Redox Reactions : Oxidation state, Equivalent weight, n-factor, Gram equivalent, Normality, Balancing of Redox reactions

Oxidation Number

Hypothetical or real charge on an atom in its combined or free state.

Oxidation number of :

1. group-I elements in periodic table is +1
   and group-II elements is +2

2. H = -1 with metals and +1 with non-metals

3. Oxygen in general = -2
   but can have +2, +1, 0, -1, -2 as well

4. F = -1 (always)
   Cl, Br, I in general = -1
   but can have +1, +3, +5, +7

5. Elements in their standard state = 0
   Example : Ca_(s), Na_(s), I_2(s), Br_2(l), F_2(g)

6. Ion = charge on ion
   Fe⁺³ = +3
   O^-2 = -2

7. Alloys O.N = 0
   Sodium amalgam (NaHg)

8. Oxidation number can not be fractional
   but average oxidation number can be fractional
   Fe₃O₄ = $\stackrel{+2}{\mathrm{Fe}}\mathrm{O}+\stackrel{+3}{{\mathrm{Fe}}_2}{\mathrm{O}}_3$
   average O.N of Fe₃O₄ : $\left[\begin{array}{rcl}3x-8& =& 0\\ x& =& \frac{8}{3}\end{array}\right]$

n-factor

n-factor - Total number of moles of e- gain or loss by oxidation or reduction of 1 mole of a compound.

1. Atoms = Valancy

2. Ions = |Charge|

3. Salts = |Charge on cation/anion|

4. Acid = Basicity or number of H+ ion donated

5. Base = Acidity or number of OH- ion donated

6. Reduction rxn. = dec. in O.N or number of e- gained

7. Oxidation rxn. = inc. in O.N or number of e- loss

8. Redox rxn. : 1 atom of a molecule oxidise or reduce
   $\stackrel{+\mathrm{p}}{{\mathrm{A}}_{\mathrm{x}}}{\mathrm{B}}_{\mathrm{y}}\to \stackrel{+\mathrm{q}}{{\mathrm{A}}_{{\mathrm{x}}_1}}{\mathrm{B}}_{\mathrm{y}}$
   $\text{n-factor of}{\mathrm{A}}_{\mathrm{x}}{\mathrm{B}}_{\mathrm{y}}=\mathrm{x\mid p}-\mathrm{q\mid }$
   Example - $\stackrel{+3}{{\mathrm{C}}_2}{\mathrm{O}}_4\to \stackrel{+4}{\mathrm{C}}{\mathrm{O}}_2$
   $\text{n-factor of}{\mathrm{C}}_2{\mathrm{O}}_4=2\mathrm{\mid }3-4\mathrm{\mid }=2$

9. more than 1 atom of a molecule oxidise or reduce
   $\stackrel{+\mathrm{p}}{{\mathrm{A}}_{\mathrm{x}}}\stackrel{+\mathrm{r}}{{\mathrm{B}}_{\mathrm{y}}}\to \stackrel{+\mathrm{q}}{{\mathrm{A}}_{{\mathrm{x}}_1}}\stackrel{+\mathrm{s}}{{\mathrm{B}}_{{\mathrm{y}}_1}}$
   $\text{n-factor of}{\mathrm{A}}_{\mathrm{x}}{\mathrm{B}}_{\mathrm{y}}=\mathrm{x\mid p}-\mathrm{q\mid }+\mathrm{y\mid r}-\mathrm{s\mid }$
   Example - $\stackrel{+1}{{\mathrm{Fe}}_2}\stackrel{-2}{\mathrm{S}}+\stackrel{0}{{\mathrm{O}}_2}\to \stackrel{+3}{{\mathrm{Fe}}_2}{\mathrm{O}}_3+\stackrel{+4}{\mathrm{S}}{\mathrm{O}}_2$
   $\text{n-factor of}{\mathrm{Fe}}_2\mathrm{S}=2\mathrm{\mid }1-3\mathrm{\mid }+1\mathrm{\mid }-2-4\mathrm{\mid }=10$

