Oxidation number of S in \({ H }_{ 2 }{ SO }_{ 5 }\)is  2(+1) + x + 5(-2) = 0 or x = +8
But oxidation number of S cannot be more than 6 because S has only 6 valence electrons. This fallacy is removed by calculating oxidation number of S by chemical bonding method.
Let us consider the structure of \({ H }_{ 2 }{ SO }_{ 5 }\)

In \({ H }_{ 2 }{ SO }_{ 5 }\), two oxygen atoms are in -1 oxidation state.
Let the oxidation number of S be x.
\(\underset { For \ H }{ 2(+1) } +\underset { For \ three \ O }{ 3(-2) } +x+\underset { For \ O-O }{ 2(-1) } =0\)
\( 2+(-6)+x+(-2)=0\)
\( x=+6\)
Therefore, the oxidation number of S in \({ H }_{ 2 }SO_{ 5 }\)is +6.
(ii) Oxidation number of Cr in \({ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }\)is 
2x + (-2)7 = -2
2x = +12
x = +6
Let us consider the structure of \({ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }\) ion

Let the oxidation number of each Cr-atom be x.
4(-2) + (-2) + 1(-2) + 2x = 0
-8-2-2 + 2x = 0
2x = +12, x = +6
Oxidation number of Cr in \({ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }\) is same i.e+6 whether it is calculated by conventional method or by chemical bonding method. Hence, there is no fallacy.
(iii) Oxidation number of N in \(NO_{ 3 }^{ - }\)be x
[x + 3(-2) = -1 or x = +5]
Let us consider the structure of \(NO_{ 3 }^{ - }\)ion

Let the oxidation number of N be x
\(\underset { For \ \overset { - }{ O } }{ 1(-1) } +x+\underset { For \ (=O) }{ 1(-2) } +\underset { For(\rightarrow O) }{ 1(-2) } =0\)
\(x=+5\)
Oxidation number of N in \({ NO }_{ 2 }^{ - }\)ion is same, i.e+5 whether it is calculated by conventional method or by chemical bonding method hence, there is no fallacy.

Step 1 : The skeletal ionic equation is
\(Mn{ O }_{ 4 }^{ - }(aq)+{ Br }^{ - }(aq)\longrightarrow Mn{ O }_{ 2 }(s)+Br{ O }_{ 3 }^{ - }(aq)\)
Step 2 :  Assign oxidation numbers for Mn and Br
\(\overset { +7 }{ Mn{ O }_{ 4 }^{ - } } (aq)+\overset { -1 }{ { Br }^{ - } } (aq)\longrightarrow \overset { +4 }{ Mn{ O }_{ 2 } } (s)+\overset { +5 }{ Br{ O }_{ 3 }^{ - } } (aq)\)
this indicates that permanganate ion is the oxidant and bromide ion is the reductant.
Step 3: Calculate the increase and decrease of oxidation number, and make the increase equal to the decrease.
\(2\overset { +7 }{ Mn{ O }_{ 4 }^{ - } } (aq)+\overset { -1 }{ { Br }^{ - } } (aq)\longrightarrow \overset { +4 }{ Mn{ O }_{ 2 } } (s)+\overset { +5 }{ Br{ O }_{ 3 }^{ - } } (aq)\)
Step 4: As the reaction occurs in the basic medium, and the ionic charges are not equal on both sides, add 2 OH– ions on the right to make ionic charges equal.
\(Mn{ O }_{ 4 }^{ - }(aq)+{ Br }^{ - }(aq)\longrightarrow Mn{ O }_{ 2 }(s)+Br{ O }_{ 3 }^{ - }(aq)+2OH^{ - }(aq)\)
Step 5: Finally, count the hydrogen atoms and add appropriate number of water molecules (i.e. one H2O molecule) on the left side to achieve balanced redox change.
\(Mn{ O }_{ 4 }^{ - }(aq)+{ Br }^{ - }(aq)+{ H }_{ 2 }O(l)\longrightarrow Mn{ O }_{ 2 }(s)+Br{ O }_{ 3 }^{ - }(aq)+2OH^{ - }(aq)\)

(a) Oxidised substance → C₆H₆O₂  Reduced substance → AgBrO xidising agent → AgBr  Reducing agent → C₆H₆O₂
(b) Oxidised substance → HCHO  Reduced [Ag(NH₃)₂]+ substance →  Oxidising agent → [Ag(NH₃)₂]+ Reducing agent → HCHO 
(c) Oxidised substance → HCHO  Reduced substance → Cu²⁺ Oxidising agent → Cu²⁺  Reducing agent → HCHO 
(d) Oxidised substance → N₂H₄  Reduced substance → H₂O₂  Oxidising agent → H₂O₂  Reducing agent → N₂H₄ 
(e) Oxidised substance → Pb  Reduced substance → PbO₂  Oxidising agent → PbO₂  Reducing agent → Pb 

(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.

