Redox Reaction
REDOX REACTIONS
INTRODUCTION
Molecular Equations: $2 \mathrm{FeCl}{3}+\mathrm{SnCl}{2} \rightarrow 2 \mathrm{FeCl}{2}+\mathrm{SnCl}{4}$
The reactants and products have been written in molecular forms; thus, the equation is termed as molecular equation.
Ionic Equations: The reactions in which the reactants and products are present in the form of ions are called ionic reactions.
For example: $2 \mathrm{Fe}^{3+}+6 \mathrm{Cl}^{-}+\mathrm{Sn}^{2+}+2 \mathrm{Cl}^{-} \rightarrow 2 \mathrm{Fe}^{2+}+4 \mathrm{Cl}^{-}+\mathrm{Sn}^{4+}+4 \mathrm{Cl}^{-}$
$$ \text { Or } 2 \mathrm{Fe}^{3+}+\mathrm{Sn}^{2+} \rightarrow 2 \mathrm{Fe}^{2+}+\mathrm{Sn}^{4+} $$
Phenomenon of Oxidation and Reduction:
Oxidation or de-electronation is a process which liberates electrons.
Reduction or electronation is a process which gains electrons.
| Oxidation | Reduction |
|---|---|
| a. $\mathrm{M} \longrightarrow \mathrm{M}^{\mathrm{n+}}+\mathrm{ne}^{-}$ | $\mathrm{M}^{\mathrm{n}^{+}}+\mathrm{ne}^{-} \longrightarrow \mathrm{M}$ |
| b. $\mathrm{M}^{\mathrm{n}{1}+} \longrightarrow \mathrm{M}^{\mathrm{n}{2}+}+\left(\mathrm{n}{2}-\mathrm{n}{1}\right) \mathrm{e}^{-} \quad\left(\mathrm{n}{2}>\mathrm{n}{1}\right)$ | $\mathrm{M}^{\mathrm{n}{2}+}+\left(\mathrm{n}{2}-\mathrm{n}{1}\right) \mathrm{e}^{-} \longrightarrow \mathrm{M}^{\mathrm{n}{1}+} \quad\left(\mathrm{n}{2}>\mathrm{n}{1}\right)$ |
| c. $\mathrm{A}^{\mathrm{n-}} \longrightarrow \mathrm{A}+\mathrm{ne}^{-}$ | $\mathrm{A}+\mathrm{ne}^{-} \longrightarrow \mathrm{A}^{\mathrm{n-}}$ |
| d. $\mathrm{A}^{\mathrm{n}{1}-} \longrightarrow \mathrm{A}^{\mathrm{n}{2}-}+\left(\mathrm{n}{1}-\mathrm{n}{2}\right) \mathrm{e}^{-}$ | $\mathrm{A}^{\mathrm{n}{2}^{-}}+\left(\mathrm{n}{1}-\mathrm{n}{2}\right) \mathrm{e}^{-} \longrightarrow \mathrm{A}^{\mathrm{n}{2}}$ |
Note: M may be an atom or a group of atoms; A may be atom or a group of atoms.
Oxidizing and Reducing Agent:
(a) If an element is in its highest possible oxidation state in a compound, it can function as an oxidizing agent, e.g. $\mathrm{KMnO}{4}, \mathrm{~K}{2} \mathrm{Cr}{2} \mathrm{O}{7}, \mathrm{HNO}{3}, \mathrm{H}{2} \mathrm{SO}{4}, \mathrm{HClO}{4}$ etc.
(b) If an element is in its lowest possible oxidation state in a compound, it can function as a reducing agent, e.g. $\mathrm{H}{2} \mathrm{~S}, \mathrm{FeSO}{4}, \mathrm{Na}{2} \mathrm{~S}{2} \mathrm{O}{3}, \mathrm{SnCl}{2}$ etc.