10. Disproportionation rxn. = Same atom of a molecule oxidise and reduce
    $\stackrel{+\mathrm{p}}{{\mathrm{A}}_{\mathrm{x}}}{\mathrm{B}}_{\mathrm{y}}\to \stackrel{+\mathrm{q}}{{\mathrm{A}}_{{\mathrm{x}}_1}}+\stackrel{+\mathrm{r}}{{\mathrm{A}}_{{\mathrm{x}}_2}}+{\mathrm{B}}_{\mathrm{y}}$
    $\mathrm{n}=\mathrm{x\mid p}-\mathrm{q\mid };\mathrm{m}=\mathrm{x\mid p}-\mathrm{r\mid }$
    $\text{n-factor of}{\mathrm{A}}_{\mathrm{x}}{\mathrm{B}}_{\mathrm{y}}=\frac{\mathrm{n}\times \mathrm{m}}{\mathrm{m}+\mathrm{n}}$
    Example - ${\mathrm{H}}_2\stackrel{-1}{{\mathrm{O}}_2}\to {\mathrm{H}}_2\stackrel{-2}{\mathrm{O}}+\stackrel{0}{{\mathrm{O}}_2}$
    $\text{n-factor of}{\mathrm{H}}_2{\mathrm{O}}_2=\frac{2\mathrm{\mid }-1-\left(-2\right)\mathrm{\mid }\times 2\mathrm{\mid }-1-0\mathrm{\mid }}{2+2}=\frac{4}{4}=1$
    For disproportionation reaction an element must have atleast 3 variable oxidation number.

Equivalent weight

Number of equivalents (n_eq)

Normality

Law of equivalence

In titration : ${\left({\mathrm{n}}_{\text{eq}}\right)}_{\text{acid}}={\left({\mathrm{n}}_{\text{eq}}\right)}_{\text{base}}$

Example : $2\mathrm{A}+3\mathrm{B}\to 5\mathrm{C}$

Balancing of Redox reactions

• Step-I : Inc. in O.N = Dec. in O.N
• Step-II : Balance Charge by adding H⁺ in acidic medium
  and OH^- in basic medium
• Step-III : Balance H & OH by adding H₂O

Q. Balance ${\mathrm{MnO}}_4^{-}+{\mathrm{H}}_2\mathrm{S}\to \mathrm{S}+{\mathrm{Mn}}^{+2}$ in acidic medium.

Decrease in O.N (n-factor) of Mn = 1 x |7 - 2| = 5
Increase in O.N (n-factor) of S = 1 x |-2 - 0| = 2

• Step-I : Inc. in O.N of S x 5 = Dec. in O.N Mn x 2
  $2{\mathrm{MnO}}_4^{-}+5{\mathrm{H}}_2\mathrm{S}\to 5\mathrm{S}+2{\mathrm{Mn}}^{+2}$
• Step-II : Right side has +4 charge and left side has -2, to balance charge
  $6{\mathrm{H}}^{+}+2{\mathrm{MnO}}_4^{-}+5{\mathrm{H}}_2\mathrm{S}\to 5\mathrm{S}+2{\mathrm{Mn}}^{+2}$
• Step-III : Left side has 16 H and 8 Oxygen, Right side does not, to balance
  $6{\mathrm{H}}^{+}+2{\mathrm{MnO}}_4^{-}+5{\mathrm{H}}_2\mathrm{S}\to 5\mathrm{S}+2{\mathrm{Mn}}^{+2}+8{\mathrm{H}}_2\mathrm{O}$

Q. Balance ${\mathrm{I}}_2+{\mathrm{S}}_2{\mathrm{O}}_3^{-2}\to {\mathrm{I}}^{-}+{\mathrm{S}}_4{\mathrm{O}}_6^{-2}$ in basic medium.

Decrease in O.N (n-factor) of I = 2 x |0 - (-1)| = 2
Increase in O.N (n-factor) of S = 2 x |2 - 2.5| = 1

• Step-I : Inc. in O.N of S x 2 = Dec. in O.N Iodine x 1
  $1{\mathrm{I}}_2+2{\mathrm{S}}_2{\mathrm{O}}_3^{-2}\to 2{\mathrm{I}}^{-}+{\mathrm{S}}_4{\mathrm{O}}_6^{-2}$
• Step-II : Charges are already balanced
• Step-III : Hydrogen & Oxygen are also balanced already