chlorine is an oxidant and K₂MnO₄ is reductant.

(a) No, because used of a bleach can also destroy a cloth.
(b) Bleaching powder, CaOCl₂ is actually Ca(OCl) Cl, in which the oxidation state of one chlorine (Cl) is -1 and that of other in parenthesis (Cl is +10.
⇒ -2 + x = -1 ⇒ (x =+ 1)
(c) You can use vanish to remove the spot.
(d) Values associated with Mrs. Sharma are helpful but unknowledgeable and not aware about side effects of bleach.

Suppose the oxidation number of N in Y is x
(N^-2)₂\(\rightarrow\)(2N)^x + 10e^-
(as N₂H₄\(\rightarrow\)Y + 10e^-)
Therefore, 2x - 10 = -4, which gives x = +3. Hence, oxidation number of H in Y = 3

Since in the above reaction the compound formed is an ionic compound, which may also be represented as \(\mathrm{Na}^{+} \mathrm{H}^{-}\) ^(s), this
suggests that one half reaction in this process is :
\(2 \mathrm{Na}(\mathrm{s}) \rightarrow 2 \mathrm{Na}^{+}(\mathrm{g})+2 \mathrm{e}^{-}\) 
and the other half reaction is: 
\(\mathrm{H}_{2}(\mathrm{~g})+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}^{-}(\mathrm{g})\)
This splitting of the reaction under examination into two half reactions automatically reveals that here sodium is oxidised and hydrogen is reduced, therefore, the complete reaction is a redox change.

In the manufacture of benzoic acid from toluene, alcoholic potassium permanganate is used as an oxidant because of the following reasons.
(i) In a neutral medium, OH^– ions are produced in the reaction itself. As a result, the cost of adding an acid or a base can be reduced.
(ii) KMnO₄ and alcohol are homogeneous to each other since both are polar. Toluene and alcohol are also homogeneous to each other because both are organic compounds.
Reactions can proceed at a faster rate in a homogeneous medium than in a heterogeneous medium.
Hence, in alcohol, KMnO4 and toluene can react at a faster rate.
The balanced redox equation for the reaction in a neutral medium is give as below:

 

When conc. H2SO4 is added to an inorganic mixture containing bromide, initially HBr is produced.
HBr, being a strong reducing agent reduces H₂SO₄ to SO₂ with the evolution of red vapour of bromine.
2NaBr+2H₂SO₄⟶2NaHSO₄+2HBr₂
HBr+H₂SO₄⟶Br₂+SO₂+2H₂O (red vapour)  
But, when conc. H₂SO₄ is added to an inorganic mixture containing chloride, a pungent smelling gas (HCl) is evolved. HCl, being a weak reducing agent, cannot reduce H₂SO₄ to SO₂
2NaCl+2H₂SO₄⟶2NaHSO₄+2HCI

(a) \(CuO(s)+{ H }_{ 2 }(g)\rightarrow Cu(s)+{ H }_{ 2 }O(g)\)
Assign oxidation numbers of each atom above its symbol.
\(\overset { +2-2 }{ CuO } (s)+\overset { 0 }{ H_{ 2 } } (g)\rightarrow \overset { 0 }{ Cu } (s)+\overset { +1 }{ H_{ 2 } } \ \overset { -2 }{ 0 } \ (g)\)
Oxidation number of Cu in Cuo is +2.It decreases from +2 to zero in Cu.While oxidation number of hydrogen increases from 0(in H₂) to +1 (in H₂O)
This shows the CuO is reduced to Cu but H₂ is oxidised to H₂O.Hence,it is an example of redox reaction.
(b) \(\overset { +3 }{ { Fe }_{ 2 } } \ { 0 }_{ 3 }(s)+3\overset { +2-2 }{ CO } (g)\longrightarrow \overset { 0 }{ 2F } \ e(s)+\overset { +4-2 }{ CO_{ 2 } } (g)\)
Oxidation number of Fe decreases from +3(in Fe₂O₃) to zero (in Fe) and oxidation number of C increases from +2 (in CO) to +4
 (in CO₂ ). This shows that Fe₂O₃ is reduced to F_e and CO is oxidised to CO₂. Hence, it is a redox reaction.
(c) \(\overset { +3-1 }{ 4BCl_{ 3 } } (g)+\overset { +1+3-1 }{ 3LiAl{ H }_{ 4 } } (s)\longrightarrow \overset { -3+1 }{ 2{ B }_{ 2 }{ H }_{ 6 } } (g)+\overset { +1-1 }{ 3LiCl } (s)+3\overset { +3-1 }{ AlC{ l }_{ 3 } } (s)\) 
Oxidation number of B decreases from +3(in BCl₃) to -3(in B₂H₆) and oxidation number of H increases from -1(in LiAlH₄) to +1 (in B₂H₆ ). This shows that BCl₃ is reduced to B₂H₄ and LiAlH₄ is oxidised. Hence, it is redox reaction.
(d) \(\overset { 0 }{ 2k(s) } +\overset { 1 }{ { P }_{ 2 } } \longrightarrow 2\overset { +1-1 }{ K{ F } } (s)\)
Oxidation number of K increases from zero(in K) to +1 (in KF) and oxidation number of F reduces from zero(in F₂) to -1(in KF).This shows that K is oxidised and F₂ is reduced.Hence it is a redox reaction
(e) \(4\overset { -3-1 }{ N{ H }_{ 3 } } (g)+5\overset { 0 }{ { O }_{ 2 } } (g)\longrightarrow 4\overset { +2-1 }{ NO } (g)+6\overset { +1-2 }{ H_{ 2 }O } (g)\)
Oxidation number of N increases from -2(in NH₃) to +2(in NO) and oxidation number of O decreases from zero (in O₂) to-2
(in NO and H₂O). This shows that NH₃ is oxidised and O₂ is reduced. Hence, it is a redox reaction.

