LITERATUR ELKTROKIMIA

            Electrochemistry is the study of the relationships that exist between chemical reactions and the flow of electricity. Included here are electrolysis reactions, in which non-spontaneous changes are forced to occur by the passage of electricity through chemical systems. Also included are spontaneous oxidation-reduction reactions (redox reactions) that are able to supply electricity.

            Electrochemical reactions are of great practical importance in the study of chemistry and also in our day-to-day activities. In the chemistry lab, they allow us to gain information about the energy changes involved in chemical reactions and help in the analysis of chemical systems. This has proven to be so useful that other sciences (biology, for example) have adapted electrochemical techniques to the study of systems of particular interest to them.

1. Concept of Redox reaction

Oxidation is the reactions that gain of oxygen, loss of hydrogen, loss of electrons, or undergo increase in oxidation number by a substance. Reduction is the reactions that loss of oxygen, gain of hydrogen, gains of electrons, or undergoes decrease in oxidation number by a substance. Consider the oxidation of magnesium:

2 Mg(s) + O₂(g) à 2 MgO(s)

The product MgO is ionic and contains the ions Mg²⁺ and O^2-, which are formed by the transfer of electrons from magnesium to oxygen. If we use the symbol e- to stand for an electron, the loss of electrons by magnesium can be written

Mg à Mg²⁺ + 2e^- (oxidation)

The change is identified as oxidation because magnesium loss electron and undergo an increase in oxidation number (0 à +2).

            For oxygen in this reaction, we can write

O₂ + 4e^- à 2O^2- (reduction)

This time the change is marked as a reduction because oxygen gains electrons and undergo a decrease in oxidation number (0 à -2).

2. Balancing Redox Reaction

            Redox reaction can be balanced by ion-electron method and oxidation number method.

Ion-Electron Method in Acid Solution

            For this method, we consider the reaction of HCl with KMnO₄. In this reaction Cl^- is oxidized to Cl₂ and MnO₄- is reduced to Mn²⁺. The skeleton equation, which does contain either H⁺ or H₂O, is Cl^- + MnO₄^- à Cl₂ + Mn²⁺ (acidic solution)

Step 1.    Divide the equation into half reactions.

Cl^- à Cl₂

MnO₄^- à Mn²⁺

Step 2.    Balance atoms other than H and O. In the first half-reaction of 2 in front of Cl^-. We don’t have to do anything to the second half-reaction.

2Cl^- à Cl₂

MnO₄^- à Mn²⁺

Step 3.    Balance oxygens by adding H₂O that side that needs O. In the second half-reaction, there are 4 oxygens on the left and none on the right. We therefore have to add 4 H₂O to the right side. Then the oxygens will balance.

2Cl^- à Cl₂

MnO₄^- à Mn²⁺ + H₂O

Step 4.    Balance hydrogens by adding H⁺ to the side that needs H. The right side of the second half-reaction has a total of 8 hydrogens; there is none on the left. Therefore, we add 8H+ to the left side.

2Cl^- à Cl₂

 H⁺ + MnO₄^- à Mn²⁺ + H₂O

Step 5.    Balance the charge by adding electrons. In this step we make the net charge the same on the both side. In the first half-reaction we have to add 2 electrons to the right side. In the second half-reaction we must add 5 electrons to the left side.

2Cl^- à Cl₂ + 2e^-

5e^-+ H⁺ + MnO₄^- à Mn²⁺ + H₂O

Step 6.    Make the number of electrons gained equal the number lost. This can be accomplished by multiplying the first half-reaction through by 5 and the second through by 2. There will then be 10e- lost and gained.

5(2Cl^- à Cl₂ + 2e^-)

2(5e^-+ H⁺ + MnO₄^- à Mn²⁺ + H₂O)

Step 7.    Add the two half-reactions. When we do this, we will not bother to bring down the electrons, because we know they will cancel. They have to-we went to the trouble to make them the same in each half-reaction.

5(2Cl^- à Cl₂ + 2e^-)

2(5e^-+ H⁺ + MnO₄^- à Mn²⁺ + H₂O)

                                     

                                        10Cl^- + 16H⁺ + 2MnO₄^- à 5Cl₂ + 2Mn²⁺ + 8H₂O

Ion-Electron Method in Basic Solution

            In a basic solution the dominant species are H₂O and OH^-, so these are the species that should be used to achieve material balance. Although you can use H₂O and OH^- directly, the simplest technique is to first balance the reactions as if occurred in acidic solution and then perform the conversion described below to adjust it to conform to conditions in basic solution. Consider for the oxidation of plumbite ion, Pb(OH)₃^-, to lead dioxide by hypochlorite ion in basic solution.

