JEE Syllabus
Electrochemistry:
Electrochemical cells and cell reactions;
Electrode potentials;
Nernst equation and its relation to DG;
Electrochemical series,
emf of galvanic cells;
Faraday's laws of electrolysis;
Electrolytic conductance, specific, equivalent and molar conductance,
Kohlrausch's law;
Concentration cells.
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Main topic in TMH Book Chapter
Section I: ELECTROLYSIS
FARADAY'S LAWS OF ELECTROLYSIS
Section II: ELECTROLYTIC CONDUCTION
Section III: GALVANIC CELLS
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Electrochemistry:
In eletrolytic conduction, ions carry electric charge. An increase in temperature increases electrolytic conductance because as the temperature is increased, the ions move faster.
In electrolytic cell, electric energy is used to cause a chemical reaction to take place.
For example, in a eletrolytic cell molten sodium chloride is the electrolyte. Two electrodes anode and cathode are placed into the molten sodium chloride and the an electri current passed through it.
The cation (Na+) moves toward the electrode called the cathode. At this electrode Na+ ions accept electrons to form sodium metal.
Na+ + e- → Na
This is reduction (gain of electrons)
At the other electrode, called the anode, the anion (Cl-) loses electrons to form chlorine gas.
2Clˉ → Cl-2 + 2eˉ
Therefore, oxidation (loss of electrons) takes place at the anode.
In the cell, electrons move from the anode through the external circuit to the cathode.
The reaction at anode and the reaction at cathode separately are called half-reactions.
Various types of half reactions are possible at anode (but all are oxidation only)
1. oxidation of an anion to an element
2Clˉ → Cl-2 + 2eˉ
2. Oxidation of an anion or cation in solution to an ion of higher oxidation state
Fe^2+ → Fe^3+ + e‾
3. Oxidation of a metal anode,
Cu(s) → Cu^2+ + 2e‾
4. Oxidation of H-2O to produce oxygen:
2H-2O → O-2 +4H^+ +4e‾
Similarly at Cathode various types of reactions occur (But all are reduction reactions)
1. Reduction of a cation to a metal
2. Reduction of an anion or cation in solution to an ion of lower oxidation state.
3. Reduction of a nonmetal to an anion.
4. Reduction of water to produce hydrogen.
Faraday's laws of electrolysis:
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Quantitative Relationships in Electrolytic Cells
Determining the amount of electrical energy necessary for accumulating a given amount material from the electrolytic cell.
Faraday's laws of electrolysis:
First law: It states that the amount of any substance that is liberated at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.
W α Q (w = weight of substance deposited and Q is charge = ampere * time)
Second law: It states tht when the same quantity of electricity is passed through different electrolytes amount of different substances liberated or deposited at the different electrodes are directly proportional to the chemical equivalents9i.e., equivalent weight) of substances.
One faraday (F) is the amount of electrical energy required for flow of 1 mole of electrons.
To three significant digits, 1 faraday equals 96,500 coulombs(coul).
Current flow is measured in amperes (A)which is coulombs/seconds or coul/s,
The equivalent weight, or gram-equivalent weight, of a substance is the amss in grams that is equivalent ot 1 mole of electrons. We can think it as gram-weight of a substance that releases one mole of electrons or reacts with one mole of electrons.
To determine equitdvalent weight of a substance it is necessary to know how many moles of electrons are transferred per mole of substance. To calculate the equivalent weight of a substance the formula weight is divided by the number of moles of electrons transferred.
Aluminium always loses three electrons to form Al^3+. Its formula weight or atomic mass is 27.
Hence it equivalent weight is 27/3 = 9.0 g.
Problem:
How long would it take a current of 100 A to deposit 10 g of Fe from a solution of FeCl-2?
Solution:
The equation for Fe in FeCl-2 is Fe^2+ + 2e‾
Formula weight is 55.8 g
Hence equivalent weight = 55.8/2 = 27.9
Electrical energy required = 10.0 g Fe* (1 equiv Fe/27.9 g Fe)* (96,500 coul/1 equiv Fe)
= 34588 coul
As 100 A or 100 coul/sec is given as current
Time taken = 34588/100 = 345.8 s.
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Voltaic or Galvanic Cells
In this cell, a chemical reaction produces electrical energy.
In this cell, the electrons being transferred from the reducing agent to the oxidizing agent travel through a wire and thus provide an elctric current.
One example of this cell type, is zinc metal in a solution containing ZnSo-4 and Cu in solution of CuSo-4. Zinc electrode acts as the cathode and Cu electrode acts as the anode.
At zinc electrode Zn → Zn^2+ +2e¯ (oxidation)
At Cu electrode: Cu^2+ +2e¯ → Cu
The above cell is represented as
Zn|ZnSO-4║CuSO-4|Cu In this symbol additionally On zinc side as it is a cathode a - sign is placed in O and on Cu side as it is anode a + sign is placed in O.
In oxidation reaction, electrons are evolved and hence this reaction can be used to produce electricity from chemical reaction.
Emf produced by the cell depends upon (a) temperature and (b) concentration of CuS0-4 and ZnSO-4.
When ZnSO-4 and CuSO-4 are i molar(1 mole per litre), emf produced is 1.1 volt.
Half cells: Every cell is made of two parts called half cell or electrodes. a metal in contact with own ions is called a half cell.
The reaction taking place at electrode is called electrode reaction or half cell reaction. All electrode reactions are remembered for oxidation.
Standard Hydrogen Electrode
An electrode in which pure dry hydrogen gas is bubbled at 1 atm and 298K about a platinized platinum plate through a solution containing H^+ ions ( for example - HCl solution)
The emf produced is taken as zero volts. All other potential are expressed with SHE potential as zero.
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Nernst equation
In electrochemistry, the Nernst equation gives the electrode potential (E), relative to the standard electrode potential, (E^0), of the electrode couple or, equivalently, of the half cells of a battery.
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web sites
Electrolysis, batteries and fuel cells
http://www.docbrown.info/page01/ExIndChem/ExtraElectrochem.htm
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