Chemistry - Electrochemistry Question with Solution | TestHub

The reaction involved is, $\mathrm{NaCl}+{\mathrm{H}}_2\mathrm{O}\stackrel{\text{ Electrolysis }}{\to }\frac{1}{2}{\mathrm{H}}_2+{\mathrm{Cl}}_2+\mathrm{NaOH}$ Amount of NaCl present in 0.5 L of $1.0\mathrm{M},\mathrm{NaCl}=0.5$ mol Quantity of electricity passed $=965\times 5$ coulombs. $\therefore 4825$ coulombs will decompose NaCl . $\begin{array}{rl}& =\frac{4825}{96500}\mathrm{mol}\\ & =0.05\mathrm{mol}\end{array}$ NaOH formed in the solution will also be 0.05 mol . Volume of solution $=0.5\mathrm{L}$ $\therefore$ Molarity of NaOH in the solution $\begin{array}{rl}& =\frac{0.05\mathrm{moles}}{0.5\mathrm{L}}=0.1\mathrm{M}\\ & =10^{-1}\mathrm{M}\end{array}$ $\begin{array}{rlrl}& \therefore & \mathrm{pOH}& =1\\ \text{ or }& \mathrm{pH}& =14-1=13& \end{array}$