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Chapter 17: Problem 80

An electrochemical cell consists of a nickel metal electrode immersed in a solution with \(\left[\mathrm{Ni}^{2+}\right]=1.0 \mathrm{M}\) separated by a porous disk from an aluminum metal electrode. a. What is the potential of this cell at \(25^{\circ} \mathrm{C}\) if the aluminum electrode is placed in a solution in which \(\left[\mathrm{Al}^{3+}\right]=7.2 \times\) \(10^{-3} M ?\) b. When the aluminum electrode is placed in a certain solution in which \(\left[\mathrm{Al}^{3+}\right]\) is unknown, the measured cell potential at \(25^{\circ} \mathrm{C}\) is \(1.62 \mathrm{V}\). Calculate \(\left[\mathrm{Al}^{3+}\right]\) in the unknown solution. (Assume Al is oxidized.)

Short Answer

Expert verified
The potential of this cell at \(25°C\), with given concentrations of \(\mathrm{Ni^{2+}}\) and \(\mathrm{Al^{3+}}\), is approximately \(1.47\,\text{V}\) (part a). The concentration of \(\mathrm{Al^{3+}}\) in the unknown solution when the measured cell potential at \(25°C\) is \(1.62\,\text{V}\) is approximately \(6.5 \times 10 ^{-4}\,\text{M}\) (part b).

Step by step solution

01

Identify the half-reactions and standard electrode potentials

To find the potential of this electrochemical cell, first, we need to identify the half-reactions occurring at the nickel and aluminum electrodes and their standard electrode potentials (E°). From a table of standard electrode potentials, we have: For Nickel: \[Ni^{2+} + 2e^- \rightleftharpoons Ni(s); E°_{Ni} = -0.26\,\text{V}\] For Aluminum: \[ Al^{3+} + 3e^- \rightleftharpoons Al(s); E°_{Al} = -1.66\,\text{V}\]
02

Calculate the standard cell potential

In order to calculate the potential of this cell under standard conditions, we need to calculate the cell potential using the standard electrode potentials we found in step 1. The cell potential under standard conditions, \(E°_{cell}\) is the difference between the standard electrode potentials of the reduction and oxidation half-reactions: \[E°_{cell} = E°_{cathode} - E°_{anode}\] As aluminum has the more negative standard reduction potential, it will be oxidized, and nickel will be reduced. \[E°_{cell} = (-0.26\,\text{V}) - (-1.66\,\text{V})\]
03

Calculate the actual cell potential at given concentrations

To find the actual cell potential at given concentrations of \(\mathrm{Ni^{2+}}\) and \(\mathrm{Al^{3+}}\), we use the Nernst equation: \[E_{cell} = E°_{cell} - \frac{RT}{nF} \ln \frac{[\text{Reductant}]}{[\text{Oxidant}]}\] At \(25°C\), \(R = 8.314\,\text{J mol}^{-1}\,\text{K}^{-1}\), \(T = 298.15\,\text{K}\), and \(F = 96,500\,\text{C mol}^{-1}\). Using the Nernst Equation for part (a), \[E_{cell} = E°_{cell} - \frac{(8.314)(298.15)}{6(96500)} \ln \frac{ \left(1.0\,\text{M} \right)}{\left(7.2 \times 10^{-3}\,\text{M} \right)}\]
04

Evaluate and display the results for part (a)

Now, we will calculate the cell potential, \[E_{cell} = (1.4\,\text{V}) - \frac{(8.314)(298.15)}{6(96500)} \ln \frac{(1.0\,\text{M})}{(7.2 \times 10^{-3}\,\text{M})} = 1.47\,\text{V}\] The potential of this cell at \(25°C\), with given concentrations of \(\mathrm{Ni^{2+}}\) and \(\mathrm{Al^{3+}}\), is approximately \(1.47\,\text{V}\).
05

Calculate the concentration of \(\mathrm{Al^{3+}}\) for part (b)

Given a cell potential of \(1.62\,\text{V}\), we need to find the concentration of \(\mathrm{Al^{3+}}\). We can rearrange the Nernst equation from step 3 to solve for \(\mathrm{[Al^{3+}]}\): \[[\mathrm{Al^{3+}}] = \frac{[\mathrm{Ni^{2+}}]}{\text{exp} \left[ \frac{nF (E_{cell} - E°_{cell})}{RT} \right]}\] Plugging in the given values, \[[\mathrm{Al^{3+}}] = \frac{1\,\text{M}}{\text{exp} \left[ \frac{6(96500)(1.62\,\text{V} - 1.4\,\text{V})}{(8.314)(298.15)} \right]} = 6.5 \times 10^{-4}\,\text{M}\] The concentration of \(\mathrm{Al^{3+}}\) in the unknown solution when the measured cell potential at \(25°C\) is \(1.62\,\text{V}\) is approximately \(6.5 \times 10 ^{-4}\,\text{M}\).

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Most popular questions from this chapter

How can one construct a galvanic cell from two substances, each having a negative standard reduction potential?

The overall reaction in the lead storage battery is $$\mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{HSO}_{4}^{-}(a q) \longrightarrow 2 \mathrm{PbSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ a. For the cell reaction \(\Delta H^{\circ}=-315.9 \mathrm{kJ}\) and \(\Delta S^{\circ}=\) \(263.5 \mathrm{J} / \mathrm{K} .\) Calculate \(\mathscr{E}^{\circ}\) at \(-20 .^{\circ} \mathrm{C} .\) Assume \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. b. Calculate \(\mathscr{E}\) at \(-20 .^{\circ} \mathrm{C}\) when \(\left[\mathrm{HSO}_{4}^{-}\right]=\left[\mathrm{H}^{+}\right]=4.5 \mathrm{M}\) c. Consider your answer to Exercise \(71 .\) Why does it seem that batteries fail more often on cold days than on warm days?

Under standard conditions, what reaction occurs, if any, when each of the following operations is performed? a. Crystals of \(\mathrm{I}_{2}\) are added to a solution of NaCl. b. \(\mathrm{Cl}_{2}\) gas is bubbled into a solution of NaI. c. A silver wire is placed in a solution of \(\mathrm{CuCl}_{2}\) d. An acidic solution of \(\mathrm{FeSO}_{4}\) is exposed to air. For the reactions that occur, write a balanced equation and calculate \(\mathscr{E}^{\circ}, \Delta G^{\circ},\) and \(K\) at \(25^{\circ} \mathrm{C}\)

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The electrolysis of \(\mathrm{BiO}^{+}\) produces pure bismuth. How long would it take to produce \(10.0 \mathrm{g}\) Bi by the electrolysis of a \(\mathrm{BiO}^{+}\) solution using a current of \(25.0 \mathrm{A} ?\)
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