Question

(a) What quantity of charge (in coulombs) is a fully charged 12-V lead–acid storage battery theoretically capable of furnishing if the spongy lead available for reaction at the anodes weighs 10 kg and there is excess PbO2?

(b) What is the theoretical maximum amount of work (in joules) that can be obtained from this battery?

Short answer

a) The total quantity of charge of lead-acid battery is ${\text{9.3×10}}^{\text{6}}\text{C}$.

b) The maximum amount of work that can be obtained from the lead-acid battery is ${\text{1.1×10}}^{\text{6}}\text{J}$.

Step-by-step solution

a) Calculate the cell potential of the lead-acid storage battery.

Given data:

${\text{E}}_{\text{Cell}}\text{=12V}$

Mass of anode = 10 kg

The molar mass of the lead (Pb) is 207.2 g/mol.

Now, calculate the number of moles of lead (Pb) we get,

$\begin{array}{l}\text{=10kg×}\frac{\text{1000g}}{\text{1kg}}\text{×}\frac{\text{1molPb}}{\text{207.2g}}\\ \text{=10×1000×0.004826}\end{array}$

$\text{=48.3molPb}$

Therefore, the number of moles of lead is48.3 moles.

The anode half-cell reaction:

$\text{Pb}\left(\text{s}\right){\text{+SO}}_{\text{4}}^{\text{-2}}\left(\text{aq}\right)\to {\text{PbSO}}_{\text{4}}\left(\text{s}\right){\text{+2e}}^{\text{-}}$

There aretwomoles of electronsare required tooxidizeonemoleoflead(Pb).

Now, calculate the number of moles of electrons required to oxidize 48 moles of lead (Pb). We get

$\begin{array}{l}\text{=}\left(\text{48.3molPb}\right)\text{×}\left(\frac{{\text{2mole}}^{\text{-}}}{\text{1molPb}}\right)\\ {\text{=96.6mole}}^{\text{-}}\end{array}$

Hence, the number of moles of electrons required to oxidize 48 moles of lead (Pb) is96.6mol${\text{e}}^{\text{-}}$.

Calculate the charge of the lead-acid storage battery.

The charge is carried by one mole of electrons.

${\text{=96485C/mole}}^{\text{-}}$

Calculate the total charge

$\begin{array}{c}\left(\text{Q}\right){\text{=96.6mole}}^{\text{-}}\text{×}\frac{\text{96485C}}{{\text{mole}}^{\text{-}}}\\ {\text{=9.3×10}}^{\text{6}}\text{C}\end{array}$

Hence, the total quantity of charge of the lead-acid storage battery is${\text{9.3×10}}^{\text{6}}\text{C}$

Part b Step 4: b) Calculate the maximum amount of work obtained from the lead-acid battery.

We know that the cell potential of the lead-acid storage battery is

${\text{E}}_{\text{cell}}\text{=12V}$

The total quantity of charge present in a lead-acid storage battery is

${\text{9.3×10}}^{\text{6}}\text{C}$

Here calculate the maximum amount of work obtained from the battery,

${\text{W}}_{\text{elec}}{\text{=itE}}_{\text{cell}}$

$\begin{array}{c}{\text{W}}_{\text{elec}}{\text{=QE}}_{\text{cell}}\\ \text{=}\left({\text{9.3×10}}^{\text{6}}\text{C}\right)\text{×}\left(\text{12V}\right)\\ \text{=}\left({\text{9.3×10}}^{\text{6}}\text{C}\right)\text{×}\left(\text{12}\frac{\text{J}}{\text{C}}\right)\end{array}$

${\text{=1.1×10}}^{\text{8}}\text{J}$

Therefore, the maximum amount of work obtained from the battery is${\text{1.1×10}}^{\text{8}}\text{J}$