Chemistry Balancing of redox reactions, Redox reactions as the basis for titrations, Limitation of concept of oxidation number

### Topics to be covered

=> Balancing of redox reactions
=> Redox reactions as the basis for titrations
=> Limitation of concept of oxidation number

### Balancing of Redox Reactions

=> Two methods are used to balance chemical equations for redox processes.
•One of these methods is based on the change in the oxidation number of reducing agent and the oxidising agent
•The other method is based on splitting the redox reaction into two half reactions — one involving oxidation and the other involving reduction.

Both these methods are in use and the choice of their use rests with the individual using them.

(a) color{green}("Oxidation Number Method:") In writing equations for oxidation-reduction reactions, just as for other reactions, the compositions and formulas must be known for the substances that react and for the products that are formed. The oxidation number method is now 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 assigning 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 number so that these become equal. (If you realise that two substances are reduced and nothing is oxidised or vice-versa, something is wrong. Either the formulas of reactants or products are wrong or the oxidation numbers have not been assigned properly).

Step 4: Ascertain the involvement of ions if the reaction is taking place in water, add color{red}(H^+) or color{red}(OH^–) ions 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 color{red}(H^+) ions in the equation; if in basic solution, use color{red}(OH^–) ions.

Step 5 : Make the numbers of hydrogen atoms in the expression on the two sides equal by adding water color{red}(H_2O) molecules to the 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.

color{green}("Let us now explain the steps involved in the method with the help of a few problems given below:")

(b) color{green}("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 color{red}(Fe^(2+)) ions to color{red}(Fe^(3+)) ions by dichromate ions color{red}((Cr_2O_7)^(2–)) in acidic medium, wherein, color{red}(Cr_2O_7^(2–)) ions are reduced to color{red}(Cr^(3+)) ions. The following steps are involved in this task.

Step 1:color{green}("Produce unbalanced equation for the reaction in ionic form :")

color{red}(Fe^(2+)(aq) + Cr_2O_7^(2–) (aq) → Fe^(3+) (aq) + Cr^(3+)(aq)) ............(8.50)

Step 2: color{green}("Separate the equation into halfreactions :")

color{green}("Oxidation half :") color{red}(overset(+2)(Fe^(2+))(aq) → overset(+3)(Fe^(3+))(aq)) .................(8.51)

color{green}("Reduction half :") color{red}(overset(+6)(Cr_2 ) overset(-2)(O_7^(2-)) (aq) → overset(+3)(Cr^(3+)) (aq))  .............(8.52)

Step 3: Balance the atoms other than color{red}(O) and color{red}(H) in each half reaction individually. Here the oxidation half reaction is already balanced with respect to color{red}(Fe) atoms. For the reduction half reaction, we multiply the color{red}(Cr^(3+)) by 2 to balance color{red}(Cr) atoms.

color{red}(Cr_2O_7^(2-)(aq) → 2Cr^(3+)(aq)) .......................(8.53)

Step 4: For reactions occurring in acidic medium, add color{red}(H_2O) to balance color{red}(O) atoms and color{red}(H^+) to balance color{red}(H) atoms.

color{green}("Thus, we get :")

color{red}(Cr_2O_7^(2-)(aq) +14H^(+)(aq) → 2Cr^(3+)(aq) +7H_2O(l)) .................(8.54)

Step 5: Add electrons to one side of the half reaction to balance the charges. If need be, make the number of electrons equal in the two half reactions by multiplying one or both half reactions by appropriate number.

color{green}("The oxidation half reaction is thus rewritten to balance the charge:")

color{red}(Fe^(2+)(aq) → Fe^(3+)(aq) + e^(-)) ...........................(8.55)

Now in the reduction half reaction there are net twelve positive charges on the left hand side and only six positive charges on the right hand side. Therefore, we add six electrons on the left side

color{red}(Cr_2O_7^(2-)(aq) +14H^(+)(aq) +6e^(-) → 2Cr^(3+)(aq) + 7H_2O(l))  ....................(8.56)

color{green}("To equalise the number of electrons in both the half reactions, we multiply the oxidation half reaction by 6 and write as :")

color{red}(6Fe^(2+) (aq) → 6Fe^(3+)(aq) +6e^(-)) ...........................(8.57)

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 :

color{red}(6Fe^(2+) (aq) +Cr_2O_7^(2-)(aq) +14H^(+) (aq) → 6Fe^(3+)(aq) +2Cr^(3+)(aq) +7H_2O(l))  ...............(8.58)

Step 7: Verify that the equation contains the same type and number of atoms and the same charges on both sides of the equation. This last check reveals that the equation is fully balanced with respect to number of atoms and the charges.

