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Chapter 20: Problem 21

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: \(TiCl_4 + 2Mg \rightarrow Ti + 2MgCl_2\). (b) Titanium is being reduced, and magnesium is being oxidized. (c) Magnesium is the reductant and titanium tetrachloride is the oxidant.

Step by step solution

01

(a) Write a balanced equation

First, we need to identify the reactants and products: Reactants: Titanium tetrachloride (TiCl4) vapor and molten magnesium metal (Mg) Products: Solid titanium metal (Ti) and molten magnesium chloride (MgCl2) Now, we can write the unbalanced equation: TiCl4 + Mg -> Ti + MgCl2 We need to balance the equation by ensuring equal number of atoms of each element on both sides: TiCl4 + 2Mg -> Ti + 2MgCl2 Now, the equation is balanced with an equal number of atoms on both sides.
02

(b) Identify what is being oxidized and what is being reduced

To determine which species is being oxidized and which is being reduced, we need to find their oxidation states before and after the reaction: 1. Titanium (Ti) in TiCl4: +4 (since Cl has an oxidation number of -1) 2. Magnesium (Mg) in Mg: 0 (element in its elemental state has an oxidation number of 0) 3. Titanium (Ti) in Ti: 0 (element in its elemental state has an oxidation number of 0) 4. Magnesium (Mg) in MgCl2: +2 (since Cl has an oxidation number of -1) Now we can identify the changes in oxidation numbers: - Ti undergoes a reduction in its oxidation state from +4 to 0. - Mg undergoes an increase in its oxidation state from 0 to +2. So, titanium is being reduced, and magnesium is being oxidized.
03

(c) Identify the reductant and the oxidant

Based on the changes in the oxidation states, we can now identify the reductant and the oxidant: Reductant: The reductant is the substance that causes the reduction of another species by giving away electrons. In this reaction, magnesium (Mg) is the reductant, as it gives away two electrons causing the reduction of titanium. Oxidant: The oxidant is the substance that causes the oxidation of another species by gaining electrons. In this reaction, titanium tetrachloride (TiCl4) is the oxidant, as it gains two electrons causing the oxidation of magnesium.

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Most popular questions from this chapter

For a spontaneous reaction \(\mathrm{A}(a q) \rightarrow \mathrm{A}^{-}(a q)+\) \(\mathrm{B}^{+}(a q),\) answer the following questions: (a) If you made a voltaic cell out of this reaction, what half-reaction would be occurring at the cathode, and what half-reaction would be occurring at the anode? (b) Which half-reaction from (a) is higher in potential energy? (c) What is the sign of \(E_{\text { cell }}^{\circ} ?[\) Section 20.3\(]\)

(a) How many coulombs are required to plate a layer of chromium metal 0.25 \(\mathrm{mm}\) thick on an auto bumper with a total area of 0.32 \(\mathrm{m}^{2}\) from a solution containing \(\mathrm{CrO}_{4}^{2-}\) ? The density of chromium metal is 7.20 \(\mathrm{g} / \mathrm{cm}^{3} .\) (b) What current flow is required for this electroplating if the bumper is to be plated in 10.0 s? (c) If the external source has an emf of \(+6.0 \mathrm{V}\) and the electrolytic cell is 65\(\%\) efficient, how much electrical power is expended to electroplate the bumper?

From each of the following pairs of substances, use data in Appendix E to choose the one that is the stronger oxidizing agent: $$ \begin{array}{l}{\text { (a) } \mathrm{Cl}_{2}(g) \text { or } \mathrm{Br}_{2}(l)} \\ {\text { (b) } \mathrm{Zn}^{2+}(a q) \text { or } \mathrm{Cd}^{2+}(a q)} \\ {\text { (c) } \mathrm{Cl}^{-}(a q) \text { or } \mathrm{ClO}_{3}(a q)} \\ {\text { (d) } \mathrm{H}_{2} \mathrm{O}_{2}(a q) \text { or } \mathrm{O}_{3}(\mathrm{g})}\end{array} $$

(a) Write the half-reaction that occurs at a hydrogen electrode in acidic aqueous solution when it serves as the cathode of a voltaic cell.(b) Write the half-reaction that occurs at a hydrogen electrode in acidic aqueous solution when it serves as the anode of a voltaic cell. (c) What is standard about the standard hydrogen electrode?

Given the following half-reactions and associated standard reduction potentials: $$ \begin{array}{c}{\text { AuBr }_{4}^{-}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s)+4 \mathrm{Br}^{-}(a q)} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad {E_{\mathrm{red}}^{\circ}=-0.86 \mathrm{V}} \\ {\mathrm{Eu}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Eu}^{2+}(a q)} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad {E_{\mathrm{red}}^{\circ}=-0.43 \mathrm{V}}\end{array} $$ $$ \begin{array}{r}{\mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{I}^{-}(a q)+2 \mathrm{OH}^{-}(a q)} \\\ {E_{\mathrm{red}}^{\circ}=+0.49 \mathrm{V}}\end{array} $$ (a) Write the equation for the combination of these half-cell reactions that leads to the largest positive emf and calculate the value. (b) Write the equation for the combination of half-cell reactions that leads to the smallest positive emf and calculate that value.
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