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Chapter 18: Problem 36

Sketch the galvanic cells based on the following overall reactions. Show the direction of electron flow, the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation. Assume that all concentrations are \(1.0 M\) and that all partial pressures are \(1.0 \mathrm{~atm}\). a. \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{Fe}^{2+}(a q) \rightleftharpoons \mathrm{Fe}^{3+}(a q)+\mathrm{I}_{2}(a q)\) b. \(\mathrm{Zn}(s)+\mathrm{Ag}^{+}(a q) \rightleftharpoons \mathrm{Zn}^{2+}(a q)+\mathrm{Ag}(s)\)

Short Answer

Expert verified
a. For the reaction \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{Fe}^{2+}(a q) \rightleftharpoons \mathrm{Fe}^{3+}(a q)+\mathrm{I}_{2}(a q)\), the anode is where oxidation of \(\mathrm{Fe}^{2+}\) to \(\mathrm{Fe}^{3+}\) occurs, while the cathode is where reduction of \(\mathrm{IO}_3^{-}\) to \(\mathrm{I}_2\) takes place. Electrons flow from the anode to the cathode. Anions in the salt bridge migrate towards the anode and cations towards the cathode to maintain neutrality. The overall balanced equation is \(2\mathrm{IO}_{3}^{-}(a q) + 12\mathrm{H}^{+}(a q) + 6\mathrm{Fe}^{2+}(a q) \rightleftharpoons 6\mathrm{Fe}^{3+}(a q) + 2\mathrm{I}_{2}(a q) + 6\mathrm{H}_{2}\mathrm{O}(l)\). b. For the reaction \(\mathrm{Zn}(s)+\mathrm{Ag}^{+}(a q) \rightleftharpoons \mathrm{Zn}^{2+}(a q)+\mathrm{Ag}(s)\), the anode is where oxidation of Zn(s) to \(\mathrm{Zn}^{2+}\) occurs, while the cathode is where reduction of \(\mathrm{Ag}^{+}\) to Ag(s) takes place. Electrons flow from the anode to the cathode. Anions in the salt bridge migrate towards the anode and cations towards the cathode to maintain neutrality. The overall balanced equation is \(\mathrm{Zn}(s) + 2\mathrm{Ag}^{+}(a q) \rightleftharpoons \mathrm{Zn}^{2+}(a q) + 2\mathrm{Ag}(s)\).

Step by step solution

01

a. Identifying Half Reactions

In order to set up the galvanic cell, we need to identify the half-reactions for the given overall reactions. For the first reaction: \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{Fe}^{2+}(a q) \rightleftharpoons \mathrm{Fe}^{3+}(a q)+\mathrm{I}_{2}(a q)\) The two half-reactions are: Reduction: \(\mathrm{IO}_{3}^{-}(a q) + 6\mathrm{e}^{-} + 6\mathrm{H}^{+}(a q) \rightleftharpoons \mathrm{I}_{2}(a q) + 3\mathrm{H}_{2}\mathrm{O}(l)\) Oxidation: \(\mathrm{Fe}^{2+}(aq) \rightleftharpoons \mathrm{Fe}^{3+}(aq) + \mathrm{e}^{-}\)
02

a. Identifying Anode and Cathode

In a galvanic cell, oxidation occurs at the anode and reduction occurs at the cathode. In this case, oxidation of \(\mathrm{Fe}^{2+}\) to \(\mathrm{Fe}^{3+}\) occurs at the anode, and \(\mathrm{IO}_3^{-}\) is reduced to \(\mathrm{I}_2\) at the cathode.
03

a. Direction of Electron Flow and Ion Migration

In a galvanic cell, electrons flow from the anode to the cathode through the external circuit. Ions in the salt bridge migrate to maintain electrical neutrality. In this reaction: Electron flow: from \(\mathrm{Fe}^{2+}\) (anode) to \(\mathrm{IO}_{3}^{-}\) (cathode). Ion migration in the salt bridge: - Anions (e.g., \(\mathrm{NO}_3^-\)) migrate towards the anode (Fe) to maintain electrical neutrality. - Cations (e.g., \(\mathrm{K}^+\)) migrate towards the cathode (IO3) to maintain electrical neutrality.
04

a. Overall Balanced Equation

From the half-reactions, we can write the overall balanced equation for this cell: \(2\mathrm{IO}_{3}^{-}(a q) + 12\mathrm{H}^{+}(a q) + 6\mathrm{Fe}^{2+}(a q) \rightleftharpoons 6\mathrm{Fe}^{3+}(a q) + 2\mathrm{I}_{2}(a q) + 6\mathrm{H}_{2}\mathrm{O}(l) \)
05

b. Identifying Half Reactions

For the second reaction: \(\mathrm{Zn}(s) + \mathrm{Ag}^{+}(a q) \rightleftharpoons \mathrm{Zn}^{2+}(a q)+\mathrm{Ag}(s)\) The two half-reactions are: Reduction: \(\mathrm{Ag}^{+}(a q) + \mathrm{e}^{-} \rightleftharpoons \mathrm{Ag}(s)\) Oxidation: \(\mathrm{Zn}(s) \rightleftharpoons \mathrm{Zn}^{2+}(a q) + 2\mathrm{e}^{-}\)
06

b. Identifying Anode and Cathode

In this galvanic cell, oxidation of Zn(s) to \(\mathrm{Zn}^{2+}\) occurs at the anode, and \(\mathrm{Ag}^{+}\) is reduced to Ag(s) at the cathode.
07

b. Direction of Electron Flow and Ion Migration

In this reaction: Electron flow: from Zn(s) (anode) to \(\mathrm{Ag}^+\)(cathode). Ion migration in the salt bridge: - Anions (e.g., \(\mathrm{NO}_3^-\)) migrate towards the anode (Zn) to maintain electrical neutrality. - Cations (e.g., \(\mathrm{K}^+\)) migrate towards the cathode (Ag) to maintain electrical neutrality.
08

b. Overall Balanced Equation

From the half-reactions, we can write the overall balanced equation for this cell: \(\mathrm{Zn}(s) + 2\mathrm{Ag}^{+}(a q) \rightleftharpoons \mathrm{Zn}^{2+}(a q) + 2\mathrm{Ag}(s)\)

