Q22.45P - Chemistry: Molecular Nature Of Matter And Change

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 ${\mathbf{\text{1.0 mol of H}}}_{\mathbf{\text{2 }}}$ 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 ${\mathbf{\text{500. mol of H}}}_{\mathbf{\text{2}}}$ if electricity is $0.06per $\mathbf{kilowatt}\mathbf{·}\mathbf{hour}$$\left(\text{1 watt}·\text{second = 1 joule}\right)\mathbf{\text{.}}$

Step-by-Step Solution

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 ${\text{500. mol of H}}_{\text{2}}$ is $$\text{2.991.}$

Step 1: 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.

Step 2: 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:

$\mathrm{\omega }=-\mathrm{n}\times \mathrm{F}\times {\mathrm{E}}_{\mathrm{cell}}$ 

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 $3589244.2 \text{J}$

Step 3: 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 $89.73\%.$

Step 4: 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