Question

Calculate the mass of Mg and the volume of Chlorine gas at NTP produced during the electrolysis of molten MgCl₂ by the passage of 2 amperes of current for 1 hour. Molar masses of Mg and Cl₂ are 24 g mol^-1 and 71 g mol^-1 respectively.

Answer 1

Given : 

Electric current = I = 2A 

Time = t = 1 hr = 1 × 60 × 60 s = 3600 s 

Molar mass of Mg = 23 g mol^-1 

Molar mass of Cl₂ = 71 g mol^-1 

Mass of Mg produced = ? 

Volume of Cl₂ at NTP produced = ?

Reactions during electrolysis :

(i) Mg²⁺ + 2e^- → Mg (Reduction half reaction) 

(ii) 2Cl^- → Cl_2(g)  + 2e^- (Oxidation half reaction)

Quantity of electricity passed = Q = I × t 

= 2 × 3600 

= 7200 C

∵ 1 Faraday = 1 mol electrons 

∴ Number of moles of electrons passed

= \(\frac{Q}{F}\) = \(\frac{7200}{96500}\) = 0.07461 mol

From half reaction (i),

∵ 2 moles of electrons deposit 1 mole of Mg 

∴ 0.07461 moles of electrons will deposit,

\(\frac{0.07461\times 1}{2}\) = 0.037305 mol Mg

Mass of Mg = Moles of Mg × Molar mass of Mg

= 0.037305 × 24 

= 0.8953 g Mg

From half reaction (ii),

∵ 2 moles of electrons produce 1 mol Cl₂ gas 

∴ 0.07461 moles of electrons will produce,

\(\frac{0.07461}{2}\) = 0.037305 mol Cl₂

∵ 1 mole of Cl₂ occupies 22.4 dm at NTP

∴ 0.037305 mole of Cl₂ will occupy,

22.4 × 0.037305 = 0.8356 dm³

∴ Volume of Cl₂ gas produced 

= 0.8356 dm³ 

= 0.8356 × 10³ cm³ 

= 835.6 cm³

∴ Mass of Mg produced = 0.8953 g

Volume of Cl_2(g) at NTP produced = 835.6 cm³