In the production of Al metal by electrolysis, approximately 15 kW⋅h of electrical energy are used to produce each kilogram of Al from an Al³⁺. Assuming that all the charge passed is used in producing aluminum, at what voltage does the cell operate?
To determine the voltage at which the cell operates, we can use the equation:
\[ \text{Electrical energy} = \text{Voltage} \times \text{Electrical charge} \times \text{Time} \]
Given that approximately 15 kWh (kilowatt-hours) of electrical energy are used to produce each kilogram of aluminum, we need to convert this value to joules (J) for further calculation.
1 kilowatt-hour (kWh) = 3.6 million joules (MJ)
Therefore, 15 kWh = 15 x 3.6 MJ = 54 MJ
The electrical charge can be calculated using Faraday's law, which states that 1 Faraday (F) of charge is equal to 96,485 coulombs (C). Since we want to find the voltage at which the cell operates, we need to calculate the charge in coulombs.
The charge (Q) can be calculated using the equation:
\[ Q = \frac{54 \times 10^6}{96,485} \, \text{C} \]
Now that we have the charge, we can rearrange the equation to solve for the voltage:
\[ \text{Voltage} = \frac{\text{Electrical energy}}{\text{Charge}} \]
Substituting the values into the equation:
\[ \text{Voltage} = \frac{54 \times 10^6 \text{ J}}{\frac{54 \times 10^6}{96,485} \, \text{C}} \]
Simplifying the equation, we get:
\[ \text{Voltage} = 96,485 \, \text{V} \]
Therefore, the cell operates at approximately 96,485 volts.