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

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) Writing the On on each atom above its symbol, then
\(3\overset { +1 }{ H } \overset { -1 }{ Cl } (aq)+\overset { +1 }{ H } \overset { +5 }{ N } \overset { -2 }{ { O }_{ 3 } } (aq)\longrightarrow \overset { 0 }{ { Cl }_{ 2 }(g)+ } \overset { +3 }{ N } \overset { -2 }{ O } \overset { -1 }{ Cl(g)+ } \overset { +1 }{ { 2H }_{ 2 } } \overset { -2 }{ O } (l)\)
Here, the On of Cl increases from -1 in HCl to O in Cl₂ , therefore, Cl^- is oxidised and hence, HCl acts as the reducing agent. The ON of N decreases from +5 in HNO₃ to +3 in NOCL, therefore, HNO₃ acts as the oxidising agent. Thus this reaction is a redox reaction.
(b) Writing the ON of each atom above its symbol, we have,
\(\overset { +2 }{ Hg } \overset { -1 }{ { Cl }_{ 2 } } (aq)+\overset { +1 }{ 2K } \overset { -1 }{ I } (aq)\longrightarrow \overset { +2 }{ Hg } { \overset { -1 }{ I } }_{ 2 }(s)+2\overset { +1 }{ K } \overset { -1 }{ { Cl }^{ - }(aq) } \)
Here, the On of none of the atoms undergo a change, therefore, this reaction is not a redox reaction.
(c) \(\overset { +3 }{ { Fe }_{ 2 } } { \overset { -2 }{ O } }_{ 3 }(s)+3\overset { +2 }{ C } \overset { -2 }{ O } (g)\ \overset { \Delta }{ \longrightarrow } \ 2\overset { 0 }{ Fe } (s)+3\overset { +4 }{ C } \overset { -2 }{ { O }_{ 2 } } (g)\)
Here, On of decreases from +3 in Fe₂O₃ to 0 in Fe, therefore, Fe₂O₃ acts as an oxidising agent. Further, On of C increases from +2 in CO to +4 in CO₂, therefore, CO acts as a reducing agent. Thus, this reaction is an example of redox reaction.
(d) Writing the ON of each atom above its symbol, then
\(\overset { +3 }{ P } { \overset { -1 }{ Cl } }_{ 3 }(l)+3\overset { +1 }{ { H }_{ 2 } } \overset { -2 }{ O } (l)\quad \longrightarrow \quad 3\overset { +1 }{ H } \overset { -1 }{ Cl } (aq)+\overset { +1 }{ { H }_{ 3 } } \overset { +3 }{ P } \overset { -2 }{ { O }_{ 3 } } (aq)\)
Here, On of none of the atoms undergo a change, therefore, this reaction is not a redox reaction.
(e) Writing the ON of each atom above its symbol, then
\(4\overset { -3 }{ N } \overset { +1 }{ { H }_{ 3 } } (aq)+3\overset { 0 }{ { O }_{ 2 } } (g)\quad \longrightarrow \quad 2\overset { 0 }{ { N }_{ 2 } } (g)+6\overset { +1 }{ { H }_{ 2 } } \overset { -2 }{ O } (l)\)

Follow the following steps to solve out such problems.
Step I Obtain the value of \({ E }_{ red }^{ 0 }\\ \) for both the redox couples (if not given) Since the given E⁰ values show the oxidation electrode potential, so invert this value to get \({ E }_{ red }^{ 0 }\\ \) (reduction electrode potential)i.e 
\({ E }_{ Zn,{ /Zn }^{ 2+ } }^{ 0 }=-0.76V\)
\({ E }_{ Cd,/{ Cd }^{ 2+ } }^{ 0 }=-0.40V\)
Step II Find cathode and anode for the cell,
Here Zn/Zn²⁺ couple is anode and Cd²⁺/Cd couple is cathode [If cell is not given, the couple having more negative E_cell or E⁰ value, acts as anode and the other one acts as cathode]
step III Calculate the cell potential or emf using the expression
\({ E }_{ red }^{ 0 }={ E }_{ cathode }^{ 0 }-{ E }_{ anode }^{ 0 }\)
\({ E }_{ cell }^{ 0 }=-0.40-(-0.76)\)
\( =-0.40+0.76=+0.36V\)
 step IV If \({ E }_{ cell }^{ 0 }\)is positive the reaction is feasible and the cell exists and if is\({ E }_{ cell }^{ 0 }\) negative or zero, the cell does not exist.
Since We get positive,\({ E }_{ cell }^{ 0 }\) so the cell will exist.

Like NaH, sodium oxide is also an ionic compound and have a formula (Na⁺)₂O^2-. Thus, oxidation and reduction half reactions can be written as
I. 2Na(s)\(\rightarrow \)2Na⁺(g)+2e-  ............(i)
Loss of 2e^-   (oxidation half reaction)
II. O₂(g)+4e^-\(\rightarrow \)20^2-(g)    ................(ii)
Gain of 4e^-  (Reduction half reaction)
To equalise number of electrons, equation (i) is multiplied with 2, and hence, we get 
III. 4Na(s) \(\rightarrow \)4Na⁺(g)+4e^-
On adding equation II and III, we get
4Na(s)+O₂(g)\(\rightarrow \)4Na⁺+2O^2-
OR
4Na(s)+O₂(g)\(\rightarrow \)2(Na⁺)₂2O^2-(s)
Since, the above reaction involve oxidation and reduction as its two half reactions, so it is a redox reaction
Note
NaCl and Na₂S being iconic compounds can also be represented as Na⁺Cl^- and (Na⁺)₂S^2-

(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

Lower the electrode potential, better is the reducing agent. Since the electrode potentials increase in the oder; K⁺/K = -2.93V,  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.

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