Any reaction in which electrons are exchanged between atoms or ions, represents simultaneous process of oxidation and reduction and is called a Redox reaction.

Oxidation is

(i) the gain of oxygen

or (ii) the loss of hydrogen

or (iii) the loss of electrons (de-electronation)

or (iv) the increase in oxidation number

Reduction is (the reverse of oxidation)

(i) the loss of oxygen

or (ii) the gain of hydrogen

or (iii) the gain of electron (electronation)

or (iv) the decrease in oxidation number

Electronic concept of Oxidation and Reduction:

Loss of electrons is called Oxidation

Gain of electrons is called Reduction

Redox Reaction

A reaction in which oxidation and reduction are carried out simultaneously is called redox reaction. One substance undergoes oxidation and other substance undergoes reduction i.e. reaction between a reducing agent and oxidising agent is termed as redox reaction.

Reducing agent $\underset{\text{ reduction }}{\stackrel{\text{ oxidation }}{\rightleftharpoons }}$ Oxidizing agent + Electrons

Redox Reactions are divided into two main types :

(i) Intermolecular redox reaction: In such a redox reaction, molecule of one reactant is oxidised whereas the molecule of other reactant gets reduced.

Oxidation Number of Elements in Covalent Compound and in Ions:

Rules for Assigning Oxidation Number (O.N.)

Oxidation Number of one particular element in a covalent compound or ion is determined by taking the Oxidation Number of other elements according to certain rules which are summarised below

$\begin{array}{ll}\text{Group I (Alkali Elements)}& +1\\ \text{Group II (Alkaline Earth Elements)}& +2\\ \text{Group III (Boron Family)}& +3\\ \text{Hydrogen}& +1\\ & -1\text{ (in metal hydride }{\mathrm{Na}}^{+}{\mathrm{H}}^{-}\text{)}\\ \text{Oxygen}& -2\\ & -1\text{ in peroxides }({\mathrm{BaO}}_2)\\ & +2\text{ in }{\mathrm{OF}}_2\\ \text{Nitrogen}& -3\text{ in }{\mathrm{NH}}_3\text{ and in nitrides}\\ & \text{varies when in combination with oxygen}\\ \text{Halogens}& -1\text{ in direct combination with metals; varies}\\ & \text{in combination with oxygen and in inter-}\\ & \text{halogen compounds such on }\mathrm{ICl}\end{array}$

Wherever there is doubt, the more electro-negative atom takes the lower oxidation number.

In neutral molecules, sum of oxidation numbers is zero.

In ions, sum of oxidation numbers is equal to charge.

While naming the ion or compound, it is desirable to put oxidation number per atom (i.e. oxidation state) in roman figures within brackets. (anion ends with ate)

Oxidation number of an atom is also called its oxidation state

If central atom is in the highest oxidation state in a compound it behaves as an oxidising agent (eg. ${\mathrm{H}}_2{\mathrm{SO}}_4$ ) and if it is in the lowest oxidation state the compound behaves as a reducing agent (eg. ${\mathrm{H}}_2\mathrm{S}$ ) If it is in its intermediate oxidation state, the compound behaves as both an oxidising as well as a reducing agent (eg. ${\mathrm{SO}}_2$ ).

Note: In an organic compound containing carbon, hydrogen and oxygen, the oxidation number of carbon can be calculated as,

Oxidation number of carbon $=\frac{2\times n_0-n_H}{n_C}$

Where, ${\mathrm{n}}_0=$ Number of oxygen atoms

${\mathrm{n}}_{\mathrm{H}}=$ Number of hydrogen atoms

${\mathrm{n}}_{\mathrm{c}}=$ Number of carbon atoms

Examples:

(I). $\mathbf{S}$ in ${\mathbf{H}}_2{\mathbf{S}}_2{\mathbf{O}}_8$ :

Here two 0 -atoms form peroxide linkage, i.e.,

$$\begin{array}{rl}& \therefore 2\times 1+2a+6\times (-2)+2\times (-1)=0\\ & a=+6\end{array}$$

(ii) $NinN{H}_{4}N{O}_3$

$2\times a+4\times 1+3\times (-2)=0$

$a=+1\text{ (wrong) }$

No doubt there are two $\mathrm{N}$-atoms in ${\mathrm{NH}}_4{\mathrm{NO}}_3$, but one $\mathrm{N}$-atom has negative oxidation number (attached to $\mathrm{H}$ ) and the other has positive oxidation number (attached to 0 ). Therefore, evaluation should be done separately as

Oxidation Number of ${\mathrm{Nin}}^{+}{\mathrm{NH}}_4^{+}$ $$ and $$ Oxidation number of ${\mathrm{NinNO}}_3^{-}$

$a+4x(+1)=+1$ $a+3\times (-2)=-1$

$a=-3$ $\therefore a=+5$

(iii) $Iin{\mathrm{KI}}_3:$

$1+3\times (\mathrm{a})=0;\therefore \mathrm{a}=-1/3$

$\therefore {\mathrm{KI}}_3$ is $\mathrm{KI}+{\mathrm{I}}_2$

$\therefore$ I has two oxidation numbers -1 and 0 .

