Electrochemistry Test - 7

molar conductivity = k x Volume in cm³ containing 1 mol of electrolyte.
so Λm = k/c × 1000
= 0.0248 × 1000/0.2 = 124Scm²/mol

Conductivity k = 1/R ×cell constant.
cell constant l/A = conductivity × R = 0.146 × 10^-3 × 1500 = 0.219 cm^-1

Λm = k/c × 1000 = 7.896 x 10^-5 ××1000/0.00241 = 32.76Scm²/mol

and α = Λm / Λ⁰m = 32.76/390.5 = 0.084

and K = Cα²/1-α = 0.00241× (0.084)²/1-0.084 = 1.85 ×10^-5 

For reduction of 1 mol of Al³⁺ to Al , Al³⁺ + 3e gives Al ,

, 3 mol of electrons are required to deposit 1 mol of Al .so total charge required will be 3F.

For reduction of 1 mol of Cu²⁺ to Cu,

2 mol of electrons are required to deposit 1 mol of copper . so total charge required will be 2F.

Ca²⁺ + 2e = Ca . At mass of Calcium is 40g.
So to deposit 40g (1mol) of calcium , 2F of current is required . Hence for 20g of Calcium to be deposited, 0.5 mol of electron or 1F current is required.

number of Moles of Al = mass/ atomic mass = 40/27 = 1.48 mol.

Al³⁺ +3e = Al . For 1 mol of Al deposition 3F charge is required.

therefore for 1.48mol of Al the current in faraday required will be = 1.48× 3 = 4.44F

For 1 mol H₂O to O₂ 2 mol of electrons are required.
H₂O = 2H⁺ + 1/2O₂ + 2e
So 2F = 2 × 96500 = 193000C

For converting FeO to Fe₂O₃ 1mol of electrons are required.