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Chapter 18: Problem 77

An electrochemical cell consists of a standard hydrogen electrode and a copper metal electrode. If the copper electrode is placed in a solution of \(0.10 \mathrm{M} \mathrm{NaOH}\) that is saturated with \(\mathrm{Cu}(\mathrm{OH})_{2}\), what is the cell potential at \(25^{\circ} \mathrm{C} ?\left[\right.\) For \(\mathrm{Cu}(\mathrm{OH})_{2}, K_{\text {sp }}=1.6 \times 10^{-19}\) ]

Short Answer

Expert verified
The cell potential of the given electrochemical cell at \(25^{\circ}\mathrm{C}\) is approximately \(0.65 V\).

Step by step solution

01

Calculate the concentration of \(Cu^{2+}\)

First, the solubility product expression for the reaction \(Cu(OH)_2 \rightleftharpoons Cu^{2+} + 2OH^-\) can be written as: \[K_{sp}=[Cu^{2+}][OH^-]^2\] Since the solubility product constant, \(K_{sp} = 1.6 \times 10^{-19}\), and the concentration of OH- ions is given as \(0.10 M "\mathrm{NaOH}"\), we can solve for the concentration of \(Cu^{2+}\): \[1.6 \times 10^{-19} = [Cu^{2+}] (0.10)^2\] \[ [Cu^{2+}]= \frac{1.6\times10^{-19}}{0.01}\] \[ [Cu^{2+}]= 1.6\times10^{-17} M\]
02

Calculate the reduction potential of copper electrode using Nernst equation

For the redox reaction: \[Cu^{2+}(aq) + 2e^- → Cu(s)\] We can use the Nernst equation to find the cell potential. The Nernst equation is given as: \[E_{cell}=E_{cell}^0 - \frac{0.0592}{n}\log{Q}\] Our reaction has \(n = 2\) electrons, and the standard cell potential for the reduction of Cu2+ to Cu is \(E_{cell}^0 = 0.34 V\). The reaction quotient, Q, can be formulated as: \[Q=[Cu^{2+}]\] Now, we can substitute the values calculated earlier in the Nernst Equation: \[E_{cell} = 0.34 - \frac{0.0592}{2}\log{1.6\times10^{-17}}\] Evaluating the expression gives us: \[E_{cell} \approx 0.65 V\]
03

Determine the overall cell potential

Now we can determine the overall cell potential as follows: \[E_{overall} = E_{copper} - E_{hydrogen}\] Since the standard hydrogen electrode has a reduction potential of 0 V: \[E_{overall} = E_{copper}\] Therefore, the overall cell potential at \(25^{\circ}\mathrm{C}\) is: \[E_{overall} \approx 0.65 V\]

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Most popular questions from this chapter

A fuel cell designed to react grain alcohol with oxygen has the following net reaction: $$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(I)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)$$ The maximum work that 1 mole of alcohol can do is \(1.32 \times\) \(10^{3} \mathrm{~kJ} .\) What is the theoretical maximum voltage this cell can achieve at \(25^{\circ} \mathrm{C} ?\)

An unknown metal \(\mathrm{M}\) is electrolyzed. It took \(74.1 \mathrm{~s}\) for a current of \(2.00 \mathrm{~A}\) to plate out \(0.107 \mathrm{~g}\) of the metal from a solution containing \(\mathrm{M}\left(\mathrm{NO}_{3}\right)_{3}\). Identify the metal.

A factory wants to produce \(1.00 \times 10^{3} \mathrm{~kg}\) barium from the electrolysis of molten barium chloride. What current must be applied for \(4.00 \mathrm{~h}\) to accomplish this?

The overall reaction and equilibrium constant value for a hydrogen-oxygen fuel cell at \(298 \mathrm{~K}\) is $$2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(I) \quad K=1.28 \times 10^{83}$$ a. Calculate \(\mathscr{C}^{\circ}\) and \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\) for the fuel cell reaction. b. Predict the signs of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the fuel cell reaction. c. As temperature increases, does the maximum amount of work obtained from the fuel cell reaction increase, decrease, or remain the same? Explain.

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