The above oxidation number of I in ${\mathrm{KI}}_3$ is an average value, i.e. [I-I-I] ${}^{-}$

Average 0.N. $=\frac{1\times (-1)+2\times (0)}{3}=-\frac{1}{3}$

(iv) $\mathbf{S}$ in ${\mathrm{Na}}_2{\mathrm{S}}_2{\mathbf{O}}_3:$

$2\times 1+2\times a+3\times (-2)=0$

$\therefore a=+2$

This is the average oxidation number:

The structure of ${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3$ is

The O.N. of $S^{-}$atom is -1 and of central $S=+5$

(v) ${\mathrm{Cr}}^2{\mathrm{K}}_3{\mathrm{CrO}}_8:$

$3\times 1+\mathrm{a}+8\times (-1)=0$

$\therefore a=+5$

${\left[{\mathrm{CrO}}_8\right]}^{3-}$ exists as ${\left[\mathrm{Cr}{\left({\mathrm{O}}_2\right)}_4\right]}^{3-}$ having four peroxide bonds, i.e., tetraperoxo ion

Exceptional cases of calculation of oxidation numbers

The rules described earlier are usually helpful in determination of the oxidation number of a specific atom in simple molecules but these rules fail in following cases. In these cases, the oxidation numbers are evaluated using the concepts of chemical bonding involved.

Type I. In molecules containing peroxide linkage in addition to element-oxygen bonds. For example,

(i) Oxidation number of $\mathrm{S}$ in ${\mathrm{H}}_2{\mathrm{SO}}_5$

(Permonosulphuric acid or Caro’s acid)

By usual method; ${\mathrm{H}}_2{\mathrm{SO}}_5$

$2\times 1+a+5\times (-2)=0$ or $a=+8$

But this cannot be true as maximum oxidation number for $S$ cannot exceed +6 , Since $S$ has only 6 electrons in its valence shell. This exceptional value is due to the fact that two oxygen atoms in ${\mathrm{H}}_2{\mathrm{SO}}_5$ shows peroxide linkage as shown below,

Therefore the evaluation of $0.N$. of sulphur here should be made as follows,

$2x(+1)+a+3x(-2)+2x(-1)$

(for $\mathrm{H}$ ) (for $\mathrm{S}$ ) (for 0 ) (for $0-0$ )

or $2+a-6-2=0$ or $a=+6$

(ii) Oxidation number of $\mathrm{S}$ in ${\mathrm{H}}_2{\mathrm{S}}_2{\mathrm{O}}_8$

(Peroxidisulphuric acid or Marshall’s acid)

By usual method: ${\mathrm{H}}_2{\mathrm{S}}_2{\mathrm{O}}_8$

$1\times 2+2a+8(-2)=0$

$2\mathrm{a}=+16-2=14$ or $\mathrm{a}=+7$

Similar to Caro’s acid, Marshall’s acid also has a peroxide linkage and $S$ shows +6 oxidation state in it.

Therefore the evaluation of oxidation state of sulphur should be made as follows,

$\frac{2\times (+1)}{(\text{ for }\mathrm{H})}+\frac{2\times (\mathrm{a})}{(\text{ for }\mathrm{S})}+\frac{6\times (-2)}{(\text{ for }0)}+\frac{2\times (-1)}{(\text{ for }0-0)}=0$

or $2+2a-12-2=0$ or $a=+6$

(iii) Oxidation number of ${\mathrm{Cr}}^2{\mathrm{CrO}}_5$

(Blue perchromate)

By usual method ${\mathrm{CrO}}_5;\mathrm{a}-10=0$ or $\mathrm{a}=+10$

This cannot be true as maximum O.N. of Cr cannot be more than +6 , since $\mathrm{Cr}$ has only five electrons in $3d$ orbitals and one electron in $4\mathrm{s}$ orbital. This exceptional value is due to the fact that four oxygen atoms in ${\mathrm{CrO}}_5$ are in peroxide linkage.

The chemical structure of ${\mathrm{CrO}}_5$ is

Therefore, the evaluation of $0.N$. of Cr should be made as follows

$a+1\times (-2)+4(-1)=0$

(for Cr) (for 0) (for 0-0)

or $\mathrm{a}-2-4=0$ or $\mathrm{a}=+6$

Type II. In molecules containing covalent and coordinate bonds, following rules are used for evaluating the oxidation numbers of atoms.

(i) For each covalent bond between dissimilar atoms the less electronegative element is assigned the oxidation number of +1 while the atom of the more electronegative element is assigned the oxidation number of -1 .

(ii) In case of a coordinate-covalent bond between similar or dissimilar atoms but the donor atom is less electronegative than the acceptor atom, an oxidation number of +2 is assigned to the donor atom and an oxidation number of -2 is assigned to the acceptor atom.

Conversely, if the donor atom is more electronegative than the acceptor atom, the contribution of the coordinate bond is neglected. Examples.

(a) Oxidation number of $\mathrm{C}$ in $\mathrm{HC}\equiv \mathrm{N}$ and $\mathrm{HN}\equiv \mathrm{C}$

The evaluation of oxidation number of $\mathrm{C}$ cannot be made directly by usual rules since no standard rule exists for oxidation numbers of $\mathrm{N}$ and $\mathrm{C}$.

In such cases, evaluation of oxidation number should be made using indirect concept or by the original concepts of chemical bonding.

(i) Oxidation number of carbon in $\mathrm{H}-\mathrm{N}\equiv \mathrm{C}$

The contribution of coordinate bond is neglected since the bond is directed from a more electronegative $\mathrm{N}$ atom (donor) to a less electronegative carbon atom (acceptor).

