Chapter 19: Problem 92
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
Step by step solution
Write Half-Reactions
Identify the Anode and Cathode
Calculate Standard Cell Potential, \( E^{\circ}_{\text{cell}} \)
Diagram of the Galvanic Cell
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Standard Cell Potential
To compute \(E^{\circ}_{\text{cell}}\), we use the standard reduction potential values of the two half-cells. This is achieved with the formula:
- \(E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}\)
Anode and Cathode Identification
In the exercise concerning magnesium and cadmium:
- Magnesium has a reduction potential of \(-2.37 \text{ V}\), which is more negative, signifying that it acts as the anode.
- Cadmium with a reduction potential of \(-0.40 \text{ V}\) plays the role of the cathode because it is less negative.
Half-Reaction Equations
For the magnesium-cadmium cell:
- Magnesium half-reaction (anode, oxidation): \[ \text{Mg}_{(s)} \rightarrow \text{Mg}^{2+}_{(aq)} + 2e^- \]
- Cadmium half-reaction (cathode, reduction): \[ \text{Cd}^{2+}_{(aq)} + 2e^- \rightarrow \text{Cd}_{(s)} \]
Electron Flow Direction
In our magnesium-cadmium cell:
- Electrons are released from the magnesium metal at the anode.
- These electrons travel through an external circuit to the cadmium cathode.