Redox Reactions Class 11 Chemistry NCERT Solutions Ch 7

(i) When emf is positive, the reaction is feasible.
E^o_{? cell} = E^o_{? Fe}? −E^o_{? I2}? =0.77−0.54=0.23V
Reaction is feasible.

(ii) E^o_? cell? =E^o_{? Ag}? −E^o_{? Cu}? =0.80−0.34=0.46V
Reaction is feasible.

(iii) E^o_? cell? =E^o_{? Fe}? −E^o_{? Cu}? =0.77−0.34 =0.43 V
Reaction is feasible.

(iv) E^o_? cell? =E^o_{? Fe}? −E^o_{? Ag}? =0.77−0.80=−0.03 V
Reaction is not feasible.

(v) E^o_? cell? =E^o_{? Br2?} ? −E_Fe? =1.09−0.77=0.32 V 
Reaction is feasible.

(a) Toluene can be oxidised to benzoic acid in acidic, basic and neutral media according to the following redox equations:

In the laboratory, benzoic acid is usually prepared by alcoholic KMnO₄ oxidation of toluene. However, in industry alcoholic KMnO₄ is preferred over acidic or alkaline KMnO₄ because of the following reasons:
(i) The cost of adding an acid or the base is avoided because in the neutral medium, the base (OH^- ions) are produced in the reaction itself.
(ii) Since reactions occur faster in homogeneous medium than in heterogeneous medium, therefore, alcohol helps in mixing the two reactants, i.e., KMnO₄  (due to its polar nature) and toluene (because of its being an organic compound).

(b) KBr + H₂? SO₄? → KHSO₄? + HBr
HBr is a strong reducing agent which further reduces H₂? SO₄?  as follows
2HBr + H₂? SO₄? → Br₂? + SO₂? + H₂? O

But in case of mixture containing Chloride the reaction ends up on forming HCl as HCl is a weak reducing agent which can't further reducethe sulphuric acid further.
KCl+H₂? SO₄? →KHSO₄? +HCl? (colourless pungent smell gas)

(a) Oxidation number method:
The oxidation number of P decreases from 0 to -3 and increases from 0 to +2. Hence,  P₄?  is oxidizing as well as reducing agent.
During reduction, the total decrease in the oxidation number for 4 P atoms is 12.
During oxidation, total increase in the oxidation number for 4 P atoms is 4.

The increase in the oxidation number is balanced with decrease in the oxidation number by multiplying H_2? PO₂⁻? with 3. 

P₄? (s) + OH⁻ (aq) → PH₃? (g) + 3H₂? PO₂⁻? (aq)

To balance O atoms, multiply OH⁻ ions by 6.

P₄? (s) + 6OH⁻ (aq) → PH₃? (g) + 3H₂? PO₂⁻? (aq)

To balance H atoms, 3 water molecules are added to L.H.S and 3 hydroxide ions on R.H.S.

P₄? (s) + 6OH⁻ (aq) + 3H₂? O (l) → PH₃? (g) + 3H₂? PO₂⁻? (aq) + 3OH⁻ (aq)
Subtract 3 hydroxide ions from both sides.

P_4? (s) + 3OH⁻ (aq) + 3H₂? O (l) → PH₃? (g) + 3H₂? PO₂⁻? (aq)

Ion electron method:
The oxidation half reaction is 

P₄? (s) → H₂? PO₂⁻? (aq).
The P atom is balanced.

P₄? (s) → 4H₂? PO₂⁻? (aq)

The oxidation number is balanced by adding 4 electrons on RHS.

P₄? (s) → 4H₂? PO₂⁻? (aq) + 4e⁻

The charge is balanced by adding 8 hydroxide ions on LHS.

P₄? (s)+8OH⁻ (aq)→4H₂? PO₂⁻? (aq)
The O and H atoms are balanced.

The reduction half reaction is P₄? (s)→PH₃? (g).
The oxidation number is balanced by adding 12 electrons on LHS.