Therefore the oxidation number of $\mathrm{N}$ in $\mathrm{H}-\mathrm{N}=\mathrm{C}$ remains -3 as it has three covalent bonds.

$1\times (+1)+1\times (-3)+a=0$

(for $\mathrm{H}$ ) (for $\mathrm{N}$ ) (for $\mathrm{C}$ )

or $1+a-3=0$ or $a=+2$.

(ii) Oxidation number of carbon in $\mathrm{HC}\equiv \mathrm{N}$

In $\mathrm{HC}\equiv \mathrm{N},\mathrm{N}$ is more electronegative than carbon, each bond gives an oxidation number of -1 to $\mathrm{N}$. There are three covalent bonds, the oxidation number of $\mathrm{N}$ in $\mathrm{HC}\equiv \mathrm{N}$ is taken as -3 .

Now $\mathrm{HC}\equiv \mathrm{N}\therefore +1+\mathrm{x}-3=0\Rightarrow \mathrm{x}=+2$

Type III. In a molecule containing two or more atoms of same or different elements in different oxidation states.

(i) Oxidation number of $\mathrm{S}$ in ${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3$

By usual method

$2x(+1)+2xa+3(-2)=0$

or $\mathrm{a}=2$

But this is unacceptable as the two sulphur atoms in ${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3$ cannot have the same oxidation number because on treatment with dil. ${\mathrm{H}}_2{\mathrm{SO}}_4$, one sulphur atom is precipitated while the other is oxidised to ${\mathrm{SO}}_2$.

${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{Na}}_2{\mathrm{SO}}_4+{\mathrm{SO}}_2+\mathrm{S}+{\mathrm{H}}_2\mathrm{O}$

In this case, the oxidation number of sulphur is revaluated, from concepts of chemical bonding. The chemical structure of ${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3$ is

Due to the presence of a co-ordinate bond between two sulphur atoms, the acceptor sulphur atom has oxidation number of -2 whereas the other $S$ atom gets oxidation number of +2 .

$\frac{2\times (+1)}{(\text{ for Na) }}+\frac{3\times (-2)}{(\text{ for }0)}+\frac{\mathrm{ax}1}{(\text{ for }S)}+\frac{1\times (-2)}{(\text{ for coordinated }S)}=0$

or $+2-6+a-2=0$ or $a=+6$

Thus two sulphur atoms in ${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3$ have oxidation numbers of -2 and +6 .

(ii) Oxidation number of chlorine in ${\mathrm{CaOCl}}_2$ (bleaching powder)

In bleaching powder $\mathrm{Ca}(\mathrm{OCI})(\mathrm{Cl})$, the two $\mathrm{Cl}$ atoms are in different oxidation states i.e., one atom as ${\mathrm{Cl}}^{-}$having oxidation number of -1 and the other as ${\mathrm{OCl}}^{-}$having oxidation number of +1 .

(iii) Oxidation number of ${\mathrm{Nin}}^2{\mathrm{NH}}_4{\mathrm{NO}}_3$

By usual method $2\mathrm{a}+4\mathrm{x}(+1)+3\mathrm{x}(-1)=0$

$2a+4-3=0$ or $2a=+1$ (wrong)

No doubt ${\mathrm{NH}}_4{\mathrm{NO}}_3$ has two nitrogen atoms but one $\mathrm{N}$ has negative oxidation number (attached to $\mathrm{H}$ ) and the other has positive oxidation number (attached to 0 ). Hence the evaluation should be made separately for ${\mathrm{NH}}_4^{+}$and ${\mathrm{NO}}_3^{-}$

${\mathrm{NH}}_4^{+}$a $+4x(+1)=+1$ or $\mathrm{a}=-3$

${\mathrm{NO}}_3^{-}$a $+3(-2)=-1$ or $\mathrm{a}=+5$

(iv) Oxidation number of ${\mathrm{Fe}}^2{\mathrm{Fe}}_3{\mathrm{O}}_4$

In ${\mathrm{Fe}}_3{\mathrm{O}}_4$, $\mathrm{Fe}$ atoms are in two different oxidation states. ${\mathrm{Fe}}_3{\mathrm{O}}_4$ can be considered as an equimolar mixture of $\mathrm{FeO}$ [iron (II) oxide] and ${\mathrm{Fe}}_2{\mathrm{O}}_3$ [iron (III) oxide]. Thus in one molecule of ${\mathrm{Fe}}_3{\mathrm{O}}_4$, two $\mathrm{Fe}$ atoms are in +3 oxidation state and one $\mathrm{Fe}$ atom is in +2 oxidation state.

(v) Oxidation number of $\mathrm{S}$ in sodium tetrathionate $\left({\mathrm{Na}}_2{\mathrm{S}}_4{\mathrm{O}}_6\right)$

Its structure can be represented as follows,

The two S-atoms which are linked to each other have oxidation number zero. The oxidation number of other $S$-atoms can be calculated as follows

Let oxidation number of $\mathrm{S}=\mathrm{a}$.

$\therefore 2a+2\times 0+6\times (-2)=-2$

(for S) (for S-S) (for 0 )

$a=+5$

APPLICATIONS OF OXIDATION STATE

If an element is in its highest possible oxidation state in a compound, it can act only as an oxidising agent, for example, ${\mathrm{KMnO}}_4,{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7,{\mathrm{HNO}}_3,{\mathrm{H}}_2{\mathrm{SO}}_4,{\mathrm{HClO}}_4$ etc.