For the reaction in a basic medium, first balance the atoms as is done in acidic medium. Then for each color{red}(H^+) ion, add an equal number of color{red}(OH^–) ions to both sides of the equation. Where color{red}(H^+) and color{red}(OH^–) appear on the same side of the equation, combine these to give color{red}(H_2O).

Q 3111080820

Write the net ionic equation for the reaction of potassium dichromate(VI), K_2Cr_2O_7 with sodium sulphite, Na_2SO_3, in an acid solution to give chromium(III) ion and the sulphate ion.

Solution:

Step 1: The skeletal ionic equation is:

Cr_2O_7^(2-)(aq) +SO_3^(2-) (aq) → Cr^(3+) (aq) +SO_4^(2-)(aq)

Step 2: Assign oxidation numbers for Cr and S

overset(+6 )(Cr_2) overset(-2)(O_7^(2-))(aq) + overset(+4)(S) overset(-2)(O_3^(2-))(aq) → overset(+3)(C)r(aq) + overset(+6)(S) overset(-2)(O_4^(2-))(aq)

This indicates that the dichromate ion is the oxidant and the sulphite ion is the reductant.

Step 3: Calculate the increase and decrease of oxidation number, and make them equal:

overset(+6)(Cr_2) overset(-2)(O_7^(2-))(aq) + overset(+4)(3S) overset(-2)(O_3^(2-))(aq) → overset(+3)(2Cr^(3+)) (aq) + overset(+6)(3S) overset(-2)(O_4^(2-)) (aq)

Step 4: As the reaction occurs in the acidic medium, and further the ionic charges are not equal on both the sides, add 8H^+ on the left to make ionic charges equal

Cr_2O_7^(2-) (aq) +3SO_3^(2-) (aq) +8H^(+) → 2Cr^(3+) (aq) +3SO_4^(2-) (aq)

Step 5: Finally, count the hydrogen atoms, and add appropriate number of water molecules (i.e., 4H_2O) on the right to achieve balanced redox change.

Cr_2O_7^(2-) (aq) +3SO_3^(2-) (aq) +8H^(+) (aq) → 2Cr^(3+) (aq) +3SO_4^(2-) (aq) +4H_2O (l)
Q 3121080821

Permanganate ion reacts with bromide ion in basic medium to give manganese dioxide and bromate ion. Write the balanced ionic equation for the reaction.

Solution:

Step 1 : The skeletal ionic equation is : MnO_4^(–)(aq) + Br^–(aq) → MnO_2(s) + BrO_3^– (aq)

Step 2 : Assign oxidation numbers for Mn and Br

overset(+7)(Mn)O_4^(-) (aq) + overset(-1)(Br^-) (aq) → overset(+4)(Mn)O_2 (s) + overset(+5)(Br)O_3^(-) (aq) this indicates that permanganate ion is the oxidant and bromide ion is the reductant.

Step 3: Calculate the increase and decrease of oxidation number, and make the increase equal to the decrease.

overset(+7)(2Mn)O_4^(-) (aq) + overset(-1)(Br^(-)) (aq) → overset(+4)(2Mn)O_2(s) + overset(+5)(Br)O_3^(-)(aq)

Step 4: As the reaction occurs in the basic medium, and the ionic charges are not equal on both sides, add 2 OH^– ions on the right to make ionic charges equal.

2MnO_4^(-)(aq) + Br^(-) (aq) → 2MnO_2 (s) +Br_3^(-) (aq) +2OH^(-) (aq)

Step 5: Finally, count the hydrogen atoms and add appropriate number of water molecules (i.e. one H_2O molecule)
on the left side to achieve balanced redox change.

2MnO_4^(-) (aq) + Br^(-) (aq) +H_2O(l) → 2MnO_2 (s) + BrO_3^(-) (aq) +2 OH^(-) (aq)
Q 3131080822

Permanganate(VII) ion, MnO_4^(–) in basic solution oxidises iodide ion, I^– to produce molecular iodine (I_2) and manganese (IV)
oxide (MnO_2). Write a balanced ionic equation to represent this redox reaction.