P_4? (s)+12e⁻→PH₃? (g)
The charge is balanced by adding 12 hydroxide ions on RHS.

P₄? (s)+12e⁻→PH_3? (g)+12OH⁻
The oxidation half reaction is multiplied by 3 and the reduction half reaction is multiplied by 2.
The half reactions are then added to obtain balanced chemical equation.

P_4? (s)+3OH⁻ (aq)+3H₂? O (l)→PH₃? (g)+3H₂? PO₂⁻? (aq)

(b) The oxidation number of N increases from – 2 in N₂H₄ to + 2 in NO and the oxidation number of Cl decreases from + 5 in CIO⁻₃ to – 1 in Cl^–. Hence, in this reaction,  N₂H₄ is the reducing agent and CIO⁻₃ is the oxidizing agent.
Ion - electron method:
The oxidation half equation is:

−2               +2

N₂H₄ (l)? NO (g)
The N atoms are balanced as:

N₂H₄ (l)?2NO (g)+8e⁻
The charge is balanced by adding 8 OH^– ions as:

N₂H₄ (I)+8OH⁻ (aq)?2NO (g)+8e⁻
The O atoms are balanced by adding 6H2O as:

N₂H₄ (I)+8OH⁻ (aq)?2NO (g)+6H₂O (I)+8e⁻. (i)
The reduction half equation is:

+5            

(a) The balanced half reaction equations are:
Oxidation half equation:

I⁻ (aq) → I₂ (s)       - (i)

Reduction half reaction equation:

MnO₄⁻ (aq) → MnO₂ (aq) - (ii)

Balance I atoms and charges in the oxidation half reaction.

2I⁻ (aq) → I₂ (s) + 2e⁻

In the reduction half reaction, the oxidation number of Mn changes from +7 to +4. Hence, add 3 electrons to reactant side of the reaction.

MnO₄⁻ (aq) + 3e⁻→ MnO₂ (aq)

Balance charge in the reduction half reaction by adding 4 hydroxide ions to product side.

MnO4⁻ (aq) + 3e⁻ → MnO₂ (aq)+4OH⁻

To balance O atoms, add 2 water molecules to reactant side.

MnO4⁻ (aq)+ 3e⁻ +2H₂O→MnO₂ (aq) + 4OH⁻

To equalize the number of electrons, multiply the oxidation half reaction by 3 and multiply the reduction half reaction by 2.

6I⁻ (aq) → 3I₂ (s)+6e⁻

2MnO₄⁻ (aq) + 6e⁻+4H₂O → 2MnO₂ (aq)+8OH⁻

Add two half-cell reactions to obtain the balanced equation.

2MnO₄⁻? (aq) + 6I⁻ (aq) + 4H₂? O₂? (l) → 2MnO₂? (s) + 3I₂? (s) + 8OH⁻

(b) The balanced half reaction equations are:
Oxidation half equation:

SO₂ (g) + 2H₂O (l) → HS0₄^–  (aq) + 3H⁺ (aq) +2e^–                        … (i)
Reduction half equation:

MnO₄^– (aq) + 8H⁺ (aq) + 5e^– → Mn²⁺ (aq) + 4H₂O (l) ………. (ii)
Multiply Eq. (i) by 3 and Eq. (ii) by 2 and add, we have,

2MnO₄^– (aq) + 5S0₂ (g) + 2H₂0 (l) + H⁺ (aq) → 2Mn²⁺ (aq) + 5HSO₄^– (aq)

(c) Oxidation half equation: Fe²⁺ (aq) → Fe³⁺ (aq) + e^– … (i)
Reduction half equation: H₂O₂ (aq) + 2H⁺ (aq) + 2e^– → 2H₂O (l) … (ii)
Multiply Eq. (i) by 2 and add it to Eq. (ii), we have,