If element is in its lowest oxidation state in a compound, it can act only as a reducing agent. For example, ${\mathrm{H}}_2\mathrm{S},{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4,{\mathrm{FeSO}}_4,{\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3,{\mathrm{SO}}_2,{\mathrm{SnCl}}_2$, many metals etc.

The strength of oxyacids of chlorine decrease in the order. ${\mathrm{HClO}}_4>{\mathrm{HClO}}_3>{\mathrm{HClO}}_2>\mathrm{HClO}$

If an element is in intermediate oxidation state in a compound, it can act both as an oxidising as well as a reducing agent. For example, ${\mathrm{H}}_2{\mathrm{O}}_2,{\mathrm{H}}_2{\mathrm{SO}}_3,{\mathrm{HNO}}_3,{\mathrm{SO}}_2$ etc.

Difference between Valency and Oxidation Number :

Types of redox reactions:

1. Combination reactions:

$\stackrel{0}{2\mathrm{Mg}}+\stackrel{0^{}}{{\mathrm{O}}_2}⟶\stackrel{+2-2}{2\mathrm{MgO}}$

$\stackrel{0}{{\mathrm{H}}_2}+{\stackrel{0}{\mathrm{Cl}}}_2⟶2\stackrel{+1-1}{\mathrm{HCl}}$

2. Decomposition Reactions : (Reverse of combination reaction)

$2\stackrel{+1-2}{\mathrm{H}}{\mathrm{L}}_2^{\mathrm{O}}\stackrel{\mathrm{\Delta }}{\to }2{\stackrel{0}{\mathrm{H}}}_2+{\stackrel{0}{\mathrm{O}}}_2$

$2{\mathrm{KClO}}_3⟶+5+2$

3. Displacement reactions

$\mathrm{A}+\mathrm{BC}⟶\mathrm{AC}+\mathrm{B}$

Metal displacement:

$\stackrel{+3-2}{{\mathrm{Cr}}_2{\mathrm{O}}_3}+\stackrel{0}{2\mathrm{Al}}⟶\stackrel{+3-2}{{\mathrm{Al}}_2{\mathrm{O}}_3}+\stackrel{0}{2\mathrm{Cr}}$

Non-metal displacement

$2\stackrel{0}{\mathrm{Na}}+2{\mathrm{H}}_2^{+1-2}\mathrm{O}⟶\stackrel{+1-2+1+}{\mathrm{NaOH}}+{\mathrm{H}}_2$

4. Disproportionation:

It is a special type of redox reaction in which an element in a particular oxidation state gets simultaneously oxidised and reduced.

$2\stackrel{+1-1}{2{\mathrm{H}}_2}{\mathrm{O}}_2⟶2{\mathrm{H}}_2\mathrm{O}+\stackrel{+0}{{\mathrm{O}}_2}$

oxidation state of 0 changes from -1 to 0 and -2 state.

5. Auto-oxidation

Certain materials such as terpentine, olefinic compounds, phosphorus, metals like zinc and lead etc. can absorb oxygen from the air in the presence of water and water gets converted to hydrogen peroxide refers to the phenomenon of auto-oxidation. The material which absorbs oxygen and activates this is called the activator. The addition compound of activator and oxygen is termed as auto-oxidator. This reacts with water or some other acceptor so as to oxidise the latter.

BALANCING OXIDATION-REDUCTION REACTIONS BY OXIDATION NUMBER METHOD :

The steps involved are:

(i) The skeleton equation of the chemical change is written.

(ii) Oxidation numbers are assigned to atoms in the equation. The atoms in which change in oxidation number has taken place are selected and two half reactions involving oxidation and reduction are selected.

(iii) Change in oxidation numbers in both the equations is made equal by multiplying with suitable integers and then both the equations are added.

(iv) First of all, those substances are balanced which have undergone change in oxidation number and then other atoms except hydrogen and oxygen. Finally hydrogen and oxygen are balanced. In ionic equations, the net charges on both sides are made equal. ${\mathrm{H}}^{+}$ions in acidic reactions and ${\mathrm{HO}}^{-}$ions in basic reactions are used to balance the charge and number of hydrogen and oxygen atoms.

ION-ELECTRON METHOD FOR BALANCING REDOX REACTIONS

The method for balancing redox reactions by ion electron method was developed by Jette and LaMev in 1927. It involves the following steps:

(i) Write down the redox reaction in ionic form.

(ii) Split the redox reaction into two half reactions, one for oxidation and the other for reduction.

(iii) Balance each half reaction for the number of atoms of each element. For this purpose:

(a) Balance the atoms other than $\mathrm{H}$ and $\mathrm{O}$ for each half reaction using simple multiples.

(b) Add water molecules to the side deficient in oxygen and ${\mathrm{H}}^{+}$to the side deficient in hydrogen. This is done in acidic or neutral solutions.

(c) In alkaline solution, for each excess of oxygen, add one water molecule to the same side and $2{\mathrm{OH}}^{-}$ions to the other side. If hydrogen is still unbalanced, add one ${\mathrm{OH}}^{-}$ ion for each excess hydrogen on the same side and one water molecule to the other side.

(iv) Add electrons to the side deficient in electrons as to equalise the charge on both sides.

(v) Multiply one or both the half reactions by a suitable number so that the numbers of electrons become equal in both the equations.