\(\overset { +8 }{ Xe } { O }_{ 6 }^{ 4- }(aq)+\overset { -1 }{ { 2F }^{ - } } (aq)+6{ H }^{ + }(aq)\longrightarrow \overset { +6 }{ Xe } { O }_{ 3 }(g)+\overset { 0 }{ { F }_{ 2 } } +3{ H }_{ 2 }O(l)\)
In the above reaction, oxidation number of Xe in  \(​​XeO_{ 6 }^{ 4- }\) and oxidation number of F increases from -1(in F^-) to zero(in F₂ ).
Hence \(​​XeO_{ 6 }^{ 4- }\) or Na₄XeO₆ is reduced and F^-  is oxidised.
This reaction occur because Na₄XeO₆ or \(​​XeO_{ 6 }^{ 4- }\) is a stronger oxidising agent than fluorine.

Suppose that the oxidation number of P in \({ HPO }_{ 3 }^{ 2- }\) be x.
Theb, 1 + x + 3 (-2) = -2
or x + 1 - 6 = -2 or x = +3

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.

Since, Mg²⁺(aq)/Mg electrode = - 2.36V is at lower potential than Al³⁺(aq)/AI electrode = - 1.66 V, therefore, Mg²⁺ (aq)/Mg electrode acts the anode and AI³⁺(aq)/Mg electrode acts as the anode and AI³⁺(aq) /AI acts as the cathode.In other words, Mg loses electrons and AI³⁺ ion accepts electrons.Thus, the cell reaction is
 3Mg +2 AI³⁺ \(\rightarrow\) 3Mg²⁺ + 2AI and  E^o_cell = E^o_{AL³⁺ I AI} - E_{Mg²⁺ I Mg} = -1.66-(-2.36)
= +0.70 V

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⁺_.

 \(CI_{ 2 }O_{ 7 }(g)+4H_{ 2 }O_{ 2 }(aq)+2OH^{ - }(aq)\longrightarrow 2CIO_{ 2 }^{ - }(aq)+5H_{ 2 }O(I)+4O_{ 2 }(g)\)

KCI cannot be used as electrolyte in the salt bridge because CI^- ions will combine with Ag⁺ ions to form white precipitates of AgCI

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

In a redox reaction, the total number of electrons lost by reducing agent must be equal to the number of electrons gained by the oxidising agent.

Pb₃O₄ is actually a stoichiometric mixture of 2 moles of PbO and 1 mole of PbO₂ i.e(2 PbO.PbO₂). In PbO₂, lead is present in +4 oxidation state, whereas the stable oxidation state of lead in PbO is +2. PbO₂ thus can act as an oxidant (oxidising agent) and therefore, can oxidise Cl^- ion of HCl into chlorine. We may also keep in mind that PbO is a basic oxide. Therefore, the reaction.
Pb₃O₄ + 8HCl\(\rightarrow\)3PbCl₂ + Cl₂ + 4H₂O
can be splitted into two reactions namely:
2PbO + 4HCl\(\rightarrow\)2PbCl₂ + 2H₂O
(acid-base reaction)
\(\overset { +4 }{ Pb{ O }_{ 2 } } +4\overset { -1 }{ HCl } \rightarrow \overset { +4 }{ Pb } \ { Cl }_{ 2 }+\overset { 0 }{ { Cl }_{ 2 } } +{ 2 }H_{ 2 }O\)
(redox reaction)
Since, HNO₃ itself is an oxidising agent, therefore it is unlikely that the reaction may occur between PbO₂ and HNO₃ . However, the acid-base reaction occurs between PbO and HNO₃ as
2PbO + 4HNO₃\(\rightarrow\)2Pb(NO₃)₂ + 2H₂O
It is the passive nature of PbO₂ against HNO₃ that makes the reaction different from the follows with HCl.