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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(NO鈧)鈧. Therefore, the two half-cell reactions involved, the standard hydrogen electrode (SHE) and the copper half-cell, are: SHE: \(2H鈦(aq) + 2e鈦 \rightarrow H鈧(g)\) Copper half-cell: \(Cu^{2+}(aq) + 2e鈦 \rightarrow Cu(s)\)
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掳) and concentrations of the ions involved in the cell reaction. For this cell, the Nernst equation can be written as: \(E = E掳 - \frac{0.0592}{n} \log{\frac{[Cu^{2+}]}{[H^+]^2}}\) In this expression, E掳 is the standard cell potential, n is the number of electrons transferred in the redox reaction (in this case, n = 2), and [Cu虏鈦篯 and [H鈦篯 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 [Cu虏鈦篯. To do this, we will rewrite the Nernst equation as follows: \(E = E掳 - \frac{0.0592}{2} \log{\frac{[Cu^{2+}]}{[H^+]^2}}\) We can take \(E掳 - \frac{0.0592}{2} \log[1/{H^+]^2}\) 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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Hydrogen Electrode
The Standard Hydrogen Electrode (SHE) serves as a universal reference for measuring electrode potentials in electrochemical cells. It is based on the half-reaction where diatomic hydrogen gas is bubbled through a solution saturated with hydrogen ions. This reaction can be represented as \[2H^+(aq) + 2e^- \rightarrow H_2(g)\]The SHE is assigned a potential of 0 volts under standard conditions, which include a pressure of 1 atm and a temperature of 25掳C (298 K). Its role is crucial in allowing scientists to determine the potential of other cells by providing a consistent reference point. In the context of a galvanic cell, the SHE acts as one of the two electrodes, working together with a secondary electrode to facilitate a redox reaction.
Nernst Equation
The Nernst Equation is a fundamental tool in electrochemistry, used to relate the cell potential to the concentrations of the reacting species. It provides a way to calculate the cell potential \(E\) for any conditions by using the standard cell potential \(E^掳\) and adjusting for concentrations and temperature.For our exercise, the Nernst Equation takes the form:\[E = E^掳 - \frac{0.0592}{n} \log{\frac{[Cu^{2+}]}{[H^+]^2}}\]Where:- \(E^掳\) is the standard cell potential.- \(n\) is the number of electrons transferred in the redox reaction (in this case, \(n=2\)).- \([Cu^{2+}]\) is the concentration of copper ions.- \([H^+]\) is the concentration of hydrogen ions.This equation enables us to calculate the cell potential at non-standard conditions, which is essential for understanding how different ion concentrations affect the overall cell behavior.
Copper Ion Concentration
Copper ions play a central role in the function of the galvanic cell used in this exercise. These ions interact with the electrode to facilitate the cell's redox reactions. The copper half-cell reaction can be written as:\[Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)\]The concentration of \([Cu^{2+}]\) affects the cell potential, as illustrated by the Nernst Equation. When the copper ions accept electrons, they get reduced to solid copper, which deposits on the electrode. The concentration of copper ions is therefore crucial in determining the cell's efficacy and potential.It's important to note that any changes in \([Cu^{2+}]\) directly influence the cell's potential, providing a direct link between concentration and electrical output. This forms the basis for constructing a calibration curve in this exercise.
Cell Potential Calibration
Calibration of cell potential involves establishing a relationship between the measured potential and known concentrations of reactants. This allows chemists to determine unknown concentrations based on cell potential readings. In the context of the given exercise, we are constructing a calibration curve for the copper ions in the solution. By plotting the cell potential \(E\) against the logarithm of the copper ion concentration \(\log[Cu^{2+}]\), we can draw a straight line. The Nernst Equation, when simplified, provided us the form:\[E = A + 0.0296 \log[Cu^{2+}]\]Here, \(A\) represents a constant from the combined terms of the Nernst equation. The slope of 0.0296 reflects how the cell potential changes for every tenfold change in \(\log[Cu^{2+}]\). Through this method, chemists can predict the copper ion concentration based on observable cell potential values effectively.

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Most popular questions from this chapter

The overall reaction and equilibrium constant value for a hydrogen-oxygen fuel cell at 298 \(\mathrm{K}\) is $$2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \quad K=1.28 \times 10^{83}$$ a. Calculate \(8^{\circ}\) 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.

Consider the electrolysis of a molten salt of some metal. What information must you know to calculate the mass of metal plated out in the electrolytic cell?

A solution containing \(\mathrm{Pt}^{4+}\) is electrolyzed with a current of 4.00 \(\mathrm{A} .\) How long will it take to plate out 99\(\%\) of the platinum in 0.50 \(\mathrm{L}\) of a \(0.010-M\) solution of \(\mathrm{Pt}^{4+} ?\)

You have a concentration cell with Cu electrodes and \(\left[\mathrm{Cu}^{2+}\right]=1.00 M(\text { right side })\) and \(1.0 \times 10^{-4} M(\text { left side })\) a. Calculate the potential for this cell at \(25^{\circ} \mathrm{C}\) b. The \(\mathrm{Cu}^{2+}\) ion reacts with \(\mathrm{NH}_{3}\) to form \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) by the following equation: $$\begin{aligned} \mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) & \\\ & K=1.0 \times 10^{13} \end{aligned}$$ Calculate the new cell potential after enough \(\mathrm{NH}_{3}\) is added to the left cell compartment such that at equilibrium \(\left[\mathrm{NH}_{3}\right]=2.0 \mathrm{M} .\)

When copper reacts with nitric acid, a mixture of \(\mathrm{NO}(g)\) and \(\mathrm{NO}_{2}(g)\) is evolved. The volume ratio of the two product gases depends on the concentration of the nitric acid according to the equilibrium $$2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}(g) \rightleftharpoons 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$$ Consider the following standard reduction potentials at \(25^{\circ} \mathrm{C} :\) $$3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathscr{E}^{\circ}=0.957 \mathrm{V}$$ $$\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathscr{E}^{\circ}=0.775 \mathrm{V}$$ a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a NO and NO \(_{2}\) mixture with only 0.20\(\% \mathrm{NO}_{2}\) (by moles) at \(25^{\circ} \mathrm{C}\) and 1.00 atm? Assume that no other gases are present and that the change in acid concentration can be neglected.

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