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CBSE Gujarat Board Haryana Board

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Class 10 Class 12
Distinguish between: (a) Electrolytes and non-electrolytes, (b) Reduction potential and oxidation potential (c) Primary cells and secondary cells, (d) Specific conductivity and molar conductivity.




An electrolyte dissociates in solution and thus produce ion.


nonelectrolyte does not dissociate at all in solution and therefore does not produce any ions.

Electrolytes are ionic  substance that dissolve in water

Nonelectrolytes are typically polar covalent substances that do dissolve in water as molecules instead of ions.



Primary cell


Lower initial cost.

Higher Initial Cost

Higher life-cycle cost ($/kWh).

Lower life-cycle cost ($/kWh) if charging in convenient and inexpensive


Regular maintenance required.

Typically lighter and smaller  thus traditionally more suited for portable applications.

Traditionally less suited for portable applications, although recent advances in Lithium battery technology have lead to the development of smaller/lighter secondary batteries.


Molar conductivity

Specific conductivity

Molar Conductivity of a solution at a given concentration is the conductance of the volume of solution containing one mole of electrolyte kept between two electrodes with area of cross section 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


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 and temperature.


Can you store copper sulphate solutions in a Zinc pot?


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)

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


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.

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


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

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


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
Eocell = Eo Mg|Mg2+
Eo Mg|Mg2+= Eocell

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


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 .