N₂S₄O₆;
+ 2 + 4x - 12 = 0
4x = 10
x = +2,5
Na₂SO₃
+ 2 + x - 6 = 0
x = +4

The middle'S' atoms have zero oxidation number and other two'S' atoms have +5 each
Average oxiIdan on number is \(\frac{5+5}{4}\)=2.5

A redox couple consists of oxidised and reduced form of the same substance taking part in the oxidation or reduction half reaction.

Oxidation half equation  Fe²⁺ (aq)\(\rightarrow\)Fe³⁺(aq) + e^-.......(i)
Reduction half equation : H2O2 (aq) + 2H⁺(aq) + 2e^- \(\rightarrow\) 2H₂O(l)......(ii)
Multiply Eq, (i) 2 and add it to Eq. (ii), we have
H₂O₂(aq) + 2Fe²⁺ (aq) + 2H⁺ (aq) \(\rightarrow\) 2F³⁺ (aq) + 2H₂O(l)

A) In the given reaction, the compound nitric oxide is formed by the combination of the elemental substances, nitrogen and oxygen. Therefore, this is an example of combination redox reaction
B) The given reaction involves the breaking down of lead nitrate into three components. Therefore, this is categorized under Decomposition redox reaction
C) Hydrogen of water has been displaced by hydride ion into dihydrogen gas. Therefore, this may be called as Displacement redox reaction.
D) This reaction involves disproportion of NO₂ (+4 state) into NO⁻₂ (+3 state) and NO⁻₃ (+5 state). Therefore, this reaction is an example of disproportion redox reaction.

At the anode, loss of electrodes takes place, ie. oxidation takes place, whereas cathode, a gain of electrodes takes place, ie. reduction takes place.Therefore, the cathode is called reduction electrode and anode is called oxidation electrode.

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

Oxidation number method

(Multiply Cr³⁺ by 2 because there are 2Cr atoms in \({ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }\)ion).
Balance increase and decrease in oxidation number
\({ 6Fe }^{ 2+ }+{ H }^{ + }+{ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }\longrightarrow { 2Cr }^{ 3+ }+{ 6Fe }^{ 3+ }+{ H }_{ 2 }O\)
Balance charge by multiplying H⁺ by 14.
\({ 6Fe }^{ 2+ }+{ 14H }^{ + }+{ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }\longrightarrow { 2Cr }^{ 3+ }+{ 6Fe }^{ 3+ }+{ H }_{ 2 }O\)
Balance H and O-atoms by multiplying H₂O by 7.
\({ 6Fe }^{ 2+ }+{ 14H }^{ + }+{ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }\longrightarrow { 2Cr }^{ 3+ }+{ 6Fe }^{ 3+ }+{ 7H }_{ 2 }O\)
This represents a balanced redox reaction.

Step I : The skeletal ionic equation is:
eg.  \(C{ r }_{ 2 }{ O }_{ 7 }^{ 2- }(aq)+{ SO }_{ 3 }^{ 2- }(aq)\longrightarrow { Cr }^{ 3+ }(aq)+{ SO }_{ 4 }^{ 2- }(aq)\)
Step II : Assign oxidation numbers for Cr and S
\(\overset { +6 }{ C{ r }_{ 2 } } \overset { -2 }{ { O }_{ 7 }^{ 2- } } (aq)+\overset { +4-2 }{ { SO }_{ 3 }^{ 2- } } (aq)\longrightarrow \overset { +3 }{ { Cr }^{ 3+ } } (aq)+\overset { +6 }{ S } \overset { -2 }{ { O }_{ 4 }^{ 2- }(aq) } \)
This indicates that the dichromate ion is the oxidant and the sulphite ion is the reductant.
Step 3 : Calculate the increase and decrease of oxidation number, and make them equal: from step-2 we can notice that there is change in oxidation state of chromium and sulphur. Oxidation state of chromium changes form +6 to +3. There is decrease of +3 in oxidation state of chromium on right hand side of the equation. Oxidation state of sulphur changes from +4 to +6. There is an increase of +2 in the oxidation state of sulphur on right hand side. To make the increase and decrease of oxidation state equal, place numeral 2 before cromium ion on right hand side and numeral 3 before sulphate ion on right hand side and balance the chromium and sulphur atoms on both the sides of the equation. Thus we get
+6  -2                +4    -2                    +3                    +6    -2
\(Cr_{ 2 }O_{ 7 }^{ 2- }(aq)+{ 3SO }_{ 3 }^{ 2- }(aq)\longrightarrow 2Cr^{ 3+ }(aq)+3{ SO }_{ 4 }^{ 2- }(aq)\)
Step 4: As the reaction occurs in the acidic medium, and further the ionic charges are not equal on both the sides, add 8H⁺ on the left to make ionic charges equal
\(Cr_{ 2 }O_{ 7 }^{ 2- }(aq)+{ 3SO }_{ 3 }^{ 2- }(aq)\longrightarrow 2Cr^{ 3+ }(aq)+3{ SO }_{ 4 }^{ 2- }(aq)\)
Step 5: Finally, count the hydrogen atoms, and add appropriate number of water molecules (i.e., 4H₂O) on the right to achieve balanced redox change.
\(Cr_{ 2 }O_{ 7 }^{ 2- }(aq)+{ 3SO }_{ 3 }^{ 2- }(aq)+{ 8H }^{ + }(aq)\longrightarrow 2C{ r }^{ 3+ }(aq)+{ 3SO }_{ 3 }^{ 2- }(aq)+4{ H }_{ 2 }O\)

\(\begin{array}{r} \mathrm{H}^{+}+\mathrm{MnO}_{4}^{-}+\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{Mn}^{2+}+\mathrm{H}_{2} \mathrm{O} {[\text { Acidic medium }]} \\ \mathrm{MnO}_{4}^{-} \longrightarrow \mathrm{Mn}^{2+} \\ 5 \mathrm{e}^{-}+8 \mathrm{H}^{+}+\mathrm{MnO}_{4}^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O} \quad \ldots(i) \\ {\left[\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{e}\right] \times 5 \quad \ldots(i i)} \\ \hline 8 \mathrm{H}^{+}+5 \mathrm{Fe}^{2+}+\mathrm{MnO}_{4}^{-} \longrightarrow 5 \mathrm{Fe}^{3+}+\mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O} \end{array}\)

Here, I^- is oxidised to I₂ and O₂(g) is reduced to OH^-.
\(4 \mathrm{I}+\mathrm{O}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 4 \mathrm{OH}^{-}+2 \mathrm{I}_{2}\)

a) (i) It completes the internal circuit.
 (ii) It maintains the electrical neutrality
b) Fe²⁺ + Cr₂\({ O }_{ 7 }^{ 2- }\) + H⁺ \(\rightarrow\) Fe³⁺ + Cr³⁺ + H₂O

The given redox reaction is Zn(S) + 2Ag⁺ (aq) \(\rightarrow\) 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) Since oxidation occurs at the zinc electrode, therefore, electrons accumulate on the zinc electrode and hence, zinc electrode is negatively charged.
(ii) The ions carry current. The electrons flow from Zn to Ag electrode while the current flows from Ag to Zn electrode.
(iii) The reactions occurring at the two electrodes are
Zn(s) \(\rightarrow\) Zn²⁺ (aq) + 2e^-
Ag⁺ (aq) \(\rightarrow\) + e^- Ag(s)