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Chapter 18: Problem 36

Which of the following species will be oxidized by \(1 \mathrm{MHCl}\) ? (a) \(\mathrm{Au}\) (b) \(\mathrm{Mg}\) (c) Cu (d) \(\mathrm{F}^{-}\)

Short Answer

Expert verified
Answer: (b) Mg

Step by step solution

01

Determine the oxidation potential of H+

First, determine the oxidation potential of H+. The standard oxidation potential of H+ in the reaction is given as: $$ 2\mathrm{H}^+ + 2\mathrm{e}^- \rightleftharpoons\mathrm{H}_{2} $$ Since the standard reduction potential of this reaction (reverse of the reaction) can be easily found in a table of standard reduction potentials, it has the value of \(0 \mathrm{V}\). The standard oxidation potential of H+ will be the negative of this value, which is also \(0\mathrm{V}\).
02

Determine the standard oxidation potential of the given species

Now, look up the standard oxidation potentials of the given species (Au, Mg, Cu, and F-). In any standard reduction potential table, you can find the reduction potentials of these species' respective half-cell reactions and switch their signs to get the oxidation potentials. The species with the highest oxidation potential will have the greatest tendency to lose electrons and be oxidized. Here are the oxidation potentials of the given species: - Au: \(1.50\mathrm{V}\) - Mg: \(-2.37\mathrm{V}\) - Cu: \(0.34\mathrm{V}\) - F-: \(2.87\mathrm{V}\)
03

Compare the oxidation potential of the given species with H+

Compare the oxidation potentials of the given species to the oxidation potential of H+. If a species has a higher oxidation potential than H+, it will have a greater tendency to lose electrons and be oxidized. - Au (\(1.50\mathrm{V}\)) has a higher oxidation potential than H+ (\(0\mathrm{V}\)), meaning it is less likely to be oxidized than H+. - Mg (\(-2.37\mathrm{V}\)) has a lower oxidation potential than H+ (\(0\mathrm{V}\)), meaning it is more likely to be oxidized than H+. - Cu (\(0.34\mathrm{V}\)) has a higher oxidation potential than H+ (\(0\mathrm{V}\)), meaning it is less likely to be oxidized than H+. - F- (\(2.87\mathrm{V}\)) has a higher oxidation potential than H+ (\(0\mathrm{V}\)), meaning it is less likely to be oxidized than H+.
04

Determine the species that will be oxidized

After comparing the oxidation potentials of the given species, we find that Mg has the lowest oxidation potential, meaning it is most likely to lose electrons and be oxidized by 1M HCl. Therefore, the answer is (b) Mg.

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Key Concepts

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

Reduction Potential
Reduction potential is a vital concept in understanding how reactions occur in electrochemistry. It refers to the tendency of a species to gain electrons and be reduced. Every substance has a characteristic reduction potential, which can be measured in volts (V). This value helps in predicting whether a particular element will undergo reduction when placed in a solution or electrochemical cell. The reduction potential is typically found by comparing it to the standard hydrogen electrode (SHE), which is set at 0.00 V. For example, if a metal like copper has a standard reduction potential of 0.34 V, it means it has a higher tendency to gain electrons compared to hydrogen ions. When you analyze reduction potential:
  • If the potential is positive, the substance has a higher likelihood of being reduced.
  • If negative, the substance is less likely to accept electrons.
These potentials are crucial in determining the feasibility of redox reactions, where one species gets reduced, and another gets oxidized.
Electron Transfer
Electron transfer is the movement of electrons from one molecule or atom to another. This process is the foundation of all redox (reduction-oxidation) reactions. During a redox reaction, one element loses electrons (oxidation), while another gains electrons (reduction). The driving force behind electron transfer is the difference in potential between the two species, which creates a flow of electrons from the reducing agent (electron donor) to the oxidizing agent (electron acceptor). To visualize this, consider an electrochemical cell where:
  • Just like water flowing downhill due to gravity, electrons flow from high potential energy to low potential energy.
  • Redox reactions can occur spontaneously if the potential for transfer (or reduction potential) is favorable.
Understanding electron transfer is fundamental to grasp how batteries work or how biological processes like photosynthesis and respiration occur. These reactions power everything from phones to the cells in your body!
Standard Reduction Potential
Standard reduction potential is a measure of the inherent ability of a chemical species to be reduced, or gain electrons, under standard conditions (25°C, 1 atm pressure, and 1M concentration). These potentials are tabulated using the standard hydrogen electrode as a reference point. In practical terms:
  • A higher standard reduction potential indicates a greater tendency for the species to gain electrons.
  • Ranking different species by their standard reduction potentials allows chemists to predict the direction of electron flow in galvanic cells.
For example, in our initial exercise, magnesium had a low reduction potential, which explains its higher oxidation likelihood since it donates electrons more readily. The applicability of standard reduction potentials is vast, including prediction of reaction spontaneity and designing electrochemical cells. Understanding this concept helps you determine which substances are better electron donors or acceptors, guiding numerous chemical and industrial processes.

