Question

Calculate the amounts of Na and Chlorine gas produced during the electrolysis of fused NaCl by the passage of 1 ampere current for 25 minutes. Molar masses of Na and Chlorine gas are 23 g mol^-1 and 71 g mol^-1 respectively.

Answer 1

Given : 

Electric current = I = 1 ampere 

Time = t = 25 minutes = 25 × 60 s = 1500 s 

Molar mass of Na = 23 g mol^-1 

Molar mass of Cl₂ = 71 g mol^-1

Mass of Na produced = ?, 

Mass of Cl produced = ? 

Reactions during electrolysis :

Quantity of electricity = Q = I × t = 1 × 1500 = 1500 C

Number of moles of electrons passed = \(\frac{Q}{F}\) = \(\frac{1500}{96500}\) 0.01554

From half reaction (i),

∵ 2 moles of electrons deposit 2 moles of Na 

∴ 0.01554 moles of electrons will deposit,

\(\frac{0.01554\times 2}{2}\) = 0.01554 mol Na

Mass of Na = Moles of Na × Molar mass of Na

= 0.01554 × 23 = 0.3572 g Na 

From half reaction (ii),

∵ 2 moles of electrons produce 1 mole Cl₂

∴ 0.01554 moles of electrons will produce,

\(\frac{0.01554\times 1}{2}\) = 7.77 × 10^-3 × 71

∴ Mass of Cl₂ gas = Moles of Cl₂ × Molar mass 

= 7.77 × 10^-3 × 71 

= 0.5518 g

∴ Mass of Na deposited = 0.3572 g

Mass of Cl₂ liberated = 0.5518 g