Question

Calculate the potential of a lead–acid cell if all reactants and products are in their standard states. What will be the voltage if six such cells are connected in a series?

Short answer

The potential of the lead (Pb) acid cell is2.041 V, and the voltage of the lead acid battery having six cells connected in series is12.246 V.

Step-by-step solution

Half-cell reaction at anode and cathode.

The lead (Pb) acid cell is arechargeable cell. The anode of the lead acid cell is made up ofspongy lead(Pb), which is oxidized${\text{Pb}}^{\text{2+}}$during discharging.

The solid lead dioxide$\left({\text{PbO}}_{\text{2}}\right)$acts as a cathode and is reduced${\text{Pb}}^{\text{4+}}$to${\text{Pb}}^{\text{2+}}$. Thesulfuric acidsolution$\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right)$is anelectrolytein the lead acid cell.

Thehalf-cell reaction at the anodeis shown below,

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

Thehalf-cell reaction at the cathodeis shown below,

${\text{PbSO}}_{\text{2}}\left(\text{s}\right){\text{+SO}}_{\text{4}}^{\text{2-}}{\text{+4H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{+2e}}^{\text{-}}\to {\text{PbSO}}_{\text{4}}\left(\text{s}\right){\text{+6H}}_{\text{2}}\text{O}\left(\text{l}\right)$

Therefore, in the electrochemical reactions, the standard state of all the reactants and products implies that the concentration of each reactant and product is 1M.

The lead (Pb) acid cell potential.

All cell species are in their standard states; the cell's potential is known asstandard cell potential.It is denoted by${\text{ΔE}}_{\text{cell}}^{\text{o}}$.

The standard cell potential is calculated by using the following formula,

${\text{ΔE}}_{\text{cell}}^{\text{o}}{\text{=E}}_{\text{cathode}}^{\text{o}}{\text{-E}}_{\text{anode}}^{\text{o}}$--- (a)

Here,

${\text{E}}_{\text{cathode}}^{\text{o}}$- is the standard reduction potential of cathode half-cell reaction.

${\text{E}}_{\text{anode}}^{\text{o}}$- is the standard reduction potential of anode half-cell reaction.

For lead-acid cells;

${\text{E}}_{\text{cathode}}^{\text{o}}\text{=1.685V}$

${\text{E}}_{\text{anode}}^{\text{o}}\text{=-0.356V}$

Put the values in equation (a);

${\text{ΔE}}_{\text{cell}}^{\text{o}}\text{=1.685V-}\left(\text{-0.356V}\right)$

${\text{ΔE}}_{\text{cell}}^{\text{o}}\text{=2.041V}$

Hence, the potential of the lead acid cell is2.041 V.

The potential of the lead acid battery has six cells in series

When n number of cells are connected, called a battery when cells are connected in a series, the potential of each cell adds up to give the overall potential of the battery.

Consider the six lead acid cells are connected in a series;

${\text{ΔE}}_{\text{battery}}^{\text{o}}{\text{=addingΔE}}_{\text{cell}}^{\text{o}}\text{6times}$

${\text{ΔE}}_{\text{battery}}^{\text{o}}{\text{=6×ΔE}}_{\text{cell}}^{\text{o}}$ --- (b)

Here, the subscript battery stands for the potential of the battery.

Now, put${\text{ΔE}}_{\text{battery}}^{\text{o}}\text{=2.041V}$in the equation (b);

${\text{ΔE}}_{\text{battery}}^{\text{o}}\text{=6×2.041V}$

${\text{ΔE}}_{\text{battery}}^{\text{o}}\text{=12.246V}$

Thus, the lead acid battery's potential, which has six cells in series, is12.246 V.