Lesson 1

What is Electrochemistry?

•  Is the branch of chemistry that deals with electricity and its relation to chemical reactions. Some reactions proceed spontaneously and produce a small amount of energy in the form of electromotive force while some nonspontaneous reactions require their energy for the reactions to push through.

Some Significant Persons in the Field of Electrochemistry

J           John Daniell & Michael Faraday

-          known as the “Founders of Electrochemistry”

Otto von Guericke

-          A German physicist to create the first electric generator

Georg Ohm

-          A German physicist that proved that there is a direct proportionality between the potential difference applied across a conductor and the resultant electric current. This relationship is now known as Ohm's law.

Svante Arrhenius

-          A Swedish scientist, devised the Acid-Base Theory which is defined by Arrhenius that “ acid-base reactions are characterized by Arrhenius acids, which dissociate in aqueous solution to form hydrogen ions (H⁺), and Arrhenius bases, which form hydroxide (OH⁻) ions”

REDOX

-          Stands for Reduction – Oxidation. These are electrochemical processes involving electron transfer to or from a molecule or ion changing its oxidation state.

Oxidation

-          It is the he loss of electrons or an increase in oxidation state by a molecule, atom, or ion.

Reduction

-           is the gain of electrons or a decrease in oxidation state by a molecule, atom, or ion.

 Oxidation Number/ Oxidation States

is an integer value assigned to each element in a compound or ion that allows us to keep track of electrons associated with each other

Rules for assigning oxidation number:

1.      All elements in their free state (uncombined with other elements) have an oxidation number of zero.

Example: C, Na, Ca, Mg, N₂, O₂, Cl₂

2.      H is +1 except in metal hydrides where it is -1.

Example: HNO₃  ->   H= +1

                 CaH  ->   H= -1

3.      O is -2 except in peroxides where it is -1 and in OF₂ where it is +2.

4.      The metallic element in an ionic compound has positive oxidation number.

Example: I- A    ->    +1

                II- A   ->   +2

                Aluminum  ->  +3

5.      Fluorine is always -1.

6.      In covalent compounds, the negative oxidation number of is assigned to the most electromagnetic atom.

7.      The algebraic sum of the oxidation number of the elements in a polyatomic ion is equal to the charge of ion.

Balancing Redox Reactions

Electrochemical reactions in water are better understood by balancing redox reactions using the ion-electron method where H⁺ , OH^– ion, H₂O and electrons (to compensate the oxidation changes) are added to cell's half-reactions for oxidation and reduction.

*Remember:

To know if the half reaction is a oxidation or reduction reaction we just need to remember two things

OIL – Oxidation is Loss of electrons

        RIG – Reduction is Gain of electrons

       

*Balancing in Acidic Mediums

In acid medium H⁺ ions and water are added to half-reactions to balance the overall reaction.

 For example, when Manganese (M) reacts with Sodium Bismuthate (NaBiO₃).

Mn²⁺(aq) + NaBiO₃(s) → Bi³⁺(aq) + MnO₄^–(aq)

Oxidation:

4 H₂O(l) + Mn²⁺(aq) → MnO₄^–(aq) + 8 H⁺(aq) + 5 e^–

Reduction:

2 e^– + 6 H⁺(aq) + BiO₃^–(s) → Bi³⁺(aq) + 3 H₂O(l)

Finally, the reaction is balanced by multiplying the number of electrons from the reduction half reaction to oxidation half reaction and vice versa and adding both half reactions, thus solving the equation.

8 H₂O(l) + 2 Mn²⁺(aq) → 2 MnO₄^–(aq) + 16 H⁺(aq) + 10 e^–

10 e^– + 30 H⁺(aq) + 5 BiO₃^–(s) → 5 Bi³⁺(aq) + 15 H₂O(l)

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14 H⁺(aq) + 2 Mn²⁺(aq) + 5 NaBiO₃(s) → 7 H₂O(l) + 2 MnO₄^–(aq) + 5 Bi³⁺(aq) +      5 Na⁺(aq)

*Balancing in Basic Mediums

In basic medium OH^– ions and water are added to half reactions to balance the overall reaction.

For example, on reaction between Potassium Permanganate (KMnO₄)and Sodium Sulfite (Na₂SO₃).

KMnO₄ + Na₂SO₃ + H₂O → MnO₂ + Na₂SO₄ + KOH

Oxidation:

3 e^– + 2 H₂O + MnO₄^– → MnO₂ + 4 OH^–              

Reduction:

 2 OH^– + SO₃^2– → SO₄^2– + H₂O + 2 e^–

The same procedure as followed on acid medium by multiplying electrons to opposite half reactions solve the equation thus balancing the overall reaction.

6 e^– + 4 H₂O + 2 MnO₄^– → 2 MnO₂ + 8 OH^–

6 OH^– + 3 SO₃^2– → 3 SO₄^2– + 3 H₂O + 6e

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2 KMnO₄ + 3 Na₂SO₃ + H₂O → 2 MnO₂ + 3 Na₂SO₄ + 2 KOH