Electrochemistry can be define as the effect  of electricity on chemicals.

Electrology is the practice of electrical epilation to permanently remove human hair……

The Electrolysis of Water

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Chemical Concepts Demonstrated: Electrolysis of water, decomposition of water

Demonstration:

The Hoffman electrolysis apparatus is filled with Na₂SO₄ solution containing universal indicator and is started.
Observations:

The solution turns blue at the cathode (basic) and red at the anode (acidic).  Twice as much gas is evolved at the cathode as at the anode. When the solutions in the two electrodes compartments are mixed, the indicator turns green (neutral).

Explanations (including important chemical equations):

Two reactions can take place at the cathode.

Two reactions can also occur at the anode.

It is easier to reduce H₂O that Na⁺ ions at the cathode and it is easier to oxidize H₂O than SO₄ ^2- at the anode. Combining the two half reactions so that electrons are conserved, we obtain the following equations.

6 H₂O (l)  —> 2 H₂ (g) + O₂ (g) + 4 H ⁺ (aq) + 4 OH^– (aq) or

2 H₂O (l)  —> 2 H₂ (g) + O₂(g)

The decomposition of water produces twice as much hydrogen gas as oxygen gas. Since the number of protons and hydroxide ions formed in the reaction are the same, when the solutions are combined at the end of the electrolysis the indicator has the characteristic green color of a neutral solution.

Electrolysis of Water, Version II

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Chemical Concept Demonstrated: Effects of electrolysis on the acidity of a solution

Demonstration:

Two crystallizing dishes are filled with Na2SO4; a salt bridge and an attachment to a power supply connect them. Both dishes contain a universal indicator: red, acidic; blue, basic; green, neutral. The current is allowed to run through. The two dishes are combined in a beaker.

Observations:

The water turns red at the anode and blue at the cathode. When the dishes are mixed together, the water turns green.

Explanations (including important chemical equations):

anode:    2 H₂O (l)      —> O₂ (g) + 4 H⁺ (aq) + 4 e ^–     E� = -1.229 V
(solution turns from green to red)

cathode: 2 H₂O (l) + 2 e ^– —> H₂(g) + 2 OH ^– (aq)        E� = -0.828 V
(solution turns from green to blue)

Electrolysis of water produces equivalent amounts of acid and base.  Notice that the cathode equation should be multiplied by two in order to conserve the charge in the chemical reaction.

When the dishes are mixed together, the solution is neutral again.   Balancing the above equations with respect to the electrons also balances it with respect to the water’s ions, so the overall sum of the two dishes is an electrically neutral and pH neutral solution.

Electrodes

They are conductors in form of rod, Plates, whereby electric current enters or leave the cell.

Talking about the anode and the cathode

Anodes: Positive electrodes by which conventional currents enters the electrolyte or by which electrons leaves the electrolyte.

Cathode: Negative electrodes by which the conventional currents leaves the electrolyte or by which electrons enters the electrolyte.

Galvanic or Voltaic Cells

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Chemical Concepts Demonstrated:  Voltaic/galvanic cells, relative half-cell potentials

Demonstration:

• One of the dishes is filled with ZnSO₄ and the other with HCl.
• A strip of Zn metal is attached at one end to the posts of the electrochemistry template and is placed at the other end into the dish filled with Zn²⁺.
• The hydrogen electrode is attached, placed into the HCl solution, and H₂ gas is bubbled in. Insert the salt bridge.
• The Zn²⁺/Zn half-cell is replaced with a Cu²⁺/Cu half-cell.
• The H⁺/H₂ half-cell is replaced with a Zn²⁺/Zn. (picture 2)
  

Observations:

The potential in the absence of the salt bridge is 0.00 V.  After the salt bridge is inserted, the potential of
the first set up is around + 0.76 V and the cell is a galvanic or voltaic cell.  The Zn²⁺/Zn half-cell is the anode the H⁺/H₂ is the cathode.

In the second set up, both the magnitude and the sign of the potential change.   The potential is now roughly – 0.34 V.

Picture 2 shows the third set up.  The potential is now – 1.10 V. If the leads are changed, the cell potential becomes + 1.10 V and the cell becomes a galvanic or voltaic cell.

Explanations (including important chemical equations):

With the leads connected so as to produce a cell potential of + 0.76 V, the half reactions are:

anode:	Zn (s) —> Zn 2+ (aq) + 2 e–	Eo = 0.76 V
cathode:	2 H + (aq) + 2 e – —> H2 (g)	Eo = 0.00 V
	Zn (s) + 2 H +  (aq) —> Zn 2+ (aq) + H2 (g)	Eo cell = 0.76 V

If the standard-state potential for the H⁺/H₂ half-cell is assumed to be 0.00 V, and the potential for the anode half-reaction is equal in magnitude but opposite in sign to the standard-state potential for the Zn²⁺/Zn couple, then the standard-state reduction potential for the Zn²⁺/Zn half-cell must be – 0.76 V.

