

HNO₃ is a neutral molecule.

Sum of oxidation number of all atoms of HNO₃ = 0

∴ Oxidation number of H + oxidation number of N +3(oxidation number of O) = 0

∴ (+1) + x + 3(—2) = 0

∴ x = + 5

∴ Oxidation number of N = +5

H₃P0₃ is a neutral molecule.

Sum of oxidation number of all atoms of H₃PO₃ = 0

∴ 3(oxidation number of H) + oxidation number of P + 3(oxidation number of O) = 0

∴ 3(+1) + x + 3(—2)

∴ x = + 3

∴ Oxidation number of P = + 3

K₂C₂O₄ is a neutral molecule.

Sum of oxidation number of all atoms of K₂C₂O₄ = 0

∴ 2(oxidation number of K) +2(oxidation number of C) + 4(oxidation number of O) = 0

∴ 2(+1) + 2x + 4(—2) = 0

∴ x = + 3

∴ Oxidation number of C = + 3

H₂S₄O₆ is a neutral molecule.

Sum of oxidation number of all atoms of O H₂S₄O₆ = 0

∴ 2(oxidation number of H) + 4(oxidation number of S) + 6(oxidation number of O) = 0

∴ 2(+1) + 4(x) + 6(—2) = 0

∴ x = + 2.5

∴ Oxidation number of S = + 2.5

Cr₂O₇^2— is an ionic species carrying charge —2

Sum of oxidation number of all atoms 0f Cr₂O₇^2— = —2

∴ 2(oxidation number of Cr) + 7(oxidation number of O) = — 2

∴ 2x + 7(—2) = -2

∴ x = +6

∴ Oxidation number of Cr = + 6

NaH₂PO₄ is a neutral molecule.

Sum of oxidation number of all atoms of NaH₂PO₄ = 0

∴ Oxidation number of Na + 2(oxidation number of H) + oxidation number of P + 4(oxidation number of O) = 0

∴  (+1) + 2(+1) + x + 4(—2) = 0

∴ x = + 5

∴ Oxidation number of P = +5

(i) The reaction is a redox reaction

(ii) Cu is reduced from + 1 to zero

(iii) S is oxidized from — 2 to + 4

(iv) Cu₂O is an oxidant

(v) Cu₂S is a reductant

Since there is no change in the oxidation number of any element it is not a redox reaction.

(i) The reaction is a redox reaction

(ii) I is reduced from zero to — 1.

(iii) S is oxidized from + 2 to + 2.5

(iv) I₂ is an oxidant.

(v) S₂O₃^2- Is a reductant.

Oxidation is defined as:

• Addition of oxygen.
• Addition of electronegative element.
• Removal of hydrogen.
• Removed of electropositive element.
• Loss of electrons by any species.

The species in the oxidized state

(a) Mg²⁺ (b) Cu²⁺ (c) O₂ (d) Cl₂

(i) The reaction is a redox reaction.

(ii) S_(S) is an oxidizing agent.

(iii) Na_(S) is a reducing agent.

(iv) Na_(S) is oxidizing and S_(S) is reduced

Cr(OH)₄ is an ionic species carrying net charge —1.

Sum of oxidation number of all atoms of Cr(OH)₄ = —1

∴ x + 4(oxidation number of O) + 4(oxidation number of H) = — 1

∴ x + 4(—2) + 4(+1) = —1

∴ x = + 3

∴ Oxidation number of Cr = + 3

Na₂S₂O₃ is a neutral molecule.

Sum of oxidation number of all atoms of Na₂S₂O₃ = 0

∴ 2(oxidation number of Na) + 2(oxidation number of S) + 3(oxidation number of O) = 0

∴ 2(+1) + 2(x) + 3(—2) = 0

∴ x = + 2

∴ Oxidation numbers of S = + 2

H₃BO₃, is a neutral molecule.

