Chemistry - Electrochemistry Question with Solution | TestHub

$\left[{\mathrm{H}}^{+}\right]=10^{-12}\mathrm{M}$
So$\left[\mathrm{O}{\mathrm{H}}^{-}\right]=10^{-2}\mathrm{M}$
Number of moles of NaOH in 50 mL solution
$=\frac{\mathrm{M}\mathrm{V}}{1000}=\frac{10^{-2}\times 50}{1000}$
$=5\times 10^{-4}$
Mass of NaOH in the solution
$=5\times 10^{-4}\times 40=2\times 10^{-2}\mathrm{g}$
$\mathrm{W}=\frac{\mathrm{I}\mathrm{t}\mathrm{E}}{96500}$
$2\times 10^{-2}=\frac{0.5\times \mathrm{t}\times 40}{96500}$
$\mathrm{t}=96.5$sec.