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From each of the following pairs of substances, use data in Appendix E to choose the one that is the stronger oxidizing agent: $$ \begin{array}{l}{\text { (a) } \mathrm{Cl}_{2}(g) \text { or } \mathrm{Br}_{2}(l)} \\ {\text { (b) } \mathrm{Zn}^{2+}(a q) \text { or } \mathrm{Cd}^{2+}(a q)} \\ {\text { (c) } \mathrm{Cl}^{-}(a q) \text { or } \mathrm{ClO}_{3}(a q)} \\ {\text { (d) } \mathrm{H}_{2} \mathrm{O}_{2}(a q) \text { or } \mathrm{O}_{3}(\mathrm{g})}\end{array} $$

Short Answer

Expert verified
The stronger oxidizing agents for each pair of substances are: (a) Cl鈧(g), (b) Cd虏鈦(aq), (c) ClO鈧冣伝(aq), and (d) O鈧(g).

Step by step solution

01

(a) Cl鈧(g) or Br鈧(l)

To compare the oxidizing abilities of Cl鈧(g) and Br鈧(l), we will check their standard reduction potentials in Appendix E. The reaction for chlorine gas is: \[ \mathrm{Cl}_{2}(g) + 2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-}(a q) \] The \( E掳 \) for this reaction is \( +1.36 \ \text{V} \). The reaction for bromine liquid is: \[ \mathrm{Br}_{2}(l) + 2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-}(a q) \] The \( E掳 \) for this reaction is \( +1.07 \ \text{V} \). Since Cl鈧(g) has a higher standard reduction potential than Br鈧(l), we can conclude that Cl鈧(g) is the stronger oxidizing agent.
02

(b) Zn虏鈦(aq) or Cd虏鈦(aq)

For this comparison, we will check the standard reduction potentials for Zn虏鈦(aq) and Cd虏鈦(aq). The reaction for zinc ions is: \[ \mathrm{Zn}^{2+}(a q) + 2 \mathrm{e}^{-} \rightarrow \mathrm{Zn}(s) \] The \( E掳 \) for this reaction is \( -0.76 \ \text{V} \). The reaction for cadmium ions is: \[ \mathrm{Cd}^{2+}(a q) + 2 \mathrm{e}^{-} \rightarrow \mathrm{Cd}(s) \] The \( E掳 \) for this reaction is \( -0.40 \ \text{V} \). Comparing the standard reduction potentials, Cd虏鈦(aq) has a higher value than Zn虏鈦(aq), which means Cd虏鈦(aq) is the stronger oxidizing agent.
03

(c) Cl鈦(aq) or ClO鈧冣伝(aq)

Let's compare the standard reduction potentials for Cl鈦(aq) and ClO鈧冣伝(aq). We already found the reaction for Cl鈦(aq) in part (a). However, oxidation is required in this case, so we need to reverse the reaction and change the sign of \( E掳 \): \[ \mathrm{2Cl}^{-}(a q) \rightarrow \mathrm{Cl}_{2}(g) + 2 \mathrm{e}^{-} \] with \( E掳 = -1.36 \ \text{V} \). The reaction for ClO鈧冣伝(aq) is: \[ \mathrm{2ClO}_{3}^{-}(a q) + 12 \mathrm{H}^{+}(a q) + 10 \mathrm{e}^{-} \rightarrow \mathrm{Cl}_{2}(g) + 6 \mathrm{H}_{2}\mathrm{O}(l) \] The \( E掳 \) for this reaction is \( +1.50 \ \text{V} \). Since ClO鈧冣伝(aq) has a higher standard reduction potential than Cl鈦(aq), we can conclude that ClO鈧冣伝(aq) is the stronger oxidizing agent.
04

(d) H鈧侽鈧(aq) or O鈧(g)

Lastly, we will compare the standard reduction potentials for H鈧侽鈧(aq) and O鈧(g). The reaction for hydrogen peroxide is: \[ \mathrm{H}_{2} \mathrm{O}_{2}(a q) + 2 \mathrm{H}^{+}(a q) + 2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2}\mathrm{O}(l) \] The \( E掳 \) for this reaction is \( +1.77 \ \text{V} \). The reaction for ozone is: \[ \mathrm{2O}_{3}(g) + 2 \mathrm{e}^{-} \rightarrow 3 \mathrm{O}_{2}(g) \] The \( E掳 \) for this reaction is \( +2.07 \ \text{V} \). Comparing the standard reduction potentials, O鈧(g) has a higher value than H鈧侽鈧(aq), so we can conclude that O鈧(g) is the stronger oxidizing agent.

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Key Concepts

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

Standard Reduction Potential
Standard reduction potential is a measurement that tells us how easily a substance gains electrons in comparison to a standard hydrogen electrode (SHE). This potential is represented by the symbol \( E^掳 \), and is measured in volts (V). A higher \( E^掳 \) value indicates a greater tendency for a substance to gain electrons, thus being a stronger oxidizing agent.
For example, in the case of chlorine gas \( \mathrm{Cl}_{2}(g) \), it has an \( E^掳 \) of \(+1.36 \ \text{V}\), meaning it is more likely to gain electrons and act as an oxidizing agent compared to bromine liquid \( \mathrm{Br}_{2}(l) \) with an \( E^掳 \) of \( +1.07 \ \text{V}\).