(vi) Add the two balanced half reactions and cancel any term common to both sides.

The following solved problems illustrate the various steps in oxidation number method.

1. Balance the given reaction by using oxidation number method.

${\mathrm{MnO}}_4^{-}+{\mathrm{Fe}}^{2+}\to {\mathrm{Mn}}^{2+}+{\mathrm{Fe}}^{3+}+{\mathrm{H}}_2\mathrm{O}$ (Acidic medium)

The balancing is done in the following steps:

Write the Oxidation Number of each atoms in the skeleton equation.

Identify the atoms which undergo change in oxidation number.

Calculate the increase and decrease in oxidation number w.r.t. reactant atoms.

Equate the increase and decrease in oxidation number on the reactant side.

${\mathrm{MnO}}_4^{-}+5{\mathrm{Fe}}^{2+}\to {\mathrm{Mn}}^{2+}+{\mathrm{Fe}}^{3+}+{\mathrm{H}}_2\mathrm{O}$

Balance Mn and Fe atoms on both sides of the equation.

${\mathrm{MnO}}_4^{-}+5{\mathrm{Fe}}^{2+}\to {\mathrm{Mn}}^{2+}+5{\mathrm{Fe}}^{3+}+{\mathrm{H}}_2\mathrm{O}$

As the reaction is carried in the acidic medium, to balance 0 atoms, add three ${\mathrm{H}}_2\mathrm{O}$ molecules on the product side.

${\mathrm{MnO}}_4^{-}+5{\mathrm{Fe}}^{2+}\to {\mathrm{Mn}}^{2+}+5{\mathrm{Fe}}^{3+}+4{\mathrm{H}}_2\mathrm{O}$

$$ In order to balance $\mathrm{H}$ atoms, add $8{\mathrm{H}}^{+}$on the reactant side.

$8{\mathrm{H}}^{+}+{\mathrm{MnO}}_4^{-}+5{\mathrm{Fe}}^{2+}\to {\mathrm{Mn}}^{2+}+5{\mathrm{Fe}}^{3+}+4{\mathrm{H}}_2\mathrm{O}$

$$ The final equation is balanced w.r.t. charge also.

2. Balance the given reaction in basic medium.

Reduction

• Write the Oxidation number of each atom

$+3-2+1+1-1+6-2+1-2$

${[\mathrm{Cr}(\mathrm{OH}{)}_4]}^{-}+{\mathrm{H}}_2{\mathrm{O}}_2\to {\left({\mathrm{CrO}}_4\right)}^{2-}+{\mathrm{H}}_2\mathrm{O}$

Calculate the increase and decrease in oxidation number.

Equate the increase and decrease in oxidation number on the reactant side

$2{[\mathrm{Cr}(\mathrm{OH}{)}_4]}^{-}+3{\mathrm{H}}_2{\mathrm{O}}_2\to {\left({\mathrm{CrO}}_4\right)}^{2-}+{\mathrm{H}}_2\mathrm{O}$

Balance the number of $\mathrm{Cr}$ atoms in the equation.

$2{[\mathrm{Cr}(\mathrm{OH}{)}_4]}^{-}+3{\mathrm{H}}_2{\mathrm{O}}_2\to 2{\left({\mathrm{CrO}}_4\right)}^{2-}+{\mathrm{H}}_2\mathrm{O}$

In order to balance the number of oxygen atoms, add five ${\mathrm{H}}_2\mathrm{O}$ molecules on the product side

$2[\mathrm{Cr}(\mathrm{OH}{)}_4]+3{\mathrm{H}}_2{\mathrm{O}}_2\to 2{\left({\mathrm{CrO}}_4\right)}^{2-}+6{\mathrm{H}}_2\mathrm{O}$

As the reaction is carried in the basic medium, in order to balance the number of negative charges add two ${\mathrm{OH}}^{-}$ions on the reactant sides and two ${\mathrm{H}}_2\mathrm{O}$ molecules on the product side.

$2{[\mathrm{Cr}(\mathrm{OH}{)}_4]}^{-}+3{\mathrm{H}}_2{\mathrm{O}}_2+2{\mathrm{OH}}^{-}\to 2{\left({\mathrm{CrO}}_4\right)}^{2-}+6{\mathrm{H}}_2\mathrm{O}+2{\mathrm{H}}_2\mathrm{O}$

or $2[\mathrm{Cr}(\mathrm{OH}{)}_4]+3{\mathrm{H}}_2{\mathrm{O}}_2+2{\mathrm{OH}}^{-}\to 2{\left({\mathrm{CrO}}_4\right)}^{2-}+8{\mathrm{H}}_2\mathrm{O}$

Example for balancing using lon electron method

$\mathrm{Cr}(\mathrm{OH}{)}_4^{-}+{\mathrm{H}}_2{\mathrm{O}}_2+{\mathrm{OH}}^{-}\to {\mathrm{CrO}}_4^{2-}+{\mathrm{H}}_2\mathrm{O}$

Oxidation half reaction.

