Calculate the current that must be passed into a solution of aluminum salt for 1 hr 30 minutes in order to deposit 1.5 g of Aluminum. $(Al = 27)$
Answer (2)
Calculate the time in seconds: t = 90 × 60 = 5400 seconds.
Calculate the number of moles of Aluminium: n = 27 1.5 = 0.0556 moles.
Calculate the number of moles of electrons: n e = 3 × 0.0556 = 0.1667 moles.
Calculate the current: I = 5400 0.1667 × 96500 = 2.978 Amperes.
The required current is 2.978 Amperes .
Explanation
Understanding the Problem We are given the mass of aluminum deposited, the atomic weight of aluminum, and the time taken for the deposition. We need to find the current required for this process.
Calculating Time in Seconds First, let's calculate the total time in seconds: t = 90 × 60 = 5400 seconds
Calculating Moles of Aluminum Next, we determine the number of moles of Aluminum deposited: n = atomic weight mass = 27 1.5 = 0.0556 moles
Calculating Moles of Electrons Since Aluminium ion has a charge of +3 ( A l 3 + ), 3 moles of electrons are required to deposit 1 mole of Aluminium. Therefore, the number of moles of electrons required is: n e = 3 × n = 3 × 0.0556 = 0.1667 moles
Calculating Total Charge Now, we calculate the total charge required using Faraday's constant (F = 96500 C/mol): Q = n e × F = 0.1667 × 96500 = 16083.33 Coulombs
Calculating the Current Finally, we calculate the current using the formula: I = t Q = 5400 16083.33 = 2.978 Amperes
Final Answer Therefore, the current that must be passed into the solution is approximately 2.978 Amperes.
Examples
Electrolysis, the process described in this problem, is used in various industrial applications. For example, it's used to refine metals like copper and aluminum, extract pure elements from their compounds, and electroplate objects with a thin layer of metal for protection or decoration. Understanding the relationship between current, time, and the amount of substance deposited is crucial in optimizing these processes for efficiency and cost-effectiveness. For instance, in aluminum refining, controlling the current allows precise deposition of aluminum, ensuring high purity and minimizing waste. The principles used here are also applicable in designing batteries and fuel cells, where controlling ion flow is essential for energy storage and conversion.
To deposit 1.5 g of aluminum in 1 hour and 30 minutes, a current of approximately 2.978 Amperes is required. This was calculated using the total charge needed and the time in seconds. Considering aluminum's ionization state, 3 moles of electrons are necessary to deposit 1 mole of aluminum.
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