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Most popular questions from this chapter

Consider the reaction below at \(25^{\circ} \mathrm{C}\) : $$ 3 \mathrm{SO}_{4}{ }^{2-}(a q)+12 \mathrm{H}^{+}(a q)+2 \mathrm{Cr}(s) \longrightarrow 3 \mathrm{SO}_{2}(g)+2 \mathrm{Cr}^{3+}(a q)+6 \mathrm{H}_{2} \mathrm{O} $$ Use Table \(18.1\) to answer the following questions. Support your answers with calculations. (a) Is the reaction spontaneous at standard conditions? (b) Is the reaction spontaneous at a \(\mathrm{pH}\) of \(3.00\) with all other ionic species at \(0.100 \mathrm{M}\) and gases at \(1.00\) atm? (c) Is the reaction spontaneous at a pH of \(8.00\) with all other ionic species at \(0.100 \mathrm{M}\) and gases at \(1.00 \mathrm{~atm}\) ? (d) At what \(\mathrm{pH}\) is the reaction at equilibrium with all other ionic species at \(0.100 \mathrm{M}\) and gases at \(1.00 \mathrm{~atm}\) ?

For the following half-reactions, answer the questions below. $$ \begin{array}{cc} \mathrm{Co}^{3+}(a q)+e^{-} \longrightarrow \mathrm{Co}^{2+}(a q) & E^{\circ}=+1.953 \mathrm{~V} \\ \mathrm{Fe}^{3+}(a q)+e^{-} \longrightarrow \mathrm{Fe}^{2+}(a q) & E^{\circ}=+0.769 \mathrm{~V} \\ \mathrm{I}_{2}(a q)+2 e^{-} \longrightarrow 2 \mathrm{I}^{-}(a q) & E^{o}=+0.534 \mathrm{~V} \\ \mathrm{~Pb}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Pb}(s) & E^{\circ}=-0.127 \mathrm{~V} \\ \mathrm{Cd}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Cd}(s) & E^{\circ}=-0.402 \mathrm{~V} \\ \mathrm{Mn}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Mn}(s) & E^{\circ}=-1.182 \mathrm{~V} \end{array} $$ (a) Which is the weakest reducing agent? (b) Which is the strongest reducing agent? (c) Which is the strongest oxidizing agent? (d) Which is the weakest oxidizing agent? (e) Will \(\mathrm{Pb}(s)\) reduce \(\mathrm{Fe}^{3+}(a q)\) to \(\mathrm{Fe}^{2+}(a q) ?\) (f) Will \(\mathrm{I}^{-}(a q)\) reduce \(\mathrm{Pb}^{2+}(a q)\) to \(\mathrm{Pb}(s) ?\) (g) Which ion(s) can be reduced by \(\mathrm{Pb}(s)\) ? (h) Which if any metal(s) can be oxidized by \(\mathrm{Fe}^{3+}(a q)\) ?

I Suppose \(E_{\text {red }}^{\circ}\) for \(\mathrm{H}^{+} \longrightarrow \mathrm{H}_{2}\) were taken to be \(0.300 \mathrm{~V}\) instead of \(0.000 \mathrm{~V}\). What would be (a) \(E_{\text {ox }}^{o}\) for \(\mathrm{H}_{2} \longrightarrow \mathrm{H}^{+}\) ? (b) \(E_{\text {red }}^{\circ}\) for \(\mathrm{Br}_{2} \longrightarrow \mathrm{Br}^{-} ?\) (c) \(E^{\circ}\) for the cell in \(24(\mathrm{c})\) ? Compare your answer with that obtained in \(24(\mathrm{c})\).

The electrolysis of an aqueous solution of \(\mathrm{NaCl}\) has the overall equation $$ 2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g)+2 \mathrm{OH}^{-}(a q) $$ During the electrolysis, \(0.228\) mol of electrons passes through the cell. (a) How many electrons does this represent? (b) How many coulombs does this represent? (c) Assuming \(100 \%\) yield, what masses of \(\mathrm{H}_{2}\) and \(\mathrm{Cl}_{2}\) are produced?

For the cell $$ \mathrm{Zn}\left|\mathrm{Zn}^{2+}\right| \mathrm{Cu}^{2+} \mid \mathrm{Cu} $$ \(E^{\circ}\) is \(1.10 \mathrm{~V}\). A student prepared the same cell in the lab at standard conditions. Her experimental \(E^{\circ}\) was \(1.0 \mathrm{~V}\). A possible explanation for the difference is that (a) a larger volume of \(\mathrm{Zn}^{2+}\) than \(\mathrm{Cu}^{2+}\) was used. (b) the zinc electrode had twice the mass of the copper electrode. (c) \(\left[\mathrm{Zn}^{2+}\right]\) was smaller than \(1 M\). (d) \(\left[\mathrm{Cu}^{2+}\right]\) was smaller than \(1 M\). (e) the copper electrode had twice the surface area of the zinc electrode.
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