Explain the important distinctions between each pair of terms: (a) half- reaction and overall cell reaction; (b) voltaic cell and electrolytic cell; (c) primary battery and secondary battery; (d) \(E_{\text {cell }}\) and \(E_{\text {cell }}^{\circ}\).

Short Answer

Expert verified
Half-reaction is a partial reaction, either oxidation or reduction, while overall cell reaction combines both. A voltaic cell generates current from a spontaneous reaction, while an electrolytic cell drives a non-spontaneous reaction. Primary batteries are non-rechargeable, converting chemical energy into electricity. Secondary batteries are rechargeable, with reversed chemical processes. \(E_{\text {cell}}\) is cell potential under non-standard conditions, while \(E_{\text {cell}}^{\circ}\) is under standard conditions.

Step by step solution

01

Half-reaction and Overall Cell Reaction

A half-reaction is either the oxidation or reduction that occurs in a cell, split into two parts for ease of analysis. In contrast, the overall cell reaction is the combination of the oxidation half-reaction and the reduction half-reaction. It represents the net chemical process in an electrochemical cell.
02

Voltaic Cell and Electrolytic Cell

A voltaic cell, also known as a galvanic cell, generates an electrical current produced by spontaneous redox reactions. On the other hand, an electrolytic cell uses electrical energy from an external source to drive a non-spontaneous redox reaction.
03

Primary Battery and Secondary Battery

A primary battery, also known as a disposable battery, can convert chemical energy into electrical energy directly until its reactants are exhausted. It’s not rechargeable. A secondary battery, on the other hand, is a rechargeable battery. It can be reused multiple times as the chemical processes inside them are reversible.
04

\(E_{\text {cell}}\) and \(E_{\text {cell}}^{\circ}\)

\(\(E_{\text {cell}}\)\) is the cell potential under non-standard conditions. It can be calculated using the Nernst equation. \(\(E_{\text {cell}}^{\circ}\)\), on the other hand, is the cell potential under standard conditions, which means a temperature of 298K, a pressure of 1 atm for gases and concentrations of 1 M for solutions.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For the reaction \(2 \mathrm{Cu}^{+}(\mathrm{aq})+\mathrm{Sn}^{4+}(\mathrm{aq}) \longrightarrow\) \(2 \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Sn}^{2+}(\mathrm{aq}), E_{\mathrm{cell}}^{\circ}=-0.0050 \mathrm{V}\) (a) can a solution be prepared at \(298 \mathrm{K}\) that is \(0.500 \mathrm{M}\) in each of the four ions? (b) If not, in which direction will a reaction occur?

The electrolysis of \(\mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) is conducted in two separate half-cells joined by a salt bridge, as suggested by the cell diagram \(\mathrm{Pt}\left|\mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})\right|\left|\mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})\right| \mathrm{Pt}\) (a) In one experiment, the solution in the anode compartment becomes more acidic and that in the cathode compartment, more basic during the electrolysis. When the electrolysis is discontinued and the two solutions are mixed, the resulting solution has \(\mathrm{pH}=7\). Write half-equations and the overall electrolysis equation. (b) In a second experiment, a 10.00 -mL sample of an unknown concentration of \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) and a few drops of phenolphthalein indicator are added to the \(\mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) in the cathode compartment. Electrolysis is carried out with a current of \(21.5 \mathrm{mA}\) (milliamperes) for 683 s, at which point, the solution in the cathode compartment acquires a lasting pink color. What is the molarity of the unknown \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) ?\)

Use the data in Appendix D to calculate the standard cell potential for each of the following reactions. Which reactions will occur spontaneously? (a) \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{F}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{F}^{-}(\mathrm{aq})\) (b) \(\mathrm{Cu}(\mathrm{s})+\mathrm{Ba}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Ba}(\mathrm{s})\) (c) \(3 \mathrm{Fe}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{Fe}(\mathrm{s})+2 \mathrm{Fe}^{3+}(\mathrm{aq})\) (d) \(\mathrm{Hg}(1)+\mathrm{HgCl}_{2}(\mathrm{aq}) \longrightarrow \mathrm{Hg}_{2} \mathrm{Cl}_{2}(\mathrm{s})\)

The theoretical \(E_{\text {cell }}^{\circ}\) for the methane-oxygen fuel cell is \(1.06 \mathrm{V} .\) What is \(E^{\circ}\) for the reduction half-reaction \(\mathrm{CO}_{2}(\mathrm{g})+8 \mathrm{H}^{+}(\mathrm{aq})+8 \mathrm{e}^{-} \longrightarrow \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(1) ?\)

Calculate the quantity indicated for each of the following electrolyses. (a) the mass of \(\mathrm{Zn}\) deposited at the cathode in 42.5 min when 1.87 A of current is passed through an aqueous solution of \(\mathrm{Zn}^{2+}\) (b) the time required to produce \(2.79 \mathrm{g} \mathrm{I}_{2}\) at the anode if a current of \(1.75 \mathrm{A}\) is passed through \(\mathrm{KI}(\mathrm{aq})\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free