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Chapter 20: Problem 20

Indicate whether the following balanced equations involve oxidation-reduction. If they do, identify the elements that undergo changes in oxidation number. $$ \begin{array}{l}{\text { (a) } 2 \mathrm{AgNO}_{3}(a q)+\mathrm{CoCl}_{2}(a q) \longrightarrow 2 \mathrm{AgCl}(s)+} \\\ {\quad\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}(a q)} \\ {\text { (b) } 2 \mathrm{PbO}_{2}(s) \longrightarrow 2 \mathrm{PbO}(s)+\mathrm{O}_{2}(g)} \\\ {\text { (c) } 2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NaBr}(s) \rightarrow \mathrm{Br}_{2}(l)+\mathrm{SO}_{2}(g)+} \\ {\quad \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)}\end{array} $$

Short Answer

Expert verified
(a) The balanced equation \(2 AgNO_3(aq) + CoCl_2(aq) \longrightarrow 2 AgCl(s) + Co(NO_3)_2(aq)\) does not involve an oxidation-reduction reaction, as no oxidation numbers change. (b) The balanced equation \(2 PbO_2(s) \longrightarrow 2 PbO(s) + O_2(g)\) involves an oxidation-reduction reaction. \(Pb\) is reduced from +4 to +2, and \(O\) is oxidized from -2 to 0. (c) The balanced equation \(2 H_2SO_4(aq) + 2 NaBr(s) \rightarrow Br_2(l) + SO_2(g) + Na_2SO_4(aq) + 2 H_2O(l)\) involves an oxidation-reduction reaction. \(S\) is reduced from +6 to +4, and \(Br\) is oxidized from -1 to 0.

Step by step solution

01

Identifying Oxidation states for Equation (a)

For the given balanced equation: \(2 AgNO_3(aq) + CoCl_2(aq) \longrightarrow 2 AgCl(s) + Co(NO_3)_2(aq)\) We can start by assigning oxidation numbers to each element: In \(AgNO_3\), \(Ag\) has an oxidation state of +1, \(N\) has an oxidation state of +5, and \(O\) has an oxidation state of -2. In \(CoCl_2\), \(Co\) has an oxidation state of +2, and \(Cl\) has an oxidation state of -1. In \(AgCl\), \(Ag\) has an oxidation state of +1, and \(Cl\) has an oxidation state of -1. In \(Co(NO_3)_2\), \(Co\) has an oxidation state of +2, \(N\) has an oxidation state of +5, and \(O\) has an oxidation state of -2. Now that we have assigned oxidation numbers to all the elements in the reactants and products, we can determine if any element has undergone a change in oxidation number. Since no oxidation number has changed in the above equation, no redox reaction occurs.
02

Identifying Oxidation states for Equation (b)

For the given balanced equation: \(2 PbO_2 (s) \longrightarrow 2 PbO (s) + O_2 (g)\) We can start by assigning oxidation numbers to each element: In \(PbO_2\), \(Pb\) has an oxidation state of +4, and \(O\) has an oxidation state of -2. In \(PbO\), \(Pb\) has an oxidation state of +2, and \(O\) has an oxidation state of -2. In \(O_2\), \(O\) has an oxidation state of 0. Comparing the oxidation numbers of the reactants and products, we can see that there has been a change in the oxidation numbers: \(Pb\) has been reduced from +4 to +2, and \(O\) has been oxidized from -2 to 0. Thus, this is an oxidation-reduction reaction.
03

Identifying Oxidation states for Equation (c)

For the given balanced equation: \(2 H_2SO_4(aq) + 2 NaBr(s) \rightarrow Br_2(l) + SO_2(g) + Na_2SO_4(aq) + 2 H_2O(l)\) We can start by assigning oxidation numbers to each element: In \(H_2SO_4\), \(H\) has an oxidation state of +1, \(S\) has an oxidation state of +6, and \(O\) has an oxidation state of -2. In \(NaBr\), \(Na\) has an oxidation state of +1, and \(Br\) has an oxidation state of -1. In \(Br_2\), \(Br\) has an oxidation state of 0. In \(SO_2\), \(S\) has an oxidation state of +4, and \(O\) has an oxidation state of -2. In \(Na_2SO_4\), \(Na\) has an oxidation state of +1, \(S\) has an oxidation state of +6, and \(O\) has an oxidation state of -2. In \(H_2O\), \(H\) has an oxidation state of +1, and \(O\) has an oxidation state of -2. Comparing the oxidation numbers of the reactants and products, we can see that there has been a change in the oxidation numbers: \(S\) has been reduced from +6 to +4, and \(Br\) has been oxidized from -1 to 0. Thus, this is an oxidation-reduction reaction.

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Key Concepts

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

Oxidation States
Understanding oxidation states, sometimes referred to as oxidation numbers, is crucial for analyzing chemical reactions, especially to identify a particular type of reaction known as a redox process. Oxidation states are hypothetical charges that an atom would have if all bonds to atoms of different elements were completely ionic. These assigned numbers are used as a bookkeeping tool to keep track of electron transfer during chemical reactions.

Oxidation numbers are usually an integer which can be positive, negative, or zero, and there are a set of rules used to allocate them to atoms in a molecule or ion:
  • Elements in their natural state have an oxidation number of 0.
  • For monoatomic ions, the oxidation number is equal to the charge of the ion.
  • Oxygen almost always has an oxidation number of -2, except in peroxides or when bonded to fluorine.
  • Hydrogen typically has a +1 oxidation state when bonded with nonmetals and -1 with metals.
  • The algebraic sum of oxidation states for all atoms in a neutral molecule or a polyatomic ion equals the total charge of the molecule or ion.
Correctly assigning oxidation states can reveal which elements are oxidized and which are reduced—critical for characterizing redox reactions.
Redox Reaction Identification
Redox reactions, or oxidation-reduction reactions, are processes that involve the transfer of electrons between two substances. These reactions can be identified by tracing the changes in oxidation states of the elements through the course of the reaction.

