Question
The energy consumption in electrolysis depends on the product of the charge and the voltage [volt $\times \text { coulomb }=\mathrm{V} \cdot \mathrm{C}=\mathrm{J}(\text { joules })] .$ Determine the theoretical energy consumption per $1000 \mathrm{kg} \mathrm{Cl}_{2}$ produced in a diaphragm chlor-alkali cell (page 902 ) that operates at 3.45 V. Express this energy in (a) $\mathrm{kJ}$;(b) kilowatt-hours, kWh.
Step 1
This can be done by multiplying the given mass by 1000, since 1 kg = 1000 g. So, $1000 \, \text{kg} \, \text{Cl}_2 = 1000 \times 1000 \, \text{g} \, \text{Cl}_2 = 1 \times 10^6 \, \text{g} \, \text{Cl}_2$. Show more…
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The energy consumption in electrolysis depends on the product of the charge and the voltage [volt $\times \text { coulomb }=\mathrm{V} \cdot \mathrm{C}=\mathrm{J} \text { (joules) }]$. Determine the theoretical energy consumption per $1000 \mathrm{kg} \mathrm{Cl}_{2}$ produced in a diaphragm chlor-alkali cell (page 906 ) that operates at 3.45 V. Express this energy in (a) $\mathrm{kJ}$; (b) kilowatt-hours, kWh.
The production of $\mathrm{H}_{2}$ gas by the electrolysis of water typically requires about $400 \mathrm{~kJ}$ of energy per mole. (a) Use the relationship between work and cell potential (Section 21.4 to calculate the minimum work needed to form $1.0 \mathrm{~mol}$ of $\mathrm{H}_{2}$ gas at a cell potential of $1.24 \mathrm{~V}$ (b) What is the energy efficiency of the cell operation? (c) Find the cost of producing 500 . mol of $\mathrm{H}_{2}$ if electricity is 50.06 per kilowatt-hour ( 1 watt-second $=1$ joule).
The production of $\mathrm{H}_{2}$ gas by the electrolysis of water typically requires about 400 $\mathrm{kJ}$ of energy per mole. (a) Use the relationship between work and cell potential (Section 21.4$)$ to calculate the minimum work needed to form 1.0 mol of $\mathrm{H}_{2}$ gas at a cell potential of 1.24 $\mathrm{V}$ . (b) What is the energy efficiency of the cell operation? (c) Find the cost of producing 500 . mol of $\mathrm{H}_{2}$ if electricity is $\$ 0.06$ per kilowatt-hour ( 1 watt-second $=1$ joule).
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