$\mathrm{Cr}(\mathrm{OH}{)}_4^{-}\to {\mathrm{CrO}}_4^{2-}$ Balance $\mathrm{H}$

$\mathrm{Cr}(\mathrm{OH}{)}_4^{-}+4{\mathrm{OH}}^{-}\to {\mathrm{CrO}}_4^{2-}+4{\mathrm{H}}_2\mathrm{O}$ Balance electrons

$\mathrm{Cr}(\mathrm{OH}{)}_4^{-}+4{\mathrm{OH}}^{-}\to {\mathrm{CrO}}_4^{2-}+4{\mathrm{H}}_2\mathrm{O}+3{\mathrm{e}}^{-}]\times 2$

Reduction half reaction

${\mathrm{H}}_2{\mathrm{O}}_2\to {\mathrm{H}}_2\mathrm{O}$ Balance Oxygen

${\mathrm{H}}_2{\mathrm{O}}_2+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{H}}_2\mathrm{O}+2{\mathrm{OH}}^{-}$

${\mathrm{H}}_2{\mathrm{O}}_2+{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2\mathrm{O}+2{\mathrm{OH}}^{-}$

${\mathrm{H}}_2{\mathrm{O}}_2+2{\mathrm{e}}^{-}\to 2{\mathrm{OH}}^{-}]\times 3$

Equate electrons lost and gained and add

$2{[\mathrm{Cr}(\mathrm{OH}{)}_4]}^{-}+2{\mathrm{OH}}^{-}+3{\mathrm{H}}_2{\mathrm{O}}_2\to 2{\mathrm{CrO}}_4{}^{2-}+8{\mathrm{H}}_2\mathrm{O}$

PRACTICE QUESTIONS

1. Balance the following equations:

(i) ${\mathrm{KMnO}}_4+\mathrm{KCl}+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{MnSO}}_4+{\mathrm{K}}_2{\mathrm{SO}}_4+{\mathrm{H}}_2{\mathrm{SO}}_4+{\mathrm{H}}_2\mathrm{O}+{\mathrm{Cl}}_2$ (acidic)

(ii) ${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+{\mathrm{I}}^{-}+{\mathrm{H}}^{+}\to {\mathrm{Cr}}^{3+}+{\mathrm{I}}_2+{\mathrm{H}}_2\mathrm{O}$ (acidic)

(iii) $\mathrm{Cu}+{\mathrm{NO}}_3^{-}+\dots \to {\mathrm{Cu}}^{2+}+{\mathrm{NO}}_2+\dots$ (acidic)

(iv) ${\mathrm{Cl}}_2+{\mathrm{IO}}_3^{-}+{\mathrm{OH}}^{-}\to {\mathrm{IO}}_4^{-}+\dots +{\mathrm{H}}_2\mathrm{O}$ (basic)

(v) ${\mathrm{Br}}^{-}+{\mathrm{BrO}}_3^{-}+{\mathrm{H}}^{+}\to {\mathrm{Br}}_2+{\mathrm{H}}_2\mathrm{O}$ (acidic)

(vi) ${\mathrm{H}}_2\mathrm{S}+{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+{\mathrm{H}}^{+}\to {\mathrm{Cr}}_2{\mathrm{O}}_3+{\mathrm{S}}_8+{\mathrm{H}}_2\mathrm{O}$ (acidic)

(vii) $\mathrm{Au}+{\mathrm{NO}}_3^{-}+{\mathrm{Cl}}^{-}+{\mathrm{H}}^{+}\to {\mathrm{AuCl}}_4^{-}+{\mathrm{NO}}_2+{\mathrm{H}}_2\mathrm{O}$ (acidic)

(viii) ${\mathrm{I}}_2+{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+{\mathrm{H}}^{+}\to {\mathrm{Cr}}^{3+}+{\mathrm{IO}}_3^{-}+{\mathrm{H}}_2\mathrm{O}$ (acidic)

(ix) ${\mathrm{Cu}}_2\mathrm{O}+{\mathrm{H}}^{+}+{\mathrm{NO}}_3^{-}\to {\mathrm{Cu}}^{2+}+\mathrm{NO}+{\mathrm{H}}_2\mathrm{O}$ (acidic)

(x) ${\mathrm{Cl}}_2+{\mathrm{I}}_2\to {\mathrm{IO}}_3^{-}+{\mathrm{Cl}}^{-}$(acidic)

PRACTICE QUESTIONS

1. In the reaction, $4\mathrm{Fe}+3{\mathrm{O}}_2\to 4{\mathrm{Fe}}^{3+}+6{\mathrm{O}}_2^{-}$which of the following statements is incorrect?

(a) It is a redox reaction.

(b) Metallic iron is a reducing agent.

(c) ${\mathrm{Fe}}^{3+}$ is an oxidising agent.

(d) Metallic iron is reduced to ${\mathrm{Fe}}^{3+}$.

2. The oxidation number of chlorine in $\mathrm{HOCl}$ is:

(a) -1

(b) zero

(c) +1

(d) +2

3. For the redox reaction,

${\mathrm{MnO}}_4^{-}+{\mathrm{C}}_2{\mathrm{O}}_4^{2-}+{\mathrm{H}}^{+}\to {\mathrm{Mn}}^{2+}+{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}$

the correct coefficients of the reactants for the balanced equation are :

$\begin{array}{llll}& {\mathrm{MnO}}_4^{-}& {\mathrm{C}}_2{\mathrm{O}}_4^{2-}& {\mathrm{H}}^{+}\\ \text{(a)}& 2& 5& 16\\ \text{(b)}& 16& 5& 2\\ \text{(c)}& 5& 16& 2\\ \text{(d)}& 2& 16& 5\end{array}$

