Tuesday, February 15, 2011
Faraday's laws of electrolysis:
Michael Faraday, a pioneer in the properties of electric currents, formulated two basic laws of electrolysis:
FARADAY'S FIRST LAW may be stated as follows:
|Faraday's First Law of Electrolysis|
|"The amount of any substance deposited, evolved, or dissolved at an electrode is directly proportional to the amount of electrical charge passing through the circuit."|
The amount of electricity passing through the circuit in a given time is the number of moles of electrons passing through the circuit in that time, and the charge Q is related to the current I by
The charge on the electron is 1.602 x 10-19 C, and Avogadro's number is 6.023 x 1023. It follows that one mole of electrons has a charge of 9.65 x 104 C. This quantity is known as the FARADAY or FARADAY'S CONSTANT (F).
FARADAY'S SECOND LAW may be stated as follows:
|Faraday's Second Law of Electrolysis|
|"The mass of different substances produced by the same quantity of electricity are directly proportional to the molar masses of the substances concerned, and inversely proportional to the number of electrons in the relevant half-reaction."|
This means that z moles of electrons are needed to discharge an ion Xz+ or Xz-.
In the apparatus below,
1 Faraday will discharge 9 g Al (1/3 mole), 20 g Ca (1/2 mole) and 23 g Na (1 mole).
The relevant half reactions are:
Al3+ + 3e- Al
Extraction of aluminium:
Aluminium is obtained by the electrolytic reduction of its molten oxide, alumina (Al2O3). Because alumina has a very high melting point (2045 ºC), the mineral cryolite (Na3AlF6) is added to lower the melting point in order that the electrolysis may be carried out at about 950 ºC. The electrolytic cell has carbon anodes and a carbon cathode (which forms the lining of the tank in which the electrolysis takes place). Carbon dioxide is formed at the anodes, and aluminium at the cathode. It is heavier than the molten alumina/cryolite mixture, and sinks to the bottom of the cell, where it is tapped off. The procedure is known as the Hall-Héroult process.
Aluminium extraction is very demanding on electrical current (typically, 3-5 V and 100 000 A), and is economical only where power is cheap.
When copper is first obtained by reduction of its ores, it is cast as impure slabs or ingots, called blister copper. In the electrorefining process, the blister ingots are used as anodes in an electrolytic cell, where an acid solution of copper (II) sulphate is used as electrolyte. Initially, the cathodes consist of thin sheets of pure copper.
During electrolysis, copper passes into solution from the anodes, (leaving the impurities, normally containing silver, gold and platinum) as an anode slime, which sinks to the bottom of the cell. The anode reaction is
At the cathode, copper (II) ions are discharged and the pure copper sheet becomes coated with an increasingly thick layer of very pure copper:
Electroplating consists of depositing a thin layer of a metal on another, either for protection or for the sake of appearance. Typically, a brass or nickel object is coated with a layer of silver by making use of electrolysis of a silver solution, using the object to be coated as the cathode:
The anode consist of pure silver, and the cathode is the object to be plated. The electrolyte is a mixure of silver nitrate with potassium cyanide.
The reactions are:
At the anode: Ag → Ag+ + e-
At the cathode: Ag+ + e- → Ag
The cyanide ensures a low concentration of silver ions, a condition for providing the best plating results.
During the process, the concentration of silver in the electrolyte remains constant, as the rate of reduction at the cathode (which is the rate of deposition of silver on the object) is the same as the rate of reduction at the anode (which is the rate of rate of dissolution of the silver anode).
Electrolytic Cell V/s Galvanic Cell
The main points of difference between an electrolytic cell and a galvanic cell (electrochemical cell) are:
Difference between an electrolytic cell and a galvanic cell
|Electrochemical cell (Galvanic Cell)||Electrolytic cell|
|A Galvanic cell converts chemical energy into electrical energy.||An electrolytic cell converts electrical energy into chemical energy.|
|Here, the redox reaction is spontaneous and is responsible for the production of electrical energy.||The redox reaction is not spontaneous and electrical energy has to be supplied to initiate the reaction.|
|The two half-cells are set up in different containers, being connected through the salt bridge or porous partition.||Both the electrodes are placed in a same container in the solution of molten electrolyte.|
|Here the anode is negative and cathode is the positive electrode. The reaction at the anode is oxidation and that at the cathode is reduction.||Here, the anode is positive and cathode is the negative electrode. The reaction at the anode is oxidation and that at the cathode is reduction.|
|The electrons are supplied by the species getting oxidized. They move from anode to the cathode in the external circuit.|| The external battery supplies the electrons. They enter through the cathode and come out through the anode. |
Electrolysis of water:
Water may be electrolyzed in the apparatus shown below. Pure water is however a very poor conductor of electricity, and one has to add dilute sulphuric acid in order to have a significant current flow.
The electrodes consist of platinum foil. The electrolyte is dilute sulphuric acid. Hydrogen gas is evolved at the cathode, and oxygen at the anode.
The ratio, by volume, of hydrogen to oxygen, is exactly 2:1.