H₂O₂ (aq) +2Fe²⁺ (aq) +2H⁺ (aq) → 2Fe³⁺ (aq) + 2H₂O (l)

(d)Oxidation half equation:

SO₂ (g) + 2H₂O (l) → SO₄^2- (aq) + 4H⁺ (aq) + 2 e^–  … (i)
Reduction half equation:

Cr₂O₇^2– (aq) + 14H⁺ (aq) + 6e^– → 2Cr³⁺ (aq) + 7H₂O (l) … (ii)
Multiply Eq. (i) by 3 and add it to Eq. (ii), we have,

Cr₂O₇^2– (aq) + 3SO₂ (q) + 2H⁺ (aq) → 2Cr³⁺ (aq) + 3SO₄^2- (aq) + H₂O (l)

In AgF₂,  oxidation state of Ag is +2 which is very unstable. Therefore, it quickly accepts an electron to form the more stable +1 oxidation state.
Ag²⁺ + e^– →Ag⁺
Therefore, AgF₂, if formed, will act as a strong oxidising agent.

Lower the electrode potential, better is the reducing agent. Since the electrode potentials increase in the order: K⁺/K (-2.93 V), Mg²⁺/Mg (-2.37 V), Cr³⁺/Cr (-0.74 V), Hg²⁺/Hg (0.79 V), Ag⁺/Ag (0.80 V), therefore, reducing power of metals decreases in the same order, i.e., K, Mg, Cr, Hg, Ag.

(i) An aqueous solution of AgNO₃?  with silver electrodes.
At cathode: Silver ions have lower discharge potential than hydrogen ions. Hence, silver ions will be deposited in preference to hydrogen ions.

At anode: Silver anode will dissolve to form silver ions in the solution.
Ag → Ag⁺ + e⁻

(ii) An aqueous solution of AgNO_3?  with platinum electrodes.

At cathode: Silver ions have lower discharge potential than hydrogen ions. Hence, silver ions will be deposited in preference to hydrogen ions.

At anode: Hydroxide ions having lower discharge potential will be discharged in preference to nitrate ions. Hydroxide ions will decompose to give oxygen.
4OH⁻ (aq) → 2H₂? O (l) + O₂? (g) + 4e⁻

(iii) A dilute solution of H₂? SO₄?  with platinum electrodes.

At cathode: 2H⁺ + 2e⁻ → H₂? (g)

At anode: Hydroxide ions having lower discharge potential will be discharged in preference to sulphate ions. Hydroxide ions will decompose to give oxygen.
4OH⁻ (aq) → 2H₂? O (l) + O₂? (g) + 4e⁻

(iv) An aqueous solution of CuCl₂?  with platinum electrodes

At cathode: Cupric ions will be reduced in preference to protons
Cu²⁺ + 2e⁻ → Cu

At anode: Chloride ions will be oxidized in preference to hydroxide ions
2Cl⁻ → Cl₂? + 2e⁻

The three examples are:

(i) When excess P_4?   (reducing agent) reacts with F_2? (oxidizing agent),  PF_3?  is produced in which P has +3 oxidation number.
                        P_4? (excess) + F₂? → PF₃?
But if fluorine is in excess,  PF₅?  is formed in which P has oxidation number of +5.
                        P_4? ? + F₂? (excess) → PF₅?

(ii) Oxidizing agent is oxygen and reducing agent is K. When excess K reacts with oxygen,  K_2? O is formed in which oxygen has oxidation number of -2.
                        4K (excess) + O₂? → 2K₂? O

But if oxygen is in excess, then K_2? O₂?  is formed in which O has oxidation number of -1.
                        2K + O₂? (excess) → K₂? O₂?

