Redox reaction : Balancing of Redox-Reaction

Balancing of oxidation-reduction reactions

Though there are a number of methods for balancing oxidation – reduction reactions, two methods are very important. These are, (1) Oxidation number method,  (2)  Ion – electron method

(1)     Oxidation number method : The method for balancing redox reactions by oxidation number change method was developed by Johnson. In a balanced redox reaction, total increase in oxidation number must be equal to the total decrease in oxidation number. This equivalence provides the basis for balancing redox reactions. This method is applicable to both molecular and ionic equations. The general procedure involves the following steps,

(i)      Write the skeleton equation (if not given, frame it) representing the chemical change.

(ii)     Assign oxidation numbers to the atoms in the equation and find out which atoms are undergoing oxidation and reduction. Write separate equations for the atoms undergoing oxidation and reduction.

(iii)    Find the change in oxidation number in each equation. Make the change equal in both the equations by multiplying with suitable integers. Add both the equations.

(iv)    Complete the balancing by inspection. First balance those substances which have undergone change in oxidation number and then other atoms except hydrogen and oxygen. Finally balance hydrogen and oxygen by putting H2O molecules wherever needed.

The final balanced equation should be checked to ensure that there are as many atoms of each element on the right as there are on the left.

(v)     In ionic equations the net charges on both sides of the equation must be exactly the same. Use H+ ion/ions in acidic reactions and OH ion/ions in basic reactions to balance the charge and number of hydrogen and oxygen atoms.

                    The following example illustrate the above rules,

Step : I      Cu + HNO3 → Cu(NO3)2 + NO2 + H2O (Skeleton equation)

Step: II     Writing the oxidation number of all the atoms. 

\[\overset{0}{\mathop{Cu}}\,\,\,+\,\overset{+1}{\mathop{H}}\,\overset{+5}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{3}}\,\,\to \,\,\overset{+2}{\mathop{Cu}}\,{{(\overset{+5}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{3}})}_{2}}+\overset{+4}{\mathop{N}}\,{{\overset{-2}{\mathop{O}}\,}_{2}}\,\,+\,\,{{\overset{+1}{\mathop{H}}\,}_{2}}\overset{-2}{\mathop{O}}\,     

Step: III    Change in oxidation number has occurred in copper and nitrogen.

\[\overset{0}{\mathop{Cu}}\,\to \overset{+2}{\mathop{Cu}}\,{{(N{{O}_{3}})}_{2}}                 ……(i)

\[H\overset{+5}{\mathop{N}}\,{{O}_{3}}\to \overset{+4}{\mathop{N}}\,{{O}_{2}}               ……(ii)

Increase in oxidation number of copper = 2 units per molecule Cu

Decrease in oxidation number of nitrogen = 1 unit per molecule HNO3

Step: IV    To make increase and decrease equal, equation (ii) is multiplied by 2.

                   Cu + 2HNO3 → Cu(NO3)2 + 2NO2 + H2

Step: V      Balancing nitrate ions, hydrogen and oxygen, the following equation is obtained.

                   Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + 2H2                         

This is the balanced equation.

(2)     Ion-electron method (half reaction method)

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 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 H and O for each half reaction using simple multiples.

(b)     Add water molecules to the side deficient in oxygen and 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 2OH ions to the other side. If hydrogen is still unbalanced, add one 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 number 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 example illustrate the above rules

Step: I       I2 + OH → IO3 + I + H2O (Ionic equation)

Step: II     Splitting into two half reactions,

I2 + OH → IO3 + H2O ; I2 → I

                   (Oxidation half reaction)   (Reduction half reaction)

Step: III    Adding OH ions, \[{{I}_{2}}+12O{{H}^{-}}\to 2IO_{3}^{-}+6{{H}_{2}}O

Step: IV    Adding electrons to the sides deficient in electrons,

\[{{I}_{2}}+12O{{H}^{-}}\to 2IO_{3}^{-}+6{{H}_{2}}O+10{{e}^{-}}\,\,;\,\,\,\,{{I}_{2}}+2{{e}^{-}}\to 2{{I}^{-}}

Step: V      Balancing electrons in both the half reactions.

