Calculate the quantity of electricity required to reduct `24.6g` of nitrobenzene to aniline if the current efficienty is `75%`. If the potential drop across the cell is `4.0V`, how much energy is consumed `(Mw` of `C_(6)H_(5)NO_(2)=123g mol^(-1))`

Explanation

Half`-` cell reaction for the reduction of nitrobenzene to aniline is as below `:`
`C_(6)H_(5)NO_(2)+6H^(o+)+6e^(-)rarrC_(6)H_(5)NH_(2)+2H_(2)O`
Moles of nitro benzene`=24.6//123=0.2 mol`
`6e^(-)=6F-=1 mol ` of `C_(6)H_(5)NH_(2)` ltbr. `0.2 mol` of `C_(6)H_(5)NH_(2)=6xx0.2=1.2F`
Hence,` 1.2 f` of electricity is used to reduce `24.6 g` of nitrobenzene if the current efficientcy is `100%`.
But if the current efficiency is `75%` , then
Number of Faradays required `=(1.2Fxx100)/(75)`
`=1.6F=1.6xx96500C`
Now potential difference`=4.0V`
Energy `(E)` consumed `=` Charge `xx` Potential difference
`E=(1.6xx96500C)xx4.0V=617600J=617.6kJ`.