Electrolysis of a solution of HSO₄^- ions produces S₂O₈^2– . Assuming 75% current efficiency, what current should be employed to achieve a production rate of 1 mole of S₂O₈^{- -} per hour ?

writting the reaction as:

2HSO4-...............+......>S2O₈^2–+ 2H+.......+2e-
Let charge on elctron=q
from the equation we can see that 1 mole of S2O82- invoves two moles of electron.
So for 1 hour, the current should be = 2*NA*q/60*60 Ampere= I
but the efiiciency is 75% only, o to get this much actual amount the total current nedded=I’(let)
Hence 75% of I’=I
i.e., I’*75/100=I
Now please solve it.