Sum of oxidation number of all atoms of H₃BO₃ = 0

∴ 3(oxidation number of H) + (oxidation number of B) + 3(oxidation number of O) = 0

∴ 3(+1) + x + 3(-2) = 0

∴ x = + 3

∴ Oxidation number of B = +3.

(a) Since the standard reduction potential, E° of C1₂ (+ 1.36 V) is higher than E° of Br₂ (1.09 V) the stronger oxidizing agent in the redox couple is Cl₂.

(b) Since standard reduction potential E° for MnO₄^— (1.51 v) is higher than 15° for Cr₂O₇^2— (1.33 V), MnO₄^— will be a stronger oxidizing agent than Cr₂O₇^2—, in the redox couple.

(a) Since the standard reduction potential, E° of Li(— 3.05 V) is less or more negative than E° of Mg (— 2.36 V) lithium will be a stronger reducing agent than. Mg in the redox couple.

(b) Since the standard reduction potential of Zn(— 0.76 V) is less or more negative than E° of Fe (— 0.44 V) zinc will be a stronger reducing agent than Fe in the redox couple.

(Note : Cr₂O₂^2— is given in text book is changed to Cr₂O₇^2—)

Step 1: Assign oxidation number to each atom in the given reaction.

Step 2: Identify the atoms that have changed oxidation numbers.

Step 3: Find the total decrease in oxidation number of the reduced atom Cr and total increase in oxidation number of the oxidized atom S.

Total decrease in oxidation number = (+6) — (+3) = 3

Total increase in oxidation number = (+6) — (+4) = 2

Step 4: To balance the net change in oxidation numbers, multiply Cr₂O₇^2- by 1 and Cr³⁺ by 2 and multiply SO₃^2- and SO₄^2- by 3.

Cr₂O₇^2-_(aq) + 3SO₃^2-_(aq) → 2Cr³⁺_(aq) + 3SO₄^2-_(aq)

Step 5: Find net charges on left hand and right hand side of the equation.

Total change on left hand side

= (—2) + 3(-2)= — 8

Total change on right hand side

= 2(+3) + 3(—2) = 0

Since there is net — 8 change on left hand side, add 8H⁺ to it since the reaction is in acidic medium.

Hence balanced equation is,

Step 1: Assign oxidation number to each atom in the equation.

Step 2: Identify the atoms that have changed oxidation numbers.

Step 3: Find the total decrease in oxidation number of reduced atom Mn and total increase in oxidation number of oxidized atom Br.

Total decrease in oxidation number

= (+7) — (+4) = 3

Total increase in oxidation number

= (+5) — (—1) = 6

Step 4: To balance the net change in oxidation numbers, multiply MnO₄^— and MnO₂ by 2.

2MnO₄^—_(aq) + Br^—_(aq) →  2MnO_2(aq) + BrO₃^—_(aq)

Step 5: Since there are 3 negative charges on left hand side and 1 negative charge on right hand side of the equation add 1H₂O on left and 2OH^— on right hand side.

This is a balanced equation.

2MnO₄^—_(aq) + Br^—_(aq) + H₂O_(l) →  2MnO_2(aq) + BrO₃^—_(aq) + 2OH^—_(aq)

Step 1: Assign oxidation number to each atom in the given equation.

Step 2: Identify the atoms that have changed oxidation numbers.

Step 3: Find the total decrease in oxidation number of reduced atom S and total increases in oxidation number of the oxidized atom C.

Total decrease in oxidation number

= (+6) — (+4) = 2

Total increase in oxidation number

= (+ 4) — (9) = 4

Step 4: To balance the net charge in oxidation numbers, multiply H₂SO₄ and SO₂ by 2.

2H₂SO_{4 (aq)} + C _(s) → CO_{2 (g)} + 2SO_{2 (g)} + H₂O _(l)

Since there are 4H on left side and 2H on right side, multiply H₂O by 2.

∴ 2H₂SO_{4 (aq)} + C _(s) → CO_{2 (g)} + 2SO_{2 (g)} + 2H₂O _(l)

Step 5: Since there is no net charge on left and right hand sides, the above equation is balanced.