Solution:

Step 1: First we write the skeletal ionic equation, which is
MnO_4^(–) (aq) + I^– (aq) → MnO_2(s) + I_2(s)

Step 2: The two half-reactions are:

Oxidation half : overset(-1)(I^(-))(aq) → overset(0)(I_2) (s)

Reduction half: overset(+7)(Mn)O_4^(-) (aq) → overset(+4)(Mn)O_2(s)

Step 3: To balance the I atoms in the oxidation half reaction, we rewrite it as:

2I^(-) (aq) → I_2 (s)

Step 4: To balance the O atoms in the reduction half reaction, we add two water molecules on the right:

MnO_4^(-) (aq) → MnO_2(s) +2H_2O (l)

To balance the H atoms, we add four H^+ ions on the left:

MnO_4^(-) (aq) +4H^(+) (aq) → MnO_2(s) +2H_2O (l)

As the reaction takes place in a basic solution, therefore, for four H^+ ions, we add four OH^– ions to both sides of the
equation:

MnO_4^(-) (aq) +4H^(+) +4OH^(-) (aq) → MnO_2(s) +2H_2O(l) +4OH^(-) (aq)

Replacing the H^+ and OH^– ions with water, the resultant equation is:
MnO_4^– (aq) + 2H_2O (l) → MnO_2 (s) + 4 OH^– (aq)

Step 5 : In this step we balance the charges of the two half-reactions in the manner depicted as:

2I^(-) (aq) → I_2 (s) +2e^(-)

MnO_4^(-) (aq) +2H_2O(l) +3e^(-) → MnO_2(s) +4OH^(-) (aq)

Now to equalise the number of electrons, we multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2.

6I^(-) (aq) → 3I_2 (s) +6e^(-)

2MnO_4^(-) (aq) +4H_2O(l) +6e^(-) → 2MnO_2 (s) +8 OH^(-) (aq)

Step 6: Add two half-reactions to obtain the net reactions after cancelling electrons on both sides.

6I^(-) (aq) +2MnO_4^(-)(aq) +4H_2O(l) → 3I_2 (s) +2MnO_2(s) +8 OH^(-) (aq)

Step 7: A final verification shows that the equation is balanced in respect of the number of atoms and charges on both sides.

### Redox Reactions as the Basis for Titrations

=> In acid-base systems we come across with a titration method for finding out the strength of one solution against the other using a pH sensitive indicator.

=> Similarly, in redox systems, the titration method can be adopted to determine the strength of a reductant/oxidant using a redox sensitive indicator.

=> color{red}("The usage of indicators in redox titration is illustrated below:")

(i) In one situation, the reagent itself is intensely coloured, e.g., permanganate ion, color{red}(MnO_4^(–)) . Here color{red}(MnO_4^(–)) acts as the self indicator. The visible end point in this case is achieved after the last of the reductant (color{red}(Fe^(2+)) or color{red}(C_2O_4^(2–))) is oxidised and the first lasting tinge of pink colour appears at color{red}(MnO_4^(–)) concentration as low as color{red}(10^(–6) mol dm^(–3)) (color{red}(10^(–6) mol L^(–1))). This ensures a minimal ‘overshoot’ in colour beyond the equivalence point, the point where the reductant and the oxidant are equal in terms of their mole stoichiometry.

(ii) If there is no dramatic auto-colour change (as with color{red}(MnO_4 ^(–)) titration), there are indicators which are oxidised immediately af ter the last bit of the reactant is consumed, producing a dramatic colour change. The best example is afforded by color{red}(Cr_2O_7^(2–)), which is not a self-indicator, but oxidises the indicator substance diphenylamine just after the equivalence point to produce an intense blue colour, thus signalling the end point.

(iii) There is yet another method which is interesting and quite common. Its use is restricted to those reagents which are able to oxidise I– ions, say, for example, color{red}(Cu(II)):

color{red}(2Cu^(2+)(aq) +4I^(-)(aq) → Cu_2I_2(s) +I_2(aq)) ...................................................(8.59)

This method relies on the facts that iodine itself gives an intense blue colour with starch and has a very specific reaction with thiosulphate ions (color{red}(S_2O_3^(2–))), which too is a redox reaction.

color{red}(I_2(aq) +2S_2O_3^(2-)(aq) → 2I^(-) (aq) +S_4O_6^(2-)(aq)) ..................... (8.60)

color{red}(I_2), though insoluble in water, remains in solution containing color{red}(KI) as color{red}(KI_3).

On addition of starch after the liberation of iodine from the reaction of color{red}(Cu^(2+)) 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 easily be tracked and the rest is the stoichiometric calculation only.

### Limitations of Concept of Oxidation Number

As you have observed in the above discussion, the concept of redox processes has been evolving with time. This process of evolution is continuing. In fact, in recent past the oxidation process is visualised as a decrease in electron density and reduction process as an increase in electron density around the atom(s) involved in the reaction.