\[{{I}_{2}}+12O{{H}^{-}}\to 2IO_{3}^{-}+6{{H}_{2}}O+10{{e}^{-}}\,\,;\,\,5\,[{{I}_{2}}+2{{e}^{-}}\to 2{{I}^{-}}]

Step: VI    Adding both the half reactions.

\[6{{I}_{2}}+12O{{H}^{-}}\to 2IO_{3}^{-}+6{{H}_{2}}O+10{{I}^{-}}

Dividing by 2,  \[3{{I}_{2}}+6O{{H}^{-}}\to IO_{3}^{-}+5{{I}^{-}}+3{{H}_{2}}O

 

Autoxidation

(1)     Turpentine and numerous other olefinic compounds, phosphorus and certain metals like Zn and Pb can absorb oxygen from the air in presence of water. The water is oxidised to hydrogen peroxide. This phenomenon of formation of H2O2 by the oxidation of H2O is known as autoxidation. The substance such as turpentine or phosphorus or lead which can activate the oxygen is called activator. The activator is supposed to first combine with oxygen to form an addition compound, which acts as an autoxidator and reacts with water or some other acceptor so as to oxidise the latter. For example;

\[\underset{(\text{activator})}{\mathop{Pb}}\,+{{O}_{2}}\to \,\,\underset{\text{(autoxidator)}}{\mathop{Pb{{O}_{2}}}}\,         \[Pb{{O}_{2}}+\underset{(\text{acceptor})}{\mathop{{{H}_{2}}O}}\,\to PbO+{{H}_{2}}{{O}_{2}}                             

(2)     The turpentine or other unsaturated compounds which act as activators are supposed to take up oxygen molecule at the double bond position to form unstable peroxide called moloxide, which then gives up the oxygen to water molecule or any other acceptor.

2KI + H2O2 → 2KOH + I2

The evolution of iodine from KI solution in presence of turpentine can be confirmed with starch solution which turns blue.

(3)     The concept of autoxidation help to explain the phenomenon of induced oxidation. Na2SO3 solution is oxidised by air but Na3AsO3 solution is not oxidised by air. If a mixture of both is taken, it is observed both are oxidised. This is induced oxidation.

Na2SO3 + O2 → Na2SO5

Moloxide

Na2SO5 + Na3AsO3 → Na3AsO4 + Na2SO4

Na2SO3 + Na3AsO3 + O2 → Na2SO4 + Na3AsO4   

 

Disproportionation

One and the same substance may act simultaneously as an oxidising agent and as a reducing agent with the result that a part of it gets oxidised 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.

Following are the some examples of disproportionation,

 

(1)           

(2)

(3)    

(4)    

 

Important applications of redox-reactions

Many applications are based on redox reactions which are occuring in environment. Some important examples are listed below;

(1)     Many metal oxides are reduced to metals by using suitable reducing agents. For example Al2O3 is reduced to aluminium by cathodic reduction in electrolytic cell. Fe2O3 is reduced to iron in a blast furnace using coke.

(2)     Photosynthesis is used to convert carbon dioxide and water by chlorophyll of green plants in the presence of sunlight to carbohydrates.

\[6C{{O}_{2}}(g)+6{{H}_{2}}O(l)\xrightarrow[Sunlight]{Chlorophyll}C{{ & }_{6}}{{H}_{12}}{{O}_{6}}(aq)+6{{O}_{2}}(g)         

In this case, CO2 is reduced to carbohydrates and water is oxidised to oxygen. The light provides the energy required for the reaction.

(3)     Oxidation of fuels is an important source of energy which satisfies our daily need of life.

Fuels + O2 → CO2 + H2O + Energy

In living cells, glucose (C6H12O6) is oxidised to CO2 and H2O in the presence of oxygen and energy is released, 

C6H12O6 (aq) + 6O2(g) → 6CO2(g) + 6H2O(l) + Energy

(4)     The electrochemical cells involving reaction between hydrogen and oxygen using hydrogen and oxygen electrodes in fuel cells meet our demand of electrical energy in space capsule.

(5)     Respiration in animals and humans is also an important application of redox reactions.