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Chapter 18: Problem 17

Define oxidation and reduction in terms of both change in oxidation number and electron loss or gain.

Short Answer

Expert verified
Oxidation is a process involving the loss of electrons and an increase in oxidation number, whereas reduction involves the gain of electrons and a decrease in oxidation number. In a redox reaction, oxidation and reduction occur simultaneously. For example, in the reaction \(2Na + Cl_2 \rightarrow 2NaCl\), the sodium atom undergoes oxidation as its oxidation number increases from 0 to +1 due to electron loss, while the chlorine molecule undergoes reduction as its oxidation number decreases from 0 to -1 due to electron gain.

Step by step solution

01

1. Definition of Oxidation and Reduction

Oxidation and reduction are two processes that occur hand in hand, forming an essential part of any redox (reduction-oxidation) reaction. Oxidation is the process that involves a loss of electrons while reduction is the process that involves a gain of electrons.
02

2. Oxidation in terms of change in oxidation number

In chemical reactions, the oxidation numbers of the elements in a compound or an ion can change. When the oxidation number of an element increases, that is called oxidation. Oxidation occurs when an atom loses one or more electrons in a reaction, leading to a more positive oxidation state. For example, consider the following reaction: \(2Na + Cl_2 \rightarrow 2NaCl\) Here, the sodium atom (Na) goes from an oxidation number of 0 to +1, indicating that it has lost one electron, which results in its oxidation.
03

3. Reduction in terms of change in oxidation number

Reduction, on the other hand, is the process where the oxidation number of an element decreases. This happens when an atom gains one or more electrons in a reaction, leading to a less positive or more negative oxidation state. For example, in the same reaction as mentioned above: \(2Na + Cl_2 \rightarrow 2NaCl\) The chlorine molecule (Cl_2) goes from an oxidation number of 0 to -1 in the chloride ion (Cl−), indicating that it has gained one electron, which leads to its reduction.
04

4. Oxidation in terms of electron loss

Oxidation is the process in which an atom or ion loses one or more electrons from its valence shell (outermost shell). This results in an increase in the atom's positive charge or a decrease in its negative charge. As mentioned earlier, when electron loss takes place, the oxidation number of the element involved in the reaction increases.
05

5. Reduction in terms of electron gain

Reduction is the process in which an atom or ion gains one or more electrons in its valence shell. This results in a decrease in the atom's positive charge or an increase in its negative charge. As mentioned earlier, when electron gain takes place, the oxidation number of the element involved in the reaction decreases. In summary, oxidation and reduction involve changes in oxidation numbers, with oxidation being characterized by an increase in oxidation number due to electron loss, while reduction is characterized by a decrease in oxidation number due to electron gain. These two processes always occur simultaneously in a redox reaction.

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Most popular questions from this chapter

Combine the equations $$\Delta G^{\circ}=-n F \mathscr{E}^{\circ} \text { and } \Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$$ to derive an expression for \(\mathscr{E}^{\circ}\) as a function of temperature. Describe how one can graphically determine \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) from measurements of \(\mathscr{E}^{\circ}\) at different temperatures, assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. What property would you look for in designing a reference half-cell that would produce a potential relatively stable with respect to temperature?

A chemist wishes to determine the concentration of \(\mathrm{CrO}_{4}^{2-}\) electrochemically. A cell is constructed consisting of a saturated calomel electrode (SCE; see Exercise 111\()\) and a silver wire coated with \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) . The \(\mathscr{E}^{\circ}\) value for the following half-reaction is 0.446 \(\mathrm{V}\) relative to the standard hydrogen electrode: $$\mathrm{Ag}_{2} \mathrm{CrO}_{4}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Ag}+\mathrm{CrO}_{4}^{2-}$$ a. Calculate \(\mathscr{E}_{\text { cell } \text { and }} \Delta G\) at \(25^{\circ} \mathrm{C}\) for the cell reaction when \(\left[\mathrm{CrO}_{4}^{2-}\right]=1.00 \mathrm{mol} / \mathrm{L}\) . b. Write the Nernst equation for the cell. Assume that the SCE concentrations are constant. c. If the coated silver wire is placed in a solution (at $25^{\circ} \mathrm{C} )\( in which \)\left[\mathrm{CrO}_{4}^{2-}\right]=1.00 \times 10^{-5} M,$ what is the expected cell potential? d. The measured cell potential at \(25^{\circ} \mathrm{C}\) is 0.504 \(\mathrm{V}\) when the coated wire is dipped into a solution of unknown \(\left[\mathrm{CrO}_{4}^{2-}\right] .\) What is \(\left[\mathrm{CrO}_{4}^{2-}\right]\) for this solution? e. Using data from this problem and from Table \(18.1,\) calculate the solubility product \(\left(K_{\mathrm{sp}}\right)\) for $\mathrm{Ag}_{2} \mathrm{CrO}_{4}$.

The solubility product for \(\operatorname{CuI}(s)\) is \(1.1 \times 10^{-12}\) . Calculate the value of \(\mathscr{E}^{\circ}\) for the half-reaction $$\mathrm{CuI}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s)+\mathrm{I}^{-}(a q)$$

Consider the galvanic cell based on the following halfreactions: $$\begin{array}{ll}{\mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}} & {\mathscr{E}^{\circ}=-0.76 \mathrm{V}} \\ {\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}} & {\mathscr{E}^{\circ}=-0.44 \mathrm{V}}\end{array}$$ a. Determine the overall cell reaction and calculate $\mathscr{E}_{\text { cell }}$ b. Calculate \(\Delta G^{\circ}\) and \(K\) for the cell reaction at $25^{\circ} \mathrm{C}$ c. Calculate \(\mathscr{E}_{\text { cell }}\) at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{Zn}^{2+}\right]=0.10 M\) and $\left[\mathrm{Fe}^{2+}\right]=1.0 \times 10^{-5} \mathrm{M} .$

A galvanic cell is based on the following half-reactions at $25^{\circ} \mathrm{C} :$ $$\begin{array}{c}{\mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}} \\\ {\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}}\end{array}$$ Predict whether \(\mathscr{E}_{\text{cell}}\) is larger or smaller than \(\mathscr{E}^{\circ}_{\text{cell}}\) for the following cases. a. [Ag1] 5 1.0 a. $\left[\mathrm{Ag}^{+}\right]=1.0 M,\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]=2.0 M,\left[\mathrm{H}^{+}\right]=2.0 \mathrm{M}$ b. $\left[\mathrm{Ag}^{+}\right]=2.0 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]=1.0 M,\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-7} \mathrm{M}$
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