Thursday, April 16, 2020

Electrochemistry :Faraday's Law &Product of electrolysis.Class -12


Faraday’s First Law of Electrolysis

Faraday’s first law of electrolysis states that the mass of a substance altered at an electrode during electrolysis is directly proportional to the quantity of electric charge transferred at that electrode.  As charge is the product of strength of current and time, the mass (m) of the substance altered at an electrode is directly proportional to the strength of the current (i) and the time (t) for which the current passes, i.e., 

∝ it. 

Or

m = Zit,                                                                  ------------------(2)

where Z is the proportionality constant, and is characteristic of the substance or the element of which the electrode is made. Z is called electro-chemical equivalent (ECE) of the substance. From equation (2), we have

Z = m /( it)                                                                ----------------(3).

Electrochemical Equivalent (ECE)

The electrochemical equivalent (ECE) of an element is defined as the mass of its ions liberated at the electrode when one coulomb of electricity is passed through the electrolyte. We know that the quantity of electricity, i.e., the amount of charge q = it. Therefore, equation (3) becomes 

Z = m / q                                                                  ----------------(4).

Since m is in gram and q in coulomb, the unit for Z is gram per coulomb. The electrochemical equivalents of some common elements are given in the table here.

Faraday’s Second Law of Electrolysis

Faraday’s second law of electrolysis states that when the same quantity of electricity passes through different electrolytes, the mass of an elemental material altered at an electrode is directly proportional to the element's chemical equivalent. Chemical equivalent is also referred to as equivalent weight. Chemical equivalent is the ratio of the atomic weight (A) of an element to its valency (V). 

Consider three electrolytes, say copper sulphate, silver nitrate and zinc sulphate, in three different voltameters connected in series with the respective electrodes in them, as shown here.

Let E1, E2, E3 be the chemical equivalents of copper, silver and zinc, respectively. Also, let Z1, Z2 and Z3 be their respective electro chemical equivalents, and m1, m2, m3 be their respective masses liberated when the same quantity (q) of electricity is passed through their respective electrolytes. According to Faraday’s second law,

(m
1 / E1) = (m2 / E2) = (m3 / E3            ----------------(5).

Or

m1 : m2 : m3 = E: E2 E3,                     ----------------(6)

where charge q is the same for all. This is one form of Faraday’s second law of electrolysis. Applying the first law to these electrolytes, we have

m1 = Z1q,                  m2 = Z2q            and             m3 = Z3q           ----------------(7).

In other words,

m1 : m2 : m3 = Z1 : Z2 : Z3,                            ----------------(8)

where charge, q is the same for all. From equations (6) and (8),

Z1 : Z2 : Z3 = E1 : E2 : E3.----------------(9)

Equation (9) represents another form of Faraday’s second law of electrolysis.It shows that the ratio of the electrochemical equivalents of elements is equal to the ratio of their chemical equivalents.

Activity to Verify Faraday’s First Law of Electrolysis

Repeat the electrolysis with the copper sulphate electrolyte. Pass a current i1 for a time t1Let the mass of copper deposited on the cathode after t1 second be m1Then 

Z = m1/( i×t1).

Change the position of the rheostat again, and allow a different current i2 to pass through the electrolyte for a time t2Determine the mass m2 deposited at the cathode C. Then 

Z = m2 / (i2 × t2).

We will find that the value of Z is the same. This verifies Faraday’ first law of electrolysis.

Verification of Faraday’s Second law

In addition to the copper voltameter containing copper sulphate electrolyte, take two more voltameters, one a silver voltameter containing silver nitrate electrolyte, and the other a zinc voltameter containing zinc sulphate electrolyte. Connect the three voltameters in series. In each of the three voltameters, the electrodes A and C are made of the metals of the respective solutions. That is, the electrodes in the copper sulphate solution are made of copper, in the silver nitrate solution of silver, and in the zinc sulphate solution of zinc. 

