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Chapter 4: Problem 21

Differentiate between the following terms. a. species reduced versus the reducing agent b. species oxidized versus the oxidizing agent c. oxidation state versus actual charge

Short Answer

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In summary: a. The species reduced gains electrons, and its oxidation state decreases. The reducing agent donates electrons and undergoes oxidation, increasing its oxidation state. For example, in \(2H_2 + O_2 \rightarrow 2H_2O\), \(O_2\) is reduced, and \(H_2\) is the reducing agent. b. The species oxidized loses electrons, and its oxidation state increases. The oxidizing agent accepts electrons and undergoes reduction, decreasing its oxidation state. For example, in \(2H_2 + O_2 \rightarrow 2H_2O\), \(H_2\) is oxidized, and \(O_2\) is the oxidizing agent. c. The oxidation state is a hypothetical charge assigned to describe redox reactions, while the actual charge is the real charge of an ion or molecule. For example, in \(H_2O\), the oxidation state of oxygen is -2, but the actual charge is neutral. As an oxide ion, \(O^{2-}\), the actual charge is -2.

Step by step solution

01

1. Species Reduced vs. Reducing Agent

The species reduced is the chemical species that gains one or more electrons during a redox (reduction-oxidation) reaction. This means that its oxidation state decreases during the redox process. On the other hand, the reducing agent is the chemical species that donates electrons to another species, causing the other species to be reduced. In a redox reaction, the reducing agent itself undergoes oxidation, which means that its oxidation state increases. Example: In the redox reaction \(2H_2 + O_2 \rightarrow 2H_2O\), the oxygen molecule (\(O_2\)) is the species reduced as its oxidation state decreases from 0 in \(O_2\) to -2 in \(H_2O\). The hydrogen molecule (\(H_2\)) is the reducing agent because it donates electrons to the oxygen molecule and its oxidation state increases from 0 in \(H_2\) to +1 in \(H_2O\).
02

2. Species Oxidized vs. Oxidizing Agent

The species oxidized is the chemical species that loses one or more electrons during a redox (reduction-oxidation) reaction. This means that its oxidation state increases during the redox process. On the other hand, the oxidizing agent is the chemical species that accepts electrons from another species, causing the other species to be oxidized. In a redox reaction, the oxidizing agent itself undergoes reduction, which means that its oxidation state decreases. Example: In the redox reaction \[2H_2 + O_2 \rightarrow 2H_2O\], the hydrogen molecule (\(H_2\)) is the species oxidized as its oxidation state increases from 0 in \(H_2\) to +1 in \(H_2O\). The oxygen molecule (\(O_2\)) is the oxidizing agent because it accepts electrons from the hydrogen molecule, and its oxidation state decreases from 0 in \(O_2\) to -2 in \(H_2O\).
03

3. Oxidation State vs. Actual Charge

The oxidation state is a hypothetical charge assigned to an atom in a molecule or a single ion to help describe redox reactions. This value is determined by the distribution of electrons between bonded atoms, assuming that all electrons in a bond are assigned to the more electronegative atom. On the other hand, the actual charge is the real charge of an atom in a molecule or an ion when electrons are not equally shared between atoms as they form a chemical bond. In other words, the actual charge reflects the charge of an ion or molecule in a specific situation. Example: In a water molecule (\(H_2O\)), the oxidation state of oxygen is -2, and the oxidation state of each hydrogen atom is +1. These oxidation states help describe the redox behavior of water molecules in chemical reactions. However, in reality, a water molecule has no overall actual charge (neutral) because electrons are shared between hydrogen and oxygen atoms, which does not make them ions. If an oxygen atom was present as an oxide ion (\(O^{2-}\)), then its actual charge would be -2 due to the loss of 2 electrons. Meanwhile, a hydrogen ion (\(H^+\)) has an actual charge of +1 due to the loss of 1 electron.

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Most popular questions from this chapter

A solution is prepared by dissolving \(10.8 \mathrm{~g}\) ammonium sulfate in enough water to make \(100.0 \mathrm{~mL}\) of stock solution. A \(10.00\) mL sample of this stock solution is added to \(50.00 \mathrm{~mL}\) of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.

A student mixes four reagents together, thinking that the solutions will neutralize each other. The solutions mixed together are \(50.0 \mathrm{~mL}\) of \(0.100 M\) hydrochloric acid, \(100.0 \mathrm{~mL}\) of \(0.200 M\) of nitric acid, \(500.0 \mathrm{~mL}\) of \(0.0100 \mathrm{M}\) calcium hydroxide, and \(200.0 \mathrm{~mL}\) of \(0.100 M\) rubidium hydroxide. Did the acids and bases exactly neutralize each other? If not, calculate the concentration of excess \(\mathrm{H}^{+}\) or \(\mathrm{OH}^{-}\) ions left in solution.

You made \(100.0 \mathrm{~mL}\) of a lead(II) nitrate solution for lab but forgot to cap it. The next lab session you noticed that there was only \(80.0 \mathrm{~mL}\) left (the rest had evaporated). In addition, you forgot the initial concentration of the solution. You decide to take \(2.00 \mathrm{~mL}\) of the solution and add an excess of a concentrated sodium chloride solution. You obtain a solid with a mass of \(3.407 \mathrm{~g}\). What was the concentration of the original lead(II) nitrate solution?

Zinc and magnesium metal each react with hydrochloric acid according to the following equations: $$ \begin{array}{l} \mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g) \\ \mathrm{Mg}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{MgCl}_{2}(a q)+\mathrm{H}_{2}(g) \end{array} $$ A \(10.00-\mathrm{g}\) mixture of zinc and magnesium is reacted with the stoichiometric amount of hydrochloric acid. The reaction mixture is then reacted with \(156 \mathrm{~mL}\) of \(3.00 M\) silver nitrate to produce the maximum possible amount of silver chloride. a. Determine the percent magnesium by mass in the original mixture. b. If \(78.0 \mathrm{~mL}\) of \(\mathrm{HCl}\) was added, what was the concentration of the \(\mathrm{HCl}\) ?

The vanadium in a sample of ore is converted to \(\mathrm{VO}^{2+}\). The VO \(^{2+}\) ion is subsequently titrated with \(\mathrm{MnO}_{4}^{-}\) in acidic solution to form \(\mathrm{V}(\mathrm{OH})_{4}{ }^{+}\) and manganese(II) ion. The unbalanced titration reaction is \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{VO}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) $$ \mathrm{V}(\mathrm{OH})_{4}^{+}(a q)+\mathrm{Mn}^{2+}(a q)+\mathrm{H}^{+}(a q) $$ To titrate the solution, \(26.45 \mathrm{~mL}\) of \(0.02250 \mathrm{M} \mathrm{MnO}_{4}^{-}\) was required. If the mass percent of vanadium in the ore was \(58.1 \%\), what was the mass of the ore sample? Hint: Balance the titration reaction by the oxidation states method.
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