(c) If an element is in its intermediate oxidation state in a compound, it can function both as an oxidizing agent as well as reducing agent, e.g. $\mathrm{H}{2} \mathrm{O}{2}, \mathrm{H}{2} \mathrm{SO}{3}, \mathrm{HNO}{3}, \mathrm{SO}{2}$ etc.
(d) If highly electronegative element is in its higher oxidation state in a compound, that compound can function as a powerful oxidizing agent, e.g. $\mathrm{KClO}{4}, \mathrm{KClO}{3}, \mathrm{KIO}_{3}$ etc.
(e) If an electronegative element is in its lowest possible oxidation state in a compound or in free state, it can function as a powerful reducing agent, e.g. $\mathrm{I}^{-}, \mathrm{Br}^{-}, \mathrm{N}_{3}^{-}$ etc.
MODERN CONCEPT OF OXIDATION AND REDUCTION
According to the modern concept, loss of electrons is oxidation whereas gain of electrons is reduction. Oxidation and reduction can be represented in a general way as shown below:
Figure 1.1: Oxidation and Reduction
ION ELECTRON METHOD FOR BALANCING REDOX REACTIONS
This method involves the following steps:
(a) Divide the complete equations into two half reactions
(i) One representing oxidation
(ii) The other representing reduction
(b) Balance the atoms in each half reaction seperately according to the following steps:-
(i) Balance all atoms other than oxygen and hydrogen
(ii) To balance oxygen and hydrogen
(c) Acidic Medium
(i) Add $\mathrm{H}_{2} \mathrm{O}$ to the side which is oxygen deficient to balance oxygen atoms
(ii) Add $\mathrm{H}^{+}$ to the side which is hydrogen deficient to balance $\mathrm{H}$ atoms
(d) Basic Medium
(i) Add ${\mathrm{OH}}$ to the side which has less negative charge
(ii) Add $\mathrm{H}_{2} \mathrm{O}$ to the side which is oxygen deficient to balance oxygen atoms
(iii) Add $\mathrm{H}^{+}$ aromatic compounds to the side which is hydrogen deficient
OXIDATION STATE AND OXIDATION NUMBER
Oxidation State
It is defined as the charge (real or imaginary) which an atom appears to have when it is in combination. In the case of electrovalent compounds, the oxidation number of an element or radical is the same as the charge on the ion.
Oxidation Number
PYQ-2024-Periodic_Table_and_Periodicity-Q2,
(a) Oxidation number of an element in a particular compound represents the number of electrons lost or gained by an element during its change from free state into that compound or Oxidation number of an element in a particular compound represent the extent of oxidation or reduction of an element during its change from free state into that compound.
(b) Oxidation number is given positive sign if electrons are lost. Oxidation number is given negative sign if electrons are gained.
(c) Oxidation number represent real change in case of ionic compounds. However, in covalent compounds it represents imaginary charge.
Rules for Calculation of Oxidation Number:
Following rules have been arbitrarily adopted to decide oxidation number of elements on the basis of their periodic properties.
(a) In uncombined state or free state, oxidation number of an element is zero.
(b) In combined state oxidation number of-
(i) $\mathrm{F}$ is always -1
(ii) $\mathrm{O}$ is -2 . In peroxide it is -1 , in superoxides it is $-1 / 2$. However in $\mathrm{F}_{2} \mathrm{O}$ it is +2 .
(ii) $\mathrm{H}$ is +1 . In ionic hydrides it is -1 . (i.e., IA, IIA and IIIA metals).
(iv) Halogens as halide is always -1 .
**(v)**Sulphur as sulphide is always -2 .
(vi) Metal is always +ve.
(vii) Alkali metals (i.e., IA group - $\mathrm{Li}, \mathrm{Na}, \mathrm{K}, \mathrm{Rb}, \mathrm{Cs}, \mathrm{Fr}$ ) is always +1 .
(viii) Alkaline earth metals (i.e., IIA group $-\mathrm{Be}, \mathrm{Mg}, \mathrm{Ca}, \mathrm{Sr}, \mathrm{Ba}, \mathrm{Ra}$ ) is always +2 .
