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Chapter 18: Problem 15

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

Short Answer

Expert verified
Oxidation is a chemical process involving electron loss, leading to an increase in the oxidation number of an atom or ion. Reduction, on the other hand, is a chemical process involving electron gain, resulting in a decrease in the oxidation number of an atom or ion.

Step by step solution

01

Define Oxidation in Terms of Electron Loss

Oxidation is a chemical process that occurs when a species loses electrons. In other words, it is the increase in the oxidation number of an atom or ion.
02

Define Reduction in Terms of Electron Gain

Reduction is a chemical process that occurs when a species gains electrons. In other words, it is the decrease in the oxidation number of an atom or ion.
03

Define Oxidation in Terms of Change in Oxidation Number

When oxidation occurs, the oxidation number of the species increases. This increase in the oxidation number indicates that the atom or ion has lost electrons, leading to a positive or less negative charge.
04

Define Reduction in Terms of Change in Oxidation Number

When reduction occurs, the oxidation number of the species decreases. This decrease in the oxidation number indicates that the atom or ion has gained electrons, leading to a negative or less positive charge.

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Most popular questions from this chapter

Consider the standard galvanic cell based on the following halfreactions: $$\begin{array}{r}\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu} \\ \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}\end{array}$$ The electrodes in this cell are \(\mathrm{Ag}(s)\) and \(\mathrm{Cu}(s)\). Does the cell potential increase, decrease, or remain the same when the following changes occur to the standard cell? a. \(\operatorname{CuSO}_{4}(s)\) is added to the copper half-cell compartment (assume no volume change). b. \(\mathrm{NH}_{3}(a q)\) is added to the copper half-cell compartment. [Hint: \(\mathrm{Cu}^{2+}\) reacts with \(\mathrm{NH}_{3}\) to form \(\left.\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) .\right]\) c. \(\mathrm{NaCl}(s)\) is added to the silver half-cell compartment. [Hint: \(\mathrm{Ag}^{+}\) reacts with \(\mathrm{Cl}^{-}\) to form \(\left.\mathrm{AgCl}(s) .\right]\) d. Water is added to both half-cell compartments until the volume of solution is doubled. e. The silver electrode is replaced with a platinum electrode. $$\mathrm{Pt}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt} \quad \mathscr{E}^{\circ}=1.19 \mathrm{~V}$$

Nerve impulses are electrical "signals" that pass through neurons in the body. The electrical potential is created by the differences in the concentration of \(\mathrm{Na}^{+}\) and \(\mathrm{K}^{+}\) ions across the nerve cell membrane. We can think about this potential as being caused by a concentration gradient, similar to what we see in a concentration cell (keep in mind that this is a very simple explanation of how nerves work; there is much more involved in the true biologic process). A typical nerve cell has a resting potential of about \(-70 \mathrm{mV}\). Let's assume that this resting potential is due only to the \(\mathrm{K}^{+}\) ion concentration difference. In nerve cells, the \(\mathrm{K}^{+}\) concentration inside the cell is larger than the \(\mathrm{K}^{+}\) concentration outside the cell. Calculate the \(\mathrm{K}^{+}\) ion concentration ratio necessary to produce a resting potential of \(-70 . \mathrm{mV}\). $$\frac{\left[\mathrm{K}^{+}\right]_{\text {inside }}}{\left[\mathrm{K}^{+}\right]_{\text {outside }}}=?$$

Sketch the galvanic cells based on the following overall reactions. Show the direction of electron flow, the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation. Assume that all concentrations are \(1.0 M\) and that all partial pressures are \(1.0 \mathrm{~atm}\). a. \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{Fe}^{2+}(a q) \rightleftharpoons \mathrm{Fe}^{3+}(a q)+\mathrm{I}_{2}(a q)\) b. \(\mathrm{Zn}(s)+\mathrm{Ag}^{+}(a q) \rightleftharpoons \mathrm{Zn}^{2+}(a q)+\mathrm{Ag}(s)\)

Consider the following half-reactions: $$\begin{array}{cc}\mathrm{IrCl}_{6}{ }^{3-}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Ir}+6 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.77 \mathrm{~V} \\ \mathrm{PtCl}_{4}{ }^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.73 \mathrm{~V} \\ \mathrm{PdCl}_{4}{ }^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pd}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.62 \mathrm{~V} \end{array}$$ A hydrochloric acid solution contains platinum, palladium, and iridium as chloro-complex ions. The solution is a constant \(1.0 \mathrm{M}\) in chloride ion and \(0.020 \mathrm{M}\) in each complex ion. Is it feasible to separate the three metals from this solution by electrolysis? (Assume that \(99 \%\) of a metal must be plated out before another metal begins to plate out.)

In making a specific galvanic cell, explain how one decides on the electrodes and the solutions to use in the cell.
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