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In your own words, define the following symbols or terms: (a) \(E^{\circ} ;\) (b) \(F ;\) (c) anode; (d) cathode.

Short Answer

Expert verified
The term \(E^{\circ}\) represents the standard electrode potential, reflecting the potential of a reversible electrode at standard state. \(F\) represents Faraday’s constant, the total charge carried by one mole of electrons. An anode is the electrode where oxidation (loss of electrons) occurs, while a cathode is the electrode where reduction (gain of electrons) occurs.

Step by step solution

01

Term (a) \(E^{\circ}\)

\(E^{\circ}\) is known as the standard electrode potential. It is a measure of the individual potential of a reversible electrode at standard state, with ion concentrations at 1molar, pressure at 1 bar, and a defined temperature, usually 298K. The electrode potential involves a redox reaction with electrons moving from the electrode into the solution or vice versa.
02

Term (b) \(F\)

\(F\) represents the Faraday’s constant. This is essentially the total amount of electric charge carried by one mole of electrons. It is defined as the magnitude of charge per mole of electrons, with an approximate value of ~96485 C/mol.
03

Term (c) Anode

An anode in the context of electrochemistry is the electrode where oxidation occurs. During this process, electrons are lost from the chemical species. In a galvanic (battery) cell, the anode is considered negative, while in an electrolytic cell (recharging a battery), it is considered positive.
04

Term (d) Cathode

A Cathode is the electrode where reduction occurs, which means gain of electrons by the chemical species. In a galvanic cell, the cathode is the positive electrode while in an electrolytic cell, it is the negative electrode. It is the opposite of the anode in terms of electron flow.

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Key Concepts

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

Standard Electrode Potential
Imagine you have two half-cells, each containing a metal and its ions, all lined up for a high school prom dance-off. The standard electrode potential, denoted by the symbol \(E^\circ\), is akin to the zest and charisma each contestant brings to the floor. More scientifically, it is a measure of a half-cell's potential to either gain or lose electrons under standard conditions, where solutions are at 1 molar, pressures at 1 bar, and the temperature fixed at 25 degrees Celsius (298K).

Poll dance to electron dance, the standard electrode potential is like a leaderboard, showcasing which metals are the rock stars of electron attraction (and repulsion), with the hydrogen electrode set as the zero-point reference. When these metals partake in the redox shuffle, their \(E^circ\) scores help predict the voltage of the full cell and determine the reaction’s spontaneity, or whether it will even occur without external force. This concept plays a starring role in designing batteries, understanding corrosion, and many other electrochemical applications.
Faraday's Constant
Now, zoom into the dance floor where the electrons are grooving through wire 'dance tunnels' between our metal 'contestants'. Here, Faraday's constant, symbolized as \(F\), is the head DJ, setting the rhythm of the electron flow; it's the number that tells us how many coulombs of charge are in a single mole of these electron party-goers. Specifically, it's approximately 96485 coulombs per mole.

Every beat (or electron) that Faraday's constant mixes on the decks is fundamental for understanding how much energy is being transferred in electrochemical cells. This figure is not only crucial for calculating the amount of substance being oxidized or reduced during electrolysis but is also instrumental in deducing the power of electroplating, battery capacity, and the overall dynamics of electric currents in chemical processes. If electrochemistry were music, Faraday’s constant would be its tempo, giving rhythm to the entire electro-dance.
Anode and Cathode
The dance partners in our electrochemical prom are the anode and cathode. The anode is like the grand entrance to the dance floor — it's where the chemical species lose their electronic dance partners through oxidation, waving goodbyes as electrons are given off to the circuit. In the redox reaction tango, the anode has different moods. In a galvanic cell party, it's the introverted negative spot, whereas at an electrolytic cell bash, it's the life-of-the-party positive zone.

The Spotlight on Cathode

The cathode, conversely, is the cool refreshment corner where the chemical species get to gain new electron pals through reduction. It's the place of warm welcomes. In our ballroom analogy, the team at a galvanic cell event sees the cathode as the VIP positive area, but switch over to an electrolytic cell rave, and it's the chill negative lounge. Distinguishing the anode from the cathode based on charge and role in electron transfer is elementary for unraveling the flow of current and the direction of redox reactions in different types of electrochemical cells.
Oxidation and Reduction
It's time for the main event, the redox ballroom dance off. Oxidation and reduction are two halves of this chemistry dance-off, where atoms, ions, or molecules change partners (electrons). In oxidation, picture a dancer (say, an atom) with a propensity to lose partners, shedding electrons in a process that increases its charge — think the electrifying solo dance moves drawing the crowd's attention as they lose their inhibitions (and electrons).