(c) The algebraic sum of the oxidation number of all the atoms in a compound is equal to zero. e.g. $\mathrm{KMnO}_{4}$.
Ox. no. of $\mathrm{K}+\mathrm{Ox}$. no. of $\mathrm{Mn}+(\mathrm{Ox}$. no. of $\mathrm{O}) \times 4=0$
$(+1)+(+7)+4 x(-2)=0$
(d) The algebraic sum of all the oxidation no. of elements in a radical is equal to the net charge on the radical. e.g. $\mathrm{CO}_{3}^{-2}$.
Oxidation no. of $\mathrm{C}+3 \times$ (Oxidation no. of $\mathrm{O})=-2(4)+3 \times(-2)=-2$
(e) Oxidation number can be zero, $+\mathrm{ve},-\mathrm{ve}$ (integer or fraction)
(f) Maximum oxidation no. of an element is $=$ Group no. (Except $\mathrm{O}$ and $\mathrm{F}$ )
Minimum oxidation no. of an element is = Group no. -8 (Except metals)
Redox reactions involve oxidation and reduction both. Oxidation means loss of electrons and reduction means gain of electrons. Thus redox reactions involve electron transfer and the number of electrons lost are same as the number of electrons gained during the reaction. This aspect of redox reaction can serve as the basis of a pattern for balancing redox reactions.
Oxidation number of $\mathbf{M n}$ in $\mathbf{K M n O}_{4}$ :
Let the oxidation number of $\mathrm{Mn}$ be $\mathrm{x}$. Now we know that the oxidation numbers of $\mathrm{K}$ is +1 and that of $\mathrm{O}$ is -2 .
Now to the sum of oxidation numbers of all atoms in the formula of the compound must be zero, i.e. $+1+x-8=0$. Hence, the oxidation number of $\mathrm{Mn}$ in $\mathrm{KMnO}_{4}$ is +7 .
$$ \begin{aligned} & 2 x=+14-2=+12 \ & x=+\frac{12}{2}=+6 \text { Hence, oxidation number of } \mathrm{Cr} \text { in is }+6 . \end{aligned} $$
Balancing of Redox Reactions by Oxidation State Method
This method is based on the fact that the number of electrons gained during reduction must be equal to the number of electrons lost during oxidation. Following steps must be followed while balancing redox equations by this method.
(a) Write the skeleton equation (if not given, frame it) representing the chemical change.
(b) With the help of oxidation number of elements, find out which atom is undergoing oxidation/reduction, and white separate equations for the atom undergoing oxidation/reduction.
(c) Add the respective electrons on the right for oxidation and on the left for reduction equation. Note that the net charge on the left and right side should be equal.
(d) Multiply the oxidation and reduction reactions by suitable integers so that total electrons lost in one reaction is equal to the total electrons gained by other reaction.
(e) Transfer the coefficients of the oxidizing and reducing agents and their products as determined in the above step to the concerned molecule or ion.
(f) By inspection, supply the proper coefficient for the other formulae of substances not undergoing oxidation and reduction to balance the equation.
Example:
$\mathrm{Cr}{2} \mathrm{O}{7}^{2-}+\mathrm{I}^{-}+\mathrm{H}^{+} \longrightarrow \mathrm{Cr}^{3+}+\mathrm{I}_{2}$
Solution: (i) Find the oxidation state of atoms undergoing redox change
$$ \stackrel{+6 \times 2}{\mathrm{Cr}{2}} \mathrm{O}{7}^{2-}+\mathrm{I}^{-1} \longrightarrow \stackrel{+3}{\mathrm{Cr}^{3+}}+\stackrel{0}{\mathrm{I}_{2}} $$
(ii) Balance the number of atoms undergoing redox change.