(iii) The oxidizing agent is oxygen and the reducing agent is C. When an excess of C reacts with oxygen, CO is formed in which C has +2 oxidation number.
                        C (excess) + O_2? → CO

When excess of oxygen is used,  CO₂?  is formed in which C has +4 oxidation number.
                        C + O₂? (excess) → CO₂

The given redox reaction can be depicted as

Zn (s) + 2Ag⁺ (aq) → Zn²⁺ (aq) + 2Ag (s)
Since Zn gets oxidised to Zn²⁺ ions, and Ag⁺ gets reduced to Ag metal, therefore,

Oxidation occurs at the zinc electrode and reduction occurs at the silver electrode. Thus, galvanic cell corresponding to the above redox reaction may be depicted as:

Zn|Zn²⁺ (aq) | Ag⁺ (aq) | Ag

(i) Zinc electrode is negatively charged because oxidation occurs at the zinc electrode (i.e. electrons accumlulate on the zinc electrode)

(ii) The ions carry current. The electrons flow from Zn to Ag electrode while the current flows from Ag to Zn electrode.

(iii) Reaction occurring at each electrode are:

Zn (s) → Zn²⁺ (aq) + 2e^-

Ag⁺ (aq) + e^- → Ag (s)

Reactions (a) and (b) indicate that H₃PO₂  (hypophosphorous acid) is a reducing agent and thus reduces both AgNO₃ and CuSO₄ to Ag and Cu respectively. Conversely, both AgNO₃ and CuSO₄ act as oxidising agent and thus oxidise H₃PO₂to H₃PO₄  (orthophosphoric acid) Reaction (c) suggests that [Ag (NH₃)₂]⁺ oxidises C₆H₅CHO (benzaldehyde) to C₆H₅COO^–  (benzoate ion) but reaction (d) indicates that Cu²⁺ ions cannot oxidise C₆H₅CHO to C₆H₅COO^–. Therefore, from the above reactions, we conclude that Ag⁺ ion is a strong deoxidising agent than Cu²⁺ ion.

XeO₆⁴⁻? oxidizes F⁻ and F⁻ reduces XeO₆⁴⁻?
Hence, the given reaction occurs.
The oxidation number of Xe decreases from +8 to +6. The oxidation number of F increases from -1 to 0.
Thus,  Na₄? XeO₆? is a stronger oxidising agent than F⁻.

The average O.N. of S in S₂O₃^2- is +2 while in S₄O₆^2- it is + 2.5. The O.N. of S in SO₄^2- is +6. Since Br₂ is a stronger oxidising agent than I₂, it oxidises S of S₂O₃^2- to a higher oxidation state of +6 and hence forms SO₄^2- ion. I₂, however, being weaker oxidising agent oxidises S of S₂O₃^2- ion to a lower oxidation of +2.5 in S₄O₆^2- ion. It is because of this reason that thiosulphate reacts differently with Br₂ and I₂.

Fluorine oxidizes chloride ion to chlorine, bromide ion to bromine and iodide ion to iodine respectively.
F₂? + 2Cl⁻ → 2F⁻ + Cl₂?
F₂? + 2Br⁻ → 2F⁻ + Br₂?
F_2? + 2I⁻ → 2F⁻ + I₂?

Chlorine oxidizes bromide ion to bromine and iodide ion to iodine.
Cl_2? + Br⁻ → 2Cl⁻ + Br₂?
Cl₂? + I⁻ → 2Cl⁻ + I₂?

Bromine oxidizes iodide ion to iodine.
Br₂? + I⁻ → 2Br⁻ + I₂?

But bromine and chlorine cannot oxidize fluoride to fluorine. Hence, fluorine is the best oxidizing agent amongst the halogens. The decreasing order of the oxidizing power of halogens is F₂? >Cl₂? >Br₂? >I₂?  

HI and HBr can reduce sulphuric acid to sulphur dioxide but HCl and HF cannot. Thus,  HI and HBr are stronger reducing agents than HCl and HF.
2HI+H₂? SO₄? →I₂? +SO₂? +2H₂? O
2HBr+H₂? SO₄? →Br₂? +SO₂? +2H₂? O

Iodide ion can reduce Cu (II) to Cu (I) but bromide cannot.
4I⁻+2Cu²⁺→Cu₂? I₂? +I₂?

