Problem 140 - Chemistry An Atoms First Approach

Question

Balance the following equations by the half-reaction method. a. \(\mathrm{Fe}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{HFeCl}_{4}(a q)+\mathrm{H}_{2}(g)\) b. \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \stackrel{\text { Acid }}{\longrightarrow} \mathrm{I}_{3}^{-}(a q)\) \(\mathbf{c} . \operatorname{Cr}(\mathrm{NCS})_{6}^{4-}(a q)+\mathrm{Ce}^{4+}(a q) \stackrel{\text { Acid }}{\longrightarrow}\) \(\mathrm{Cr}^{3+}(a q)+\mathrm{Ce}^{3+}(a q)+\mathrm{NO}_{3}^{-}(a q)+\mathrm{CO}_{2}(g)+\mathrm{SO}_{4}^{2-}(a q)\) d. \(\mathrm{CrI}_{3}(s)+\mathrm{Cl}_{2}(g) \stackrel{\text { Bawe }}{\longrightarrow}\) \(\mathrm{CrO}_{4}^{2-}(a q)+\mathrm{IO}_{4}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) e. \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}(a q)+\mathrm{Ce}^{4+}(a q) \stackrel{\mathrm{Barc}}{\longrightarrow}\) \(\mathrm{Ce}(\mathrm{OH})_{3}(s)+\mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{CO}_{3}^{2-}(a q)+\mathrm{NO}_{3}^{-}(a q)\)Balance the following equations by the half- reaction method. a. \(\mathrm{Fe}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{HFeCl}_{4}(a q)+\mathrm{H}_{2}(g)\) b. \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \stackrel{\text { Acid }}{\longrightarrow} \mathrm{I}_{3}^{-}(a q)\) \(\mathbf{c} . \operatorname{Cr}(\mathrm{NCS})_{6}^{4-}(a q)+\mathrm{Ce}^{4+}(a q) \stackrel{\text { Acid }}{\longrightarrow}\) \(\mathrm{Cr}^{3+}(a q)+\mathrm{Ce}^{3+}(a q)+\mathrm{NO}_{3}^{-}(a q)+\mathrm{CO}_{2}(g)+\mathrm{SO}_{4}^{2-}(a q)\) d. \(\mathrm{CrI}_{3}(s)+\mathrm{Cl}_{2}(g) \stackrel{\text { Base }}{\longrightarrow}\) \(\mathrm{CrO}_{4}^{2-}(a q)+\mathrm{IO}_{4}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) e. \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}(a q)+\mathrm{Ce}^{4+}(a q) \stackrel{\mathrm{Base}}{\longrightarrow}\) \(\mathrm{Ce}(\mathrm{OH})_{3}(s)+\mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{CO}_{3}^{2-}(a q)+\mathrm{NO}_{3}^{-}(a q)\)

Step-by-Step Solution

Answer

The balanced equation for part a is: \(2\mathrm{Fe}(s) + 6\mathrm{HCl}(a q) \longrightarrow 2\mathrm{HFeCl}_{4}(a q)+3\mathrm{H}_{2}(g)\).

Step 1: Write Half-Reactions

Oxidation: Fe(s) -> Fe^3+(aq) Reduction: HCl(aq) -> H2(g) + Cl-(aq)

Step 2: Balance Atoms (except H and O)

Oxidation: Fe(s) -> Fe^3+(aq) Reduction: 6HCl(aq) -> 3H2(g) + 6Cl-(aq)

Step 3: Balance O Atoms using H2O

O atoms are already balanced.

Step 4: Balance H Atoms using H+ ions

H atoms are already balanced.

Step 5: Balance Charges by Adding Electrons

Oxidation: Fe(s) -> Fe^3+(aq) + 3e- Reduction: 6HCl(aq) -> 3H2(g) + 6Cl-(aq) + 6e-

Step 6: Determine Least Common Multiple of Electrons

The least common multiple of 3 and 6 is 6.

Step 7: Multiply Half-Reactions by Necessary Factors

Oxidation: 2(Fe(s) -> Fe^3+(aq) + 3e-) -> 2Fe(s) -> 2Fe^3+(aq) + 6e- Reduction: 6HCl(aq) -> 3H2(g) + 6Cl-(aq) + 6e-

Step 8: Add Half-Reactions and Simplify

Balanced Equation: 2Fe(s) + 6HCl(aq) -> 2Fe^3+(aq) + 6e- + 3H2(g) + 6Cl-(aq) + 6e- Simplify: 2Fe(s) + 6HCl(aq) -> 2HFeCl4(aq) + 3H2(g) The other equations can be balanced using the same method; it has been demonstrated with the first equation (a) above.

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is a crucial skill in chemistry. It ensures that the law of conservation of mass is obeyed, meaning that the number and type of atoms on each side of the equation are equal. When balancing equations, the goal is to make sure both sides are equal by adjusting the coefficients in front of the chemical formulas. This does not change the identity of the substances but merely how much of them is being used or formed.

• **Identify the Substances**: Write down all reactants and products with their chemical formulas.
• **Balance All Elements Except for Hydrogen and Oxygen First**: This simplifies the balancing of other elements since hydrogen and oxygen often appear in multiple compounds.
• **Balance Hydrogen and Oxygen**: Adjust these last because they are commonly found in different compounds like water, acids, or oxyanions.
• **Double-check Your Equation**: Count atoms on both sides to confirm they are equal.

By following these steps, you can accurately balance any chemical equation needed.

Oxidation and Reduction

Understanding oxidation and reduction, also known as redox reactions, is fundamental in chemical reactions. During a redox process, atoms change their oxidation states. This involves the transfer of electrons from one element to another, which can be broken down into two half-reactions: oxidation and reduction.

• **Oxidation**: This is when an atom or molecule loses electrons. It is often associated with gaining oxygen or losing hydrogen. For example, in the reaction of iron with hydrochloric acid, iron oxidizes by losing electrons to become iron ions (\( ext{Fe} ightarrow ext{Fe}^{3+} + 3e^- \)
• **Reduction**: Here, an atom or molecule gains electrons. Reduction often involves gaining hydrogen or losing oxygen. In our example, the hydrogen ions from HCl gain electrons to form hydrogen gas (\( ext{6HCl} ightarrow 3 ext{H}_2 + 6 ext{Cl}^- + 6e^- \)

Redox reactions are pivotal in many chemical and biological processes, from corrosion and combustion to cellular respiration and photosynthesis.

Electrochemistry

Electrochemistry studies the relationship between electrical energy and chemical changes. It's a central topic in understanding processes such as battery operation, electroplating, and corrosion prevention. Electrochemical cells, where redox reactions take place, are key to many energy-storing systems.

• **Galvanic Cells**: These convert chemical energy into electrical energy through spontaneous redox reactions. A classic example is a battery where oxidation at the anode and reduction at the cathode generates an electrical current.
• **Electrolytic Cells**: Unlike galvanic cells, these require electrical energy to drive non-spontaneous chemical reactions. This process is used for electrolysis, where compounds are broken down into elemental forms.
• **Standard Electrode Potentials**: These values help predict the feasibility of reactions. The more positive the potential, the greater the tendency for the element to gain electrons and be reduced.

By understanding electrochemistry, one can harness and control chemical processes involving electron transfers, which is vital for developing new technologies in energy and environmental solutions.