Pb(OH)₃^- + OCl^- à PbO₂ + Cl^-

Step 1.    The equation is balanced as though it occurred in acidic solution. We begin by dividing it into half-reaction.

Pb(OH)₃^-  à PbO₂ 

 OCl^- à Cl^-

Step 2.    We balance them according to atoms.

 Pb(OH)₃^-  à PbO₂ + H₂O + H⁺

H⁺ + OCl^- à Cl^- + H₂O

Step 3.    Balance the charge by adding electrons.

Pb(OH)₃^-  à PbO₂ + H₂O + H⁺ + 2 e^-

2 e^- + 2H⁺ + OCl^- à Cl^- + H₂O

Step 4.    Since number of electron lost in the first half-reaction is already equal to the number gained in the second, we can add them.

H⁺ + OCl^- + Pb(OH)₃^- à PbO₂ + 2H₂O + Cl^-

Step 5.    We perform three step conversion to basic solution. First we add to each side the same number of OH^- as there are H⁺ in the equation. H⁺ + OH^- forms H₂O.

H⁺ + OH^- + OCl^- + Pb(OH)₃^- à PbO₂ + 2H₂O + Cl^-+ OH^-

H₂O + OCl^- + Pb(OH)₃^- à PbO₂ + 2H₂O + Cl^-+ OH^- 

Step 6.    We cancel one H₂O from each side to get the final balanced equation.

OCl^- + Pb(OH)₃^- à PbO₂ + H₂O + Cl^-+ OH^- 

Oxidation State Method

            To see how these methods apply, let’s balance the equation for the reaction KMnO₄ with NaSO₃ and sulphuric acid as catalyst. The reaction was produced K₂SO₄, MnSO₄, Na₂SO₄ and H₂O.

Step 1. Write down the equation of reaction.

            KMnO₄ + Na₂SO₃ + H₂SO₄  à K₂SO₄+ MnSO₄ + Na₂SO₄ + H₂O

Step 2. Give the oxidation number each element.

+1+7(-2)₄   (+1)₂+4(-2)₃  (+1)₂+6(-2)₆  (+1)₂+6(-2)₄  +2+6(-2)₄  (+2)₂+6(-2)₄      (+1)₂-2

KMnO₄    +   Na₂SO₃     +   H₂SO₄    à   K₂SO₄        +  MnSO₄  +    Na₂SO₄    +    H₂O

Step 3.    Choose the elements that change the oxidation number. Write the oxidation reaction and reduction reaction.

Oxidation :              S⁴⁺    à     S⁶⁺ + 2e^-

Reduction :          Mn⁷⁺ + 5e^-   à    Mn²⁺

Step 4.    Make the number of electrons gained equal the number lost. This can be accomplished by multiplying the first half-reaction through by 5 and the second through by 2. There will then be 10e^- lost and gained.

Oxidation :             (S⁴⁺    à     S⁶⁺ + 2e^-)5

Reduction :          (Mn⁷⁺ + 5e^-   à    Mn²⁺)2

Step 5.    Add the two half-reactions. When we do this, we will not bother to bring down the electrons, because we know they will cancel. They have to-we went to the trouble to make them the same in each half-reaction.

           S⁴⁺    à     S⁶⁺ + 2e^-

         Mn⁷⁺ + 5e^-   à    Mn²⁺

 

                   5 S⁴⁺   +  2Mn⁷⁺ à 5S⁶⁺ + 2 Mn²⁺

Step 6.    We write the equation with the coefficient.

          2KMnO₄ + 5Na₂SO₃   +  ? H₂SO₄ à  K₂SO₄  + 2MnSO₄ + 5Na₂SO₄  + ? H₂O

Step 7.    We determine the coefficient of H₂SO₄ by the number of mole S on the both side. On the right side, there are 8 mole S, and the left side there 6 mole S. Therefore, we add 2 mole S to the left side by multiplying H₂SO₄ by 3.    

           2KMnO₄ + 5Na₂SO₃   +  3 H₂SO₄ à  K₂SO₄  + 2MnSO₄ + 5Na₂SO₄  + ? H₂O

Step 8.    We determine the coefficient of H₂O by the number of mol H or O on the both side. On the left side, there are 6 mole H and 35 mole O, and the right side there are 2 mole H and 32 mole O. Therefore, we add 4 mole S and 3 mole O to the right side by multiplying H₂O by 3.   

         2KMnO₄ + 5Na₂SO₃   +  3 H₂SO₄ à  K₂SO₄  + 2MnSO₄ + 5Na₂SO₄  + 3H₂O

3. Equivalent of redox Reaction