Calculate the number of kilowatt-hours of electricity required to produce 1.0 * 103 kg (1 metric ton) of aluminum by electrolysis of Al3+ if the applied voltage is 4.50 V and the process is 45% efficient.

Verified step by step guidance

1. Identify the electrochemical reaction for the production of aluminum: \( \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \).
2. Calculate the moles of aluminum in 1.0 \times 10^3 \text{ kg} using the molar mass of aluminum (26.98 \text{ g/mol}).
3. Determine the total charge required using Faraday's law: \( Q = n \times F \), where \( n \) is the moles of electrons and \( F \) is Faraday's constant (96485 C/mol).
4. Calculate the electrical energy required using the formula \( E = Q \times V \), where \( V \) is the applied voltage (4.50 V).
5. Adjust for the efficiency of the process (45%) to find the actual energy consumed, and convert this energy from joules to kilowatt-hours.

Key Concepts

Electrolysis

Electrolysis is a chemical process that uses electrical energy to drive a non-spontaneous reaction. In the case of aluminum production, an electric current is passed through a molten aluminum oxide solution, causing the Al3+ ions to gain electrons and form aluminum metal. Understanding the principles of electrolysis, including Faraday's laws, is essential for calculating the energy requirements for the process.

Efficiency

Efficiency in a chemical process refers to the ratio of useful output to the total input, often expressed as a percentage. In this context, the 45% efficiency indicates that only 45% of the electrical energy supplied is effectively used for the electrolysis of aluminum. This concept is crucial for determining the actual energy consumption needed to produce the desired amount of aluminum.

Kilowatt-Hour (kWh)

A kilowatt-hour (kWh) is a unit of energy equivalent to one kilowatt (1 kW) of power used for one hour. It is commonly used to measure electrical energy consumption. In this problem, calculating the total kWh required involves converting the energy needed for the electrolysis process into this unit, taking into account the voltage applied and the efficiency of the process.