Halogens have a strong tendency to accept electrons. Therefore, they are strong oxidising agents. Their relative oxidising power is, however, measured in terms of their electrode potentials. Since the electrode potentials of halogens decrease in the order: F2(+2.87V) > Cl2 (+1.36V) > Br2(+1.09V) > 12(+0.54V), therefore, their oxidising power decreases in the same order.
This is evident from the observation that F2 oxidises Cl^- to Cl₂' Br^- to Br₂, I^- - to I₂; Cl₂ oxidises Br^- to Br₂ and r to I₂ but not F- to F₂. Br₂, however, oxidises 1- to I₂but not F^- to F2^' and Cl^- to Cl₂
F2(g) + 2Cl^-(aq) \(\longrightarrow \) 2F^-(aq) + CI₂(g); F2(g) + 2Br^-(aq) \(\longrightarrow \) 2F^-(aq) + Br₂(l)
F2(g) + 2I^-(aq) \(\longrightarrow \) 2F^-(aq) + I₂(s);Cl₂(g) + 2Br^-(aq) \(\longrightarrow \) 2Cl^-(aq) + Br₂(l)
Cl₂(g) + 2l^-(aq) \(\longrightarrow \) 2Cl^-(aq) +I₂(s) and Br₂(l) + 2I^- \(\longrightarrow \) 2Br^-(aq) + I₂(s)
Thus, F2 is the best oxidant.
Conversely, halide ions have a tendency to lose electrons and hence can act as reducing agents. Since the electrode potentials of halide ions decreases in the order: F (-0.54 V) > Br" (-1.09 V) > Cl" (-1.36 V) > F (-2.87 V), therefore, the reducing power of the halide ions or their corresponding hydrohalic acids decreases in the same order: HI > HBr > HCl > HF. Thus, hydroiodic acid is the best reductant. This is supported by the following reactions. For example, HI and HBr reduce H₂SO₄ to SO₂ while HCI and HF do not.
2HBr + H₂SO₄ \(\longrightarrow \) Br₂ + SO₂ + 2H₂O; 2HI + H₂SO₄ \(\longrightarrow \) I₂+ SO₂ + 2H₂O
Further T reduces Cu²⁺ to Cu⁺ but Br" does not.
2Cu²⁺(aq) + 4I^-(aq) \(\longrightarrow \) Cu₂I₂(s)+ 12(aq);Cu2+(aq) + 2Br^- \(\longrightarrow \) No reaction. Thus, HI is a stronger reductant than HBr
Further among HCl and HF, HCl is a stronger reducing agent than HF because HCI reduces MnO₂ to Mn²⁺ but HF does not.
MnO₂(s) + 4HCl(aq) \(\longrightarrow \) MnCl₂(aq) + Cl₂(g) + 2H₂O
MnO₂(s) + 4HF(I) \(\longrightarrow \) No reaction
Thus, the reducing character of hydrohalic acids decreases in the order: HI > HBr > HCl > HF

The balanced equation for the reaction is
\(\underset { \overset { 4\quad \times \quad 17 }{ =68g } }{ 4{ NH }_{ 3 }(g) } +\underset { \overset { 5\quad \times \quad 32 }{ =160g } }{ { 5O }_{ 2 }(g) } \longrightarrow \underset { \overset { 4\quad \times \quad 30 }{ =120g } }{ { 4NO(g) } } +{ 6H }_{ 2 }O(g)\)
Here, 68 g of NH₃ will react will 02 = 160 g
\(\therefore\) 10 g of NH3 will react with 0₂  = \(\frac { 160g }{ 68g } \) x 10g = 23.6g
But the amount of 02 which is actually available is 20.0 g which is less than the amount which is needed. Therefore, 02 is the limiting reagent and hence calculations must be based upon the amount of 02 taken and not on the amount of NH3 taken. From the equation
160 g of O₂ produce NO = 120 g
\(\therefore\) 20g of O₂ will produce NO = \(\frac { 120 }{ 160 } \times 20\) = 15g