Hence, among the hydrohalic compounds, hydroiodic acid is the best reductant. The reducing power of hydrohalic acids is HF

Let x be the oxidation number to the underlined elements in the given species:

(a) NaH₂PO₄

(+1) + 2 (+1) + x + 4 (-2) = 0

x + 3 – 8 = 0

x = +5

(b) NaHSO₄

(+1) + (+1) + x + 4 (-2) = 0

x – 6 = 0

x = +6

(c) H₄P₂O₇

4 (+1) + 2x + 7 (-2) = 0

2x -10 =0

x = +5

(d) K₂MnO₄

2 (+1) + x + 4 (-2) = 0

x – 6 = 0

x = +6

(e) CaO₂

2 + 2x = 0

x = -1

(f) NaBH₄

1 + x + 4 (-1) = 0 (Since H is present as hydride ion.)

x = +3

(g) H₂S₂O₇

2 (+1) + 2x + 7 (-2) = 0

x = +6

(h) KAl (SO₄)₂.12H₂O

+1 + 3 + 2x + 8 (-2) + 12 (2 x 1 - 2) = 0

x = +6

(a) In Kl₃, since the oxidation number of K is +1, therefore, the average oxidation number of iodine = -1/3. But the oxidation number cannot be fractional. Therefore, we must consider its structure, K⁺ [I —I <— I]^–. Here, a coordinate bond is formed between I₂ molecule and I^– ion. The oxidation number of two iodine atoms forming the I₂ molecule is zero, while that of iodine forming the coordinate bond is -1. Thus, the oxidation number of the three I atoms, atoms in Kl₃ is 0, 0 and -1, respectively.

(b) By conventional method O.N. of S in H₂S₄O₆is calculated as:

2 (+1) +4x + 6) (-2) = 0

Or x = +2.5

But all the four S atoms cannot be in the same oxidation state.

So by chemical bonding method:

The O.N. of each of the S-atoms linked with each other in the middle is zero while that of each of the remaining two S-atoms is +5.

(c) By conventional method O.N. of Fe in Fe₃O₄ is calculated as:

3x + 4 (-2) = 0 or x = 8/3

By stoichiometry O.N. of Fe in Fe₃O₄ is +2 and +3.

(d) By conventional method, O.N. of C in CH3CH2OH is calculated as:

2x + 6 (+1) +1 (-2) = 0 or x = -2

(e) By conventional method, O.N. of C in CH₃COOH is calculated as:

2x + 4 - 4 = 0 or x = 0

By chemical bonding method, C2 is attached to three H-atoms (less electronegative than carbon) and one -COOH group (more electronegative than carbon).

Therefore, O.N. of C₂ = 3 (+1) + x + 1 (-1) = 0 or x = -2

But C₁ is attached to one oxygen atom by a double bond, one OH (-1) and one CH3 (+1) group, therefore, O.N. of C₁ = +1 + x+ 1 (-2) + 1 (-1) = 0 or x = +2

(a) Here, O is removed from CuO, therefore, it is reduced to Cu, while O is added to H₂ to form H₂O, therefore, it is oxidised. Further, O.N. of Cu decreases from +2 in CuO to 0 in Cu but that of H increases from 0 in H₂ to +1 in H₂0. Therefore, CuO is reduced to Cu but H₂ is oxidised to H₂O. Thus, this is a redox reaction.

(b) Here O.N. of Fe decreases from +3 in Fe₂O₃ to 0 in Fe while that of C increases from +2 in CO to +4 in CO₂. Further, oxygen is removed from Fe₂O₃ and added to CO, therefore, Fe₂O₃ is reduced while CO is oxidised. Thus, this is a redox reaction.

