For the standard cell, Cu(s) | Cu2+(aq) || Ag+ (aq) | Ag(s)
Given: E° Cu2+/Cu = 0.34V, E° = 0.80V
(a) Identify cathode and anode as the current is drawn through the cell.
(b) Write the reactions taking place at the electrodes.
(c) Calculate the standard cell potential.

a)ECu2+Cu0 <EAg+Ag0Therefore copper electorode isthe anode and silverelectrode act as a cathode. The half reaction are:anode: Cu(s) Cu2+ +2e-Cathode: 2Ag+ +2e- 2Agb) net reaction :Cu(s) +2Ag+(aq) Cu2+(aq) +2Ag(s)c) The potential of the cell is given by E =Ecell0 -0.059nlog[Cu2+][Ag+]20 =(+0.80-0.34) -0.0592log0.01[Ag+]2 log0.01(Ag+)2+ = 2 x 0.460.059 =15.593taking antilog 0.01[Ag+]2 =antilog 15.593 =3.919 x 1015[Ag+]2 =0.013.919 x1015 =2.55 x10-18[Ag+] =1.59 x 10-9M

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 For the cell Zn/Zn2+ (aq) || Cu2+(aq) | Cu, derive the relation between E°cell and Kc at 298 k.

Gibbs energy of the reaction given by:
rG = – nFE(cell)
thus the reaction 
Zn(s) + Cu2+(aq)---> Zn2+(aq) + Cu(s)

rG = – 2FE(cell)

but when we write the reaction
2 Zn (s) + 2 Cu2+----->2 Zn2+(aq) + 2Cu(s)

rG = – 4FE(cell)

If the concentration of all the reacting species is unity, then
E(cell)Ecell0
and we have
rG = – nFEcell0


Thus, from the measurement of Ecell0 we can obtain an important thermodynamic quantity, rG, standard Gibbs energy of the reaction.
From the latter we can calculate equilibrium constant by the equation:

rG = –RT ln Kc

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Define conductivity and molar conductivity for the solution of an electrolyte.

Conductivity :–
Conductivity of a solution is equal to the conductance of a solution of 1 cm length
and cross section area of 1 square cm.  it may also be define as the conductance of ine centimeter cube of the conductor . It is represented by the symbol Kappa (κ). mathematically we can write
κ = 1/ p
here ρ is resistivity
the unit of K is ohm –1 cm –1 or S cm–1
The conductivity, κ, of an electrolytic solution depends on the concentration of the electrolyte, nature of solvent andtemperature.
 
Molar conductivity:
Molar conductivity of a solution at a given concentration is the conductance of the volume V of solution containing one mole of electrolyte kept between two electrodes with area of cross section A and distance of unit length. Therefore,
Distance is unit  so l = 1
Volume          = area of base × length
So V   = A × 1  = A
Λm       =κA/l
Λm       = κV
Or
Molar conductivity increases with decrease in concentration. When concentration approaches zero, the molar conductivity is known as limiting molar conductivity and is represented by the symbol Ëom.
For strong electrolytes, Λ increases slowly with dilution and can be represented by the equation:
Λm = Ë°m° – A c ½

 
It can be seen that if we plot Λm against c1/2, we obtain a straight line with intercept equal to Ëm° and slope equal to ‘–A’. The value of the constant ‘A’ for a given solvent and temperature depends on the type of electrolyte i.e., the charges on the cation and anion produced on the dissociation of the electrolyte in solution.
 
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How the Kohlrausch’s law is used to determine the degree of ionization of weak electrolyte?

 Kohlrausch’s law of independent migration of ions states molar conductivity of an electrolyte at infinite dilution can be expressed as the sum of the contribution of individual ions. If molar conductivity of cations and anions are represented by λ+ and λ respectively.

λm  =V+λ+  + V-λ+

where v+ and v are number of cations and anions per formula of electrolyte e.g.,

Λ CaCl2 = λ (Ca2+) + 2 λ (CI)
Λ = KCl = λ (K+) + λ (CI)

Uses 1. It is used to find molar conductivity of weak electrolyte at infinite dilution which 
cannot be obtained by extrapolation.

2. It is used to calculate degree of dissociation of weak electrolyte at a particular concentration.
Degree of dissociation 
α =ΛmΛm°
where Λm is molar conductivity of weak electrolyte at a particular concentration and Λemis molar conductivity of weak electrolyte at infinite dilution.

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What do you mean by a secondary cell? Discuss the function of lead storage battery.


A secondary cell after use can be recharged by passing current through it in the opposite direction so that it can be used again. A good secondary cell can undergo a large number of discharging and charging cycles. The most important secondary cell is the lead storage battery commonly used in automobiles and invertors.
It consists of a lead anode and a grid of lead packed with lead dioxide (PbO2 ) as cathode. A 38% solution of sulphuric acid is used as an electrolyte.
The cell reactions when the battery is in use are given below:

Anode: Pb(s) + SO42–(aq)----> PbSO4(s) + 2e

Cathode: PbO2(s) + SO42–(aq) + 4H+(aq) + 2e ----> PbSO4 (s) + 2H2O (l )

i.e., overall cell reaction consisting of cathode and anode reactions is:

Pb(s) + PbO2(s) + 2H2SO4(aq)---> 2PbSO4(s) + 2H2O(l)

On charging the battery the reaction is reversed and PbSO4(s) on anode and cathode is converted into Pb and PbO2, respectively.

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