Complete and balance the following half-reactions. In each case, indicate whether the half-reaction is an oxidation or a reduction. $$ \text { (a)} \mathrm{Sn}^{2+}(a q) \longrightarrow \mathrm{Sn}^{4+}(a q) \text {(acidic solution)} \\ \text {(b)} \mathrm{TiO}_{2}(s) \longrightarrow \mathrm{Ti}^{2+}(a q) \text {(acidic solution)} \\ \text {(c)} \mathrm{ClO}_{3}^{-}(a q) \longrightarrow \mathrm{Cl}^{-}(a q) \text {(acidic solution)} \\ \text {(d)} \mathrm{N}_{2}(g) \longrightarrow \mathrm{NH}_{4}^{+}(a q) \text {(acidic solution)} \\ \text {(e)} \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{O}_{2}(g) \text {(acidic solution)} \\ \text {(f)} \operatorname{SO}_{3}^{2-}(a q) \longrightarrow \mathrm{SO}_{4}^{2-}(a q) \text {(acidic solution)} \\\\(\mathrm{g}) \mathrm{N}_{2}(g) \longrightarrow \mathrm{NH}_{3}(g) \text {(acidic solution)} $$

Short Answer

Expert verified
Following the steps for all the half-reactions, we get: (a) \( \mathrm{Sn}^{2+}(a q) - 2e^{-} \longrightarrow \mathrm{Sn}^{4+}(a q) \) (Oxidation) (b) \( \mathrm{TiO}_{2}(s) + 2e^{-} + 4\mathrm{H}^{+}(a q) \longrightarrow \mathrm{Ti}^{2+}(a q) + 2\mathrm{H}_{2}\mathrm{O}(l) \) (Reduction) (c) \( \mathrm{ClO}_{3}^{-}(a q) + 6e^{-} + 6\mathrm{H}^{+}(a q) \longrightarrow \mathrm{Cl}^{-}(a q) + 3\mathrm{H}_{2}\mathrm{O}(l) \) (Reduction) (d) \( \mathrm{N}_{2}(g) + 6\mathrm{H}^{+}(a q) + 6e^{-} \longrightarrow 2\mathrm{NH}_{4}^{+}(a q) \) (Reduction) (e) \( 2\mathrm{OH}^{-}(a q) - 4e^{-} \longrightarrow \mathrm{O}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(l) \) (Oxidation) (f) \( \mathrm{SO}_{3}^{2-}(a q) + 2\mathrm{H}^{+}(a q) - 2e^{-} \longrightarrow \mathrm{SO}_{4}^{2-}(a q) + \mathrm{H}_{2}\mathrm{O}(l) \) (Oxidation) (g) \( \mathrm{N}_{2}(g) + 6\mathrm{H}_{2}(g) \longrightarrow 2\mathrm{NH}_{3}(g) \) (Reduction)

Step by step solution

01

Determine the oxidation number of Sn

In this half-reaction, the initial oxidation state of Sn is +2, and the final oxidation state is +4.
02

Determine the number of electrons gained or lost

To go from an oxidation number of +2 to +4, two electrons are lost: \[ \mathrm{Sn}^{2+}(a q) - 2e^{-} \longrightarrow \mathrm{Sn}^{4+}(a q) \]
03

Balance the half-reaction

The half-reaction is already balanced, as there are equal numbers of charges on both sides of the equation.
04

Identify the type of half-reaction

Since the oxidation number of Sn increased and electrons were lost, this half-reaction is an oxidation process. ##Half-reaction (b)##
05

Determine the oxidation number of Ti

In TiO2, the oxidation number of Ti is +4, while in Ti^2+ the oxidation number of Ti is +2.
06

Determine the number of electrons gained or lost

To go from an oxidation number of +4 to +2, two electrons are gained: \[ \mathrm{TiO}_{2}(s) + 2e^{-} \longrightarrow \mathrm{Ti}^{2+}(a q) \]
07

Balance the half-reaction

Add four H+ ions and two H2O molecules to balance the half-reaction: \[ \mathrm{TiO}_{2}(s) + 2e^{-} + 4\mathrm{H}^{+}(a q) \longrightarrow \mathrm{Ti}^{2+}(a q) + 2\mathrm{H}_{2}\mathrm{O}(l) \]
08

Identify the type of half-reaction

Since the oxidation number of Ti decreased and electrons were gained, this half-reaction is a reduction process. Follow the same steps for the remaining half-reactions.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Oxidation and Reduction
Oxidation and reduction are two key concepts in the study of chemical reactions, particularly in the context of redox (reduction-oxidation) reactions. These processes involve the transfer of electrons between atoms, ions, or molecules.

Oxidation

Oxidation refers to the loss of electrons by a substance. When a substance undergoes oxidation, its oxidation number increases. This is seen in our exercise where Sn^{2+} loses two electrons to become Sn^{4+}, thus undergoing oxidation. A helpful mnemonic to remember oxidation is 'LEO' which stands for 'Loss of Electrons is Oxidation'.

