The following reactions are used in batteries:

$\begin{array}{c}{\mathbf{\text{I2H}}}_{\mathbf{\text{2}}}{\mathbf{\text{(g)+O}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}\mathbf{\to }{\mathbf{\text{2H}}}_{\mathbf{\text{2}}}{\mathbf{\text{O(l)E}}}_{\mathbf{\text{cell}}}\mathbf{\text{=1.23V}}\\ {\mathbf{\text{IIPb(s)+PbO}}}_{\mathbf{\text{2}}}{\mathbf{\text{(s)+2H}}}_{\mathbf{\text{2}}}{\mathbf{\text{SO}}}_{\mathbf{\text{4}}}\mathbf{\text{(aq)}}\mathbf{\to }{\mathbf{\text{2PbSO}}}_{\mathbf{\text{4}}}{\mathbf{\text{(s)+2H}}}_{\mathbf{\text{2}}}{\mathbf{\text{O(l)E}}}_{\mathbf{\text{cell}}}\mathbf{\text{=2.04V}}\\ {\mathbf{\text{III2Na(l)+FeCl}}}_{\mathbf{\text{2}}}\mathbf{\text{(s)}}\mathbf{\to }{\mathbf{\text{2NaCl(s)+Fe(s)E}}}_{\mathbf{\text{cell}}}\mathbf{\text{=2.35V}}\end{array}$

The reaction I is used in fuel cells, II in the automobile lead-acid battery, and III in an experimental high-temperature battery for powering electric vehicles. The aim is to obtain as much work as possible from a cell while keeping its weight to a minimum. (a) In each cell, find the moles of electrons transferred and$∆\text{G}$ . (b) Calculate the ratio, in kJ/g, of${\text{w}}_{\text{max}}$to mass of reactants for each of the cells. Which has the highest ratio, which is the lowest, and why? (Note: For simplicity, ignore the masses of cell components that do not appear in the cell as reactants, including electrode materials, electrolytes, separators, cell casing, wiring, etc.)

Chemical synthesis or, alternatively, chemical breakdown into two or more separate chemicals occurs when one component interacts with another to generate new material. These processes are known as chemical reactions, and they are generally irreversible until followed by other chemical reactions.

${\text{2H}}_{\text{2}}{\text{(g) +O}}_{\text{2}}\text{(g)}\to {\text{2H}}_{\text{2}}\text{O(l)}$

The reactions are then divided into half.

$$\begin{gathered} {\text{OHR:2H}}_{\text{2}}\to {\text{4H}}^{\text{+}}{\text{+ 4e}}^{\text{-}} \\ {\text{RHR:O}}_{\text{2}}{\text{+ 4e}}^{\text{-}}\to {\text{2 +O}}^{\text{2 -}} \end{gathered}$$

Four electrons are involved in this reaction having the value as ${{\text{E}}^{\text{o}}}_{\text{cell}}\text{= 1.23 V}$

The value of $∆{\text{G}}^{\text{o}}$ then can be computed as:

$\begin{array}{rcl}& & ∆{\text{G}}^{\text{o}}\text{= - nF}{{\text{E}}^{\text{o}}}_{\text{cell}}\\ & & \text{= - 4×96500×1.23}\\ & & \text{= - 474,780J}\end{array}$

The second reaction is:

${\text{Pb(s) + PbO}}_{\text{2}}{\text{(s) + 2H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)}\to {\text{2PbSO}}_{\text{4}}{\text{(s) + 2H}}_{\text{2}}\text{O(l)}$

The reactions are then divided into half.

$$\begin{gathered} {\text{OHR:Pb +H}}_{\text{2}}{\text{SO}}_{\text{4}}\to {\text{PbSO}}_{\text{4}}{\text{+ 2e}}^{\text{-}}{\text{+ 2H}}^{\text{+}} \\ {\text{RHR:PbO}}_{\text{2}}{\text{+H}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{+ 2e}}^{\text{-}}{\text{+ 2H}}^{\text{+}}\to {\text{PbSO}}_{\text{4}}{\text{+ 2H}}_{\text{2}}\text{O} \end{gathered}$$

Two electrons are involved in this reaction having the value as ${{\text{E}}^{\text{o}}}_{\text{cell}}\text{= 2.04 V}$

