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Chapter 17: Problem 140

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. \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}(a q)+\mathrm{Ce}^{4+}(a q) \stackrel{\mathrm{Base}}{\longrightarrow}\) \(\mathrm{Ce}(\mathrm{OH})_{3}(s)+\mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{CO}_{3}^{2-}(a q)+\mathrm{NO}_{3}^{-}(a q)\)

Short Answer

Expert verified
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 by step solution

01

Write Half-Reactions

Oxidation: Fe(s) -> Fe^3+(aq) Reduction: HCl(aq) -> H2(g) + Cl-(aq)
02

Balance Atoms (except H and O)

Oxidation: Fe(s) -> Fe^3+(aq) Reduction: 6HCl(aq) -> 3H2(g) + 6Cl-(aq)
03

Balance O Atoms using H2O

O atoms are already balanced.
04

Balance H Atoms using H+ ions

H atoms are already balanced.
05

Balance Charges by Adding Electrons

Oxidation: Fe(s) -> Fe^3+(aq) + 3e- Reduction: 6HCl(aq) -> 3H2(g) + 6Cl-(aq) + 6e-
06

Determine Least Common Multiple of Electrons

The least common multiple of 3 and 6 is 6.
07

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-
08

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.

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Key Concepts

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

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.

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Most popular questions from this chapter

Given the following two standard reduction potentials, $$\begin{array}{ll} \mathrm{M}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{M} & \mathscr{E}^{\circ}=-0.10 \mathrm{V} \\ \mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M} & \mathscr{E}^{\circ}=-0.50 \mathrm{V} \end{array}$$ solve for the standard reduction potential of the half-reaction $$ \mathbf{M}^{3+}+\mathbf{e}^{-} \longrightarrow \mathbf{M}^{2+} $$ (Hint: You must use the extensive property \(\Delta G^{\circ}\) to determine the standard reduction potential.)

Explain the following relationships: \(\Delta G\) and \(w,\) cell potential and \(w,\) cell potential and \(\Delta G,\) cell potential and \(Q .\) Using these relationships, explain how you could make a cell in which both electrodes are the same metal and both solutions contain the same compound, but at different concentrations. Why does such a cell run spontaneously?

Balance the following oxidation-reduction reactions that occur in acidic solution using the half-reaction method. a. \(\mathrm{I}^{-}(a q)+\mathrm{ClO}^{-}(a q) \rightarrow \mathrm{I}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) b. \(\mathrm{As}_{2} \mathrm{O}_{3}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{NO}(g)\) c. \(\mathrm{Br}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{Br}_{2}(l)+\mathrm{Mn}^{2+}(a q)\) d. \(\mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \rightarrow \mathrm{CH}_{2} \mathrm{O}(a q)+\mathrm{Cr}^{3+}(a q)\)

A galvanic cell is based on the following half-reactions: $$\begin{array}{cl} \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) & \mathscr{E}^{\circ}=0.80 \mathrm{V} \\ \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & \mathscr{E}^{\circ}=0.34 \mathrm{V} \end{array}$$ In this cell, the silver compartment contains a silver electrode and excess \(\mathrm{AgCl}(s)\left(K_{\mathrm{sp}}=1.6 \times 10^{-10}\right),\) and the copper compartment contains a copper electrode and \(\left[\mathrm{Cu}^{2+}\right]=2.0 \mathrm{M}\) a. Calculate the potential for this cell at \(25^{\circ} \mathrm{C}\) b. Assuming \(1.0 \mathrm{L}\) of \(2.0 \mathrm{M}\space \mathrm{Cu}^{2+}\) in the copper compartment, calculate the moles of \(\mathrm{NH}_{3}\) that would have to be added to give a cell potential of \(0.52 \mathrm{V}\) at \(25^{\circ} \mathrm{C}\) (assume no volume change on addition of \(\mathrm{NH}_{3}\) ). $$\begin{aligned} \mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) & \rightleftharpoons \ \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) & K=1.0 \times 10^{13} \end{aligned}$$

The blood alcohol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) level can be determined by titrating a sample of blood plasma with an acidic potassium dichromate solution, resulting in the production of \(\mathrm{Cr}^{3+}(a q)\) and carbon dioxide. The reaction can be monitored because the dichromate ion \(\left(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right)\) is orange in solution, and the \(\mathrm{Cr}^{3+}\) ion is green. The unbalanced redox equation is $$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{CO}_{2}(g)$$ If \(31.05 \mathrm{mL}\) of \(0.0600 M\) potassium dichromate solution is required to titrate \(30.0 \mathrm{g}\) blood plasma, determine the mass percent of alcohol in the blood.
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