Reduction

Reduction is the gain of electrons by a substance, causing a decrease in oxidation number. The mnemonic here is 'GER', meaning 'Gain of Electrons is Reduction'. In the exercise,TiO_{2} gains two electrons, resulting in Ti^{2+}, showcasing a typical reduction.

In a redox reaction, one species is oxidized (loses electrons), while another is reduced (gains electrons). It's essential to remember that oxidation and reduction always occur together; for every electron lost by one substance, another must gain it. This interdependence is the core of redox reactions and highlights the importance of simultaneously considering both half-reactions.
Balancing Chemical Equations
Balancing chemical equations is a fundamental skill in chemistry that ensures the law of conservation of mass is upheld. This principle states that matter cannot be created or destroyed in a chemical reaction, thus the quantity of each element must remain constant throughout the reaction.

When balancing half-reactions, we make sure that the electrons lost in the oxidation half-reaction are equal to the electrons gained in the reduction half-reaction. In our exercise, for instance, Sn^{2+} turning into Sn^{4+} involves a loss of two electrons. This balance is crucial for accurately representing the stoichiometry of the redox process.

Consideration of charge is also important. For reactions in acidic solutions, H^+ ions and water molecules are added to balance both the charge and the hydrogen and oxygen atoms. In the case of TiO_{2} to Ti^{2+}, the addition of 4H^+ and 2H_2O is required to balance the equation. Remember to confirm that both sides of the equation have the same number of each type of atom and the same overall charge.
Determining Oxidation Numbers
Determining oxidation numbers (also known as oxidation states) is a crucial step in the study of redox reactions. An oxidation number is a theoretical charge on an atom if it were in its most stable ionic form. This number can either be an actual charge for monatomic ions or, more commonly, a hypothetical charge assigned to individual atoms within a molecule or ion based on certain rules.

Some general rules for assigning oxidation numbers include:
  • Oxygen usually has an oxidation number of -2, except in peroxides where it is -1.
  • Hydrogen is typically given +1 when paired with nonmetals, and -1 with metals.
  • The oxidation number of a pure element is always zero.
  • For a molecule or ion, the sum of the oxidation numbers must equal the total charge of the molecule or ion.
In our exercise, tin (Sn) in Sn^{2+} has an oxidation number of +2, but it becomes +4 in Sn^{4+}. These numbers help in understanding the redox processes taking place, identifying whether a half-reaction involves oxidation or reduction, and ultimately are invaluable for balancing the chemical equation.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Mercuric oxide dry-cell batteries are often used where a flat discharge voltage and long life are required, such as in watches and cameras. The two half-cell reactions that occur in the battery are $$ \begin{array}{l}{\mathrm{HgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Hg}(l)+2 \mathrm{OH}^{-}(a q)} \\ {\mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{znO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-}}\end{array} $$ (a) Write the overall cell reaction. (b) The value of \(E_{\text { red }}^{\circ}\) for the cathode reaction is \(+0.098 \mathrm{V}\) . The overall cell potential is \(+1.35 \mathrm{V}\) . Assuming that both half-cells operate under standard conditions, what is the standard reduction potential for the anode reaction? (c) Why is the potential of the anode reaction different than would be expected if the reaction occurred in an acidic medium?

Hydrogen gas has the potential for use as a clean fuel in reaction with oxygen. The relevant reaction is $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) $$ Consider two possible ways of utilizing this reaction as an electrical energy source: (i) Hydrogen and oxygen gases are combusted and used to drive a generator, much as coal is currently used in the electric power industry; (ii) hydrogen and oxygen gases are used to generate electricity directly by using fuel cells that operate at \(85^{\circ} \mathrm{C}\) . (a) Use data in Appendix C to calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the reaction. We will assume that these values do not change appreciably with temperature. (b) Based on the values from part (a), what trend would you expect for the magnitude of \(\Delta G\) for the reaction as the temperature increases? (c) What is the significance of the change in the magnitude of \(\Delta G\) with temperature with respect to the utility of hydrogen as a fuel? (d) Based on the analysis here, would it be more efficient to use the combustion method or the fuel-cell method to generate electrical energy from hydrogen?

(a) In the Nernst equation, what is the numerical value of the reaction quotient, Q, under standard conditions? (b) Can the Nernst equation be used at temperatures other than room temperature?

If the equilibrium constant for a one-electron redox reaction at 298 \(\mathrm{K}\) is \(8.7 \times 10^{4}\) , calculate the corresponding \(\Delta G^{\circ}\) and \(E^{\circ} .\)

Copper corrodes to cuprous oxide, \(\mathrm{Cu}_{2} \mathrm{O},\) or cupric oxide, \(\mathrm{CuO},\) depending on environmental conditions. (a) What is the oxidation state of copper in cuprous oxide? (b) What is the oxidation state of copper in cupric oxide? (c) Copper peroxide is another oxidation product of elemental copper. Suggest a formula for copper peroxide based on its name. (d) Copper(III) oxide is another unusual oxidation product of elemental copper. Suggest a chemical formula for copper(II) oxide.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free