$$ \stackrel{(+6) \times 2}{\mathrm{Cr}{2} \mathrm{O}{7}^{2-}}+\stackrel{2 \times(-1)}{2 \mathrm{I}^{-}} \longrightarrow \stackrel{(+3) \times 2}{2 \mathrm{Cr}^{3+}}+\stackrel{0 \times 2}{\mathrm{I}_{2}} $$
(iii) Find the change in oxidation state and balance the change in oxidation states by multiplying the species with a suitable integer.
$$ \begin{aligned} & \stackrel{+12}{\mathrm{Cr}{2}} \mathrm{O}{7}^{2-}+2 \mathrm{I}^{-2} \longrightarrow 2 \mathrm{Cr}^{+6}+\mathrm{I}_{2} \ & \text { Change in Change in } \ & \text { ox. state }=6 \quad \text { ox. state }=2 \times 3 \end{aligned} $$
As the decrease in oxidation state if chromium is 6 and increase in oxidation state of iodine is 2, so we will have to multiply $\mathrm{I}^{-} / \mathrm{I}_{2}$ by 3 equalize the changes in oxidation state.
$$ \mathrm{Cr}{2} \mathrm{O}{7}^{2-}+6 \mathrm{I}^{-} \longrightarrow 2 \mathrm{Cr}^{3+}+3 \mathrm{I}_{2} $$
(iv) Find the total charges on both the sides and also find the difference of charges.
$$ \begin{aligned} & \text { Charge on LHS }=-2+6 \times(-1)=-8 \ & \text { Charge on RHS }=2 \times(+3)=+6 \ & \text { Difference in charge }=+6-(-8)=14 \end{aligned} $$
(v) Now, as the reaction is taking place in acidic medium, we will have to add the ions, to $\mathrm{H}^{+}$ the side falling short in positive charges, so we will add $14 \mathrm{H}^{+}$ and $\mathrm{LH}$ s to equalize the charges on both sides.
$$ \mathrm{Cr}{2} \mathrm{O}{7}^{2-}+6 \mathrm{I}^{-}+14 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Cr}^{3+}+3 \mathrm{I}_{2} $$
(vi) To equalize the $\mathrm{H}$ and $\mathrm{O}$ atoms, add $7 \mathrm{H}_{2} \mathrm{O}$ on $\mathrm{RHS}$
$$ \mathrm{Cr}{2} \mathrm{O}{7}^{2-}+6 \mathrm{I}^{-}+14 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Cr}^{3+}+3 \mathrm{I}{2}+7 \mathrm{H}{2} \mathrm{O} $$
$$ \begin{aligned} & \stackrel{+5}{\mathrm{IO}{3}^{-}} \longrightarrow \stackrel{+7}{\mathrm{I}} \mathrm{O}{4}^{-} \end{aligned} $$
TYPES OF REACTIONS
The redox reactions are of the following types:
(a) Combination reactions: A compound is formed by chemical combination of two or more elements. The combination of an element or compound with oxygen is called combustion. The combustion and several other combinations which involve change in oxidation state are called redox reactions.
$$ \begin{aligned} & \text { e.g., } \stackrel{-4}{C}^{+1}+2 \mathrm{O}{2} \longrightarrow \stackrel{+4}{-2} \mathrm{CO}{2}+2 \mathrm{H}{2}^{+1} \mathrm{O}^{-2} \ & \stackrel{0}{\mathrm{C}}(\mathrm{s})+\stackrel{0}{\mathrm{O}}{2}(\mathrm{~g}) \longrightarrow \stackrel{+4}{\mathrm{CO}{2}^{-2}}(\mathrm{~g}) \ & 3 \stackrel{0}{\mathrm{Mg}}+\stackrel{0}{\mathrm{~N}{2}} \longrightarrow \stackrel{+2}{\mathrm{Mg}}{3} \stackrel{-3}{\mathrm{~N}}{2} \ & \stackrel{0}{\mathrm{H}{2}}+\stackrel{0}{\mathrm{C}} \mathrm{I}{2} \longrightarrow \stackrel{+1}{2 \mathrm{HCl}} \end{aligned} $$
(b) Decomposition reactions: Decomposition is the reverse process of combination, it involves the breakdown of the compound into two or more components. The product of decomposition must contain at least one component in elemental state.
