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Chapter 18: Problem 141

Show a sketch of a galvanic concentration cell. Each compartment consists of a Co electrode in a \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}\) solution. The concentrations in the compartments are \(2.0 \mathrm{M}\) and \(0.10 \mathrm{M}\), respectively. Label the anode and cathode compartments. Show the direction of electron flow. (a) Calculate the \(E_{\text {cell }}\) at \(25^{\circ} \mathrm{C}\). (b) What are the concentrations in the compartments when the \(E_{\text {cell }}\) drops to 0.020 V? Assume volumes to remain constant at \(1.00 \mathrm{~L}\) in each compartment.

Short Answer

Expert verified
The cell potential (\(E_{\text {cell }}\)) is initially calculated to be approximately 0.0592 V using the Nernst equation. When \(E_{\text {cell }}\) drops to 0.020 V, the calculated concentrations in the compartments would no longer be 2.0 M and 0.10 M.

Step by step solution

01

Sketch the galvanic concentration cell

In a galvanic concentration cell, all parts of the cell are identical except for the concentrations of the ions in the two half-cells. A sketch of such a cell would show a system consisting of two compartments containing Co electrodes in a \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}\) solution. The concentrations of the solutions would be \(2.0 \mathrm{M}\) and \(0.10 \mathrm{M}\), respectively.
02

Label Anode and Cathode

In this cell, oxidation (loss of electrons) occurs at the electrode with the higher concentration, and reduction (gain of electrons) occurs at the electrode with the lower concentration. Thus, the anode (where oxidation occurs) has 2.0 M \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}\) and the cathode (where reduction occurs) has 0.10 M \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}\).
03

Indicate Direction Of Electron Flow

The electrons in a galvanic cell always flow from the anode (where oxidation occurs and electrons are produced) to the cathode (where reduction occurs and electrons are consumed). In this case, electrons would flow from the electrode with 2.0 M solution to the electrode with 0.10 M solution.
04

Calculate \(E_{\text {cell }}\)

The Nernst equation can be used to calculate the cell potential (E), given the concentrations of the reactants and products. Since the reaction involves the same ion on both sides, it may simplify to:\[E_{\text {cell }} = E_{\text {cell }}^0 - \frac{0.0592}{n} \log \frac{[Co^{2+}]_{cathode}}{[Co^{2+}]_{anode}}\] Since the cell involves a standard Cobalt-Cobalt couple, \(E_{\text {cell }}^0\) is 0. In this case, n, the number of electrons transferred, is 2, and the concentrations are given. Substituting these values yields an \(E_{\text {cell }}\) value.
05

Calculate The New Concentrations

When \(E_{\text {cell }}\) drops to 0.020 V, the concentrations of \(Co^{2+}\) in the anode and cathode compartment changes. By rearranging the Nernst equation, you can solve for the ratio of the concentrations. Because the volumes of the compartments remain constant, the ratio of the concentrations also represents the ratio of moles of cobalt ions. By subtracting the transferred moles from the initial moles at the anode, and adding them to the initial moles at the cathode, you can find the final concentrations.

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Most popular questions from this chapter

Consider the following half-reactions: $$\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q)+5 e^{-} \overrightarrow{\mathrm{Mn}^{2+}}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l)$$ $$\mathrm{NO}_{3}^{-}(a q)+4 \mathrm{H}^{+}(a q)+3 e^{-} \longrightarrow_{\mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)}$$

What is the function of a salt bridge? What kind of electrolyte should be used in a salt bridge?

Describe the electrolysis of an aqueous solution of \(\mathrm{KNO}_{3}\)

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?

What is the difference between a galvanic cell (such as a Daniell cell) and an electrolytic cell?
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