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Key Concepts

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

Electrochemistry
Electrochemistry is a branch of chemistry that deals with the interrelation of electrical currents and chemical reactions. This area of science is fundamental in explaining how batteries, like galvanic or voltaic cells, work. A galvanic cell, which is our focus, converts chemical energy into electrical energy through spontaneous redox reactions. The standard conditions for these reactions typically involve solute concentrations of 1.0 M and gases at 1.0 atm.

In electrochemistry, the electrical potential, or voltage, between two electrodes can be measured and is determined by the inherent tendencies of different substances to gain or lose electrons. Such tendencies are quantified through the standard electrode potential, which is central to the concept of electromotive force (EMF) of a galvanic cell.
Redox Reactions
Redox reactions, or oxidation-reduction reactions, are processes where electrons are transferred between substances. These reactions are the backbone of electrochemistry, as they are responsible for the electron movement that generates electricity. In a redox reaction, one species loses electrons (oxidation) while another gains electrons (reduction).

  • Oxidation: Loss of electrons or an increase in oxidation state by a molecule, atom, or ion.
  • Reduction: Gain of electrons or a decrease in oxidation state by a molecule, atom, or ion.
Understanding the redox processes in galvanic cells is essential to predict the flow of electrons and the direction of chemical reactions under specific conditions.
Galvanic Cell Components
A Galvanic cell consists of several important components that work together to allow for the flow of electrons and the corresponding ionic currents. These include:
  • Anode: The electrode where oxidation occurs, releasing electrons into the external circuit.
  • Cathode: The electrode where reduction takes place, receiving electrons from the external circuit.
  • Salt Bridge: This allows for the flow of ions to maintain charge balance as electrons move through the external circuit. It connects the two half-cells electrolytically but prevents the direct mixing of the different solutions that might lead to immediate reaction without electricity production.
  • External Circuit: Allows for the flow of electrons from the anode to the cathode, driving the redox reaction.
Each of these parts plays a crucial role in ensuring the continual flow of electrons and ions, and thus the successful operation of the cell.
Electron Flow in Electrochemical Cells
In a galvanic cell, the electron flow is always from the anode to the cathode through the external circuit. Electrons are pushed away from the anode because oxidation, losing electrons, occurs there. Meanwhile, they are pulled toward the cathode because reduction, gaining electrons, happens at this point.

The directional flow of ions in the salt bridge is also essential to the overall function of the cell. Anions move toward the anode to balance the positive charge left behind by the lost electrons, while cations move toward the cathode to balance the negative charge acquired by the gain of electrons. This migration maintains electrical neutrality and keeps the cell functioning.

Proper identification of electron flow direction and ion migration within the electrochemical cells is crucial for accurately sketching galvanic cell diagrams as well as understanding how these cells operate.

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Most popular questions from this chapter

When balancing reactions in Chapter 3 , we did not mention that reactions must be charge balanced as well as mass balanced. What do charge balanced and mass balanced mean? How are redox reactions charge balanced?

The blood alcohol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) level can be determined by titrating a sample of blood plasma with an acidic potassium dichromate solution, resulting in the production of \(\mathrm{Cr}^{3+}(a q)\) and carbon dioxide. The reaction can be monitored because the dichromate ion \(\left(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right)\) is orange in solution, and the \(\mathrm{Cr}^{3+}\) ion is green. The unbalanced redox equation is $$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{CO}_{2}(g)$$ If \(31.05 \mathrm{~mL}\) of \(0.0600 M\) potassium dichromate solution is required to titrate \(30.0 \mathrm{~g}\) blood plasma, determine the mass percent of alcohol in the blood.

A galvanic cell is based on the following half-reactions: $$\begin{array}{ll}\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s) & \mathscr{E}^{\circ}=-0.440 \mathrm{~V} \\ 2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}(g) & \mathscr{E}^{\circ}=0.000 \mathrm{~V} \end{array}$$ where the iron compartment contains an iron electrode and \(\left[\mathrm{Fe}^{2+}\right]=1.00 \times 10^{-3} M\) and the hydrogen compartment contains a platinum electrode, \(P_{\mathrm{H}_{2}}=1.00 \mathrm{~atm}\), and a weak acid, \(\mathrm{HA}\), at an initial concentration of \(1.00 M .\) If the observed cell potential is \(0.333 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\), calculate the \(K_{\mathrm{a}}\) value for the weak acid HA.

Gold metal will not dissolve in either concentrated nitric acid on concentrated hydrochloric acid. It will dissolve, however, in aqua regia, a mixture of the two concentrated acids. The products of the reaction are the \(\mathrm{AuCl}_{4}^{-}\) ion and gaseous NO. Write a balanced equation for the dissolution of gold in aqua regia.

The overall reaction and equilibrium constant value for a hydrogen-oxygen fuel cell at \(298 \mathrm{~K}\) is $$2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(I) \quad K=1.28 \times 10^{83}$$ a. Calculate \(\mathscr{C}^{\circ}\) and \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\) for the fuel cell reaction. b. Predict the signs of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the fuel cell reaction. c. As temperature increases, does the maximum amount of work obtained from the fuel cell reaction increase, decrease, or remain the same? Explain.
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