If the Zn²⁺/Zn half-cell is replaced with a Cu²⁺/Cu half-cell without reversing the leads to the voltmeter, the overall cell potential is – 0.34 V and the standard-state reduction potential for the Cu²⁺/Cu couple is therefore + 0.34 V.

anode:	Cu (s) —> Cu 2+ (aq) + 2 e–	Eo =  – 0.34 V
cathode:	2 H + (aq) + 2 e – —> H2 (g)	Eo =  – 0.00 V
	Cu (s) + 2 H +  (aq) —> Cu 2+ (aq) + H2 (g)	Eo cell  = -0.34 V

If the H⁺/H₂ half-cell is replaced with a Zn²⁺/Zn half-cell, the overall cell potential should be – 1.10V.

anode:	Cu (s) —> Cu 2+ (aq) + 2 e–	Eo =  – 0.34 V
cathode:	Zn 2+ (aq) + 2 e– —> Zn (s)	Eo =  – 0.76 V
	Cu (s) + Zn 2+ (aq) —> Cu 2+ (aq) + Zn (s)	Eo cell  = -1.10 V

To set up a voltaic cell using these half reactions, one would have to reverse the leads to the voltmeter.

Work From a Voltaic Cell

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Chemical Concept Demonstrated: Electrochemical work with voltaic or galvanic cells

Demonstration:

The beaker contains a solution of H2SO4.A copper wire electrode along with a magnesium electrode are inserted into the solution.The wires are connected to a flashbulb. Alternately, a normal light bulb may be used.

Observations:

The flashbulb is set off.
Explanation (including important chemical equations):

The standard-state reduction potentials are:

Mg 2+ (aq) + 2 e– —> Mg (s)	Eo = -2.37 V
Cu 2+ (aq) + 2 e– —> Cu (s)	Eo = 0.34 V

The reaction that has an overall positive cell potential is therefore

Mg (s) —> Mg 2+ (aq) + 2 e–	Eo = 2.37 V
2 H + (aq) + 2 e – —> H2 (g)	Eo =   0.00 V
Mg (s) + 2 H + (aq) —>  Mg 2+ (aq) +   H2 (g)	Eo cell  = 2.37 V

Electrons flow from the magnesium electrode to the copper electrode through the external circuit, setting off the flashbulb.

Cathodic Protection

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Chemical Concepts Demonstrated: Corrosion, cathodic protection, sacrificial anode

Demonstration:

Both dishes contain a solution of HCl, 3% H2O2, and KSCN. The first dish contains an iron nail. The second dish contains an iron nail with one end wrapped in copper wire connected to a piece of zinc metal.

Observations:

The solution in the dish containing the lone nail turns a pink color, and the intensity of the color increases over time.  In the dish containing the nail/zinc metal connection, a gas is produced on the surface of the metal, but the pink color does not appear.  When the connection between the nail and the zinc metal is severed, the solution begins to turn pink.

Explanations (including important chemical equations):

The pink coloration is due to the formation of Fe(SCN)²⁺/Fe(SCN)₂⁺ complexes.  This indicates that the iron metal is being oxidized to the Fe³⁺ ion.  This won’t occur in the second dish as long as the nail is connected to the zinc metal strip.   When the two metals are connected, the zinc undergoes oxidation to Zn ²⁺ at the same time that H ⁺ ions are reduced to H₂ on the surface of the iron nail.

Ionic Theory: Was postulated in the year 1887 by Arrhenius he said: if an electrolyte is melted or dissolved in water, it turns into freely moving  charge particles Known as IONS. 

Note: Salt are strong alkali consist of oppositely charged ions even in solid state.

Arrhenius                                     Nacl _(s) —> Na ⁺ Cl^–

Modern Ionic Theory              Na⁺ Cl^– _(s) —–> Na⁺ Cl^– _(aq)

REVIEW QUESTIONS

WHAT IS ELECTROLYSIS

EXPLAIN ELECTROLYSIS OF ACIDIFIED TETRAOXOSULPHATE(VI) ACID H2SO4

EXPLAIN:

A. ELECTROLOGY

B. ELECTROCHEMISTRY

C. ELECTRODE

D. ANOTHER AND CATHODE

E. DISCHARGE OF IONS IS GOVERNED BY —– CONDITIONS

F. IMPORTANCE OF ELECTROLYSIS

G. ELECTROLYSIS OF BRINE

INERT ELECTRODES & ELECTROLYTIC RXN.