(a) KI₃ \(\overset { +1 }{ K } \overset { x }{ { I }_{ 3 } } ;\) 1(+1) + 3x = 0 or x = -\(\frac { 1 }{ 3 } \)
Therefore, the average oxidation number of I is -\(\frac { 1 }{ 3 } \)
It is wrong because oxidation number can never be fractional. Let us consider the structure oh KI₃
\(\overset { +1 }{ K }\) (I-I\(\leftarrow \))^-1, in this structure, a coordinate bond is formed between I₂ molecule and I^- ion. Hence, the  oxidation number of three I-atoms in KI₃ are 0,0 (in I₂) and -1 respectively.
(b) By conventional method. O.N. of 5 in \(\overset { +1 }{ { H }_{ 2 } } \overset { x }{ { S }_{ 4 } } \overset { -2 }{ { O }_{ 6 } } \)
or 2 (+1) + 4x + 6 (-2) = 0 or x = +2.5 (wrong)
But it is wrong because all the four 5 atoms cannot be in the same oxidation state.
By chemical bonding method. The structure of H₂S₄0₆ is shown below;
\(H-O\overset { +5 }{ - } \overset { \overset { O }{ || } }{ \underset { \underset { O }{ || } }{ C } } -\overset { 0 }{ S } -\overset { 0 }{ S } -\overset { \overset { O }{ || } }{ \underset { \underset { O }{ || } }{ S } } \overset { +5 }{ - } OH\)
The O.N. of each of the 5-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 \(\overset { x }{ { Fe }_{ 3 } } \overset { -2 }{ { O }_{ 4 } } \)  in  or 3x + 4 (- 2) = 0 x = 8/3
By stoichiometry. Fe₃0₄ = \(\overset { +2 }{ { Fe }_{ 3 } } \overset { -2 }{ { O } } \overset { +3 }{ { Fe }_{ 2 } } \overset { -2 }{ { O }_{ 3 } } \)
\(\therefore\) O.N. of Fe in Fe₃0₄ is + 2 and + 3
(d) By conventional method. O.N. of C in CH₃CH₂OH = \(\overset { x }{ { C }_{ 2 } } \overset { +1 }{ { H }_{ 6 } } \overset { -2 }{ { O } } \)
or 2x + 6 (+ 1) + 1 (- 2) = 0 or x = -2.
(e) By conventional method. CH₃COOH = 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).
\(H\overset { 2 }{ - } \overset { \overset { H }{ I } }{ \underset { \overset { I }{ H } }{ C } } -\overset { \overset { 0 }{ II } }{ C } -OH\)
therefore, O.N. of C₂ = 3 (+1) + x + 1 (-1) = 0 or x = -2
C1 is, however, attached to one oxygen atom by a double bond, one .O.H (O.N. = -1) and one CH₃ (O.N. = +1) group, therefore, O.N. of C₁ = + 1 + x + 1
(-2) + 1 (-1) = 0 or x = +2

\(4\overset { -3-1 }{ N{ H }_{ 3 } } (g)+5\overset { 0 }{ { O }_{ 2 } } (g)\longrightarrow 4\overset { +2-1 }{ NO } (g)+6\overset { +1-2 }{ H_{ 2 }O } (g)\)
Oxidation number of N increases from -2(in NH₃) to +2(in NO) and oxidation number of O decreases from zero (in 0₂) to-2
(in NO and H₂O). This shows that NH₃ is oxidised and O₂ is reduced. Hence, it is a redox reaction.

(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.

\(\overset { 0 }{ 2k(s) } +\overset { 1 }{ { P }_{ 2 } } \longrightarrow 2\overset { +1-1 }{ K{ F } } (s)\)
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

\(\overset { +3-1 }{ 4BCl_{ 3 } } (g)+\overset { +1+3-1 }{ 3LiAl{ H }_{ 4 } } (s)\longrightarrow \overset { -3+1 }{ 2{ B }_{ 2 }{ H }_{ 6 } } (g)+\overset { +1-1 }{ 3LiCl } (s)+3\overset { +3-1 }{ AlC{ l }_{ 3 } } (s)\) 
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.

When conc. H₂SO₄is added to an inorganic mixture containing chloride, a pungent smelling gas HCl is produced because a stronger acid displaces a weaker acid from its salt
2NaCl + 2H₂SO₄ \(\rightarrow\) 2NaHS0₄ + 2HCI
Stronger acid Weaker acid
HCl + H₂SO₄ \(\rightarrow\)  Cl₂ + SO₂ + 2H₂O
Since HCI is a very weak reducing agent, it can not reduce H₂SO₄ to SO2 and hence HCI is not oxidised to Cl₂"
However, when the mixture contains bromide ion, the initially produced HBr
being a strong reducing agent than HCl reduces H₂SO₄to SO₂ and is itself oxidised to produce red vapour of Br₂
2NaBr + 2H₂SO₄ \(\rightarrow\) 2NaHSO₄ + 2HBr
2HBr + H₂SO₄ \(\rightarrow\) Br₂+ SO₂ + 2H₂O

(a) Oxidants H⁺
Reductants Zn
(b) Oxidants H₂O₂
Reductants  2[Fe(CN)₆]^4-
(c) Oxidants 2[Fe(CN)₆]^3-
Reductants H₂O₂
(d) Oxidants F₂
Reductants \(Br{ O }_{ 3 }^{ - }\)
(e) Oxidants \(NaCl{ O }_{ 3 }\)
Reductants 2I₂