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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 exidant?

Short Answer

Expert verified
The balanced equation for the reaction is \(TiCl_4 + Mg \rightarrow Ti + MgCl_2\). In this reaction, titanium tetrachloride (TiCl4) is reduced, and magnesium (Mg) is oxidized. The reductant is magnesium (Mg), and the exidant is titanium tetrachloride (TiCl4).

Step by step solution

01

Write the balanced equation.

First, we write the unbalanced equation as follows: \(TiCl_4 + Mg \rightarrow Ti + MgCl_2\) Next, we balance the equation. Since there are 2 moles of chlorine atoms on each side, the equation is already balanced: \(TiCl_4 + Mg \rightarrow Ti + MgCl_2\)
02

Determine the oxidation state of each element in the reactants and products.

We can determine the oxidation states of each element in the reactants and products, which will help us identify which substance is being oxidized and which is being reduced. Oxidation state of Ti in TiCl4: +4 Oxidation state of Cl in TiCl4: -1 Oxidation state of Mg in Mg: 0 Oxidation state of Ti in Ti: 0 Oxidation state of Mg in MgCl2: +2 Oxidation state of Cl in MgCl2: -1
03

Identify the oxidized and reduced substances.

We will now identify which substances are being oxidized and which are being reduced by comparing the oxidation states before and after the reaction. Oxidation occurs when the oxidation state of a substance increases, while reduction occurs when the oxidation state of a substance decreases. From the oxidation state changes, we can see that: - Titanium is reduced because the oxidation state of Ti decreases from +4 in TiCl4 to 0 in Ti - Magnesium is oxidized because the oxidation state of Mg increases from 0 in Mg to +2 in MgCl2 Therefore, TiCl4 is being reduced, and magnesium is being oxidized.
04

Identify the reductant and exidant.

The reductant is the substance that causes another substance to be reduced by providing electrons and gets oxidized in the process. On the other hand, the exidant causes another substance to be oxidized by taking electrons and gets reduced in the process. From the previous step, we know that: - Magnesium is oxidized (loses electrons, increases in oxidation state) - Titanium tetrachloride is reduced (gains electrons, decreases in oxidation state) Therefore, magnesium is the reductant and titanium tetrachloride is the exidant. In summary, the balanced equation for the reaction is: \(TiCl_4 + Mg \rightarrow Ti + MgCl_2\) TiCl4 is reduced, Mg is oxidized, and the reductant is Mg, while the exidant is TiCl4.

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Key Concepts

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

Oxidation State
Understanding oxidation states is crucial for analyzing redox reactions. An oxidation state, also known as oxidation number, is a conceptual charge on an atom if all bonds were purely ionic with no covalent character. It helps determine how electrons are transferred in reactions.

For example, in titanium tetrachloride (\(TiCl_4\)), titanium (Ti) has an oxidation state of +4. Chlorine (Cl) has an oxidation state of -1. Here, the positive charge on titanium indicates it has lost electrons, as chlorine's negative charge indicates it gained electrons from titanium.

Examining the changes in oxidation states can indicate whether an element is oxidized or reduced during a reaction.
  • Oxidation involves an increase in oxidation state (loss of electrons).
  • Reduction involves a decrease in oxidation state (gain of electrons).
This clear distinction is essential to correctly interpret redox reactions, as displayed in the reaction of \(TiCl_4\) with magnesium (Mg), where titanium is reduced, and magnesium is oxidized.
Chemical Equation Balancing
Balancing chemical equations is a fundamental skill in chemistry that conserves mass and charge. In a reaction, the total number of each type of atom must be equal on both sides of the equation. This reflects the law of conservation of mass.

To balance the chemical equation \(TiCl_4 + Mg \rightarrow Ti + MgCl_2\), we must ensure each atom's count remains constant:
  • Chlorine atoms: 4 in \(TiCl_4\) and 4 in \(2 \times MgCl_2\), leading to 2 \(MgCl_2\) molecules on the product side.
  • Magnesium and titanium are balanced with 1 mole each, as both switch from element to compound and back in a 1:1 ratio.
After balancing, the equation remains \(TiCl_4 + Mg \rightarrow Ti + MgCl_2\). Here, the equation is already balanced, aligning with the conservation laws.