4. Which one of the following is not a redox reaction?

(a) ${\mathrm{CaCO}}_3\to \mathrm{CaO}+{\mathrm{CO}}_2$

(b) $2{\mathrm{H}}_2+{\mathrm{O}}_2\to 2{\mathrm{H}}_2\mathrm{O}$

(c) $\mathrm{Na}+{\mathrm{H}}_2\mathrm{O}\to \mathrm{NaOH}+\frac{1}{2}{\mathrm{H}}_2$

(d) ${\mathrm{MnCl}}_3\to {\mathrm{MnCl}}_2+\frac{1}{2}{\mathrm{Cl}}_2$

5. The oxidation state of iodine in ${\mathrm{H}}_4{\mathrm{IO}}_6^{-}$is :

(a) +7

(b) -1

(c) +5

(d) +1

6. The oxide which cannot act as a reducing agent is :

(a) ${\mathrm{SO}}_2$

(b) ${\mathrm{NO}}_2$

(c) ${\mathrm{CO}}_2$

(d) ${\mathrm{ClO}}_2$

7. In a chemical reaction,

${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7+{\mathrm{xH}}_2{\mathrm{SO}}_4+{\mathrm{ySO}}_2\to {\mathrm{K}}_2{\mathrm{SO}}_4+{\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3+{\mathrm{zH}}_2\mathrm{O}$

$x,y$ and $z$ are :

(a) $1,3,1$

(b) $4,1,4$

(c) $3,2,3$

(d) $2,1,2$

8. When copper is treated with a certain concentration of nitric acid, nitric oxide and nitrogen dioxide are liberated in equal volumes according to the equation,

$\mathrm{xCu}+\mathrm{yHNO}\to \mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2+\mathrm{NO}+{\mathrm{NO}}_2+{\mathrm{H}}_2\mathrm{O}$

The coefficients $x$ and $y$ are :

(a) 2,3

(b) 2,6

(c) 1,3

(d) 3,8

9. Amongst the following, identify the species with an atom in +6 oxidation state :

(a) $\mathrm{Mn}(\mathrm{CO}{)}_4^{-}$

(b) $\mathrm{Cr}(\mathrm{CN}{)}_6{}^{3-}$

(c) ${\mathrm{NiF}}_6{}^{2-}$

(d) ${\mathrm{Cr}}_2{\mathrm{O}}_2{\mathrm{Cl}}_2$

10. The reaction, $3{\mathrm{ClO}}^{-}(\mathrm{aq})\to {\mathrm{ClO}}_3^{-}+2{\mathrm{Cl}}^{-}(\mathrm{aq})$

is an example of

(a) oxidation reaction

(b) reduction reaction

(c) disproportionation reaction

(d) decomposition reaction

11. The oxidation state of iodine in ${\mathrm{IPO}}_4$ is:

(a) +1

(b) +3

(c) +5

(d) +7

12. In alkaline medium, ${\mathrm{H}}_2{\mathrm{O}}_2$ reacts with ${\mathrm{Fe}}^{3+}$ and ${\mathrm{Mn}}^{2+}$ separately to give:

(a) ${\mathrm{Fe}}^{4+}$ and ${\mathrm{Mn}}^{4+}$

(b) ${\mathrm{Fe}}^{2+}$ and ${\mathrm{Mn}}^{2+}$

(c) ${\mathrm{Fe}}^{2+}$ and ${\mathrm{Mn}}^{4+}$

(d) ${\mathrm{Fe}}^{4+}$ and ${\mathrm{Mn}}^{2+}$

13. Which of the following chemical reactions depict the oxidising behaviour of ${\mathrm{H}}_2{\mathrm{SO}}_4$ ?

(a) $2\mathrm{HI}+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{I}}_2+{\mathrm{SO}}_2+2{\mathrm{H}}_2\mathrm{O}$

(b) $\mathrm{Ca}(\mathrm{OH}{)}_2+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{CaSO}}_4+2{\mathrm{H}}_2\mathrm{O}$

(c) $\mathrm{NaCl}+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{NaHSO}}_4+\mathrm{HCl}$

(d) $2{\mathrm{PCl}}_5+{\mathrm{H}}_2{\mathrm{SO}}_4\to 2{\mathrm{POCl}}_3+2\mathrm{HCl}+{\mathrm{SO}}_2{\mathrm{Cl}}_2$

14. Oxidation state of oxygen in ${\mathrm{KO}}_3$ and ${\mathrm{Na}}_2{\mathrm{O}}_2$ is:

(a) 3,2

(b) 1,0

(c) 0,1

(d) $-0.33,-1$

15. The reaction, ${\mathrm{P}}_4+3\mathrm{NaOH}+3{\mathrm{H}}_2\mathrm{O}\to 3{\mathrm{NaH}}_2{\mathrm{PO}}_2+{\mathrm{PH}}_3$ is an example of:

(a) disproportionation reaction

(b) neutralisation reaction

(c) double decomposition reaction

(d) pyrolytic reaction

16. Match the items in list ’ $A$ ’ with relevant items in list ’ $B$ '

REDOX TITRATIONS

There are two approaches to analyse a redox reaction

(I) mole-concept

(ii) gram or milli equivalents (meq)

Mole Concept

First balance a given redox reaction then simply apply mole-concept.

