Why is potassium usually not prepared electrolytically from one of its salts?

Short Answer

Expert verified
Potassium is not usually prepared electrolytically from its salts due to its high reactivity with water. The potassium produced during electrolysis could react violently with any water present, leading to potential dangers such as fires or explosions.

Step by step solution

01

Understanding Potassium

Potassium is a highly reactive metal. It belongs to Group 1 of the Periodic Table, the alkali metals. These metals react vigorously, and often violently, with water.
02

Understanding Electrolysis

Electrolysis is a process that uses an electric current to drive a non-spontaneous chemical reaction. It is often used to decompose chemical compounds, in a process known as electrolytic decomposition. Electrolysis can cause salts to decompose into their constituent elements.
03

The Problem with Potassium and Electrolysis

In the case of potassium, electrolytic decomposition of a potassium salt would produce potassium metal. However, because of potassium's high reactivity with water, the potassium produced would react violently with any water present. This could lead to fires or explosions. To avoid these dangers, potassium is typically prepared by other methods.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reactivity of Alkali Metals
Alkali metals such as potassium are quite fascinating due to their place in Group 1 of the Periodic Table. These metals, including lithium, sodium, and potassium, are characterized by their single valence electron, which makes them very reactive. Now, why is this single electron so important? It's because it's relatively easy for these atoms to lose that electron, turning them into positively charged ions.

When alkali metals come into contact with water, they engage in a chemical reaction that is both exothermic (releases heat) and quite dramatic. For example, when potassium meets water, it forms potassium hydroxide and hydrogen gas, and enough heat is released during this reaction to ignite the hydrogen. The reactivity increases as you move down the group, which means potassium is more reactive than sodium, which in turn is more reactive than lithium.

This high level of reactivity is why we seldom prepare potassium by electrolysis of its salts. We must handle these metals with extreme care to prevent unexpected and possibly violent reactions.
Electrolytic Decomposition
Electrolysis is akin to a magical process where an electric current compels a chemical compound to break up into its elemental parts. It's a bit like convincing otherwise content roommates (the elements) to move out on their own (decomposition). With sufficient electrical energy, compounds such as salts can be decomposed during electrolysis into their constituent elements.

In the case of potassium salts, electrolysis would strip away the elements other than potassium, leaving pure potassium metal behind. However, this is not as simple as packing up belongings and moving out. The electrolytic process needs to be carefully controlled. Unwanted reactions can occur, and in the case of highly reactive metals, these reactions can be hazardous. It is for this precise reason that alternative methods are preferred for obtaining pure alkali metals like potassium.
Chemical Safety in Potassium Handling
Handling potassium is not like handling play dough; it's not something we approach without serious precautions. Given its eagerness to react, especially with water, we need a game plan for safety. Chemical safety is paramount when dealing with such reactive substances.

For starters, working with potassium should take place in an environment free from water and moisture. A dry, non-reactive atmosphere, often using noble gases like argon, is preferred. Safety gear, including goggles, gloves, and protective clothing, should be worn to prevent contact with the skin. Also, fire extinguishers or other fire-quenching materials that are suitable for metal fires should be accessible—water can't be used because it would exacerbate a potassium fire.

In educational and professional settings, understanding and adhering to these safety practices could mean the difference between a standard procedure and a dangerous situation. Due to these stringent requirements, handling potassium is approached with the utmost respect for its reactive nature, underscoring why electrolysis is not the favored method for obtaining this metal.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The following are two reaction schemes involving magnesium. Scheme I: When magnesium burns in oxygen, a white solid (A) is formed. A dissolves in \(1 M \mathrm{HCl}\) to give a colorless solution (B). Upon addition of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) to \(\mathrm{B},\) a white precipitate is formed(C). On heating, C decomposes to \(D\) and a colorless gas is generated (E). When \(\mathrm{E}\) is passed through limewater [an aqueous suspension of \(\left.\mathrm{Ca}(\mathrm{OH})_{2}\right], \underline{\mathrm{a}}\) white precipitate appears (F). Scheme II:Magnesium reacts with \(1 M \mathrm{H}_{2} \mathrm{SO}_{4}\) to produce a colorless solution (G). Treating G with an excess of NaOH produces a white precipitate \((\mathrm{H}) . \mathrm{H}\) dissolves in \(1 M \mathrm{HNO}_{3}\) to form a colorless solution. When the solution is slowly evaporated, a white solid (I) appears. On heating I, a brown gas is given off. Identify \(\mathrm{A}-\mathrm{I}\) and write equations representing the reactions involved.

Write balanced equations for the following reactions: (a) the heating of aluminum carbonate; (b) the reaction between \(\mathrm{AlCl}_{3}\) and \(\mathrm{K} ;\) (c) the reaction between solutions of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) and \(\mathrm{Ca}(\mathrm{OH})_{2}\)

Which of the following compounds would require electrolysis to yield the free metals: \(\mathrm{Ag}_{2} \mathrm{~S}, \mathrm{CaCl}_{2}\) \(\mathrm{NaCl}, \mathrm{Fe}_{2} \mathrm{O}_{3}, \mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{TiCl}_{4} ?\)

What is wrong with the following procedure for obtaining magnesium?$$\begin{aligned}& \mathrm{MgCO}_{3} \longrightarrow \mathrm{MgO}(s)+\mathrm{CO}_{2}(g) \\\\\mathrm{MgO}(s)+\mathrm{CO}(g) & \longrightarrow \mathrm{Mg}(s)+\mathrm{CO}_{2}(g)\end{aligned}$$

A \(0.450-\mathrm{g}\) sample of steel contains manganese as an impurity. The sample is dissolved in acidic solution and the manganese is oxidized to the permanganate ion \(\mathrm{MnO}_{4}^{-}\). The \(\mathrm{MnO}_{4}^{-}\) ion is reduced to \(\mathrm{Mn}^{2+}\) by reacting with \(50.0 \mathrm{~mL}\) of \(0.0800 \mathrm{M} \mathrm{FeSO}_{4}\) solution. The excess \(\mathrm{Fe}^{2+}\) ions are then oxidized to \(\mathrm{Fe}^{3+}\) by \(22.4 \mathrm{~mL}\) of \(0.0100 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} .\) Calculate the percent by mass of manganese in the sample.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free