# Oxidising and Reducing Agent

Equivalent mass of Oxidising and Reducing Agent

The equivalent mass of an oxidising and a reducing agent may be calculated by dividing the molecular mass of the oxidising or reducing agent by change in oxidation number in the reaction. Thus,

$\text{Equi. mass of Ox. Agent} = \dfrac{\text{Mol.mass of Ox. Agent}}{\text{Decrease in Ox. No.}}$

$\text{Equi. mass of Red. Agent} = \dfrac{\text{Mol.mass of Red. Agent}}{\text{Increase in Ox. No.}}$

For example:

(i) $KMnO_4$ has different equivalent masses in different mediums (acidic, alkaline and neutral).

In acidic medium:

$\overset{7+}{2KMnO_4} + 3 H_2SO_4 \to K_2SO_4 + \overset{2+}{2MnSO_4} 3H_2O + 5O$

Equivalent mass of $KMnO_4$

$= \dfrac{\text{Mol. Mass of} KMnO_4}{\text{Decrease in Ox. No.}} = \dfrac{158}{5} = 31.6$

In alkaline medium:

$2K \overset{7+}{Mn}O_4 + 2KOH \to 2K_2\overset{6+}{Mn}O_4 + H_2O + O$

Equivalent mass of $KMnO_4$

$= \dfrac{\text{Mol. Mass of} KMnO_4}{\text{Decrease in Ox. No.}} = \dfrac{158}{1} = 158$

In neutral medium:

$\overset{7+}{2KMnO_4} + H_2O \to 2KOH + \overset{4+}{2MnO_2} + 2O$

Equivalent mass of $KMnO_4$

$= \dfrac{\text{Mol. Mass of} KMnO_4}{\text{Decrease in Ox. No}} = \dfrac{158}{3} = 52.67$

(ii) $K_2Cr_2O_7$ in acidic medium:

$K_2\overset{7+}{Cr_2}O_7 + 4H_2SO_4 \to K_2SO_4 + \overset{3+}{Cr_2}(SO_4)_3 + 4H_2O + 3O$

Equivalent mass of $K_2Cr_2O_7$

$= \dfrac{\text{Mol. Mass of} K_2Cr_2O_7}{2 \times \text{Decrease in Ox. No.}}$

(because in $K_2Cr_2O_7 \text{and} Cr_2(SO_4)_3$ two chromium atoms are present

$= \dfrac{294}{6} = 49.0$

(iii) $2K\overset{7+}{Mn}O_4 + 3H_2SO_4 +5H_2 \overset{3+}{C_2}O_4 \to K_2SO_4 + 2\overset{2+}{Mn}So_4 + 8H_2O + 10\overset{+4}{C}O_2$

In this reaction $KMnO_4$ a oxidising agent (decrease in Ox. No. +7 to +2, i.e., 5) and $H_2C_2O_4$ is reducing agent (increase in Ox. No. 3 to 4 and total increase is 2 due to 2 atoms of C).

Equivalent mass of $KMnO_4$ = 31.6 (As above)

Equivalent mass of $H_2 C_2 O_4$
= $\dfrac{\text{Molecular Mass of} H_2 C_2 O_4}{2 \times \text{Decrease in oxidation number}}$

= $\dfrac{90}{2} = 45$

Example 1. Calculate the oxidation number atoms in the following compounds or ions:

$(A) \underset{-}{Fe}O_4^{2-}, \underset{-}{Fe_3}O_4, K_4\underset{-}{Fe}(CN)_6, \underset{-}{Fe}(CO)_5 \\[3mm] (B) Na_2 \underset{-}{S_4} O_6, Na_2 \underset{-}{S_2}O_3, H_2 \underset{-}{S_2}O_8, H_2 \underset{-}{S}O_5 \\[3mm] (C) \underset{-}{H_2}O_2, \underset{-}{C_2} O^{2-}_4, \underset{-}{PH_3}, \underset{-}{C_3} H_3, Na \underset{-}{N_3} \\[3mm] (D) H_2 \underset{-}{SiF_6}, \underset{-}{Si_2}O^{6-}_7 \\[3mm] (E) Na_2 \underset{-}{B_4} O_7, (NH_4)_6 \underset{-}{Mo_7} O_{24}, Na_2H_2 \underset{-}{V_6} O_{17} Na_2 \underset{-}{Cr_3}O_{10} \\[3mm] (F) NaO \underset{-}{Cl}, \underset{-}{CI} O_3, Mg_3 \underset{-}{N_2}, \underset{-}{N_2}H_4 \\[3mm] (G) \underset{-}{P}O^{3-}_4, \underset{-}{P_2} O_5, H_2 \underset{-}{P}O_3, \underset{-}{P}OCl_{3-} \\ (H) H \underset{-}{Cl}O_4 \\[3mm] (I) H \underset{-}{N}C, H \underset{-}{N}C, \underset{-}{N}OCl \\ (J) \underset{-}{S}F_6, Ca \underset{-}{S}O_3, \underset{-}{S}O_3, H_2 \underset{-}{S}$

Solution: Let the oxidation number of the bold atoms in each be ‘x’.