Step 1: Assign oxidation number to each atom in the equation.

Step 2: Identify the atoms that have changed oxidation number.

Step 3: Find the total decrease in oxidation number of reduced atom Bi and total increase in oxidation number of oxidized atom Sn.

Total decrease in oxidation number

= (+ 3) — (0) = 3

Total increase in oxidation number

= (+4) — (+2) = 2

Step 4: To balance the net charge in oxidation numbers, multiply Bi(OH)₃ and Bi by 2 while Sn(OH)₃^— and Sn(0H)₆^2— by 3.

2Bi(OH)_{3 (g)} + 3Sn(OH)_{3 (aq)} → 2Bi _(s) + 3Sn(OH)₆^2— _(aq)

Step 5: Since there are 3 negative charges on left hand side and 6 negative charges on right hand side of the equation, add 3OH^- for basic medium on left hand side.

2Bi(OH)_{3 (g)} + 3Sn(OH)_{3 (aq)} + 3OH^— _(aq) → 2Bi _(s) + 3Sn(OH)₆^2— _(aq)

This is a balanced equation.

Step 1: Assign oxidation numbers to all atoms

Step 2: Divide the given equation into two half equations, one for oxidation and the other for reduction.

C³⁺         → CO₂  (oxidation half equation)

⁷⁺MnO^-  → Mn²⁺  (reduction half reaction)

Step 3:

Loss of electrons by C atom = (+ 4) — (+ 3) = 1

Gain of electrons by Mn atom = (+ 7) — (+ 2) = 5

Hence we can write,

H₂C₂O_{4 (aq)} → 2CO₂ + 2e^— (oxidation)

MnO₄^— _(aq) + 5e^— → Mn²⁺ (reduction)

Step 4 : T0 balance loss and gain of electrons, multiply oxidation equation by 5 and reduction equation by 2.

5H₂C₂O_{4 (aq)} → 10CO₂ + 10e^— (oxidation)

2MnO₄^— _(aq) + 10e^— → 2Mn²⁺ (reduction)

Step 5: By adding both the equations,

5H₂C₂O_{4 (aq)} + 2MnO₄^— _(aq) → 2Mn²⁺ + 10CO_{2 (g)}

Step 6: To balance the charges and H atoms add 6H⁺ on left hand side of equation.

5H₂C₂O_{4 (aq)} + 2MnO₄^— _(aq) + 6H⁺ → 2Mn²⁺ + 8H₂O  + 10CO_{2 (g)}

The above equation is a balanced equation.

Step 1 : Assign oxidation numbers to all atoms.

Step 2 : Divide the equation into two half equations one for oxidation and the other for reduction.

SnO₂^2— _(aq) → SnO₃^2— _(aq) (oxidation half reaction)

Bi(OH)_{3 (s)} → Bi _(s) (reduction half reaction)

Step 3 : Loss of electrons by Sn = ( + 4) — (+ 2) = 2

Gain of electrons by Bi = (+ 3) — (O) = 3

Hence we can write,

SnO₂^2— _(aq) → SnO₃^2— _(aq) + 2e^— (Oxidation)

Bi(OH)_{3 (s)} + 3e^— → Bi (reduction)

Step 4 : To balance loss and gain of electrons multiply oxidation equation by 3 and reduction equation by 2.

3SnO₂^2— _(aq) → 3SnO₃^2— _(aq) + 6e^— (Oxidation)

2Bi(OH)_{3 (s)} + 6e^— → 2Bi (reduction)

Step 5 : By adding both the equations,

3SnO₂^2— _(aq) + 2Bi(OH)_{3 (s)} → 3SiO₃^2— _(aq) + 2Bi

Step 6 : To balance H and O atoms add 3H₂O on right hand side.

3SnO₂^2— _(aq) + 2Bi(OH)_{3 (s)} → 3SnO₃^2— _(aq) + Bi _(s) + 3H₂O _(l)