The value of $∆{\text{G}}^{\text{o}}$ then can be computed as:

$$\begin{gathered} ∆{\text{G}}^{\text{o}}\text{= - nF}{{\text{E}}^{\text{o}}}_{\text{cell}} \\ \text{= - 2×96500×2.04} \\ =\text{- 393,720 J} \end{gathered}$$

The third reaction is:

${\text{2Na(l) + FeCl}}_{\text{2}}\text{(s)}\to \text{2NaCl(s) + Fe(s)}$

The reactions are then divided into half.

$$\begin{gathered} \text{OHR:2Na}\to {\text{2Na}}^{\text{+}}{\text{+ 2e}}^{\text{-}} \\ {\text{RHR:Fe}}^{\text{2 +}}{\text{+ 2e}}^{\text{-}}\to \text{Fe} \end{gathered}$$

Two electrons are involved in this reaction having the value as ${{\text{E}}^{\text{o}}}_{\text{cell}}\text{= 2.35 V}$

The value of $∆{\text{G}}^{\text{o}}$ then can be computed as:

$$\begin{gathered} ∆{\text{G}}^{\text{o}}\text{= - nF}{{\text{E}}^{\text{o}}}_{\text{cell}} \\ \text{= - 2×96500×2.35} \\ \text{= - 453,550J} \end{gathered}$$

Therefore, the values are:

$$\begin{gathered} {\text{I 4 electrons,∆G}}^{\text{o}}\text{= - 474,780 J} \\ {\text{II 2 electrons,∆G}}^{\text{o}}\text{= - 393,720 J} \\ {\text{III 2 electrons,∆G}}^{\text{o}}\text{= - 453,550 J} \end{gathered}$$

${\text{w}}_{\text{max}}{\text{=∆G}}^{\text{o}}$

For the first reaction we obtain:

$\begin{array}{rcl}& & {\text{mass = 2MM}}_{{\text{H}}_{\text{2}}}{\text{+ MM}}_{{\text{O}}_{\text{2}}}\\ & & {\text{= 21.016 gH}}_{\text{2}}{\text{/mol + 32.00 gO}}_{\text{2}}\text{/mol}\\ & & \text{= 36.032}\\ & & \end{array}$

$$\begin{gathered} \frac{{\text{w}}_{\text{max}}}{\text{mass}}\text{=}\frac{\text{- 474,780 J}}{\text{36.032 g}}\text{×}\frac{\text{1k J}}{\text{1000 J}} \\ \text{= - 13.2 J/g} \end{gathered}$$

For the second reaction we obtain:

$$\begin{gathered} {\text{mass = 2MM}}_{{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{+ MM}}_{{\text{PbO}}_{\text{2}}}{\text{+ MM}}_{\text{Pb}} \\ {\text{= 298.09 gH}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{/mol + 239.20 g PbO}}_{\text{2}}\text{/mol + 207.2 g Pb/mol} \\ \text{= 624.58} \end{gathered}$$

$$\begin{gathered} \frac{{\text{w}}_{\text{max}}}{\text{mass}}\text{=}\frac{\text{- 393,720 J}}{\text{624.58 g}}\text{×}\frac{\text{1k J}}{\text{1000 J}} \\ \text{= - 0.61 J/g} \end{gathered}$$

For the third reaction we obtain:

$$\begin{gathered} {\text{mass = 2MM}}_{\text{Na}}{\text{+ MM}}_{{\text{FeCl}}_{\text{2}}} \\ {\text{= 222.99 g Na/mol + 126.75 g FeCl}}_{\text{2}}\text{/mol} \\ \text{= 172.73} \end{gathered}$$

$$\begin{gathered} \frac{{\text{w}}_{\text{max}}}{\text{mass}}\text{=}\frac{\text{- 453,550 J}}{\text{172.73 g}}\text{×}\frac{\text{1k J}}{\text{1000 J}} \\ \text{= - 2.62 J/g} \end{gathered}$$

Due to the modest mass of its reactants, reaction I has the greatest ratio. Because of the huge mass of its reactants, Cell II has the lowest ratio.

Therefore, the values are:

$$\begin{gathered} \text{I 13.2 kJ/g} \\ \text{II - 0.61 kJ/g} \\ \text{III - 2.62 kJ/g} \end{gathered}$$

As the reaction I has the smallest mass, it has the greatest ratio, whereas reaction II has the lowest ratio owing to the huge mass of reactants.