The possible reaction between Ag⁺(aq) and Cu(s) is
Cu(s) + 2Ag⁺(aq) \(\rightarrow\) Cu²⁺(aq) + 2Ag(s)
The above redox reaction can be split into the following two half reactions. Writing electrode potential for each half reaction from Table 8.1, we have
Oxidation Cu(s) \(\rightarrow\) Cu²⁺(aq) + 2e^-; E^o= -0.34 V
Reduction: Ag⁺(aq) + e^- \(\rightarrow\) Ag(s)] x 2; E^o= + 0.80 V

Standard hydrogen electrode is known as reference electrode. Its electrode potential is taken as 0.000 volt.

(a) Cs exhibits only positive oxidation state because it is the most electropositive element.
(b) I exhibits both positive and negative oxidation states.Iodine exhibits -1,0,+1,3,+5 and +7 oxidation states(+3,+5 and +7 oxidation states are exhibited by I due to the presence of vacant d-orbitals.)
(c) Ne is an inert gas, so it neither exhibits negative nor positive oxidation states.

No, the oxidation number of an element in any species is an apparent charge on the atom which it appears to have acquired when all other atoms in the species are removed as ions.

In \({ MnO }_{ 4 }^{ 2- }\)the oxidation number of Mn is +6. It can increase its oxidation number(to+7) or decrease its oxidation number (to+, +4,+3,+2,0)
Hence, it undergoes disproportionation reaction in acidic medium

In  \({ MnO }_{ 4 }^{ - }\) ,Mn is in its highest oxidation state,i.e +7 .It can only decrease its oxidation number.Hence,it cannot undergo disproportional reaction.

E^o_{AI³⁺ / AI} = -.166V, E^o_{Cu²⁺ / Cu} =+ 0.34 V
E_{Fe²⁺ / Fe} = - 0.44 V, E_{Mg²⁺ / Mg} = - 2.36 V
and  E^o_{Zn²⁺ / Zn} = - 0.76 V
A metal with more negative value of E^o_red is a stronger reducing agent than those which have less negative or positive value of E^o_red .Therefore, Mg can displace all the given metals from their aqueous salt solutions.AI can displace all metals from their aqueous salt solutions.Zinc can displace Fe and Cu from their aqueous salt solutions and Fe can displace only Cu from its aqueous salt solution. Hence, the orderin which they can display each other from the solution of their salts is as follows.
 Mg, AI, Zn, Fe, Cu

Ag⁺ and Cu²⁺ act as oxidising agents in reactions (a) and (b) respectively.
In reaction (c), Ag⁺ oxidises C₆H₅CHO to C₆H₅COO^–, but in reaction (d), Cu²⁺ cannot oxidise C₆H₅CHO.
Hence, we can say that Ag⁺ is a stronger oxidising agent than Cu²⁺_.

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 netrual.

4MnO₄^-+8H⁺ +ne^- → Mn²⁺ +4H₂O
-1 + 8 + n  = +2
-1 - 2+ 8 + n = 0
n = - 5  or 5e^-