Question

In lead storage battery, how many moles of electrons will be involved in the consumption of $98g$ of sulphuric acid?

Answer 1

Lead storage battery

at anode $Pb\left(S\right)$ + $S{O}_4^{2-}$ $\to$ $PbS{O}_4\left(s\right)$ + $2e^{-}$

at cathode $Pb{O}_2+4H^{+}$ $\to S{O}_4^{2-}$+$2e^{-}$

total cell reaction $Pb\left(S\right)$+$Pb{O}_2$+$2H_{2}S{O}_4$ $\to 2PbS{O}_4$+$2H_{2}Ol$

here 2 mole e^{-} are involved in comsumption of 2 mole sulphuric acid

$i.e.,1mole{H}_{2}S{O}_4=1mole{e}^{-}$

$incomsumptionof98g{H}_{2}S{O}_4=1mole{e}^{-}areused$