Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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) TiCl4 + 2 Mg → Ti + 2 MgCl2 (b) Mg is oxidized, TiCl4 is reduced. (c) Mg is the reductant, TiCl4 is the oxidant.

Step by step solution

01

Writing the balanced chemical equation

First, let's write down the reactants and products given in the problem. Reactants: titanium tetrachloride (TiCl4) and magnesium (Mg) Products: titanium (Ti) and magnesium chloride (MgCl2) Next, we'll write down the unbalanced chemical equation: TiCl4 + Mg → Ti + MgCl2 In order to balance the equation, we need to match the number of atoms of all elements on both sides of the reaction. TiCl4 has 1 Ti atom and 4 Cl atoms. Mg has 1 Mg atom. On the product side, Ti has 1 Ti atom, and MgCl2 has 1 Mg atom and 2 Cl atoms. By observing the species present, if we double the amount of MgCl2, the Cl atoms count will balance. TiCl4 + Mg → Ti + 2 MgCl2 Now, we have a balanced chemical equation: TiCl4 + 2 Mg → Ti + 2 MgCl2
02

Identifying oxidation and reduction

Oxidation is the process of losing electrons, and reduction is the process of gaining electrons. To find out which species is being oxidized or reduced, we need to determine the changes in the oxidation states between the reactants and products. TiCl4: Ti has an oxidation state of +4 (as it is sharing its 4 electrons with the four Cl atoms) Mg: Oxidation state is 0 Ti : Oxidation state is 0 MgCl2: Mg has an oxidation state of +2 (it gave its 2 electrons to the two Cl atoms) Now, comparing the oxidation states between reactants and products: - Ti goes from +4 in TiCl4 to 0 in Ti, which means Ti gains 4 electrons (reduction) - Mg goes from 0 to +2 in MgCl2, which means Mg loses 2 electrons (oxidation)
03

Identifying the reductant and oxidant

The reductant is the substance that loses electrons and gets oxidized, while the oxidant is the substance that gains electrons and gets reduced. In this case, magnesium (Mg) is being oxidized, meaning it loses electrons. Therefore, Mg is the reductant. Titanium tetrachloride (TiCl4) is being reduced (gaining electrons), so TiCl4 is the oxidant. In summary, (a) The balanced chemical equation for the reaction is: TiCl4 + 2 Mg → Ti + 2 MgCl2 (b) Magnesium (Mg) is being oxidized, and titanium tetrachloride (TiCl4) is being reduced. (c) The reductant is magnesium (Mg), and the oxidant is titanium tetrachloride (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 Vaia!

Key Concepts

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

Balancing Chemical Equations
Balancing chemical equations is a crucial skill in chemistry that ensures the law of conservation of mass is respected, meaning that the number of atoms of each element is the same on both sides of the equation. Let's delve into the process of balancing the equation TiCl4 + Mg → Ti + 2 MgCl2.

When balancing, start by listing the number of atoms of each element in the reactants and products. In our problem, we have titanium, chlorine, and magnesium to consider. The initial equation shows that chlorine is out of balance, with 4 atoms on the reactant side and only 2 on the product side. By doubling the amount of MgCl2, you balance the chlorine atoms. The titanium and magnesium atoms are already balanced—one atom each on both sides.

Thus, the corrected equation becomes TiCl4 + 2 Mg → Ti + 2 MgCl2, with equal numbers of each type of atom on either side, reflecting the mass conservation principle.
Oxidation States
Understanding oxidation states is key to grasping redox reactions. The oxidation state, often referred to as oxidation number, is a measure of the degree of oxidation of an atom in a substance. It helps us track the transfer of electrons during the reaction.

In our reaction, TiCl4 has a titanium (Ti) atom with an oxidation state of +4 and chlorine (Cl) with -1 each, summing to zero. Magnesium (Mg) in its elemental form has an oxidation state of 0, while in MgCl2, magnesium exhibits a +2 oxidation state because it has lost two electrons, and each chlorine has an oxidation state of -1. By comparing these states, we can see that Ti is reduced as it goes from +4 to 0, and Mg is oxidized as it goes from 0 to +2, indicating that the transfer of electrons is indeed occurring.
Identifying Reductant and Oxidant
In redox chemistry, identifying the reductant and oxidant is fundamental. The reductant (also known as the reducing agent) is the substance that donates electrons, thus being oxidized itself. Conversely, the oxidant (oxidizing agent) accepts electrons, undergoing reduction.

In the given equation, magnesium's oxidation state increases from 0 to +2, indicating it loses electrons and, hence, is the reductant. Titanium tetrachloride, on the other side, gains electrons as Ti's oxidation state decreases from +4 in TiCl4 to 0 in Ti, making TiCl4 the oxidant. Understanding the role of each substance in a redox reaction is essential, as it lays the groundwork for predicting reaction outcomes and crafting further chemical equations.

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

(a) Write the half-reaction that occurs at a hydrogen electrode in acidic aqueous solution when it serves as the anode of a voltaic cell. (b) The platinum electrode in a standard hydrogen electrode is specially prepared to have a large surface area. Why is this important? (c) Sketch a standard hydrogen electrode.

If the equilibrium constant for a two-electron redox reaction at \(298 \mathrm{~K}\) is \(1.5 \times 10^{-4},\) calculate the corresponding \(\Delta G^{\circ}\) and \(E_{\text {red }}^{\circ}\).

The Haber process is the principal industrial route for converting nitrogen into ammonia: $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)$$ (a) What is being oxidized, and what is being reduced? (b) Using the thermodynamic data in Appendix \(\mathrm{C}\), calculate the equilibrium constant for the process at room temperature. (c) Calculate the standard emf of the Haber process at room temperature.

A voltaic cell utilizes the following reaction: $$\mathrm{Al}(s)+3 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Al}^{3+}(a q)+3 \mathrm{Ag}(s)$$ What is the effect on the cell emf of each of the following changes? (a) Water is added to the anode half-cell, diluting the solution. (b) The size of the aluminum electrode is increased. (c) A solution of \(\mathrm{AgNO}_{3}\) is added to the cathode half-cell, increasing the quantity of \(\mathrm{Ag}^{+}\) but not changing its concentration. (d) \(\mathrm{HCl}\) is added to the \(\mathrm{AgNO}_{3}\) solution, precipitating some of the \(\mathrm{Ag}^{+}\) as \(\mathrm{AgCl}\).

(a) Write the reactions for the discharge and charge of a nickelcadmium (nicad) rechargeable battery. (b) Given the following reduction potentials, calculate the standard emf of the cell: $$ \begin{array}{r} \mathrm{Cd}(\mathrm{OH})_{2}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cd}(s)+\begin{array}{c} 2 \mathrm{OH}^{-}(a q) \\ E_{\mathrm{red}}^{\circ}=-0.76 \mathrm{~V} \end{array} \\ \mathrm{NiO}(\mathrm{OH})(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{e}^{-} \longrightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{OH}^{-}(a q) \\ E_{\mathrm{red}}^{\circ}=+0.49 \mathrm{~V} \end{array} $$ (c) A typical nicad voltaic cell generates an emf of \(+1.30 \mathrm{~V}\). Why is there a difference between this value and the one you calculated in part (b)? (d) Calculate the equilibrium constant for the overall nicad reaction based on this typical emf value.

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