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Chapter 18: Problem 120

A galvanic cell consists of a standard hydrogen electrode and a copper electrode immersed in a Cu(NO \(_{3} )_{2}(a q)\) solution. If you wish to construct a calibration curve to show how the cell potential varies with \(\left[\mathrm{Cu}^{2+}\right],\) what should you plot to obtain a straight line? What will be the slope of this line?

Short Answer

Expert verified
To obtain a straight line, plot the cell potential (E) against the logarithm of the copper ion concentration, \(\log[Cu^{2+}]\). The slope of this line will be 0.0296.

Step by step solution

01

Understand the galvanic cell composition

The galvanic cell in this problem consists of a standard hydrogen electrode and a copper electrode immersed in a solution of Cu(NO₃)₂. Therefore, the two half-cell reactions involved, the standard hydrogen electrode (SHE) and the copper half-cell, are: SHE: \(2H⁺(aq) + 2e⁻ \rightarrow H₂(g)\) Copper half-cell: \(Cu^{2+}(aq) + 2e⁻ \rightarrow Cu(s)\)
02

Write the overall cell reaction

By combining the two half-cell reactions, we can write the overall cell reaction: \(2H^+ + Cu^{2+} + 2e^- \rightarrow H_2 + Cu + 2e^-\) Which can be simplified as: \(2H^+(aq) + Cu^{2+}(aq) \rightarrow H_2(g) + Cu(s)\)
03

Apply the Nernst Equation

The Nernst equation relates the cell potential (E) to the standard cell potential (E°) and concentrations of the ions involved in the cell reaction. For this cell, the Nernst equation can be written as: \(E = E° - \frac{0.0592}{n} \log{\frac{[Cu^{2+}]}{[H^+]^2}}\) In this expression, E° is the standard cell potential, n is the number of electrons transferred in the redox reaction (in this case, n = 2), and [Cu²⁺] and [H⁺] represent the concentrations of copper and hydrogen ions, respectively.
04

Simplify the Nernst Equation

The problem asks us to establish a linear relationship between the cell potential E and the copper ion concentration [Cu²⁺]. To do this, we will rewrite the Nernst equation as follows: \(E = E° - \frac{0.0592}{2} \log{\frac{[Cu^{2+}]}{[H^+]^2}}\) We can take \(E° - \frac{0.0592}{2} \log[1/{H^+]^2}\) as constant "A". Therefore, \(E = A + \frac{0.0296}{1} \log[Cu^{2+}]\) Now, we have a linear equation in the form of \(E = A + 0.0296\log[Cu^{2+}]\)
05

Determine the slope

From the above equation, the plot we can use for our calibration curve is the potential, E, versus the logarithm of the copper ion concentration \(\log[Cu^{2+}]\). The slope of this straight line will be equal to 0.0296 since that is the coefficient of \(\log[Cu^{2+}]\) in our equation. In conclusion, the calibration curve is a straight line when we plot cell potential E against the logarithm of the copper ion concentration, \(\log[Cu^{2+}]\), and the slope of this line is 0.0296.

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Most popular questions from this chapter

What reactions take place at the cathode and the anode when each of the following is electrolyzed? a. molten \(\mathrm{NiBr}_{2} \quad\) b. molten \(\mathrm{AlF}_{3} \quad\) c. molten \(\mathrm{MgI}_{2}\)

The overall reaction in the lead storage battery is $\mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{HSO}_{4}^{-}(a q) \longrightarrow$ $$\quad\quad\quad\quad\quad\quad\quad 2 \mathrm{PbSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ 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} \mathrm{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 69. Why does it seem that batteries fail more often on cold days than on warm days?

The following standard reduction potentials have been determined for the aqueous chemistry of indium: $$\operatorname{In}^{3+}(a q)+2 \mathrm{e}^{-} \longrightarrow \operatorname{In}^{+}(a q) \quad \mathscr{E}^{\circ}=-0.444 \mathrm{V}$$ $$\operatorname{In}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \operatorname{In}(s) \qquad \quad \mathscr{E}^{\circ}=-0.126 \mathrm{V}$$ a. What is the equilibrium constant for the disproportionation reaction, where a species is both oxidized and reduced, shown below? $$3 \ln ^{+}(a q) \longrightarrow 2 \operatorname{In}(s)+\operatorname{In}^{3+}(a q)$$ b. What is \(\Delta G_{i}^{\circ}\) for \(\operatorname{In}^{+}(a q)\) if $\Delta G_{f}^{\circ}=-97.9 \mathrm{kJ} / \mathrm{mol}\( for \)\operatorname{In}^{3+}(a q) ?$

Which of the following statement(s) is(are) true? a. Copper metal can be oxidized by \(\mathrm{Ag}^{+}\) (at standard conditions). b. In a galvanic cell the oxidizing agent in the cell reaction is present at the anode. c. In a cell using the half reactions $\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow$ Al and \(\mathrm{Mg}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mg}\) , aluminum functions as the anode. d. In a concentration cell electrons always flow from the compartment with the lower ion concentration to the compartment with the higher ion concentration. e. In a galvanic cell the negative ions in the salt bridge flow in the same direction as the electrons.

Electrolysis of an alkaline earth metal chloride using a current of 5.00 \(\mathrm{A}\) for 748 s deposits 0.471 \(\mathrm{g}\) of metal at the cathode. What is the identity of the alkaline earth metal chloride?
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