$nX+mY\to pA+qB$

from stoichiometry of reaction

n mole of $X=$ m moles of $Y=p$ moles of $A=q$ moles of $B$

millimoles $=M\times V(mL)$

or Moles $=\mathrm{MxV}(\mathrm{L})$

Question. 1 $200\mathrm{mL}$ of $\mathrm{M}/60{\mathrm{KMnO}}_4$ are used to oxidise a given sample of $\mathrm{HCl}$. How many grams of ${\mathrm{Cl}}_2$ are produced if redox rection takes place in acidic medium

GRAM EQUIVALENTS

In this method there is no need to balance a redox reaction

$\mathrm{X}+\mathrm{Y}\to \mathrm{A}+\mathrm{B}$

Reducing $$ oxidising

agent $$ agent

meq of $X=$ meq of $Y=$ meq of $A=$ meq of $B$

meq $=\frac{\text{ mass }}{\text{ eq.mass }}\times 1000$

Eq. mass $=\frac{\text{ mol. mass }}{n}$

$\mathrm{n}=$ number of electrons transferred by one molecule/ion of oxidising or reducing agent

EXAMPLES

1. ${\mathrm{FeSO}}_4$

As a reducing agent

${\mathrm{Fe}}^{+2}\to {\mathrm{Fe}}^{+3}+{\mathrm{e}}^{-}$

${\mathrm{E}}_{{\mathrm{FeSO}}_4}=152/1=152$

${\mathrm{FeSO}}_4$ (As a oxidising agent)

${\mathrm{Fe}}^{+2}+2{\mathrm{e}}^{-}\to \mathrm{Fe}$

${\mathrm{E}}_{{\mathrm{FeSO}}_4}=152/2=76$

2. ${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3$

As a reducing agent (Acidic medium)

$2{\mathrm{S}}_2{\mathrm{O}}_3^{-2}\to {\mathrm{S}}_4{\mathrm{O}}_6^{-2}+2{\mathrm{e}}^{-}$

for one mole $n=1$

${\mathrm{E}}_{{\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3}=158/1=158$

As a reducing agent (Basic medium)

${\mathrm{S}}_2{\mathrm{O}}_3^{-2}+10{\mathrm{OH}}^{-}\to 2{\mathrm{SO}}_4^{-2}+5{\mathrm{H}}_2\mathrm{O}+8{\mathrm{e}}^{-}$

$\mathrm{n}=8$

$E=158/8$

3. ${\mathrm{HNO}}_3$

As an oxidant (conc. ${\mathrm{HNO}}_3$ )

${\mathrm{NO}}_3^{-}+2{\mathrm{H}}^{+}+{\mathrm{e}}^{-}\to {\mathrm{NO}}_2+{\mathrm{H}}_2\mathrm{O}$

$\mathrm{n}=1$

$\mathrm{E}=63/1$

As an oxidant (dil ${\mathrm{HNO}}_3$ )

${\mathrm{NO}}_3^{-}+4{\mathrm{H}}^{+}+3{\mathrm{e}}^{-}\to \mathrm{NO}+2{\mathrm{H}}_2\mathrm{O}$

$\mathrm{n}=3$

${\mathrm{E}}_{{\mathrm{HNO}}_3}=63/3=21$

Note: Equivalent mass depends on dilution

Question.2 $5.5\mathrm{g}$ of a mixture of ${\mathrm{FeSO}}_4\cdot 7{\mathrm{H}}_2\mathrm{O}$ and ${\mathrm{Fe}}_2{\left({\mathrm{SO}}_4\right)}_3\cdot 9{\mathrm{H}}_2\mathrm{O}$ required $5.4\mathrm{mL}$ of $0.1\mathrm{N}{\mathrm{KMnO}}_4$ solution for complete oxidation. Calculate the moles of hydrated ferric sulphate in the mixture.

Question.3 An element $A$ in a compound $ABD$ has an oxidation number ’ $-n$ ‘. It is oxidized by ${\mathrm{Cr}}_2{\mathrm{O}}_7^{-2}$ in acid medium. In an experiment $1.68\times {10}^{-3}$ mole ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ was required for $3.26\times {10}^{-3}$ mole of the compound ABD. Calculate new oxidation state of $A$

Question.4 $20\mathrm{mL}$ of a solution containing $0.2\mathrm{g}$ of impure sample of ${\mathrm{H}}_2{\mathrm{O}}_2$ reacts with $0.316\mathrm{g}$ of ${\mathrm{KMnO}}_4$ (acidic) calculate :

(a) Purity of ${\mathrm{H}}_2{\mathrm{O}}_2$

(b) Volume of dry ${\mathrm{O}}_2$ evolved at ${27}^{\circ }\mathrm{C}$ and $750\mathrm{mm}$ pressure.

Question. 5 $1\mathrm{g}$ of ${\mathrm{AgNO}}_3$ is dissolved in $100\mathrm{mL}$ of water and is titrated with $100\mathrm{mL}$ of $\mathrm{KI}$ solution resulting in precipitation of $\mathrm{Agl}$. The $\mathrm{KI}$ left in the filtrate is then titrated with $\mathrm{M}/20{\mathrm{KIO}}_3$ in presence of $6\mathrm{M}$ $\mathrm{KCl}$. The volume of ${\mathrm{KIO}}_3$ required is $100\mathrm{mL}$. It is observed that $20\mathrm{mL}$ of same $\mathrm{KI}$ solution requires $30\mathrm{mL}$ of $\mathrm{M}/20{\mathrm{KIO}}_3$. What is the percentage of ${\mathrm{AgNO}}_3$ in the sample