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Chapter 18: Problem 77

An electrochemical cell consists of a standard hydrogen electrode and a copper metal electrode. a. What is the potential of the cell at \(25^{\circ} \mathrm{C}\) if the copper electrode is placed in a solution in which \(\left[\mathrm{Cu}^{2+}\right]=\) \(2.5 \times 10^{-4} \mathrm{M} ?\) b. The copper electrode is placed in a solution of unknown \(\left[\mathrm{Cu}^{2+}\right] .\) The measured potential at $25^{\circ} \mathrm{C}\( is 0.195 \)\mathrm{V}\( . What is \)\left[\mathrm{Cu}^{2+}\right] ?$ (Assume \(\mathrm{Cu}^{2+}\) is reduced.)

Short Answer

Expert verified
The cell potential in part a is approximately \(0.308V\), and the unknown concentration of copper ions in the solution for part b is approximately \(80M\).

Step by step solution

01

Gather given information and convert temperature to Kelvin

Given values: - Standard hydrogen electrode - Copper metal electrode - Temperature: \(25^{\circ}C = 298K\) For part a: - \([\mathrm{Cu^{2+}}] = 2.5 \times 10^{-4}M\) For part b: - Measured potential: \(0.195V\) - Unknown \([\mathrm{Cu^{2+}}]\)
02

Write down the Nernst equation

The Nernst equation is as follows: \[E = E^\circ - \frac{RT}{nF} \times \ln{Q}\] where - E: cell potential - \(E^\circ\): standard cell potential - R: universal gas constant (\(8.314 J K^{-1} mol^{-1}\)) - T: temperature in Kelvin - n: number of moles of electrons transferred - F: Faraday's constant (\(96,485\;C mol^{-1}\)) - Q: reaction quotient
03

Calculate E for part a, given the Cu(2+) concentration

Using the standard reduction potentials, we can determine the standard cell potential for the given cell configuration. Standard reduction potential for hydrogen: \(E^\circ_{H+} = 0.000V\) Standard reduction potential for copper: \(E^\circ_{Cu^{2+}/Cu} = 0.34V\) \(E^\circ_{cell} = E^\circ_{Cu^{2+}/Cu} - E^\circ_{H+} = 0.34\;V\) For the cell reaction, the reduction half-cell reaction involves two electrons: \[\mathrm{Cu^{2+}} + 2e^- \rightarrow \mathrm{Cu}\] So, \(n = 2\), and we have a concentration of \([\mathrm{Cu^{2+}}] = 2.5 \times 10^{-4}M\) Now, let's plug all these values into the Nernst equation: \[E_{cell} = 0.34 - \frac{(8.314)(298)}{(2)(96485)} \times \ln{(2.5 \times 10^{-4})}\]
04

Solve for the cell potential E for part a

\[E_{cell} = 0.34 - \frac{(8.314)(298)}{(2)(96485)} \times \ln{(2.5 \times 10^{-4})} \approx 0.308V\] So, the cell potential in part a is \(0.308V\).
05

Setup Nernst equation for part b

We already know the measured potential for part b (it's given as \(0.195V\)), so E can be replaced with this value. \[0.195 = 0.34 - \frac{(8.314)(298)}{(2)(96485)} \times \ln{[\mathrm{Cu^{2+}}]}\] Then, isolate the unknown concentration term \([\mathrm{Cu^{2+}}]\) to solve for it: \[\ln{[\mathrm{Cu^{2+}}]} = \frac{(0.34 - 0.195)(2)(96485)}{(8.314)(298)}\]
06

Solve for the unknown Cu(2+) concentration in part b

First, compute the value of the logarithm: \[\ln{[\mathrm{Cu^{2+}}]} = \frac{(0.34 - 0.195)(2)(96485)}{(8.314)(298)} \approx 4.38\] Next, exponentiate both sides to find the concentration: \[[\mathrm{Cu^{2+}}] = e^{4.38} \approx 80M\] Thus, the unknown concentration of copper ions in the solution for part b is approximately \(80M\).