When deciding which substance is a stronger oxidizing agent, always compare their standard reduction potentials. The greater the potential, the stronger the oxidizing ability.
  • Measured against the standard hydrogen electrode.
  • The higher the \( E^掳 \) value, the more potent the oxidizing agent.
  • Helps to predict the direction of redox reactions.
Redox Reactions
Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons from one substance to another. Every redox reaction is composed of two half-reactions: reduction, where a substance gains electrons, and oxidation, where a substance loses electrons.

For instance, in our exercise, when chlorine gas \( \mathrm{Cl}_{2}(g) \) gains two electrons, it is reduced to two chloride ions \( \mathrm{Cl}^{-}(aq) \). This process shows the reduction half-reaction of the redox process. The corresponding oxidation half would involve a different substance that donates electrons.

Recognizing redox reactions is critical because they are foundational for many chemical processes, such as energy production in batteries and metabolic pathways in living organisms.
  • Composed of two processes: reduction and oxidation.
  • Essential for energy transformations.
  • Involves electron transfer between compounds.
Oxidation and Reduction
Oxidation and reduction are two halves of a redox reaction. Oxidation is the loss of electrons, while reduction is the gain of electrons. Remember this by using the mnemonic: "OIL RIG" - Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons).

When looking at the oxidation of a substance in redox reactions, it is important to know which component is providing the electrons. In our example, \( \mathrm{Zn}^{2+}(aq) \) undergoes reduction by gaining electrons to become metallic zinc \( \mathrm{Zn}(s) \), while a different substance would be oxidized during this process.

The ability for a substance to oxidize or reduce another compound is defined by its standard reduction potential. Therefore, substances with a high standard reduction potential will be excellent at undergoing reduction reactions and will thus be good oxidizing agents. Understanding these processes gives insight into both natural and human-made systems.
  • Oxidation = Electron loss.
  • Reduction = Electron gain.
  • Central to processes like corrosion and respiration.

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Most popular questions from this chapter

If the equilibrium constant for a two-electron redox reaction at 298 \(\mathrm{K}\) is \(1.5 \times 10^{-4}\) , calculate the corresponding \(\Delta G^{\circ}\) and \(E^{\circ} .\)

A voltaic cell is constructed with two \(\mathrm{Zn}^{2+}-\) Zn electrodes. The two half-cells have \(\left[\mathrm{Zn}^{2+}\right]=1.8 M\) and \(\left[\mathrm{Zn}^{2+}\right]=1.00 \times 10^{-2} M,\) respectively. (a) Which electrode is the anode of the cell? (b) What is the standard emf of the cell? (c) What is the cell emf for the concentrations given? (d) For each electrode, predict whether \(\left[\mathrm{Zn}^{2+}\right]\) will increase, decrease, or stay the same as the cell operates.

A cell has a standard cell potential of \(+0.177 \mathrm{V}\) at 298 \(\mathrm{K}\) . What is the value of the equilibrium constant for the reaction ( a ) if \(n=1 ?(\mathbf{b})\) if \(n=2 ?(\mathbf{c})\) if \(n=3 ?\)

A common shorthand way to represent a voltaic cell is $$ \text {anode} | \text {anode solution} | | \text {cathode solution} | \text {cathode} $$ A double vertical line represents a salt bridge or a porous barrier. A single vertical line represents a change in phase, such as from solid to solution. (a) Write the half-reactions and overall cell reaction represented by Fel Fe \(^{2+} \| \operatorname{Ag}^{+} | A g;\) calculate the standard cell emf using data in Appendix E. (b) Write the half-reactions and overall cell reaction represented by Zn \(\left|Z \mathrm{n}^{2+}\right| \mathrm{H}^{+} | \mathrm{H}_{2} ;\) calculate the standard cell emf using data in Appendix E and use Pt for the hydrogen electrode. (c) Using the notation just described, represent a cell based on the following reaction: $$ \begin{aligned} \mathrm{ClO}_{3}^{-}(a q)+3 \mathrm{Cu}(s)+6 \mathrm{H}^{+}(a q) & \\ \longrightarrow & \mathrm{Cl}^{-}(a q)+3 \mathrm{Cu}^{2+}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$ Pt is used as an inert electrode in contact with the ClO \(_{3}^{-}\) and \(\mathrm{Cl}^{-} .\) Calculate the standard cell emf given: \(\mathrm{ClO}_{3}^{-}(a q)+\) \(6 \mathrm{H}^{+}(a q)+6 \mathrm{e}^{-} \longrightarrow \mathrm{Cl}^{-}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l); E^{\circ}=1.45 \mathrm{V}\).

A voltaic cell utilizes the following reaction: $$ 4 \mathrm{Fe}^{2+}(a q)+\mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q) \longrightarrow 4 \mathrm{Fe}^{3+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ (a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when \(\left[\mathrm{Fe}^{2+}\right]=1.3 \mathrm{M},\left[\mathrm{Fe}^{3+}\right]=\) \(0.010 \mathrm{M}, P_{\mathrm{O}_{2}}=0.50 \mathrm{atm}\) , and the \(\mathrm{pH}\) of the solution in the cathode half-cell is 3.50\(?\)

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