Calculate the emf of following cell at 298K

Mg (s) | Mg²⁺ (0.01 M) ||Ag⁺ (0.0001 M) l Ag (s)

(Given:E^° _Mg2+/Mg = -2.37V E^° _Ag+/Ag = +0.80V)

OR

The conductivity of 0.001 mol/L solution of CH₃COOH is 4.95 x 10^-5 S cm^-1. Calculate its molar conductivity and degree of dissociation (α). Given: λ^°_H+ = 349.6 S cm²/mol and λ^°_CH3COO- = 40.9 S cm²/mol

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

E_cell = 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(λ_m) = × 1000= = 49.5

Acc to kohlaursch’s law

λ^°_CH3COOH=λ^°_CH3COO- +λ^°_H+

λ^°_{CH3COOH =} 349.6 + 40.9 = 390.5 S cm₂/mol

Degree of disassociation (α) = = = 0.125