Draw curves to show how the molar conductance of strong electrolytes varies with dilution. from Chemistry Electrochemistry Class 12 Nagaland Board
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Class 10 Class 12
Draw curves to show how the molar conductance of strong electrolytes varies with dilution.

When a plot is drawn between the molar conductivity versus square root of concentration for strong electrolytes, a curve (nearly straight line) shown in the given Fig is obtained. It is clear from the figure that molar conductance has higher value and increases linearly. Slight increase in the value of Λwith dilution is due to the decrease in interionic attraction with dilution i.e., decrease in concentration.

The molar conductivity at infinite dilution (Λm) can be obtained by extrapolating the above graph of strong electrolytes to zero concentration.
Kohlrausch Law of Independent migration of ions: The law states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte. Thus, if λ°Na+ and λCl are molar limiting conductivity of the sodium and chloride ions respectively, then the limiting molar conductivity for sodium chloride is given by the equation:
Λ°NaCl = λ°Na + λCl
In general, if an electrolyte on dissociation gives v+ cations and v anions then its limiting molar conductivity is given by
Λ° = v + λ°+ + V λ°
Here, λ°+ and λ° are the limiting molar conductivities of the cation and anion respectively.
Kohlrausch’s law helps us to calculate
(i) Determination of molar conductivities of weak electrolytes at infinite dilution.
(ii) Determination of the degree of dissociation of electrolytes.
Degree of dissociation: It is ratio of molar conductivity at a specific countraction ‘C’ to the molar conductivity at infinite dilution. It is denoted by α.
i.e.,                   α = AmCAm

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Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

Answer:

For hydrogen electrode, ,
given that      pH = 10
use formula [H+] = 10–pH
so that  [H+] = 10−10 M
Electrode reaction will
H+  + e –  →1/2 H2
Use the formula


EH+/H2  = EoH+/H2 - RTnFlnH2[H+]2              = 0-8.314×2982×96500ln IH+2              = 0.05915 log [H+]              =-0.05915 pH             =-0.05915 × 10 =- 0.59 V
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Calculate the emf of the cell in which the following reaction takes place:
Ni(s) + 2Ag+ (0.002 M)  Ni2+ (0.160 M) + 2Ag(s)
Given that EoCell = 1.05 V

Answer:

Ni(s)+2Ag+(aq)Ni2+(aq)+2As(s)
Ecell = Eocell + RT2FlnAg+(aq)2Ni2+ (aq)
or      Ecell = Eocell+0.0592 logAg+(aq)2Ni2+ (aq)
 
The equation is also written as

or    Ecell = Eocell-0.0592logNi2+(aq)Ag+ (aq)2         = 1.05 V - 0.0295 log 0.1600.002  

= 1.05 V – 0.0295 x log 80

= 1.05 V – 0.0295 x 1.9031

= 1.05 V – 0.056 = 0.99 V.

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How would you determine the standard electrode potential of the system Mg2+/Mg?

Answer:

Use standard hydrogen electrode as anode and Mg2+ | Mg as a cathode we can measure the standard electrodepotential of systemMg2+ | Mg. Standard hydrogen electrode, represented by Pt(s), H2(g) (1 atm) | H+ (aq) and dip the electrode of Magnesium wire in a 1M MgSO4 solution .The standard hydrogen electrode is always zero.
Use formula
Eocell = Eo right  – Eoleft
The standard hydrogen electrode is always zero.
So that the value of
Eoleft =0
Hence
Eocell = Eo Mg|Mg2+
Or
Eo Mg|Mg2+= Eocell
2203 Views

Can you store copper sulphate solutions in a Zinc pot?

Answer:

No. Because zinc is more reactive than copper and thus holes will be developed in zinc pot.

Cu2+(aq) + Zn(s) → Zn2+ (aq) + Cu(s)
1384 Views

Consult the table of standard electrode potentials and suggest three substance that can oxidize ferrous ions under suitable conditions.

Answer:


oxidation of ferrous ion means :

Fe2+--> Fe3+ +e-EFe3+Fe2+0 = 0.77V
Any substance which standard electrode potential is more than that of Fe+3 /F+2 can oxidise ferrous ions. 
(refer to the table given in book)

The EMF of the substance whose reduction potentials greater than 0.77V will oxidised ferrous ion.
for example Br2, Cl2,and F2 .

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