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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