This process helps ensure that the reaction conforms to natural laws, illustrating the elegant balance inherent in chemical transformations.
Oxidant and Reductant Identification
In redox reactions, identifying the oxidant and reductant is a key step in understanding the electron transfer between reactants.

The oxidant, or oxidizing agent, gains electrons and is reduced in the process. Conversely, the reductant, or reducing agent, loses electrons and is oxidized. Identifying these agents relates to changes in oxidation states.

In the reaction \(TiCl_4 + Mg \rightarrow Ti + MgCl_2\):
  • Titanium tetrachloride (\(TiCl_4\)) is the oxidant, as it gains electrons (reduced) and reduces to metallic titanium (\(Ti\)).
  • Magnesium (Mg) is the reductant, as it loses electrons (oxidized) to become \(MgCl_2\).
Understanding oxidants and reductants helps unravel how energy is transferred via electron movements, enlightening areas of reaction mechanics and energetics in chemistry.

This classification can beautifully simplify complex mechanisms, shining light on the fundamental workings of redox reactions.

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Most popular questions from this chapter

Using the standard reduction potentials listed in Appendix E, calculate the equilibrium constant for each of the following reactions at \(298 \mathrm{~K}\) - (a) \(\mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Cu}^{2}+(a q)+2 \mathrm{Ag}(s)\) (b) \(3 \mathrm{Ce}^{4+}(a q)+\mathrm{Bi}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 3 \mathrm{Ce}^{3+}(a q)+\) \(\mathrm{BiO}^{+}(a q)+2 \mathrm{H}^{+}(a q)\) (c) \(\mathrm{N}_{2} \mathrm{H}_{5}^{+}(a q)+4 \mathrm{Fe}(\mathrm{CN})_{6}^{3-}(a q) \longrightarrow \mathrm{N}_{2}(\mathrm{~g})+\) \(5 \mathrm{H}^{+}(a q)+4 \mathrm{Fe}(\mathrm{CN})_{6}^{4}(a q)\)

(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 solation when it serves as the anode of a voltaic cell. (c) What is standard about the standard hydrogen electrode?

(a) What is meant by the term reduction? (b) On which side of a reduction half-reaction do the electrons appear? (c) What is meant by the term reductant? (d) What is meant by the term reducing agent?

In each of the following balanced oxidation-reduction equations, identify those elements that undergo changes in oxidation number and indicate the magnitude of the change in each case. (a) \(\mathrm{I}_{2} \mathrm{O}_{\mathrm{s}}(s)+5 \mathrm{CO}(\mathrm{g}) \longrightarrow \mathrm{I}_{2}(s)+5 \mathrm{CO}_{2}(\mathrm{~g})\) (b) \(2 \mathrm{Hg}^{2+}(a q)+\mathrm{N}_{2} \mathrm{H}_{4}(a q) \longrightarrow 2 \mathrm{Hg}(l)+\mathrm{N}_{2}(g)+4 \mathrm{H}^{+}(a q)\) (c) \(3 \mathrm{H}_{2} \mathrm{~S}(a q)+2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q) \longrightarrow 3 \mathrm{~S}(s)+\) \(2 \mathrm{NO}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\)

(a) Write the reactions for the discharge and charge of a nickel-cadmium (nicad) rechargeable battery. (b) Given the following reduction potentials, calculate the standard emf of the cell: $$ \begin{array}{r} \mathrm{Cd}(\mathrm{OH})_{2}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cd}(s)+2 \mathrm{OH}^{-}(a q) \\ E_{\mathrm{red}}^{\mathrm{e}}=-0.76 \mathrm{~V} \\ \mathrm{NiO}(\mathrm{OH})(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{e}^{-} \longrightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{OH}^{-}(a q) \\ E_{\text {red }}^{e}=+0.49 \mathrm{~V} \end{array} $$ (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.

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