Sketch a galvanic cell, and explain how it works. Look at Figs. \(18.1\) and \(18.2 .\) Explain what is occurring in each container and why the cell in Fig. \(18.2\) "works" but the one in Fig. \(18.1\) does not.

Short Answer

Expert verified
A galvanic cell consists of two half-cells, each containing an electrode and an electrolyte solution. A salt bridge is essential for maintaining electrical neutrality, as it allows the flow of ions between the half-cells. In this cell, oxidation occurs at the anode, and reduction occurs at the cathode, with electrons flowing from the anode to the cathode, generating an electric current. Figure 18.2 "works" because it includes a salt bridge, unlike Figure 18.1, which lacks one. The absence of a salt bridge in Figure 18.1 leads to charge buildup, inhibiting the redox reactions and ceasing the production of electricity.

Step by step solution

01

Sketching a galvanic cell

A galvanic cell comprises two half-cells, each containing an electrode and an electrolyte solution. Start by drawing two containers, one for each half-cell. Label the containers "Reduction Half-cell" and "Oxidation Half-cell." In each container, draw a metal electrode submerged in an electrolyte solution. Label the electrodes and solutions according to their corresponding chemical reactions. You may use any example of a redox pair, such as placing a zinc electrode (Zn) in a solution of zinc sulfate (ZnSO4) for the oxidation half-cell and a copper electrode (Cu) in a solution of copper sulfate (CuSO4) for the reduction half-cell.
02

Salt bridge

Draw a salt bridge (a U-shaped tube) connecting the two containers—a necessary component to maintain electrical neutrality. The salt bridge contains an inert electrolyte, such as potassium nitrate (KNO3), which allows the flow of ions between the half-cells.
03

External wire

Draw a wire connecting the two electrodes (one end connected to the zinc electrode and the other end connected to the copper electrode). The wire will facilitate the flow of electrons from the anode (oxidation electrode) to the cathode (reduction electrode).
04

Voltmeter

Add a voltmeter to the wire circuit between the two electrodes to measure the potential difference (voltage) generated by the galvanic cell. This will help in determining the electric current produced.
05

Explain the workings of a galvanic cell

In a galvanic cell, redox reactions occur at the electrodes, which leads to the generation of electricity. At the anode (oxidation half-cell), the metal undergoes oxidation, losing electrons that travel through the external wire towards the cathode (reduction half-cell). At the cathode, the electrons are gained by the ions in the solution, resulting in a reduction process. The flow of electrons from the anode to the cathode generates an electric current, which can be measured by the voltmeter. The salt bridge completes the circuit by maintaining electrical neutrality between the half-cells. It allows the flow of ions between the containers to balance the loss of electrons at the anode and the gain of electrons at the cathode. Now, let's analyze Figures 18.1 and 18.2.
06

Comparison between Figure 18.1 and Figure 18.2

The main difference between the two figures is the presence (or absence) of a salt bridge. In Figure 18.1, there is no salt bridge connecting the two containers, while in Figure 18.2, a salt bridge is present.
07

Explain why Figure 18.2 "works" and Figure 18.1 does not

In the absence of a salt bridge, as in Figure 18.1, the galvanic cell cannot maintain electrical neutrality, resulting in the buildup of positive and negative charges in respective half-cells. This charge accumulation opposes the flow of electrons, inhibiting the redox reactions and ceasing the production of electricity. On the other hand, Figure 18.2 "works" because it has a salt bridge that maintains electrical neutrality between the two half-cells. The presence of the salt bridge allows ions to flow between the containers, preventing charge buildup and allowing the redox reactions to proceed. This continuous flow of electrons generates electricity. In conclusion, the primary reason why the galvanic cell in Figure 18.2 "works" and the one in Figure 18.1 does not is due to the presence of a salt bridge in the former, which is essential for maintaining electrical neutrality and enabling the production of electricity.

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Most popular questions from this chapter

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 \(\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}\) \(\mathrm{Ag}^{+}(a q)+2 \mathrm{~S}_{2} \mathrm{O}_{3}{ }^{2-}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}(a q) \quad K=?\)

A zinc-copper battery is constructed as follows at \(25^{\circ} \mathrm{C}:\) $$\mathrm{Zn}\left|\mathrm{Zn}^{2+}(0.10 \mathrm{M}) \| \mathrm{Cu}^{2+}(2.50 \mathrm{M})\right| \mathrm{Cu}$$ The mass of each electrode is \(200 . \mathrm{g}\). a. Calculate the cell potential when this battery is first connected. b. Calculate the cell potential after \(10.0 \mathrm{~A}\) of current has flowed for \(10.0 \mathrm{~h}\). (Assume each half-cell contains \(1.00 \mathrm{~L}\) of solution.) c. Calculate the mass of each electrode after \(10.0 \mathrm{~h}\). d. How long can this battery deliver a current of \(10.0 \mathrm{~A}\) before it goes dead?

Given the following two standard reduction potentials, $$\mathrm{M}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{M} \quad \mathscr{E}^{\circ}=-0.10 \mathrm{~V}$$ $$\mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M} \quad \mathscr{E}^{\circ}=-0.50 \mathrm{~V}$$ solve for the standard reduction potential of the half- reaction$$\mathrm{M}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{M}^{2+}$$

A solution at \(25^{\circ} \mathrm{C}\) contains \(1.0 \mathrm{M} \mathrm{Cd}^{2+}, 1.0 \mathrm{MAg}^{+}, 1.0 \mathrm{M} \mathrm{Au}^{3+}\), and \(1.0 \mathrm{M} \mathrm{Ni}^{2+}\) in the cathode compartment of an electrolytic cell. Predict the order in which the metals will plate out as the voltage is gradually increased.

Sketch the galvanic cells based on the following half-reactions. Show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation, and determine \(\mathscr{E}^{\circ}\) for the galvanic cells. Assume that all concentrations are \(1.0 M\) and that all partial pressures are \(1.0 \mathrm{~atm} .\) \(\begin{array}{ll}\text { a. } \mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} & \mathscr{6}^{\circ}=1.78 \mathrm{~V} \\ \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}_{2} & \mathscr{6}^{\circ}=0.68 \mathrm{~V}\end{array}\) b. \(\mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn}\) \(\mathscr{6}^{\circ}=-1.18 \mathrm{~V}\) \(\mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} \quad \mathscr{6}^{\circ}=-0.036 \mathrm{~V}\)

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