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What is electrochemistry? What are redox reactions? Explain the difference between a galvanic and an electrolytic cell.

Short Answer

Expert verified
Electrochemistry is the study of chemical reactions involving electron transfer, usually in redox reactions. Redox reactions are chemical reactions where oxidation states change due to electron transfer, with one species being reduced and the other being oxidized. A galvanic cell converts chemical energy into electrical energy through a spontaneous redox reaction, whereas an electrolytic cell uses electrical energy to drive a non-spontaneous redox reaction, like in electrolysis or electroplating. The key differences are energy conversion, spontaneity, direction of electron flow, and applications.

Step by step solution

01

Define Electrochemistry

Electrochemistry is a branch of chemistry that deals with the study of chemical reactions involving the transfer of electrons between different chemical species, usually in the form of a redox reaction. It encompasses the processes and technologies in which the interconversion of electrical energy and chemical energy takes place, such as batteries and electrolysis.
02

Define Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are chemical reactions where the oxidation states of the elements involved change due to the transfer of electrons. In a redox reaction, one element or species is reduced (gains electrons) while the other is oxidized (loses electrons).
03

Explain Galvanic Cells

A galvanic cell, also known as a voltaic cell, is an electrochemical cell that converts the chemical energy produced by a spontaneous redox reaction (meaning it occurs naturally, without any external input) into electrical energy. In a galvanic cell, the reactants are two different metals or metal ions, each in contact with an electrolyte solution. The two reactants form a redox pair, and the electrons flow from the more oxidized species (anode) to the less oxidized species (cathode) through an external circuit, generating electrical current.
04

Explain Electrolytic Cells

An electrolytic cell is an electrochemical cell where electrical energy is used to drive a non-spontaneous redox reaction (meaning it does not occur naturally and requires an external energy source). In an electrolytic cell, an external voltage source (a battery or a power supply) forces the electrons to flow in the opposite direction of the natural redox reaction. Electrolysis is the most common application of electrolytic cells, and it is used to split compounds into their constituent elements or to plate one metal onto another.
05

Highlight the Differences between Galvanic and Electrolytic Cells

The main differences between galvanic and electrolytic cells are as follows: 1. Energy Conversion: A galvanic cell converts chemical energy into electrical energy, while an electrolytic cell converts electrical energy into chemical energy. 2. Spontaneity: Redox reactions in galvanic cells are spontaneous, while those in electrolytic cells are non-spontaneous and need an external energy source to occur. 3. Direction of Electron Flow: In a galvanic cell, electrons flow from the anode to the cathode (in a direction that is favourable for the redox reaction), while in an electrolytic cell, they flow in the opposite direction (forced from the cathode to the anode). 4. Applications: Galvanic cells are used in batteries and other energy storage devices, while electrolytic cells are applied in processes like electrolysis and electroplating.

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Most popular questions from this chapter

Aluminum is produced commercially by the electrolysis of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) in the presence of a molten salt. If a plant has a continuous capacity of \(1.00\) million A, what mass of aluminum can be produced in \(2.00 \mathrm{~h} ?\)

An aqueous solution of an unknown salt of ruthenium is electrolyzed by a current of \(2.50\) A passing for \(50.0 \mathrm{~min}\). If \(2.618 \mathrm{~g}\) Ru is produced at the cathode, what is the charge on the ruthenium ions in solution?

You have a concentration cell in which the cathode has a silver electrode with \(0.10 \mathrm{MAg}^{+}\). The anode also has a silver electrode with \(\mathrm{Ag}^{+}(a q), 0.050 \mathrm{M} \mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-}\), and \(1.0 \times 10^{-3} \mathrm{M} \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}\). You read the voltage to be \(0.76 \mathrm{~V}\). a. Calculate the concentration of \(\mathrm{Ag}^{+}\) at the anode. b. Determine the value of the equilibrium constant for the formation of \(\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}\) \(\mathrm{Ag}^{+}(a q)+2 \mathrm{~S}_{2} \mathrm{O}_{3}{ }^{2-}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}(a q) \quad K=?\)

When copper reacts with nitric acid, a mixture of \(\mathrm{NO}(\mathrm{g})\) and \(\mathrm{NO}_{2}(g)\) is evolved. The volume ratio of the two product gases depends on the concentration of the nitric acid according to the equilibrium \(2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}(g) \rightleftharpoons 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) Consider the following standard reduction potentials at \(25^{\circ} \mathrm{C}\) : \(3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) $$\begin{array}{r}\mathscr{E}^{\circ}=0.957 \mathrm{~V}\end{array}$$ \(\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) $${8}^{\circ}=0.775 \mathrm{~V}$$ a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a NO and \(\mathrm{NO}_{2}\) mixture with only \(0.20 \% \mathrm{NO}_{2}\) (by moles) at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) ? Assume that no other gases are present and that the change in acid concentration can be neglected.

Consider the cell described below: $$\mathrm{Zn}\left|\mathrm{Zn}^{2+}(1.00 M) \| \mathrm{Cu}^{2+}(1.00 M)\right| \mathrm{Cu}$$ Calculate the cell potential after the reaction has operated long enough for the \(\left[\mathrm{Zn}^{2+}\right]\) to have changed by \(0.20 \mathrm{~mol} / \mathrm{L}\). (Assume \(\left.T=25^{\circ} \mathrm{C} .\right)\)

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