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Chapter 19: Problem 4

Describe the basic features of a galvanic cell. Why are the two components of the cell separated from each other?

Short Answer

Expert verified
A galvanic cell, also known as a voltaic cell, is an electrochemical cell that derives electrical energy from spontaneous redox reactions occurring within the cell. It has two half-cells, each containing an electrode immersed in a solution of its own ions. The separation of the Anode and Cathode is essential to prevent a rapid, uncontrolled direct reaction between the two. A salt bridge completes the circuit, allowing the flow of ions and maintaining cell neutrality.

Step by step solution

01

Describe a Galvanic Cell

A galvanic cell, also known as a voltaic cell, is a device in which a redox reaction spontaneously occurs and produces an electric current. It consists of two half-cells, each containing an electrode (which can be a metal) immersed in a solution of its own ions.
02

Name the components of a Galvanic Cell

There are two primary components in a galvanic cell: the Anode, which is the electrode where oxidation occurs, and the Cathode, where reduction happens. It also includes an electrolyte, which allows the flow of ions, and a salt bridge that completes the circuit.
03

Discuss the separation of the cell components

The anode (negative electrode) and the cathode (positive electrode) must be separate from each other to prevent a direct reaction between the two, which would occur rapidly and not produce a controlled electrical current. The separation ensures a sustained, controlled chemical reaction leading to a steady flow of electric current between the two.
04

Explain the purpose of the salt bridge

The salt bridge completes the electrical circuit and allows the flow of ions between the half-cells. It is pivotal to maintain the cell neutrality, i.e., to prevent the solutions in the half-cells from becoming too negatively charged at the cathode or too positively charged at the anode.

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Most popular questions from this chapter

Explain why chlorine gas can be prepared by electrolyzing an aqueous solution of \(\mathrm{NaCl}\) but fluorine gas cannot be prepared by electrolyzing an aqueous solution of NaF.

Consider a Daniell cell operating under nonstandardstate conditions. Suppose that the cell's reaction is multiplied by 2 . What effect does this have on each of the following quantities in the Nernst equation? (a) \(E,\left(\right.\) b) \(E^{\circ},\) (c) \(Q,\) (d) \(\ln Q\), and (e) \(n\) ?

When \(25.0 \mathrm{~mL}\) of a solution containing both \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) ions is titrated with \(23.0 \mathrm{~mL}\) of \(0.0200 \mathrm{M}\) \(\mathrm{KMnO}_{4}\) (in dilute sulfuric acid), all of the \(\mathrm{Fe}^{2+}\) ions are oxidized to \(\mathrm{Fe}^{3+}\) ions. Next, the solution is treated with \(Z\) n metal to convert all of the \(\mathrm{Fe}^{3+}\) ions to \(\mathrm{Fe}^{2+}\) ions. Finally, \(40.0 \mathrm{~mL}\) of the same \(\mathrm{KMnO}_{4}\) solution are added to the solution in order to oxidize the \(\mathrm{Fe}^{2+}\) ions to \(\mathrm{Fe}^{3+}\). Calculate the molar concentrations of \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) in the original solution.

Industrially, copper is purified by electrolysis. The impure copper acts as the anode, and the cathode is made of pure copper. The electrodes are immersed in a \(\mathrm{CuSO}_{4}\) solution. During electrolysis, copper at the anode enters the solution as \(\mathrm{Cu}^{2+}\) while \(\mathrm{Cu}^{2+}\) ions are reduced at the cathode. (a) Write half-cell reactions and the overall reaction for the electrolytic process. (b) Suppose the anode was contaminated with \(\mathrm{Zn}\) and \(\mathrm{Ag} .\) Explain what happens to these impurities during electrolysis. (c) How many hours will it take to obtain \(1.00 \mathrm{~kg}\) of \(\mathrm{Cu}\) at a current of \(18.9 \mathrm{~A} ?\)

From the following information, calculate the solubility product of AgBr: $$ \begin{array}{ll} \mathrm{Ag}^{+}(a q)+e^{-} \longrightarrow \mathrm{Ag}(s) & E^{\circ}=0.80 \mathrm{~V} \\ \operatorname{AgBr}(s)+e^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{Br}^{-}(a q) & E^{\circ}=0.07 \mathrm{~V} \end{array} $$
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