In above example, there is no change in oxidation state of potassium. Thus, it should be noted that the decomposition does not result into change in the oxidation number of each element.
(c) Displacement reactions: The reactions in which an atom or ion in a compound is displaced by another atom or ion are called displacement reactions. The displacement reactions are of 2 types:
(i) Metal displacement: In these reactions, a metal in a compound is replace by another metal in an uncombined state. It is found that a metal with stronger reducing character can displace the other metal having a weaker reducing character.
$$ \begin{aligned} \text { e.g., } \quad & \stackrel{+3}{\mathrm{Cr}{2}} \mathrm{O}{3}^{-2}+2 \stackrel{0}{\mathrm{Al}}(\mathrm{s}) \longrightarrow \mathrm{Al}{2} \mathrm{O}{3}^{-2}(\mathrm{~s})+2 \stackrel{0}{\mathrm{Cr}}(\mathrm{s}) \ & \stackrel{+2+6}{\mathrm{CuSO}} \mathrm{O}{4}^{-2}+\stackrel{0}{\mathrm{Zn}}(\mathrm{s}) \longrightarrow \mathrm{ZnSO}{4}^{+6}(\mathrm{aq})+\stackrel{0}{\mathrm{Cu}(\mathrm{s})} \end{aligned} $$
(ii) Non-metal displacement: These displacement reactions generally involve redox reactions, where the hydrogen is displaced. Alkali and alkaline earth metals are highly electropositive, they displace hydrogen from cold water.
$$ \begin{aligned} & 2 \stackrel{0}{\mathrm{Na}}(\mathrm{s})+2 \stackrel{+1}{2}^{-1} \mathrm{O}(l) \longrightarrow 2 \stackrel{+1}{\mathrm{Na}} \stackrel{-2}{+1} \mathrm{H}(\mathrm{aq})+\stackrel{0}{\mathrm{H}}{2}(\mathrm{~g}) \ & \stackrel{0}{\mathrm{Ca}}(\mathrm{s})+2 \stackrel{+1}{\mathrm{H}}{2} \stackrel{-2}{\mathrm{O}}(\mathrm{l}) \longrightarrow \stackrel{+1}{\mathrm{Ca}}\left(\mathrm{O}^{-2+1}\right){2}(\mathrm{aq})+\stackrel{0}{\mathrm{H}{2}}(\mathrm{~g}) \end{aligned} $$
(d) Disproportionation and Oxidation-Reduction: One and the same substance may act simultaneously as an oxidizing agent with the result that a part of it gets oxidized to a higher state and rest of it is reduced to lower state of oxidation. Such a reaction, in which a substance undergoes simultaneous oxidation and reduction is called disproportionation and the substance is said to disproportionate.
The following are some of the examples of disproportionation:
(a)
(b)
(e) Oxidation state of chlorine lies between -1 to +7 ; thus out of $\mathrm{ClO}^{-}, \mathrm{ClO}{2}^{-}, \mathrm{ClO}{3}, \mathrm{ClO}{4}^{-} ; \mathrm{ClO}{4}^{-}$ does not undergo disproportionation because in this oxidation state of chlorine is highest, i.e., +7. Disproportionation of the other oxoanions are:
$$ \begin{aligned} & 3 \stackrel{+1}{\mathrm{ClO}^{-}} \longrightarrow 2 \stackrel{-1}{2 \mathrm{Cl}}+\stackrel{+5}{\mathrm{ClO}{3}^{-}} \ & 6 \mathrm{ClO}{2}^{-} \longrightarrow 4 \stackrel{+5}{\mathrm{ClO}{3}^{-}}+2 \mathrm{Cl}^{-} ; \quad 4 \mathrm{ClO}{3}^{-} \longrightarrow \mathrm{Cl}^{-}+3 \stackrel{+7}{\mathrm{ClO}_{4}^{-}} \end{aligned} $$