(c) Here, O.N. of B decreases from +3 in BrCl₃to -3 in B₂H₆ while that of H increases from -1 in LiAlH₄to +1 in B₂H₆. Therefore, BCl₃ is reduced while LiAlH₄ is oxidised. Further, H is added to BCl₃ but is removed from LiAlH₄, therefore, BC1₃ is reduced while LiAlH₄ is oxidised. Thus, it is a redox reaction.

(d) Here, each K atom as lost one electron to form K⁺ while F₂ has gained two electrons to form two F^– ions. Therefore, K is oxidised while F₂ is reduced. Thus, it is a redox reaction.

(e) Here, the oxidation number of N increases from –3 in NH₃ to +2 in NO. On the other hand, the oxidation number of O₂ decreases from 0 in O₂ to –2 in NO and H₂O i.e., O₂ is reduced. Hence, the given reaction is a redox reaction.

Writing the O.N. of each atom above its symbol, we have,          

here, the O.N. of F decreases from 0 in F₂ to -1 in HF and increases from 0 in F₂ to +1 in HOF. Therefore, F₂ is both reduced as well as oxidised. Thus, it is a redox reaction and more specifically, it is a disproportionation reaction.

(a) H_2? SO₅?  by conventional method.
Let x be the oxidation number of S
2 (+1) + x + 5 (−2) = 0
x = +8
+8 oxidation state of S is not possible as S cannot have an oxidation number more than 6. The fallacy is overcome if we calculate the oxidation number from its structure HO−S (O₂)−O−O−H.
−1+X+2 (−2)+2 (−1)+1=0
x=+6

(b) Dichromate ion
Let x be the oxidation number of Cr in dichromate ion
2x+7 (−2)=−2
x=+6
Hence the oxidation number of Cr in dichromate ion is +6. This is correct and there is no fallacy.

(c) Nitrate ion, by conventional method
Let x be the oxidation number of N in nitrate ion.
x+3 (−2)=−1
From the structure O−−N+ (O)−O−
x+1 (−1)+1 (−2)+1 (−2)=0
x=+5
Thus there is no fallacy.

(a) HgCl₂  (b) NiSO₄  (c)SnO₂  (d) Tl₂SO₄  (e) Fe₂ (S0₄)₃  (f) Cr₂O₃.

(i) In SO₂, O.N. of S is +4. In principle, S can have a minimum O.N. of -2 and maximum of +6. Therefore, S in SO₂ can either decrease or increase its O.N. and hence can act both as an oxidising as well as a reducing agent.

(ii) In H₂O₂, the O.N. of O is -1. In principle, O can have a minimum O.N. of -2 and maximum of zero (+1 is possible in O₂F₂ and +2 in OF₂). Therefore, O in H₂O₂ can either decrease its O.N. from -1 to -2 or can increase its O.N. from -1 to zero. Therefore, H₂O₂ acts both as an oxidising as well as a reducing agent.

(iii) In O₃, the O.N. of O is zero. It can only decrease its O.N. from zero to -1 or -2, but cannot increase to +2. Therefore, O₃ acts only as an oxidant.

(iv) In HNO₃, O.N. of N is +5 which is maximum. Therefore, it can only decrease its O.N. and hence it acts as an oxidant only.

(a) Therefore, it is more appropriate to write the equation for photosynthesis as (iii) because it emphasises that 12H₂O are used per molecule of carbohydrate formed and 6H₂O are produced during the process.

(b) The purpose of writing O₂ two times suggests that O₂  is being obtained from each of the two reactants.

The path of reactions (a) and (b) can be determined by using  H₂O₂¹⁸ or D₂O in reaction
(a) or by using  H₂O₂¹⁸ or O₃¹⁸in reaction (b).