Remember that electron flow in the circuit is opposite to the conventional current flow. Thus, while the conventional current flows from the positive pole through the electrolyte and ends up at the negative pole, electrons flow from the negative pole in the reverse direction. The positive pole of a battery accepta electrons from the electrolyte by means of the anode of the electrolytic cell. The reaction at the cathode (tube A) is the reduction of protons:
Oxidation takes place at the anode (tube B). There are two anions competing to give up their electrons, namely sulphate (SO42-), and hydroxide (OH-) from the ionization of water. The oxidation of OH- according to the reaction
has a standard electrode potential of -0.40V, compared to the oxidation of sulphate (-0.17V), and consequently, OH- will be oxidized preferentially. The overall reaction is therefore
and the electrons are reurned to the battery, thus completing the circuit.
In the electrolysis of aqueous solutions only one ion is involved in the selective discharge of ions at each electrode during the electrolysis. The ion which is selected for discharge at an electrode depends on a number of factors, including
- Position of the ions in the electrochemical series,
- For positive ions, the facility of discharge decrease in going from those least electropositive to those most electropositive. For example, if both copper and hydrogen ions are present in solution, it will be the copper ions which take electrons from the cathode to become copper atoms.
For negative ions, the ease of discharge decrease in going from those least electronegative to those most electronegative.
- Concentration of the Ions in Solution
- Irrespective of the position of the ions in the electrochemical series, there is a tendency to promote the discharge of the most concentrated ion present. For example, in concentrated sodium chloride solution (i.e. brine) , the two cations present are the chlorine ion and the hydroxyl ion. Although the hydroxyl ion is more easily oxidised than the chlorine ion, it is the chlorine ion which will be discharged because its concentration is much greater than that of the hydroxyl ion.
- Nature of the Electrode
- This is not as important as either of the other two factors, except in certain cases. For example in the electrolysis of molten sodium chloride using a mercury cathode, sodium ions are discharged in preference to hydrogen ions which are lower in the series.
The electrolysis of an aqueous solution of copper sulphate using copper electrodes (i.e. using active electrodes) results in transfer of copper metal from the anode to the cathode during electrolysis. The copper sulphate is ionised in aqueous solution.
CuSO4 ==> Cu(++) + SO4(-.-)
The positively charged copper ions migrate to the cathode, where each gains two electrons to become copper atoms that are deposited on the cathode.
Cu(++) + 2e(-) ==> Cu
At the anode, each copper atom loses two electrons to become copper ions, which go into solution.
Cu ==> Cu(++) + 2e(-)
The sulphate ion does not take part in the reaction and the concentration of the copper sulphate in solution does not change. The reaction is completed when the anode is completely eaten away. This process is used in electroplating.
The electrolysis of an aqueous solution of dilute sulphuric acid is often carried out in a Hofmann Voltammeter, an apparatus in which the gases evolved at the anode and cathode can be collected in separate graduated tubes. When the solution is electrolyzed hydrogen is produced at the cathode and oxygen at the anode. These gases can be shown to be present in a 2 to 1 ratio and result from the electrolysis of water under acidic conditions.
Sulphuric acid is a strong electrolyte is fully dissociated in aqueous solution.
H2SO4 ==> 2 H(+) + SO4(2 -)
Water is a weak electrolyte and is only slightly dissociated
H2O ==> H(+) + OH(-)
During electrolysis, the hydrogen ions migrates towards the cathode, and are discharged there (i.e. they gain an electron and are converted to hydrogen gas).
2 H(+) + 2 e(-) ==> H2
At the anode the concentration of hydroxyl ions is too low to maintain a reaction and the sulphate ions are not oxidized but remain on in solution at the end. Water molecules must be the species reacting at the anode.
2 H2O ==> O2 + 4 H(+) + 4 e(-)
The overall reaction is
- Cathode Reaction :
2 H(+) + 2e(-) ==> H2
4 H(+) + 4e(-) ==> 2H2
- Anode Reaction :
2 H2O ==> O2 + 4 H(+) + 4 e(-)
- Overall Cell Reaction:
4 H(+) + 2 H2O ==> 2 H2 + O2 + 4 H(+)
For every hydrogen ions discharged at the anode, another hydrogen ion is formed at the cathode. The net result is that the concentration of the sulphuric acid remains constant and this electrolysis consists of the decomposition of water with the overall reaction
2H2O ==> 2H2 + O2
Electrolysis of Molten Lead Bromide
A small quantity of solid lead bromide is taken in a silica crucible and two graphite electrodes (may be obtained from used torch cells) are inserted. A battery consisting of two dry cells is connected to the electrodes through a key and an ammeter.
Fig: 9.11 - Electrolysis of molten lead bromideWhen the key is pressed, no current flows through the system. This is because solid lead bromide does not conduct electricity. But, when the crucible containing lead bromide is heated the solid lead bromide melts. Now, on pressing the key, electricity flows through the system and a red brown gas (bromine) evolves at the anode and metallic lead deposits at the cathode.
