1.

Calculate e.m.f. of cell for the reaction : Mg_((s))+Cu^(2+)"(0.0001 M)"rarr Mg^(2+)"(0.001 M)"+Cu_((s)) Given that : E_(Mg^(2+)//Mg)^(@)=-2.37V E_(Cu^(2+)//Cu)^(@)=+0.34V

Answer»

Solution :CELL equatio :Mg(s) +` Cu^(2+) (aq) to Mg^(2+) (aq)+Cu(s) underset(n=2)`
Nernst equaton:
`E_(cell)=-0.00591/2log. [Mg^(2+)]/[Cu^(2+)]``E_(cell)=-0.00591/2log. [Mg^(2+)]/[Cu^(2+)]`
EMF of the cell,
the limiting molar PRODUCTIVITY electroytcan be represent as the sum of limiting molar conductivities of the individual CATION and anion .
pplication : kohlraush law is appliedv to calculate the limiting molar conductivityb of any electroyt from the limitinf molar conductivity lambda@ of individual ion.s.
when concentrationapproches zero,the molar conductance is KNOWN as limitiomng molar conductance.


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