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Most popular questions from this chapter

Consider a galvanic cell based on the following theoretical half-reactions: $$\begin{array}{ll}{\text {}} & { \mathscr{E}^{\circ}(\mathbf{V}) } \\\ \hline{M^{4+}+4 e^{-} \longrightarrow M} & {0.66} \\ {N^{3+}+3 e^{-} \longrightarrow N} & {0.39}\end{array}$$ What is the value of \(\Delta G^{\circ}\) and \(K\) for this cell?

A standard galvanic cell is constructed so that the overall cell reaction is $$2 \mathrm{Al}^{3+}(a q)+3 \mathrm{M}(s) \longrightarrow 3 \mathrm{M}^{2+}(a q)+2 \mathrm{Al}(s)$$ where \(\mathrm{M}\) is an unknown metal. If \(\Delta G^{\circ}=-411 \mathrm{kJ}\) for the overall cell reaction, identify the metal used to construct the standard cell.

One of the few industrial-scale processes that produce organic compounds electrochemically is used by the Monsanto Company to produce 1,4-dicyanobutane. The reduction reaction is $$2 \mathrm{CH}_{2}=\mathrm{CHCN}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{NC}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{CN}$$ The \(\mathrm{NC}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{CN}\) is then chemically reduced using hydrogen gas to $\mathrm{H}_{2} \mathrm{N}-\left(\mathrm{CH}_{2}\right)_{6}-\mathrm{NH}_{2},$ which is used to produce production of nylon. What current must be used to produce \(150 . \mathrm{kg} \mathrm{NC}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{CN}\) per hour?

Consider a concentration cell that has both electrodes made of some metal M. Solution A in one compartment of the cell contains 1.0\(M \mathrm{M}^{2+} .\) Solution \(\mathrm{B}\) in the other cell compartment has a volume of 1.00 L. At the beginning of the experiment 0.0100 mole of \(\mathrm{M}\left(\mathrm{NO}_{3}\right)_{2}\) and 0.0100 mole of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) are dissolved in solution \(\mathrm{B}\) (ignore volume changes), where the reaction $$\mathrm{M}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \rightleftharpoons \mathrm{MSO}_{4}(s)$$ occurs. For this reaction equilibrium is rapidly established, whereupon the cell potential is found to be 0.44 \(\mathrm{V}\) at \(25^{\circ} \mathrm{C} .\) Assume that the process $$\mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M}$$ has a standard reduction potential of \(-0.31 \mathrm{V}\) and that no other redox process occurs in the cell. Calculate the value of \(K_{\mathrm{sp}}\) for \(\mathrm{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C} .\)

A chemist wishes to determine the concentration of \(\mathrm{CrO}_{4}^{2-}\) electrochemically. A cell is constructed consisting of a saturated calomel electrode (SCE; see Exercise 111\()\) and a silver wire coated with \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) . The \(\mathscr{E}^{\circ}\) value for the following half-reaction is 0.446 \(\mathrm{V}\) relative to the standard hydrogen electrode: $$\mathrm{Ag}_{2} \mathrm{CrO}_{4}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Ag}+\mathrm{CrO}_{4}^{2-}$$ a. Calculate \(\mathscr{E}_{\text { cell } \text { and }} \Delta G\) at \(25^{\circ} \mathrm{C}\) for the cell reaction when \(\left[\mathrm{CrO}_{4}^{2-}\right]=1.00 \mathrm{mol} / \mathrm{L}\) . b. Write the Nernst equation for the cell. Assume that the SCE concentrations are constant. c. If the coated silver wire is placed in a solution (at $25^{\circ} \mathrm{C} )\( in which \)\left[\mathrm{CrO}_{4}^{2-}\right]=1.00 \times 10^{-5} M,$ what is the expected cell potential? d. The measured cell potential at \(25^{\circ} \mathrm{C}\) is 0.504 \(\mathrm{V}\) when the coated wire is dipped into a solution of unknown \(\left[\mathrm{CrO}_{4}^{2-}\right] .\) What is \(\left[\mathrm{CrO}_{4}^{2-}\right]\) for this solution? e. Using data from this problem and from Table \(18.1,\) calculate the solubility product \(\left(K_{\mathrm{sp}}\right)\) for $\mathrm{Ag}_{2} \mathrm{CrO}_{4}$.
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