The following reactions occur at the two electrodes:
Thus, electrolysis of lead bromide using graphite electrodes produces lead metal at the cathode and bromine gas at the anode.
An electrolytic cell has three component parts: an electrolyte and two electrodes (a cathode and an anode). The electrolyte is usually a solution of water or other solvents in which ions are dissolved. Molten salts such as sodium chloride are also electrolytes. When driven by an external voltage applied to the electrodes, the electrolyte provides ions that flow to and from the electrodes, where charge-transferring, or faradaic, or redox, reactions can take place. Only for an external electrical potential (i.e. voltage) of the correct polarity and large enough magnitude can an electrolytic cell decompose a normally stable, or inert chemical compound in the solution. The electrical energy provided undoes the effect of spontaneous chemical reactions.
Anode and cathode definitions depend on charge and discharge
Michael Faraday defined the cathode as the electrode to which cations flow (positively charged ions, like silver ions Ag+
), to be reduced by reacting with (negatively charged) electrons on the cathode. Likewise he defined the anode as the electrode to which anions flow (negatively charged ions, like chloride ions Cl−
), to be oxidized by depositing electrons on the anode. Thus positive electric current flows from the cathode to the anode. To an external wire connected to the electrodes of a battery, thus forming an electric circuit, the cathode is positive and the anode is negative.
Consider two voltaic cells, A and B, with the voltage of A greater than the voltage of B. Mark the positive and negative electrodes as cathode and anode, respectively. Place them in a circuit with anode near anode and cathode near cathode, so the cells will tend to drive current in opposite directions. The cell with the larger voltage discharges, making it a voltaic cell. Likewise the cell with the smaller voltage charges, making it an electrolytic cell. For the electrolytic cell, the external markings of anode and cathode are opposite the chemical definition. That is, the electrode marked as anode for discharge acts as the cathode while charging and the electrode marked as cathode acts as the anode while charging.
As already noted, water, particularly when ions are added (salt water or acidic water) can be electrolyzed (subject to electrolysis). When driven by an external source of voltage, H+
ions flow to the cathode to combine with electrons to produce hydrogen gas in a reduction reaction. Likewise, OH−
ions flow to the anode to release electrons and an H+
ion to produce oxygen gas in an oxidation reaction.
In molten sodium chloride, when a current is passed through the salt the anode oxidizes chloride ions (Cl−
) to chlorine gas, releasing electrons to the anode. Likewise the cathode reduces sodium ions (Na+
), which accept electrons from the cathode and deposits on the cathode as sodium metal.
NaCl dissolved in water can also be electrolyzed. The anode oxidizes chloride ions (Cl−
), and Cl2 gas is still produced. However, at the cathode, instead of sodium ions being reduced to sodium metal, water molecules are reduced to hydroxide ions (OH−
) and hydrogen gas (H2). The overall result of the electrolysis is the production of chlorine gas and aqueous sodium hydroxide (NaOH) solution.
Commercially, electrolytic cells are used in electrorefining and electrowinning of several non-ferrous metals. Almost all high-purity aluminium, copper, zinc and lead is produced industrially in electrolytic cells.
The Nernst Equation
Ecell = E0cell - (RT/nF)lnQ
Ecell = cell potential under nonstandard conditions (V)
E0cell = cell potential under standard conditions
R = gas constant, which is 8.31 (volt-coulomb)/(mol-K)
T = temperature (K)
n = number of moles of electrons exchanged in the electrochemical reaction (mol)
F = Faraday's constant, 96500 coulombs/mol
Q = reaction quotient, which is the equilibrium expression with initial concentrations rather than equilibrium concentrations
Sometimes it is helpful to express the Nernst equation differently:
Ecell = E0cell - (2.303*RT/nF)logQ
at 298K, Ecell = E0cell - (0.0591 V/n)log Q
Nernst Equation Example
A zinc electrode is submerged in an acidic 0.80 M Zn2+ solution which is connected by a salt bridge to a 1.30 M Ag+ solution containing a silver electrode. Determine the initial voltage of the cell at 298K.
(Unless you've done some serious memorizing, you'll need to consult the standard reduction potential table, which will give you the following information):
E0red: Zn2+aq + 2e- → Zns = -0.76 V
E0red: Ag+aq + e- → Ags = +0.80 V
Ecell = E0cell - (0.0591 V/n)log Q
Q = [Zn2+]/[Ag+]2
The reaction proceeds spontaneously so E0 is positive. The only way for that to occur is if Zn is oxidized (+0.76 V) and silver is reduced (+0.80 V). Once you realize that, you can write the balanced chemical equation for the cell reaction and can calculate E0:
Zns → Zn2+aq + 2e- and E0ox = +0.76 V
2Ag+aq + 2e- → 2Ags and E0red = +0.80 V
which are added together to yield:
Zns + 2Ag+aq → Zn2+a + 2Ags with E0 = 1.56 V
Now, applying the Nernst equation:
Q = (0.80)/(1.30)2
Q = (0.80)/(1.69)
Q = 0.47
E = 1.56 V - (0.0591 / 2)log(0.47)
E = 1.57 V