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. a. Solid mercuric oxide is put into a test tube and gently heated. Liquid mercury forms on the sides and in the bottom of the tube, and oxygen gas bubbles out from the test tube. b. Solid copper pieces are put into a solution of silver nitrate. Silver metal appears and blue copper(II) nitrate forms in the solution.

Short answer

a. Mercuric Oxide Reaction: Unbalanced equation: HgO → Hg + O2 Oxidation states: Hg(+2), O(-2) in HgO; Hg(0), O(0) in products Oxidation half-reaction: Hg^(+2) → Hg^(0) + 2e^(-) Reduction half-reaction: O^(−2) + 2e^(-) → O^(0) Balanced equation: 2HgO → 2Hg + O2 b. Copper and Silver Nitrate Reaction: Unbalanced equation: Cu + AgNO3 → Ag + Cu(NO3)2 Oxidation states: Cu(0), Ag(+1), N(+5), O(-2); Ag(0), Cu(+2) in products Oxidation half-reaction: Cu^(0) → Cu^(+2) + 2e^(-) Reduction half-reaction: Ag^(+1) + e^(-) → Ag^(0) Balanced equation: 2AgNO3 + Cu → 2Ag + Cu(NO3)2

Step-by-step solution

Write the unbalanced chemical equation

The reaction states that solid mercuric oxide (HgO) is heated, and we get liquid mercury (Hg) and oxygen gas (O2). So, the unbalanced chemical equation is: HgO → Hg + O2

Determine the oxidation state of each element

For HgO: - Oxygen (O) usually has an oxidation state of -2. - Therefore, mercury (Hg) must have an oxidation state of +2 to balance the charge of the compound. For Hg: - Mercury (Hg) in its elemental form has an oxidation state of 0. For O2: - Oxygen (O) has an oxidation state of 0 in its elemental form.

Write the two half-reactions

Oxidation half-reaction: Hg (in HgO, with an oxidation state of +2) loses 2 electrons to become Hg (oxidation state of 0): Hg^(+2) → Hg^(0) + 2e^(-) Reduction half-reaction: O (in HgO, with an oxidation state of -2) gains 2 electrons to become O2 (oxidation state of 0): O^(−2) + 2e^(-) → O^(0)

Write a balanced equation for the reaction

To balance the equation, we need to combine the two half-reactions we got previously. The electrons should cancel out: Hg^(+2) + O^(-2) → Hg^(0) + O2 The balanced equation is: 2HgO → 2Hg + O2 #b. Copper and Silver Nitrate Reaction#

Write the unbalanced chemical equation

Solid copper (Cu) is put into a solution of silver nitrate (AgNO3), and we observe silver (Ag) forming as well as blue copper(II) nitrate (Cu(NO3)2). The unbalanced chemical equation is: Cu + AgNO3 → Ag + Cu(NO3)2

Determine the oxidation state of each element

For Cu: - Copper (Cu) has an oxidation state of 0 in its elemental form. For AgNO3: - Nitrogen (N) has an oxidation state of +5 and Oxygen (O) has an oxidation state of -2. - Silver (Ag) has an oxidation state of +1 to balance the charge of the compound. For Ag: - Silver (Ag) has an oxidation state of 0 in its elemental form. For Cu(NO3)2: - Nitrogen (N) has an oxidation state of +5 and Oxygen (O) has an oxidation state of -2. - Copper (Cu) has an oxidation state of +2 to balance the charge of the compound.

Write the two half-reactions

Oxidation half-reaction: Cu (oxidation state of 0) loses 2 electrons to become Cu^(+2) (oxidation state of +2): Cu^(0) → Cu^(+2) + 2e^(-) Reduction half-reaction: Ag^(+1) (in AgNO3, with an oxidation state of +1) gains an electron to become Ag (oxidation state of 0) Ag^(+1) + e^(-) → Ag^(0)

Write a balanced equation for the reaction

To balance the equation, we need to combine the two half-reactions and ensure the number of electrons lost and gained are equal: 2Ag^(+1) + Cu^(0) → 2Ag^(0) + Cu^(+2) Balanced equation: 2AgNO3 + Cu → 2Ag + Cu(NO3)2

Key concepts

Oxidation State

To understand redox reactions, it's crucial to comprehend the concept of oxidation state. The oxidation state helps us determine how electrons are distributed among atoms in a molecule. It indicates the degree of oxidation or reduction of an element. Generally, the more positive the oxidation state, the more oxidized the element is. Conversely, the more negative it is, the more reduced.

Let's take mercury(II) oxide (\(\text{HgO}\)) as an example. Oxygen typically has an oxidation state of \(-2\). In order for the compound to be neutral, mercury must have an oxidation state of \(+2\). When mercury is in its elemental form (\(\text{Hg}\)), it has an oxidation state of 0.

Similarly, in the reaction of copper with silver nitrate (\(\text{Cu} + \text{AgNO}_3\)), copper starts with an oxidation state of 0 and ends with an oxidation state of \(+2\) in copper nitrate (\(\text{Cu(NO}_3\text{)}_2\)). Silver goes from an oxidation state of \(+1\) in silver nitrate to 0 in its elemental form (\(\text{Ag}\)).

Half-Reactions

Half-reactions are a way to break down a redox reaction into two parts - an oxidation reaction and a reduction reaction. By doing this, you can see the transfer of electrons clearly. Each half-reaction shows either the loss or gain of electrons.

• **Oxidation:** This involves the loss of electrons. For instance, when mercury oxide decomposes, mercury goes from \(+2\) to 0, losing electrons in the process. Similarly, when copper reacts with silver nitrate, copper goes from 0 to \(+2\).

• **Reduction:** This involves the gain of electrons. In mercury(II) oxide decomposition, oxygen goes from \(-2\) to 0, meaning it gains electrons. During the copper and silver nitrate reaction, silver goes from \(+1\) to 0 by gaining electrons.

Split a reaction into half-reactions by identifying which elements lose electrons and which elements gain electrons. This makes the process of understanding electron flow simpler.

Balancing Chemical Equations

Balancing chemical equations is necessary to satisfy the law of conservation of mass, which states that matter cannot be created or destroyed. Therefore, the same number of each type of atom must appear on both sides of a chemical equation.

In redox reactions, balancing requires accounting for both mass and charge.

Take the decomposition of mercury(II) oxide as an example:
Suppose the unbalanced equation is \(\text{HgO} \to \text{Hg} + \text{O}_2\).

1. **Balance the atoms:** You start by ensuring that you have the same number of each type of atom on both sides.
* In this example, you need two molecules of \(\text{HgO}\) to balance the oxygen on both sides.
2. **Balance the charge:** Ensure the electrons lost in oxidation are gained in reduction. In your half-reactions, lose electrons must equal gained electrons.
* Combine the balanced half-reactions to obtain the fully balanced equation: \(2\text{HgO} \to 2\text{Hg} + \text{O}_2\).

By methodically adjusting coefficients in front of compounds, you maintain the balance of atoms and charge across the equation, which is crucial for accurately representing chemical reactions.