To spot a redox reaction, look for signs of oxidation and reduction—processes where atoms lose and gain electrons, respectively. An increase in oxidation state indicates oxidation, while a decrease means reduction. Remember the mnemonic 'OIL RIG'—Oxidation Is Loss, Reduction Is Gain—referring to the loss or gain of electrons. When at least one element gets oxidized and another gets reduced in a reaction, you're witnessing a redox process.

This transfer of electrons has pivotal roles in energy production, corrosion, and even cellular respiration. Using the oxidation state method we discussed previously, we can analyze chemical equations to determine if they represent redox reactions, like in the provided step-by-step solutions.
Chemical Equations
Chemical equations are symbolic representations of chemical reactions, illustrating the substances involved and their transformation from reactants to products. Balancing these equations is an essential skill in chemistry, ensuring that the law of conservation of mass is upheld—matter can neither be created nor destroyed in a chemical reaction.

For a balanced chemical equation, the number of atoms for each element in the reactants must be equal to the number in the products. This also plays a role in identifying redox reactions. The balanced equations given in the examples demonstrate this principle, with correct stoichiometric coefficients ensuring the equality of atoms before and after the reaction.

While not every chemical equation represents a redox reaction, those that do must be balanced not only in terms of mass, but also in terms of charge. Each redox reaction has two halves: the oxidation half, where a substance loses electrons, and the reduction half, where a substance gains electrons. When combined, they must cancel each other out, resulting in a net charge of zero for the whole balanced equation.

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Most popular questions from this chapter

Disulfides are compounds that have \(S-\) S bonds, like peroxides have \(O-O\) bonds. Thiols are organic compounds that have the general formula \(R-S H,\) where \(R\) is a generic hydrocarbon. The SH \(^{-}\) is the sulfur counterpart of hydroxide, OH \(^{-} .\) Two thiols can react to make a disulfide, \(\mathrm{R}-\mathrm{S}-\mathrm{S}-\mathrm{R}\) (a) What is the oxidation state of sulfur in a thiol? (b) What is the oxidation state of sulfur in a disulfide? (c) If you react two thiols to make a disulfide, are you oxidizing or reducing the thiols? (d) If you wanted to convert a disulfide to two thiols, should you add a reducing agent or oxidizing agent to the solution? (e) Suggest what happens to the H's in the thiols when they form disulfides.

(a) Assuming standard conditions, arrange the following in order of increasing strength as oxidizing agents in acidic solution: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}, \mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{Cu}^{2+}, \mathrm{Cl}_{2}, \mathrm{O}_{2} .\) (b) Arrange the following in order of increasing strength as reducing agents in acidic solution: \(\text {Zn,}\) \(\mathrm{I}^{-}, \mathrm{Sn}^{2+}, \mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{Al} .\)

A voltaic cell consists of a strip of cadmium metal in a solution of \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) in one beaker, and in the other beaker a platinum electrode is immersed in a NaCl solution, with \(\mathrm{Cl}_{2}\) gas bubbled around the electrode. A salt bridge connects the two beakers. (a) Which electrode serves as the anode, and which as the cathode? (b) Does the Cd electrode gain or lose mass as the cell reaction proceeds? (c) Write the equation for the overall cell reaction. (d) What is the emf generated by the cell under standard conditions?

A disproportionation reaction is an oxidation-reduction reaction in which the same substance is oxidized and reduced. Complete and balance the following disproportionation reactions: $$ \begin{array}{l}{\text { (a) } \mathrm{Ni}^{+}(a q) \longrightarrow \mathrm{Ni}^{2+}(a q)+\mathrm{Ni}(s)(\text { acidic solution })} \\ {\text { (b) } \operatorname{MnO}_{4}^{2-}(a q) \longrightarrow \mathrm{MnO}_{4}^{-}(a q)+\mathrm{MnO}_{2}(s) \text { (acidic }} \\ \quad {\text { solution) }} \\\ {\text { (c) } \mathrm{H}_{2} \mathrm{SO}_{3}(a q) \longrightarrow \mathrm{S}(s)+\mathrm{HSO}_{4}^{-}(a q)(\text { acidic solution })} \\ {\text { (d) } \mathrm{Cl}_{2}(a q) \longrightarrow \mathrm{Cl}^{-}(a q)+\mathrm{ClO}^{-}(a q) \text { (basic solution) }}\end{array} $$

Using the standard reduction potentials listed in Appendix E, calculate the equilibrium constant for each of the following reactions at \(298 \mathrm{K} :\) $$ \begin{array}{l}{\text { (a) } \mathrm{Fe}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+\mathrm{Ni}(s)} \\ {\text { (b) } \mathrm{Co}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Co}^{2+}(a q)+\mathrm{H}_{2}(g)} \\ {\text { (c) } 10 \mathrm{Br}^{-}(a q)+2 \mathrm{MnO}_{4}^{-}(a q)+16 \mathrm{H}^{+}(a q) \rightarrow} \\ {\quad 2 \mathrm{Mn}^{2+}(a q)+8 \mathrm{H}_{2} \mathrm{O}(l)+5 \mathrm{Br}_{2}(l)}\end{array} $$
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