Question

Write balanced equations for the following reactions described in the chapter. (a) \(\operatorname{Sc}(\text { l) is produced by the electrolysis of } \mathrm{Sc}_{2} \mathrm{O}_{3}\) dis solved in \(\mathrm{Na}_{3} \mathrm{ScF}_{6}(1)\) (b) Cr(s) reacts with HCl(aq) to produce a blue solution containing \(\mathrm{Cr}^{2+}(\mathrm{aq})\) (c) \(\mathrm{Cr}^{2+}(\text { aq })\) is readily oxidized by \(\mathrm{O}_{2}(\mathrm{g})\) to \(\mathrm{Cr}^{3+}(\mathrm{aq})\) (d) \(\mathrm{Ag}(\mathrm{s})\) reacts with concentrated \(\mathrm{HNO}_{3}(\mathrm{aq}),\) and \(\mathrm{NO}_{2}(\mathrm{g})\) is evolved.

Short answer

The balanced equations are: (a) \(2 \ \mathrm{Sc}_{2} \mathrm{O}_{3}(a q) + 3 \ \mathrm{Na}_{3} \mathrm{ScF}_{6}(l) \rightarrow 4\ \mathrm{Sc}(l) + 18\ \mathrm{NaF}(a q) + 3\ \mathrm{O}_{2}(g)\), (b) \(\mathrm{Cr}(s) + 2 \ \mathrm{HCl}(a q) \rightarrow \mathrm{Cr}^{2+}(a q) + 2\ \mathrm{Cl}^-(a q) + \mathrm{H}_{2}(g)\), (c) \(\mathrm{4Cr}^{2+}(a q) + \mathrm{O}_{2}(g) \rightarrow \mathrm{4Cr}^{3+}(a q)\), (d) \(\mathrm{Ag}(s) + \mathrm{2HNO}_{3}(a q) \rightarrow \mathrm{Ag}^+(a q) + 2\ \mathrm{NO}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(l)\)

Step-by-step solution

- Scandium production

For the production of Scandium, we have Scandium ions in Sc2O3 form and the reaction medium as Na3ScF6 producing Scandium metal. The reaction can be represented as: \[2 \ \mathrm{Sc}_{2} \mathrm{O}_{3}(a q) + 3 \ \mathrm{Na}_{3} \mathrm{ScF}_{6}(l) \rightarrow 4\ \mathrm{Sc}(l) + 18\ \mathrm{NaF}(a q) + 3\ \mathrm{O}_{2}(g)\]

- Chromium reaction

For Chromium reaction with HCl, we have Cr as solid and HCl in aqueous state producing Cr2+ in aqueous medium. The HCl is reduced to H2 gas and the chlorine becomes chloride ions. The reaction is: \[\mathrm{Cr}(s) + 2 \ \mathrm{HCl}(a q) \rightarrow \mathrm{Cr}^{2+}(a q) + 2\ \mathrm{Cl}^-(a q) + \mathrm{H}_{2}(g)\]

- Chromium oxidation

For the oxidation of Cr2+ by O2, we are given Cr2+ is oxidized by O2 to produce Cr3+. The reaction can be represented as: \[\mathrm{4Cr}^{2+}(a q) + \mathrm{O}_{2}(g) \rightarrow \mathrm{4Cr}^{3+}(a q)\]

- Silver reaction

For Silver reacting with HNO3, we have Ag in solid state and reacting with nitric acid, where Nitrogen dioxide is evolved. The reduction of concentrated nitric acid produces NO2 and water while the silver is left as silver ions in solution. The reaction can be represented as: \[\mathrm{Ag}(s) + \mathrm{2HNO}_{3}(a q) \rightarrow \mathrm{Ag}^+(a q) + 2\ \mathrm{NO}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(l)\]

Key concepts

Balancing Chemical Equations

Understanding how to balance chemical equations is fundamental in studying chemical reactions. When we balance an equation, we ensure that the number of atoms for each element is the same on both sides of the reaction. This reflects the Law of Conservation of Mass, which states that mass is neither created nor destroyed in a chemical reaction.

In the step-by-step solution, note how the equations are balanced:

• For Scandium production, we see that the coefficient of \(2\) in front of \(\mathrm{Sc}_{2} \mathrm{O}_{3}\) ensures that there are a total of four Scandium atoms on each side of the equation.
• In the Chromium reaction with HCl, each element has the same number of atoms on both sides, balancing the equation.
<.li>Lastly, for the reaction involving Silver with HNO\(_3\), balancing is achieved by ensuring there are two Nitrogen atoms and two Hydrogen atoms resulting in the formation of two NO\(_2\) molecules and one H\(_2\)O molecule on the right-hand side.

Whenever you're asked to balance an equation, start by counting the number of atoms of each element in the reactants and products and adjust the coefficients until you have the same number of each kind of atom on both sides.

Electrolysis Reactions

Electrolysis reactions involve using electrical energy to drive a non-spontaneous chemical reaction. During electrolysis, compounds are decomposed into their constituent elements or simpler compounds. A classic example is the electrolysis of water into hydrogen and oxygen gas.

The given exercise specifically looks at the electrolysis reaction where Scandium metal is produced. This reaction occurs when an electric current passes through the compound Na\(_3\)ScF\(_6\), which results in the formation of Scandium metal. Such processes are vital in industries for extracting metals from their naturally occurring compounds. To grasp the concept, it is helpful to remember that electrolysis involves the movement of ions towards electrodes and the subsequent reactions that occur at these electrodes.

Redox Reactions

Redox reactions, or oxidation-reduction reactions, are processes where electrons are transferred between substances. They involve the simultaneous processes of oxidation (loss of electrons) and reduction (gain of electrons). The understanding of redox reactions is critical when dealing with the chemistry of metals.

In our solution, we observed a redox reaction when Chromium (Cr) metal reacted with HCl to produce Cr\(^{2+}\). Cr is oxidized while HCl is reduced. Another redox reaction occurs when Cr\(^{2+}\) ions are further oxidized by oxygen to form Cr\(^{3+}\) ions in aqueous solution. By identifying the changes in oxidation states, students can better understand the electron transfer that characterizes these reactions.

Reaction of Metals with Acids

Metals react with acids to produce salts and hydrogen gas, typically; however, with different acids and metals, the products can vary. In our exercise, the reaction of Chromium with HCl is a good example of a metal reacting with an acid, resulting in the production of a salt (Chromium chloride) and hydrogen gas.

Another interesting reaction is when Silver reacts with nitric acid (HNO\(_3\)) which is concentrated. Unlike simpler acid-metal reactions, this reaction produces a salt (Silver nitrate), water, and a gas (NO\(_2\)), but not hydrogen gas, due to the oxidizing nature of nitric acid. It's important for students to recognize that the type of acid and the metal involved can influence the products formed in such reactions.