Question

the eletrolysis of 0.5 liter of 0.1 molar molar strength of MgOH2 is carried out. determine the reactions occurring in cathode and anode. write the general reaction .number the mass of the catodes and the anodes

Answer 1

Mg(OH)₂ is considered insoluble in water as it's $K_{sp} = 1.5\times 10^{-11}$. Threfore Mg(OH)₂ does not dissosiate in water solution to give Mg²⁺ and OH^- and nor Mg²⁺ ions or OH^- take part in elecrolysis process.

Molecules of water take part in the process of electrolysis;

$$Cathode^- : 2\times (2H_2O + 2e^- \rightarrow H_2 + 2OH^-)$$ $$Anode^+ : 2H_2O \rightarrow O_2 +4H^+ +4e^-$$

General reaction

$$4H_2O + 4e^- + 2H_2O \rightarrow 2H_2 + 4OH^- + O_2 + 4H^+ +4e^-$$ $$6H_2O \rightarrow 2H_2 + O_2 + 4H_2O$$ $$2H_2O \rightarrow 2H_2 + O_2$$

Find mass of H₂ (process on cathode):

$$m=d\times V , m(H_2O) = 1 \frac{g}{cm^3}\times 500 cm^3 = 500 g$$

$$n(H_2O) = \frac{m}{M} = \frac{500 g}{18.02 \frac{g}{mol}} = 27.7 mol$$

As mole ratio $n(H_2O):n(H_2)= 2:2$ , then $n(H_2) = n(H_2O) = 27.7 mol$

$$m(H_2) = n(H_2)\times M(H_2) = 27.7 mol \times 2.02\frac{g}{mol} = 56.0 g$$

Find mass of O₂ (process on anode):

As mole ratio $n(H_2O):n(O_2) = 2:1$ , then $n(O_2)= \frac{n(H_2O)}{2} = \frac{27.7 mol}{2} = 13.85 mol$

$$m(O_2) = n(O_2)\times M(O_2) = 13.85 mol \times 32.00 \frac{g}{mol} = 443.2 g$$