Question

The production of ${\mathbf{\text{H}}}_{\mathbf{\text{2}}}$ gas by the electrolysis of water typically requires about$\mathbf{\text{400 kJ}}$ of energy per mole.

(a) Use the relationship between work and cell potential$\left(\text{Section 21.4}\right)$ to calculate the minimum work needed to form${\mathbf{\text{1.0 mol ofH}}}_{\mathbf{\text{2}}}$ gas at a cell potential of$\mathbf{\text{1.24 V.}}$

(b) What is the energy efficiency of the cell operation?

(c) Find the cost of producing ${\mathbf{\text{500. mol ofH}}}_{\mathbf{\text{2}}}$ if electricity is$0.06per$\mathbf{kilowatt}\mathbf{·}\mathbf{hour}$$\left(\text{1 watt}·\text{second = 1 joule}\right)\mathbf{\text{.}}$

Short answer

a) The minimum work needed is $3589244.2\mathrm{J}$.

b) The efficiency of the cell is $\text{89.73\%}$.

c) The cost of production of ${\text{500. mol ofH}}_{\text{2}}$is $$\text{2.991.}$

Step-by-step solution

Concept Introduction

The process in which the electricity is used to breakdown the water into two gases,oxygen gas as well as hydrogen gas is known as electrolysis of water.

Calculating the Minimum Work

a) Let us calculate the minimum work needed.

The electrical work done (minimal work, $\mathrm{\omega }$, in J) can be expressed as a product of cell potential and the flowing charge The flowing charge is expressed as the charge is then passed through the cell per mol of the electron transferred:

role="math" localid="1663324679787" $\mathrm{\omega }=-\mathrm{n}\times \mathrm{F}\times {\mathrm{E}}_{\mathrm{cell}}$

In the case of the reaction of water electrolysis, two ${\text{e}}^{\text{-}}$flow per mol of ${\text{H}}_{\text{2}}$produced

${\text{2H}}_{\text{2}}{\text{O(l) + 2e}}^{\text{-}}\to {\text{H}}_{\text{2}}{\text{(g) + 2OH}}^{\text{-}}\text{(aq)}$

If $1\mathrm{mol}\mathrm{of}{\mathrm{H}}_2$is produced at a cell potential of $1.24\mathrm{V}$,

$$\begin{gathered} \mathrm{\omega }=3{\mathrm{mole}}^{-}\times \frac{96485\mathrm{J}}{\mathrm{V}\times {\mathrm{mole}}^{-}}\times 1.24\mathrm{V} \\ \mathrm{\omega }=358924.2\mathrm{J} \end{gathered}$$

Therefore, the minimum work needed is$3589244.2\text{J}$

Calculating the Energy Efficiency of the Cell

b) Let us calculate the energy efficiency of the cell.

The efficiency of the cell then can be calculated, if $400\mathrm{kJ}$of energy per mole of ${\mathrm{H}}_2$is required,

$$\begin{gathered} \mathrm{Efficiency}=\frac{\mathrm{\omega }}{{\mathrm{\omega }}_{\max }}\times 100\% \\ \mathrm{Efficiency}=\frac{3589244.2\mathrm{J}}{400\times {10}^3\mathrm{J}}\times 100\% \\ \mathrm{Efficiency}=89.73\% \end{gathered}$$

The energy efficiency of the cell is$89.73\%.$

Calculating the Cost of Production of Hydrogen Gas

c) Let us do the calculations.

To produce $500\mathrm{mol}\mathrm{of}{\mathrm{H}}_2,$the minimum work required is,

$\begin{array}{rcl}\mathrm{\omega }& =& 500\mathrm{mol}\times 358924.2\mathrm{J}/\mathrm{mol}\\ \mathrm{\omega }& =& 179462100\mathrm{J}\\ & =& 179462.100\mathrm{kJ}\end{array}$

Converting$(\mathrm{kW}\times \mathrm{h})\mathrm{to}(\mathrm{W}\times \mathrm{s}),$

$\begin{array}{rcl}1\mathrm{kW}\times \mathrm{h}& =& 3.6\times {10}^6\\ \mathrm{W}\times \mathrm{s}& =& 3.6\times {10}^6\mathrm{J}\end{array}$

Given that the work is $179462100\mathrm{J}$and the price is$0.06\mathrm{per}\mathrm{kW}\times \mathrm{h},$

$$\begin{gathered} Cos\mathrm{t}=\frac{0.06}{3.6\times {10}^6\mathrm{J}}\times 179462100\mathrm{J} \\ Cos\mathrm{t}=2.991 \end{gathered}$$

The cost of producing $500\mathrm{mol}$of hydrogen gas is $2.991