Let the initial masses of the three cathodes in the three voltameters be m1i, m2i and m3i, respectively. Adjust the rheostat such that current i is read in the ammeter. Since the three voltameters are in series, the same current passes through all the electrolytes. Allow the current to pass for about half an hour. At the end of half an hour, let the mass of the three cathodes be m1f, m2f and m3f, respectively. The differences between the corresponding final and initial masses give the masses of copper, silver and zinc deposited on the respective cathodes. That is, m1 = m1f - m1i, m2 = m2f - m2i, and m3 = m3f - m3iWhen the ratios m1 : E1, m2 : E2 and m3 : E3 are calculated, it is found that they are equal, where charge q is the same for all. Hence

m1 : E1,                     m2 : E2                and              m3 : E3.

This verifies Faraday’s second law of electrolysis. The Faraday constant is the quantity of charge that must be passed through an electrolyte to deposit one gram equivalent (equivalent weight expressed in grams) of the substance at an electrode. The Faraday constant has a fixed value of 96,500 C which means that when a charge of 96,500 C is passed through an electrolyte (say copper sulphate), 1 gram equivalent of copper (31.77 g) is deposited at the cathode. Therefore, the chemical equivalent (E) of copper is equal to the relative atomic mass of copper divided by its valency, which is equal to 31.77.

Let us find the ECE, or (Z), of copper in the copper sulphate electrolysis activity that we just carried out. Let W1 grams be the initial mass of the cathode (C). Repeat the electrolysis activity for a duration of t second, and record the current i ampere. At the end of t seconds, the weight of the cathode will be W2 gram. The difference (W2 - W1) gives the mass m of the copper deposited on the cathode. From Faraday’s first law, m = Zit, the ECE of copper can be calculated, by rearranging the formula as Z = m/(it) gram per coulomb.

Applications of Electrolysis
  • In metallurgy, it is used for refining and extracting metals.
  • Certain metals like copper, tin, lead, gold, zinc, chromium and nickel are purified and extracted by electrolysis.
  • Electroplating is the process of coating a thin film of costlier or less corrodible metal on a base metal by electrolysis. For example, a silver film can be deposited on a copper base, or a plating of gold can be made on ornaments with a copper base. Here, the article to be electroplated forms the cathode, and the metal to be coated forms the anode. The electrolyte is a solution containing the salt of the anode material.
  • Electrotyping is a method of obtaining an exact copy of an engraved block containing letters or figures by electrolysis.

Tuesday, April 14, 2020

Conductance of Electrolytic Solutions Class 12

                                   Class 12
                                  Chemistry
                            Electrochemistry👍👍

Variation of Conductivity and Molar Conductivity with Concentration
  • They depend on the concentration of the electrolyte. The Conductivity and Molar Conductivity of both weak and strong electrolytes decreases with decrease in concentration as the number of ions per unitv olume carrying the current in a solution decreases on dilution.
Fig. Variation of electrolytic conductivity
Fig. Variation of Molar conductivity 
  • Conductivity of a solution at a specific concentration = Conductance of solution placed in between the two platinum electrodes where
  • Volume of solution = 1 unit
 Cross sectional area of electrodes = 1unit
Distance = 1unit
  • Molar conductivity of a solution at a specific concentration = Conductance of solution where
Volume of solution = V
Concentration of electrolyte = 1 mole
Cross sectional area of electrodes = A
Distance = 1unit
  • When the concentration approaches to zero, the molar conductivity is referred to as limiting molar conductivity. It is represented by the symbol Ëom.





