Problem 89 - Chemistry Matter and Change

Question

For each reaction described, write the corresponding chemical equation without putting coefficients to balance it. Next, determine the oxidation state of each element in the equation. Then, write the two half-reactions, labeling which is oxidation and which is reduction. Finally, write a balanced equation for the reaction. \begin{equation} \begin{array}{l}{\text { a. Solid mercuric oxide is put into a test tube and gently }} \\ {\text { heated. Liquid mercury forms on the sides and in the }} \\\ {\text { bottom of the tube, and oxygen gas bubbles out from }} \\ {\text { the test tube. }} \\ {\text { b. Solid copper pieces are put into a solution of silver }} \\ {\text { nitrate. Silver metal appears and blution of silver }} \\ {\text { nitrate forms in the solution. }}\end{array} \end{equation}

Step-by-Step Solution

Answer

a. HgO → Hg + O₂ (balanced: 2HgO → 2Hg + O₂), b. Cu + AgNO₃ → Ag + Cu(NO₃)₂ (balanced: Cu + 2AgNO₃ → 2Ag + Cu(NO₃)₂).

Step 1: Writing Chemical Equations

For reaction (a), the decomposition of mercuric oxide, the unbalanced chemical equation is: \[ \text{HgO(s)} \rightarrow \text{Hg(l)} + \text{O}_2(g) \] For reaction (b), where copper reacts with silver nitrate, the unbalanced equation is: \[ \text{Cu(s)} + \text{AgNO}_3(aq) \rightarrow \text{Cu}( ext{NO}_3)_2(aq) + \text{Ag(s)} \].

Step 2: Determining Oxidation States

For reaction (a), in \( \text{HgO} \), Hg has an oxidation state of +2 and O is -2. In the products, Hg is 0 in \( \text{Hg(l)} \) and O is 0 in \( \text{O}_2 \).For reaction (b), \( \text{Cu(s)} \) is 0, Ag in \( \text{AgNO}_3 \) is +1, Ag in \( \text{Ag(s)} \) is 0, and Cu in \( \text{Cu(NO}_3)_2 \) is +2.

Step 3: Writing Half-Reactions for Reaction a

The oxidation half-reaction is: \[ \text{2Hg}^{2+} \rightarrow \text{2Hg}^0 \] (reduction: Hg gains electrons). The reduction half-reaction is: \[ \text{O}^{2-} \rightarrow \text{O}_2 \] (oxidation: O loses electrons).

Step 4: Writing Half-Reactions for Reaction b

The oxidation half-reaction is: \[ \text{Cu}^0 \rightarrow \text{Cu}^{2+} \] (loses 2 electrons). The reduction half-reaction is: \[ \text{Ag}^+ + e^- \rightarrow \text{Ag}^0 \] (gains electrons).

Step 5: Balancing Reaction a

Balance the decomposition of \( \text{HgO} \):\[ 2\text{HgO (s)} \rightarrow 2\text{Hg (l)} + \text{O}_2(g) \] The mercury and oxygen atoms are balanced, as well as the charges.

Step 6: Balancing Reaction b

Balance the equation based on electron transfers:Start by balancing silver: \[ 2\text{AgNO}_3 + \text{Cu} \rightarrow 2\text{Ag} + \text{Cu(NO}_3)_2 \]Ensure both charge and mass balance.

Key Concepts

Oxidation State

The oxidation state of an element signifies the degree of oxidation of an atom. It's a useful concept for keeping track of electrons in reactions, particularly redox reactions. For elements, their oxidation state is typically zero. In covalent compounds, it's calculated based on electronegativity differences, while in ionic compounds, it reflects ion charges.
Some general rules help determine oxidation states:

• Pure elements have an oxidation state of zero. For instance, \( ext{Hg} \) in liquid mercury and \( ext{O}_2 \) in oxygen gas both have oxidation states of zero.
• Alkali metals often have an oxidation state of +1, whereas alkaline earth metals have +2.
• Oxygen is typically -2, except in peroxides like \( ext{H}_2 ext{O}_2 \) or when bonded to fluorine.
• For a neutral compound, the sum of oxidation states must equal zero; for a polyatomic ion, it equals the ion's charge.

Understanding oxidation states helps us ascertain which elements undergo reduction and oxidation in a reaction, as seen when converting \( ext{HgO} \) to \( ext{Hg} \) and \( ext{O}_2 \), or \( ext{Cu} \) to \( ext{Cu(NO}_3)_2 \).

Half-Reactions

Half-reactions break down a redox process into separate oxidation and reduction processes which help clarify electron transfers. By identifying these exchanges, it's easier to balance complex reactions.
The oxidation half-reaction represents the loss of electrons. For instance, when copper undergoes oxidation, it goes from \( ext{Cu}^0 \) to \( ext{Cu}^{2+} \), releasing two electrons. In the context of mercuric oxide, \( ext{O}^{2-} \) loses electrons to form \( ext{O}_2 \).
Conversely, the reduction half-reaction showcases the gain of electrons. During the transformation involving silver nitrate, \( ext{Ag}^+ \) ions gain electrons to become \( ext{Ag}^0 \). Meanwhile, Hg receives electrons reducing from \( ext{Hg}^{2+} \) to \( ext{Hg}^0 \).
By visualizing these separate reactions, students can better comprehend how charge conservation dictates that electrons lost in oxidation are equivalent to those gained in reduction. This process assists in balancing the overall redox reaction.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions, outlining the substances involved and their transformations. An unbalanced equation simply shows reactants and products, without indicating proportionate amounts.

• Identifying reactants and products helps streamline equation writing. For reaction (a), mercuric oxide decomposes into mercury and oxygen, while in reaction (b), copper interacts with silver nitrate yielding silver and copper nitrate.
• Chemical equations need to reflect practical observations, such as oxygen evolving as a gas or silver precipitating as a metal.
• Writing correct formulas ensures that equations depict actual chemical processes. For instance, mercuric oxide is represented as \( ext{HgO} \), while silver nitrate is \( ext{AgNO}_3 \).

Initially, equations might not fulfill the requirement for balanced chemical processes, but they present a blueprint for further manipulations during the balancing act.

Balancing Chemical Equations

Balancing chemical equations is crucial as it honors the law of conservation of mass, ensuring that matter is neither created nor destroyed during a reaction. This balance is essential for both mass and charge in ionic reactions.
In starting the balancing process, each element is examined to ensure it appears equally on both sides of the reaction. For example, in decomposing \( ext{HgO} \) into mercury and oxygen, the correct balanced equation: \( 2 ext{HgO} ightarrow 2 ext{Hg} + ext{O}_2 \) confirms twin mercury atoms balance the oxygen molecule. Additionally, in the copper and silver nitrate reaction, balancing involves setting the stoichiometry of two silver atoms for one mole of \( ext{Cu} \), forming: \( 2 ext{AgNO}_3 + ext{Cu} ightarrow 2 ext{Ag} + ext{Cu(NO}_3)_2 \).

• Always ensure atoms for each element are equal on both sides.
• Match charges across ionic species, adjusting coefficients as necessary.
• Balance complex ions like \( ext{NO}_3^- \) as singular units initially and adjust accordingly after atoms and charges are settled.

Balancing forms the heart of stoichiometry, guaranteeing reactions occur as physically feasible transformations.