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Chapter 18: Problem 99

Consider a galvanic cell consisting of a magnesium electrode in contact with \(1.0 \mathrm{M} \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) and a cadmium electrode in contact with \(1.0 \mathrm{M}\) \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} .\) Calculate \(E^{\circ}\) for the cell, and draw a diagram showing the cathode, anode, and direction of electron flow.

Short Answer

Expert verified
The \(E^{\circ}\) for the given galvanic cell is 1.97 V. The Magnesium electrode, where oxidation is taking place, is the anode and The Cadmium electrode where reduction is taking place is the cathode. The electrons flow from the anode to the cathode in the galvanic cell.

Step by step solution

01

Identify the Half-Reactions

Starting off the exercise, the half-reactions for magnesium and cadmium should be identified using standard reduction potentials. In a typical reduction reaction, the metal ion gains electrons to become a solid metal. These reactions are: \n1. \( \mathrm{Mg^{2+}} + 2 e^- \rightarrow \mathrm{Mg}, E^{\circ} = -2.37 V \) \n 2. \( \mathrm{Cd^{2+}} + 2 e^- \rightarrow \mathrm{Cd}, E^{\circ} = -0.40V \)
02

Identify the Cathode and Anode

In a galvanic cell, the anode is the electrode where oxidation occurs, i.e., where there is a loss of electrons. The cathode is the electrode where reduction occurs, i.e., where there is a gain of electrons. The substance with a higher reduction potential (less negative) will undergo reduction and hence will be the cathode. Therefore, the Cd electrode (with \(E^{\circ} = -0.40V\)) is the cathode and the Mg electrode (with \(E^{\circ} = -2.37 V\)) is the anode.
03

Calculate the Cell Potential

The potential of a galvanic cell can be determined by calculating the difference between the cathode potential and the anode potential. Given by the formula \(E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}\), substituting the given values gives: \(E^{\circ}_{cell} = -0.40 V - ( -2.37 V) = 1.97 V.\)
04

Draw the Galvanic Cell

For depicting the cathode, anode, and direction of electron flow in a galvanic cell: \n1. Anode: The Magnesium electrode, where oxidation is taking place, is the anode. Magnesium is losing electrons, transforming into Mg2+. \n2. Cathode: The Cadmium electrode, where reduction is taking place, is the cathode. Cadmium ions gain electrons to form Cd. \n3. Electron Flow: The electrons flow from the anode (where they are produced) to the cathode (where they are consumed).

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Most popular questions from this chapter

Balance the following redox equations by the ionelectron method: (a) \(\mathrm{Mn}^{2+}+\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{MnO}_{2}+\mathrm{H}_{2} \mathrm{O}\) (in basic solution) (b) \(\mathrm{Bi}(\mathrm{OH})_{3}+\mathrm{SnO}_{2}^{2-} \longrightarrow \mathrm{SnO}_{3}^{2-}+\mathrm{Bi}\) (in basic solution) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{Cr}^{3+}+\mathrm{CO}_{2}\) (in acidic solution) (d) \(\mathrm{ClO}_{3}^{-}+\mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_{2}+\mathrm{ClO}_{2}\) (in acidic solution)

Calculate the standard emf of a cell that uses the \(\mathrm{Mg} / \mathrm{Mg}^{2+}\) and \(\mathrm{Cu} / \mathrm{Cu}^{2+}\) half-cell reactions at \(25^{\circ} \mathrm{C}\) Write the equation for the cell reaction that occurs under standard-state conditions.

Compare the pros and cons of a fuel cell, such as the hydrogen-oxygen fuel cell, and a coal-fired power station for generating electricity.

Given that \(\begin{array}{ll}2 \mathrm{Hg}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Hg}_{2}^{2+}(a q) & E^{\circ}=0.92 \mathrm{~V} \\\ \mathrm{Hg}_{2}^{2+}(a q)+2 e^{-} \longrightarrow 2 \mathrm{Hg}(l) & E^{\circ}=0.85 \mathrm{~V}\end{array}\) calculate \(\Delta G^{\circ}\) and \(K\) for the following process at \(25^{\circ} \mathrm{C}\) : \(\mathrm{Hg}_{2}^{2+}(a q) \longrightarrow \mathrm{Hg}^{2+}(a q)+\mathrm{Hg}(l)\) (The preceding reaction is an example of a disproportionation reaction in which an element in one oxidation state is both oxidized and reduced.)

Explain why most useful galvanic cells give voltages of no more than 1.5 to \(2.5 \mathrm{~V}\). What are the prospects for developing practical galvanic cells with voltages of \(5 \mathrm{~V}\) or more?
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