/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 At \(900^{\circ} \mathrm{C}\) ti... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 20: Problem 17

At \(900^{\circ} \mathrm{C}\) titanium tetrachloride vapor reacts with molten magnesium metal to form solid titanium metal and molten magnesium chloride. (a) Write a balanced equation for this reaction. (b) What is being oxidized, and what is being reduced? (c) Which substance is the reductant, and which is the oxidant?

Short Answer

Expert verified
a) The balanced equation for the reaction is \(TiCl_4 (g) + 2Mg (l) \rightarrow Ti (s) + 2MgCl_2 (l)\). b) Magnesium is being oxidized and titanium is being reduced. c) The reductant is molten magnesium (Mg) and the oxidant is titanium tetrachloride vapor (TiCl4).

Step by step solution

01

Write the reaction equation

To write the balanced equation, first write down the reactants and products: Reactants: Titanium tetrachloride vapor (TiCl4) and molten magnesium metal (Mg) Products: Solid titanium metal (Ti) and molten magnesium chloride (MgCl2) Now, we need to balance the equation: TiCl4 (g) + 2Mg (l) 鈫� Ti (s) + 2MgCl2 (l) The balanced equation is: \(TiCl_4 (g) + 2Mg (l) \rightarrow Ti (s) + 2MgCl_2 (l)\) ##Step 2: Identify the oxidation states of each element before and after the reaction##
02

Determine the oxidation states

We will determine the oxidation state of each element before and after the reaction: Before the reaction: - Ti in TiCl4: +4 - Cl in TiCl4: -1 - Mg: 0 After the reaction: - Ti: 0 - Cl in MgCl2: -1 - Mg in MgCl2: +2 ##Step 3: Determine which species is being oxidized and which is being reduced##
03

Identify oxidized and reduced species

In this reaction, the oxidation states of the elements change as follows: - Ti: +4 (TiCl4) 鈫� 0 (Ti), oxidation state decreases, so Ti is being reduced. - Mg: 0 (Mg) 鈫� +2(MgCl2), oxidation state increases, so Mg is being oxidized. ##Step 4: Identify the reductant and the oxidant##
04

Determine the reductant and the oxidant

The reductant is the species that reduces the other species by losing electrons. The oxidant is the species that oxidizes the other species by gaining electrons. In our reaction: - Mg (molten magnesium) is being oxidized, so it's the reductant. - TiCl4 (titanium tetrachloride vapor) is being reduced, so it's the oxidant. In conclusion: a) The balanced equation for the reaction is \(TiCl_4 (g) + 2Mg (l) \rightarrow Ti (s) + 2MgCl_2 (l)\). b) Magnesium is being oxidized and titanium is being reduced. c) The reductant is molten magnesium (Mg) and the oxidant is titanium tetrachloride vapor (TiCl4).

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 91影视!

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

Mercuric oxide dry-cell batteries are often used where a high energy density is required, such as in watches and cameras. The two half-cell reactions that occur in the battery are \(\mathrm{HgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-}--\rightarrow \mathrm{Hg}(l)+2 \mathrm{OH}^{-}(a q)\) \(\mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ZnO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-}\) (a) Write the overall cell reaction. (b) The value of \(E_{\text {red }}^{\circ}\) for the cathode reaction is \(+0.098 \mathrm{~V}\). The overall cell potential is \(+1.35 \mathrm{~V}\). Assuming that both half-cells operate under standard conditions, what is the standard reduction potential for the anode reaction? (c) Why is the potential of the anode reaction different than would be expected if the reaction occurred in an acidic medium?

A plumber's handbook states that you should not connect a copper pipe directly to a steel pipe because electrochemical reactions between the two metals will cause corrosion. The handbook recommends you use, instead, an insulating fitting to connect them. What spontaneous redox reaction(s) might cause the corrosion? Justify your answer with standard emf calculations.

Indicate whether each of the following statements is true or false: (a) If something is oxidized, it is formally losing electrons. (b) For the reaction \(\mathrm{Fe}^{3+}(a q)+\mathrm{Co}^{2+}(a q)-\cdots\) \(\mathrm{Fe}^{2+}(a q)+\mathrm{Co}^{3+}(a q), \mathrm{Fe}^{3+}(a q)\) is the reducing agent and \(\mathrm{Co}^{2+}(a q)\) is the oxidizing agent. (c) If there are no changes in the oxidation state of the reactants or products of a particular reaction, that reaction is not a redox reaction.

(a) What is the difference between a battery and a fuel cell? (b) Can the "fuel" of a fuel cell be a solid? Explain.

Hydrazine \(\left(\mathrm{N}_{2} \mathrm{H}_{4}\right)\) and dinitrogen tetroxide \(\left(\mathrm{N}_{2} \mathrm{O}_{4}\right)\) form a self-igniting mixture that has been used as a rocket propellant. The reaction products are \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). (a) Write a balanced chemical equation for this reaction. (b) What is being oxidized, and what is being reduced? (c) Which substance serves as the reducing agent, and which as the oxidizing agent?
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Kinetics

Read Explanation

Making Measurements

Read Explanation

Inorganic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation
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.