Let x be the O.N. of C.
O.N. of C in cyanogen, (CN)₂ = 2 (x – 3) = 0 or x = +3

O.N. of C in cyanide ion, CN^- = x – 3 = -1 or x = +2

O.N. of C in cyanate ion, CNO =x-3-2 = -l or x: = +4

The four information about the reaction are:
(i) The reaction involves decomposition of cyanogen, (CN)₂ in the alkaline medium to cyanide ion, CN and cyanate ion, CNO^–.

(ii) The O.N. of C decreases from +3 in (CN)₂ to +2 in CN^–ion and increases from +3 in (CN)₂ to +4 in CNO^– ion. Thus, cyanogen is simultaneously reduced to cyanide ion and oxidised to cyanate ion.

(iii) It is an example of a redox reaction in general and a disproportionation reaction in particular.

(iv) Cyanogen is a pseudohalogen (behaves like halogens) while cyanide ion is a pseudohalide ion (behaves like halide ion).

The unbalanced chemical reaction is:
Mn³⁺ (aq) → Mn²⁺ (aq) + MnO₂? (s) + H⁺ (aq)
The oxidation half reaction is,  
Mn³⁺ (aq) → MnO₂? (s).
To balance oxidation number, one electron is added on R.H.S.
Mn³⁺ (aq) → MnO₂? (s) + e⁻
4 protons are added to balance the charge.
Mn³⁺ (aq) → MnO₂? (s) + 4H⁺ (aq) + e⁻
2 water molecules are added to balance O atoms.
The reduction half reaction is Mn³⁺ (aq) → Mn²⁺ (aq).
An electron is added to balance oxidation number.
Mn³⁺ (aq) + e⁻ → Mn²⁺ (aq)
Two half-cell reactions are added to obtain balanced chemical equation.
2Mn³⁺ (aq) + 2H₂? O (l) → Mn2+ (aq) + MnO₂? (s) + 4H⁺ (aq)

(a) F. Fluorine being the most electronegative element shows only a -ve oxidation state of -1.

(b) Cs. Alkali metals because of the presence of a single electron in the valence shell, exhibit an oxidation state of +1.

(c) I. Because of the presence of seven electrons in the valence shell, I shows an oxidation state of -1 (in compounds of I with more electropositive elements such as H, Na, K, Ca, etc.) or an oxidation state of +1 compounds of I with more electronegative elements, i.e., O, F, etc.) and because of the presence of d-orbitals it also exhibits +ve oxidation states of +3, +5 and +7.

(d) Ne. It is an inert gas (with high ionization enthalpy and high positive electron gain enthalpy) and hence it neither exhibits -ve nor +ve oxidation states.

The balanced chemical reaction for the redox reaction between chlorine and sulphur dioxide is:

Cl₂? + SO₂? + 2H₂? O → 2Cl⁻ + SO₄²⁻ ? + 4H⁺.

(a) Phosphorous, chlorine and sulphur are the non metals which can show disproportionation reaction. The reactions shown are:

P_4? + 3OH^? + 3H₂? O? PH₃? + 3H₂? PO₂^?

? Cl₂? + 2OH^? ? Cl^? + ClO^? + H₂? O

S₈? +12OH^? ?4S^2? +2S₂? O₃^2? ? +6H₂? O

(b) Copper, gallium and indium are the metals that show disproportionation reaction. 

The reactions are shown below.

2Cu⁺? Cu²⁺ + Cu

3Ga⁺? Ga³⁺ + 2Ga

3In⁺? In³⁺ + 2In

The balanced equation for the reaction is:

4NH_3? (g)+5O₂? (g)→4NO (g)+6H₂? O (g)

The molar masses of ammonia and oxygen are 17 g/mol and 32 g/mol respectively.

68 g of NH₃ will react with O₂ = 160 g.

Therefore, 10 g of NH₃ will react with O₂ = 160/68 x 10 g = 23.6 g
But the amount of O₂ which is actually available is 20.0 g which is less than the amount which is needed. Therefore, O₂ is the limiting reagent and hence calculations must be based upon the amount of O₂ taken and not on the amount of NH₃ taken. From the equation,
160 g of O₂ produce NO = 120 g
Therefore, 20 g of O₂ will produce NO =120/160 x 20 = 15 g

Based on the relative positions of these metals in the activity series, the correct order is Mg, Al, Zn, Fe, Cu.

