Chemistry - Electrochemistry Question with Solution | TestHub

The complete electrolysis reaction is as follow, ${\mathrm{B}}_2{\mathrm{H}}_6+3{\mathrm{O}}_2⟶{\mathrm{B}}_2{\mathrm{O}}_3+3{\mathrm{H}}_2\mathrm{O}$ 27.66 g of ${\mathrm{B}}_2{\mathrm{H}}_6=1\mathrm{mole}$ of ${\mathrm{B}}_2{\mathrm{H}}_6$ It requires 3 moles of ${\mathrm{O}}_2$ for complete burning ${\mathrm{H}}_2\mathrm{O}⟶{\mathrm{H}}_2+\frac{1}{2}{\mathrm{O}}_2$ $\frac{1}{2}$ mole of ${\mathrm{O}}_2$ requires 2 Faraday. $\therefore 3{\mathrm{mole}}_2$ will require $=12$ Faraday $\begin{array}{rl}12\times 96500& =i\times t\\ 12\times 96500& =100\times t\\ t& =3.2\mathrm{h}\end{array}$