Question

The pH of 1.0 L of 1.0 M NaCl after the electrolysis for 965 s using 5.0 A current (100% efficiency;log2 = 0.3), is

Answer 1

$q=i\times t=5\times 965C=4825C=\frac{4825}{96500}F=0.05F$
The moles of $O{H}^{-1}$ produced = 0.05 moles
[$O{H}^{-}$]=0.05 (as volume is 1L )
pOH = - log 0.05 = 1.3
pH = 12.70