While the discharging of a lead storage battery following reaction take place.
`PbO_(2) +Pb +4H^(+) +2SO_(4)^(-2) rarr 2PbSO_(4) +2H_(2)O ,E^(@) = 2.01`
Calculate the energy (in kJ) obtained from a lead storage battery in which `0.014` mol of lead is consumed. Assume a constant concentration of `10.M H_(@)SO_(4)(log2 = 0.3)`

Answer

6

Explanation

`E = E^(@) -(0.059)/(n)log.(1)/([H^(+)]^(4)[SO_(4)^(-2)]^(2)) =2.01 -(0.059)/(2)log.(1)/([20]^(4)[10]^(2)) = 2.22V`
Energy ` = qE`
`= 2 xx 0.014 xx 96500 xx 2.22`
`= 6000J = 6 kJ`