Question

One ampere of current is passed for 9650 seconds through molten \(\mathrm{AlCl}_{3} .\) What is the weight in grams of \(\mathrm{Al}\) deposited at cathode? (Atomic weight of \(\mathrm{Al}=27)\) (a) \(0.9\) (b) \(9.0\) (c) \(0.18\) (d) \(18.0\)

Short answer

The correct answer is (a) 0.9 grams.

Step-by-step solution

Understanding Faraday's Laws

Faraday's First Law of Electrolysis states that the mass of a substance deposited or liberated at any electrode is directly proportional to the quantity of electricity passed through the electrolyte. The formula is \[ m = \frac{E \cdot C \cdot t}{F} \]where \( m \) is the mass of the substance (in grams), \( E \) is the molar mass (also called equivalent weight), \( C \) is the current (in amperes), \( t \) is the time (in seconds), and \( F \) is the Faraday constant, \( 96500 \text{ C/mol} \).

Calculating Moles of Electrons

The balanced equation for the reduction at the cathode is \[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \]Each mole of \( \text{Al}^{3+} \) requires 3 moles of electrons to be reduced, therefore the equivalent weight of Al is given by \( \frac{M}{3} \).

Substituting Values into Faraday's Formula

Using the modified formula \[ m = \frac{M \cdot C \cdot t}{n \cdot F} \]Substitute in the values: \( M = 27 \), \( C = 1 \text{ ampere} \), \( t = 9650 \text{ seconds} \), \( n = 3 \), and \( F = 96500 \text{ C/mol} \).

Calculation

Now, calculate the mass \( m \) using:\[ m = \frac{27 \cdot 1 \cdot 9650}{3 \cdot 96500} = \frac{27 \cdot 9650}{3 \cdot 96500} \]The fraction simplifies to \( \frac{27}{30} = 0.9 \text{ grams} \).

Conclusion

The weight of \( \text{Al} \) deposited at the cathode is \( 0.9 \text{ grams} \), which matches option (a).

Key concepts

Electrochemistry

When we hear electrochemistry, it often refers to chemical reactions where electricity is involved. It's the bridge between electrical and chemical systems. In particular, electrolysis, a process where electrical energy causes a chemical change, is a prime example. Let’s explain this using the provided exercise about depositing aluminum at the cathode.

In the exercise, we dealt with aluminum chloride, \(\text{AlCl}_3\), undergoing electrolysis. This involves passing current through it to obtain aluminum metal. Electrolytic cells, which drive this process, separate compounds into their constituent elements using electricity. This conversion of energy exemplifies why electrochemistry is a critical field of study.

• Electrochemical reactions involve the transfer of electrons, facilitating the movement of ions in the solution.
• The direction of current dictates the type of reaction at each electrode; oxidation happens at the anode, while reduction occurs at the cathode.
• Quantifying electrochemical processes requires understanding concepts like Faraday's Laws, which describe the relationship between electric charge and the substance amount altered during electrolysis.

Remember that in these reactions, ions travel towards electrodes where they either gain or lose electrons, forming new substances.

Molar Mass Calculation

Molar mass calculation is a fundamental skill in chemistry, helping us relate the mass of a substance to the amount in moles. This is especially important in electrochemistry when applying Faraday's Laws. Molar mass, specially in our context, is synonymous with equivalent weight.

To calculate the mass of aluminum deposited during electrolysis, we need to understand the relationship between the molar mass of aluminum and the electrons required for its reduction. This is a straightforward calculation: \(m = \frac{E \cdot C \cdot t}{F}\), where \(E\), the molar mass (equivalent weight), ensures that the calculation aligns with the chemical equation.

• The molar mass of aluminum is 27, but since we are concerned with a single aluminum ion requiring 3 electrons, the equivalent weight equals the molar mass divided by the electron moles involved.
• This gives \(\frac{27}{3}\) or 9, correlating molar quantities with the electrons needed, making the calculation feasible without ambiguities.
• This simplification allows us to calculate the mass of aluminum precisely when a current is passed through the electrolyte.

Understanding molar mass calculations deeply is crucial for tackling similar problems efficiently.

Chemical Reactions

Chemical reactions, the heart of chemistry, involve the transformation of reactants into products through bond formation and breaking. In our exercise, the reaction focuses on aluminum ion reduction. This involves electrons congregating at the cathode, reducing \(\text{Al}^{3+}\) ions to metallic aluminum.

The balanced equation: \[\text{Al}^{3+} + 3e^- \rightarrow \text{Al}\] indicates the stoichiometry of the reaction. Aluminium ions (\(\text{Al}^{3+}\) require three electrons to become stable metallic aluminum. Thus, identification of reactants products and understanding balance equations are crucial in predicting quantities formed.

• The process is highly dependent on the movement of electrons; hence understanding oxidation-reduction reactions enables efficient application of electrolysis.
• The prior knowledge of periodic tables assists in deducing reactive possibilities.
• Mastery of balancing chemical equations allows tackling various electrochemical processes with confidence.

Through studying this reaction, you can appreciate how electricity drives chemical changes, connecting electrical applications to tangible chemical results.