Q. 12

Calculate the emf

Answer :

As the reduction potential of Ag+ is more than Mg2+, so silver will undergo reduction at cathode and magnesium wil undergo oxidation at cathode.


E°cell= E°cathode - E°anode


= 0.80-(-2.37)= 0.80 + 2.37 = 3.17V


Ecell = E°cell log


= 3.17 –


= 3.17-[(0.0295) (6)


= 3.17 – 0.1770= 2.99V


OR


Given:


Concentration (c) = 0.001 mol/L


Conductivity (κ) = 4.95 × 10-5 S/cm


Molar Conductivity(λm) = × 1000= = 49.5


Acc to kohlaursch’s law


λ°CH3COOH°CH3COO-°H+


λ°CH3COOH = 349.6 + 40.9 = 390.5 S cm2/mol


Degree of disassociation (α) = = = 0.125


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