Question

Time required to deposit 1 millimol of aluminium metal by the passage of \(9.65\) ampere of current through aqueous solution of aluminium ion, is (a) \(30 \mathrm{~s}\) (b) \(10 \mathrm{~s}\) (c) \(30,000 \mathrm{~s}\) (d) \(10,000 \mathrm{~s}\)

Short answer

The time required is 30 seconds, which is option (a).

Step-by-step solution

Understand the Problem

We need to find the time required to deposit 1 millimol of aluminium using a current of 9.65 A. The reaction involves the reduction of aluminium ions to aluminium metal, which requires 3 moles of electrons per mole of aluminium, as represented by the half-equation: \( ext{Al}^{3+} + 3e^- \rightarrow ext{Al} \).

Use Faraday's Law of Electrolysis

According to Faraday's first law of electrolysis, the mass (or moles) of a substance deposited at an electrode is directly proportional to the quantity of electricity (charge) passed through the electrolyte. The mathematical form is: \( m = \frac{Q}{F} \cdot \frac{1}{n} \cdot M \), where \( Q \) is the charge, \( F \) is Faraday's constant, \( n \) is the number of electrons required, and \( M \) is the molar mass. But, for molar calculation only \( n \) and \( F \) play a role.

Calculate the Charge Required

The molar quantity of charge required for the deposition of 1 mole of aluminium is \( 3F \), as \( n = 3 \). For depositing 1 millimol (i.e., \( 0.001 \) mol), the charge required would be \( 0.001 \times 3 \times 96485 \text{ C/mol} = 289.455 \text{ C} \).

Relate Charge to Current and Time

The relationship between charge (\( Q \)), current (\( I \)), and time (\( t \)) is given by \( Q = I \cdot t \). Rearranging this formula to find \( t \), we get \( t = \frac{Q}{I} \). Substituting the known values, \( t = \frac{289.455}{9.65} = 30 \text{ s} \).

Conclude with Answer

After calculating, the time required is found to be 30 seconds, which matches option (a).

Key concepts

Electrochemistry

Electrochemistry is a branch of chemistry that studies the relationship between electricity and chemical reactions. It involves the movement of electrons from one substance to another, usually in an electrolyte solution.

In electrochemical reactions, chemical changes produce electric current or electricity can cause chemical reactions. These reactions occur at the electrodes within an electrochemical cell.

Electrochemistry encompasses various processes, like electrolysis, which is essential for electroplating, battery technology, and metal extraction. Faraday's Law of Electrolysis is crucial in understanding how electric currents can drive chemical changes, such as in this exercise where we deposit aluminum by passing a current through an electrolyte. Understood by students, this forms the basis for manipulating and applying electrochemical processes in real-world applications.

Reduction of aluminium ions

In this exercise, the focus is on the reduction of aluminum ions. Reduction involves gaining electrons and is essential in converting ionic forms to their pure metallic form.

Aluminum ions, \(\text{Al}^{3+}\), are reduced at the cathode through the equation:

• \( \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \)

This equation shows that each aluminum ion requires three electrons (\(3e^-\)) to be reduced to aluminum metal. Understanding this process is key in electroplating and industrial aluminum production. It involves manipulating the ionic species under controlled electrochemical conditions to obtain desired metal forms.

By mastering ion reduction, students can better appreciate complex processes like electrolysis, which crucially impact industries.

Charge calculation using current

Understanding charge calculation using current is a fundamental concept in electrochemistry. The relation between charge, current, and time is given by the equation:

• \( Q = I \cdot t \)

Where \( Q \) is the charge in coulombs, \( I \) is the current in amperes, and \( t \) is the time in seconds.

In our exercise, we applied this relationship to determine how long it takes to deposit aluminum. Given \( I = 9.65 \) amperes, we calculated that depositing 1 millimol of aluminum requires 289.455 coulombs.

To find the time, we rearrange the formula:

• \( t = \frac{Q}{I} \)

Substituting the values gives us \( t = \frac{289.455}{9.65} = 30 \text{ s} \). This demonstrates how electrochemical principles link electric current with process timing, enabling practical applications like efficient material deposition.