Reduction Steals the Show

Then, reduction is that smooth dancer, gliding across the floor to gain electrons, reducing its charge, and drawing in stray partners looking for a connection. This occurs simultaneously with oxidation; as one species loses electrons, another gains. Together, they form the redox reaction, an essential chemical interplay that's fundamental to battery function, corrosion prevention, biochemical processes, and more. Each process can be identified by specific signs: an increase in oxidation state signifies oxidation, while a decrease points to reduction. Mastering the identification and understanding of these processes is crucial for analyzing reaction mechanisms and predicting the outcomes of electrochemical systems.

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Most popular questions from this chapter

Describe in words how you would calculate the standard potential of the \(\mathrm{Fe}^{2+} / \mathrm{Fe}(\mathrm{s})\) couple from those of \(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+} / \mathrm{Fe}(\mathrm{s})\).

For the reaction \(\operatorname{Co}(\mathrm{s})+\mathrm{Ni}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{Co}^{2+}(\mathrm{aq})+\) \(\mathrm{Ni}(\mathrm{s}), E_{\mathrm{cell}}^{\circ}=0.03 \mathrm{V} .\) If cobalt metal is added to an aqueous solution in which \(\left[\mathrm{Ni}^{2+}\right]=1.0 \mathrm{M},\) (a) the reaction will not proceed in the forward direction at all; (b) the displacement of \(\mathrm{Ni}(\mathrm{s})\) from the \(\mathrm{Ni}^{2+}(\mathrm{aq})\) will go to completion; (c) the displacement of \(\mathrm{Ni}(\mathrm{s})\) from the solution will proceed to a considerable extent, but the reaction will not go to completion; (d) there is no way to predict how far the reaction will proceed.

Briefly describe each of the following ideas, methods, or devices: (a) salt bridge; (b) standard hydrogen electrode (SHE); (c) cathodic protection; (d) fuel cell.

Show that for a combination of half-cell reactions that produce a standard reduction potential for a half-cell that is not directly observable, the standard reduction potential is $$E^{\circ}=\frac{\sum n_{i} E_{i}^{\circ}}{\sum n_{i}}$$ where \(n_{i}\) is the number of electrons in each half-reaction of potential \(E_{i}^{\circ} .\) Use the following half-reactions: $$ \begin{array}{c} \mathrm{H}_{5} \mathrm{IO}_{6}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{IO}_{3}^{-}(\mathrm{aq})+ \\ 3 \mathrm{H}_{2} \mathrm{O}(1) \quad E^{\circ}=1.60 \mathrm{V} \\ \mathrm{IO}_{3}^{-}(\mathrm{aq})+6 \mathrm{H}^{+}(\mathrm{aq})+5 \mathrm{e}^{-} \longrightarrow \frac{1}{2} \mathrm{I}_{2}(\mathrm{s})+3 \mathrm{H}_{2} \mathrm{O}(1) \\ E^{\circ}=1.19 \mathrm{V} \\ 2 \mathrm{HIO}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{I}_{2}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(1) \\ E^{\circ}=1.45 \mathrm{V} \\ \mathrm{I}_{2}(\mathrm{s})+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{I}^{-}(\mathrm{aq}) \quad \quad E^{\circ}=0.535 \mathrm{V} \end{array} $$ Calculate the standard reduction potential for $$ \mathrm{H}_{6} \mathrm{IO}_{6}+5 \mathrm{H}^{+}+2 \mathrm{I}^{-}+3 \mathrm{e}^{-} \longrightarrow $$ $$ \frac{1}{2} \mathrm{I}_{2}+4 \mathrm{H}_{2} \mathrm{O}=2 \mathrm{HIO} $$

Calculate the quantity indicated for each of the following electrolyses. (a) \(\left[\mathrm{Cu}^{2+}\right]\) remaining in \(425 \mathrm{mL}\) of a solution that was originally \(0.366 \mathrm{M} \mathrm{CuSO}_{4},\) after passage of \(2.68 \mathrm{A}\) for 282 s and the deposition of Cu at the cathode (b) the time required to reduce \(\left[\mathrm{Ag}^{+}\right]\) in \(255 \mathrm{mL}\) of \(\mathrm{AgNO}_{3}(\mathrm{aq})\) from 0.196 to \(0.175 \mathrm{M}\) by electrolyzing the solution between \(\mathrm{Pt}\) electrodes with a current of \(1.84 \mathrm{A}\)

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