Chapter 10

Redox Reactions

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Two methods are used for balancing chemical equations for redox reactions. The first method is based on the change in the oxidation number of the reductant and the oxidant and the second method is based on splitting the redox reaction into two half reactions one involving oxidation and another involving reduction. Both methods work very well .

a) Oxidation Number Method

In writing equations for oxidation-reduction reactions , the composition and formulas must be known for substances that react and for the products that are formed. The oxidation number method is best illustrated in the following steps :
Step 1
Write the correct formula for each reactant and product.
Step 2
Identify atoms which undergo change in oxidation number in the reaction by assiggning the oxidation number to all elements in the reaction.
Step 3
Calculate the increase or decrease in the oxidation number per atom and for the entire molecule/ion in which it occurs. If these are not equal then multiply by suitable coefficients so that these become equal.
Step 4
Ascertain the involvement of the ions if the reaction is taking place in water , add H+ or OH to the expression on the appropriate side so that the total ionic charges of reactants and products are equal. If the reaction is carried out in acidic solution , use H+ ions in the equation ; if in basic solution, use OH ions.
Step 5
Make the numbers of hydrogen atoms in the expression on two sides equal by adding water(H2O) molecules to thge reactants or products. Now , also check the number of oxygen atoms. If there are the same number of oxygen atoms in the reactants and products , the equation then represents the balanced redox reaction.

b) Half Reaction Method

In this method , the two half equations are balanced separately and then added together to give balanced equation.
Suppose we are to balance the equation showing the oxidation of Fe2+ ions by dichromate ions(Cr2O7)2− in acidic medium , wherein, Cr2O72− ions are reduced to Cr3+ ions. The following steps are involved :
Step 1
Produce the unbalanced equation for the reaction in ionic form :
                              Fe2+(aq) + Cr2O72−(aq) Fe3+(aq) + Cr3+(aq)
Step 2
Separate the equation into two half reactions :
Step 3
Balance the atoms other than O and H in each half reaction individually. Here the oxidation half reaction is already balanced with respect to Fe atoms. For the reduction half reaction , we multiply the Cr3+ by 2 to balance Cr atoms.
                              Cr2O72−(aq) 2 Cr3+(aq)
Step 4
For reactions occurring in acidic medium , add H2O to balance O atoms and H+ to balance H atoms.
Thus we get ,
                              Cr2O72−(aq) + 14 H+ → 2 Cr3+(aq) + 7 H2O
Step 5
Add electrons to one side of the half reaction to balance the charges. If needed , make the number of electrons equal in two half reactions by multiplying one or both half reactions by appropriate coefficients.
The oxidation half reactions is thus written to balance the charge:
                              Fe2+(aq) Fe3+(aq) + e
Now in the reduction half reaction there are net twelve positive charges on the left side and only six positive charges on the right hand side.
Therefore , we add six electrons on the left side.
                              Cr2O72−(aq) + 14 H+ + 6 e → 2 Cr3+(aq) + 7 H2O
To equalise the number of electrons in both half reactions , we multiply the oxidation half reaction by 6 and write as :
                              6 Fe2+(aq) 6 Fe3+(aq) + 6 e
Step 6
We add the two half reactions to achieve the overall reaction and cancel the electrons on each side. This gives the net ionic equation as :
                              6 Fe2+(aq) + Cr2O72−(aq) + 14 H+6 Fe3+(aq) + 2 Cr3+(aq) + 7 H2O
Step 7
Verify that the equation contains the same type and number of atoms and the same charges on both sides of the equation. The last check reveals that the equation is fully balanced with respect to number of atoms and the charges.
For the reaction in basic medium , first balance the atoms as is done in acidic medium. Then for each H+ ion , add an equal number of OH ions to both sides of the equation. Where H+ and OH appear on the same side of the equationb , combine these to give H2O.


In general , in a chemical reaction described by the balanced chemical equation :
                              a A + b B + …. c C + d D + …..
A, B, …. are reactants and C, D, ….. are the products. The coefficients a, b, c and d are known as stoichiometric coefficients. These indicate the number of moles of the reactants used and the number of moles of the products obtained in the reaction.
               Volumetric titrations based on redox reactions are carried out in laboratories. If we know the molarities of the reductant and oxidant and the volumes used in titration and n1 and n2 are their stoichiometric coefficients, then it is possible to use the equation :
for volumetric estimation.
For example, in the titration of FeSO4 with KMnO4 , the balanced equation is :
          10 FeSO4 + 2 KMnO4 + 8 H2SO4 5 Fe2(SO4)3 + K2SO4+ 2 MnSO4+ 8 H2O
If FeSO4 is the reagent ‘1' and KMnO4 is reagent ‘2' then this equation becomes :
and can be used to calculate the unknown quantity , if the other quantities are known.

Redox Reactions as Basis for Titrations

In redox systems , the titration method can be adopted to determine the strength of a reductant/oxidant using a redox sensitive indicator. The usage of indicators in redox titration is illustrated below :

  1. In one situation, the reagent itself is intensely coloured, e.g., permanganate ion, MnO4 . Here MnO4 acts as the self indicator. The visible end point in this case is achieved after the last of the reductant (Fe2+ or C2 O42 ) is oxidized and the first lasting tinge of pink colour appears at MnO4 concentration as low as 106 mol dm3(106 mol L1). This ensures a minimal ‘overshoot' in colour beyond the equivalence point, the point where the reductant and oxidant are equal in terms of their mol stoichiometry.
  2. If there is no dramatic auto-colour change ( as with MnO4 titration) , there are indicators which are oxidized immediately after the last bit of the reactant is consumed, producing a dramatic colour change. The best example is afforded by Cr2O72 , which is not a self indicator, but oxidizes the indicator substance diphenylamine just after the equivalence point to produce an intense blue colour, thus signaling the end point.
  3. There is another method which is used to those reagents which are able to oxidize I ions, for example , Cu( II ) :
                                  2 Cu 2+ (aq) + 4 I Cu 2 I 2 (aq) + I2 (aq)
    This method relies on the fact that iodine itself gives an intense blue colour with starch and has a very specific reaction with thiosulphate ions
    (S2O32), which too is a redox reaction.
                                  I2 (aq) + 2 S2 O32 (aq) 2 I (aq) + S4O62(aq)
    I2 though insoluble in water , remains in solution containing KIas KI3 .
    On addition of starch after liberation of iodine from the reaction of Cu2+ ions on iodide ions , an intense blue colour appears. This colour disappears as soon as the iodine is consumed by the thiosulphate ions. Thus the end point can be easily be tracked and the rest is the stiochiometric calculation only.

Limitations of Concept of Oxidation Number

According to the concept of oxidation number , oxidation means increase in oxidation number by the loss of electrons while reduction implies decrease in oxidation number by gain of electrons. However, actually during oxidation , there is a decrease in electron density around the atom (or atoms) undergoing oxidation. At the same time, there is an increase in electron density around atom(or atoms) undergoing reduction. This may be regarded as the limitations of the concept.


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