Reason: It can react both as an anode as well a cathode in an electrochemical cell.
Answer: (a) A standard hydrogen electrode is called a reversible electrode because it can react both as anode as well as cathode in an electrochemical cell.

(b) Both are correct statements but reason is not correct for the assertion. The highest oxidation number of a representative element is the group number for the first two groups and the group number minus 10 (following the long form of periodic table) for the other groups.

(d) In H₂O₂ oxidation number of O = -1 and can vary from 0 to -2 (+2 is possible in OF₂). The oxidation number can decrease or increase, because of this H₂O₂ can act both asoxidising and reducing agent.

(c) In ClO₄^–, chlorine is in maximum oxidation state of +7. So, it does not show the disproportionation reaction.

(d) None of the above

(b) Oxidation involves gain of one or more electrons by a species during a reaction.

Electrochemical cell is a device in which the redox reaction is carried indirectly and the decrease in free energy appears as electrical energy.

At cathode there is gain of electrons.
At anode there is loss of electrons.
In electrochemical cell anode is written on L.H.S while cathode is written on R.H.S.

Electrochemical series is the series of elements in which elements are arranged in decreasing order of their reduction potential.
Reducing power goes on increasing whereas oxidising power goes on decreasing down the series.

(a) Combination reactions (b) Decomposition reactions (c) displacement reactions (d) Disproportionation reactions.

Two methods are used to balance chemical equations for redox processes. One of these methods is based on the change in the oxidation number of reducing agent and the oxidising agent (i.e. oxidation number method) and the other method is based on splitting the redox reaction into two half reactions — one involving oxidation and the other involving reduction (half reaction method). Both these methods are in use and the choice of their use rests with the individual using them.

The potential associated with each electrode is known as electrode potential. If the concentration of each species taking part in the electrode reaction is unity (if any gas appears in the electrode reaction, it is confined to 1 atmospheric pressure) and further there action is carried out at 298K, then the potential of each electrode is said to be the Standard Electrode Potential. By convention, the standard electrode potential (E^? ) of hydrogen electrode is 0.00 volts. The electrode potential value for each electrode process is a measure of the relative tendency of the active species in the process to remainin the oxidised/reduced form. A negative E means that the redox couple is a stronger reducing agent than the H⁺/H₂ couple. A positive E^? means that the redox couple is a weaker reducing agent than the H⁺/H₂ couple.

Kindly go through the solution

(a)

Kindly go through the solution

(d) +6

Kindly go through the solution

(c) Al

Kindly go through the solution

(c) Li

In a chemical reaction when electrons transfer simultaneously between substances, it is called as the redox reaction.  In redox reaction, one substance gets oxidized by losing electrons and another substance reduces by gaining the electrons. These two processes together occurs in a redox reaction.

There are four types of redox reactions. These are combination, Decomposition, Displacement, and Disproportionation.
In the combination redox reaction, two or more substances form a single product, in decomposition reactions, a compound breaks down into two or more substances. When one element in a compound replaces another, it is called displacement reactions. When simultaneously, one substance is oxidized and reduced, the reaction is known as the disproportionation redox reaction.

The redox reactions are significant for some of the basic functions of life such as respiration, photosynthesis, corrosion or rusting and combustion.

The applications of redox reactions include in cellular respiration, and batteries. In industries, these are used in the chemical production, metal extraction and water treatment. 

In NEET exam, there can be 1 to 2 questions from redox reactions, carrying 4-8 marks. The weightage is around 2% to 4%. In JEE Main, the weightage of this chapter is from 3.3% to 4-6%. You can expect nearly 1 to 2 questions from this chapter.