$(A) FeS^{2-}_4 : x- 8 = -2 \\ x = 6 K_4Fe(CN)_6 : 4 + x- 6 = 0; \\ x = 2 \\ Fe_3O_4 : 3x- 8 = 0 \\ x = 8 / 3 Fe(CO)_5 : x + 0 = 0 \\ x = 0 \\[5mm] (B) Na_2 S_4O_6 : 2 + 4x- 12 = 0; \\ x = 2.5 \\ H_2S_2O^*_8 : 2 +2x- (6 \times 2 + 2 \times 1) = 0 \\ x = 0 \\ Na_2S_2O_3 : 2 + 2x- 6 = 0 \\ x = 0 \\ H_2SO^*_5 : 2 + x- (3 \times 2 + 2 \times 1) \\ x = 6 \\[5mm] (C) H_2O_2 : 2x- 2 = 0; \\ x = 1 PH_2 : x + 3 = 0 \\ x = -3 \\ NaN_3 : 1 + 3x = 0 \\ x = -1/3 \\ C_2O^{2-}_4 : 2x- 8 = -2 \\ x = 3 \\ C_3H_8 : 3x + 8 = 0 \\ x = -8/3 \\[5mm] (D) H_2SiF_6 : 2 + x- 6 = 0; \\ x = 4 \\ Si_2O^6_7 : 2x- 14 = -6 \\ x = 4$

$(E) Na_2B_4O_7 : 2 + 4x- 14 = 0; \\ x = 4 \\ Na_2H_2V_6O_{17} : 2 + 2 + 6x- 34 = 0; \\ x = 5 \\ (NH_4)_6Mo_7O_{24} : 6 7x- 48 = 0 \\ x = 6 \\ Na_2Cr_3O_{10} : 2 + 3x- 20 = 0 \\ x = 6 \\[5mm] (F) NaOCl : 1- 2 + x = 0; \\ x = 1 \\ Mg_3N_2 : 6 + 2x = 0 \\ x = -3 \\ KIO_3 : 1 + x- 6 = 0 \\ x = 5 \\ N_2H_4 : 2x + 4 = 0 \\ x = -2 \\[5mm] (G) PO^{3-}_4 : x- 8 = -3 \\ x = 5 \\ H_2PO_3 : 2 + x- 6 = 0; \\ x = 4 \\ P_2O_5 : 2x- 10 = 0 \\ x = 5 \\ POCl_3 : x- 2- 3 = 0 \\ x = 5 \\[5mm] (H) HClO_4 : 1 + x- 8 = 0 \\ x = 7 \\ (I) in HCN : H- C \equiv \equiv N \\ 1 + x + 3(-1) = 0 \\ \text{or} \hspace{20mm} x = 2$

Or three bonds on nitrogen atom and nitrogen is more electronegative element.

In $HNC : H- N \underset{=}{\rightarrow} C$

[/latex]Ox. No. of H = $+ 1 \because$

Ox. No. of N = (-2) (for covalent bond with C) + (-1) (for covalent bond with H) + 0 (no contribution for coordinate bond)

Ox. No. of N = -2 – 1 + 0 = -3

Ox. No. of C = 1 + (-3) + x = 0

x = 2

In NOCl : Cl – N = O or $NO^+Cl^-$

Ox No. of N = + 1 (for covalent bond with Cl)

Ox. No. of N = +2 (for covalent bond with O)

Ox. No. of N in NOCl = + 3

$(J) SF_6 : x + 6(-1) = 0 \\ x = 6 \\ SO_3 : x + 3(-2) = 0 \\ x = 6 \\ CaSO_3 : 2 + x + 3(-2) = 0 \\ x = 4 \\ H_2S : 2(+1) + x = 0 \\ x = -2$

*These compounds are the derivatives of hydrogen peroxide $(H_2O_2)$ hence only two atoms have -2 charge $(\text{i.e.}, O^{2-}_2, -1$ charge on each oxygen atom). Other oxygen atoms have -2 charge.

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