Thursday, April 9, 2020

                              Nernst Equation

Nernst equation is a general equation that relates the Gibbs free energy and cell potential in electrochemistry.  It is very helpful in determining cell potential, equilibrium constant etc.
Nernst Equation
It takes into account the values of standard electrode potentials, temperature, activity and the reaction quotient for the calculation of cell potential. For any cell reaction, Gibbs free energy can be related to standard electrode potential as:
ΔG =-nFE
Where, ΔG= Gibbs free energy, n = number of electrons transferred in the reaction, F = Faradays constant (96,500 C/mol) and E= cell potential. Under standard conditions, the above equation can be given as,
ΔGo =-nFEo
According to the theory of thermodynamics, Gibbs free energy under general conditions can be related to Gibbs free energy under the standard condition and the reaction quotient as:
ΔG=ΔG+ RT lnQ
Where, Q= reaction quotient, R= universal gas constant and T= temperature in Kelvin. Incorporating the value of ΔG and ΔGo, from the first two equations, we get:
-nFE = -nFE0 + RT lnQ
E = E0 – (RT/nF) lnQ
Converting natural log to log10, the above equation is known as the Nernst equation. Here, it relates the reaction quotient and the cell potential. Special cases of Nernst equation:
E =  Eo   − (2.303RT/nF)  log10Q
At standard temperature, T= 298K:
=  Eo   − (0.0592V/n) log10Q
At standard temperature T = 298 K, the 2.303RTF, term equals 0.0592 V.

Wednesday, April 8, 2020

Understand Daniel Cell Deeply

Class 12 Chemistry                              Electrochemistry                                      Daniel cell


Daniel cell
  • The cell that converts the chemical energy liberatedas a result of redox reaction to electrical energy is called a Daniel cell.
  • It has anelectrical potential of 1.1 V.
 
  • The setup for Daniel cell is as follows:
    • In a beaker a plate of zinc is dipped in a solution of zinc sulfate (ZnSO4).
    • In another beaker a plate of copper is dipped in a solution of copper (II) sulfate in another container. These plates of metal are called the electrodes of the cell.
    • These electrodes behave as terminal to hold the electrons.
    • The two electrodes are connected via wire.
    • A salt bridge is placed between the two beakers. This provides a path for the movement of ions from one beaker to the other in order to maintain electrical neutrality.
    • Zinc electrode gets oxidized and hence releases electrons that flow through the wire towards the copper electrode.
    • The copper (II) sulfate solution releases copper ions Cu2+.                                                                                                                                                                                                                                                                                                                                                                           
    • At the anode:

      Oxidation ---------------- loss of electrons.
      Zn --> Zn2+ + 2e-

      At cathode,

      Reduction -------------gain of electrons.

      Cu2+ + 2e- --> Cu
    • Zinc atoms being more reactive have a greater tendency to lose electrons than that of copper.
      The electrons in this cell moves from zinc anode to copper cathode through the wire connecting the two electrodes in the external circuit
    • A bulb placed within this circuit will glow and a voltmeter connected within this circuit will show deflection.
    • The net reaction of this cell is the sum of two half-cell reactions.
      Zn(s) + Cu2+ (aq) --> Zn2+ (aq) + Cu(s)

In a Daniel cell a salt bridge is placed between the two beakers containing a solution of zinc sulfate (ZnSO4) and a solution of copper (II) sulfate respectively. This provides a path for the movement of ions from one beaker to the other in order to maintain electrical neutrality.

Tuesday, April 7, 2020

Sadia Sheikh

Sadia Sheikh Wiki, Family, Biography & more....💖

Sadia Sheikh



👸Sadia Sheikh is an Indian teacher. She is famous for teaching Chemistry🙋 in school(2013) and for uploading educational videos on her YouTube channel.😊





                                                                                                 
                                            🙎  Wiki/Biography
Sadia Sheikh was born on 21 september in Uttar Pradesh,Meerut.
After completing her schooling from MPGS,
MPGS
💖
 she enrolled herself in DN Degree College, Meerut to pursue MSC and B.ED. Thereafter, she completed her Masters of 
science from there and Bachlers in teaching arena.😎
DN College


                       👪 Family
She was born into Sheikh Family.She used to live in Meerut,Uttar Pradesh with her Parents and relatives. She has two younger brothers. One brother is elder from her and another brother is younger.😊


                                             💁  Youtube Channel

Thursday, April 2, 2020

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Mole Concept –  Some Basic Concepts of Chemistry –  Class 11       One gram atom of any element contains the same number of atoms and on...