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Chapter 18: Problem 66

A concentration cell is constructed having Cu electrodes in two CuSO \(_{4}\) solutions \(\mathrm{A}\) and \(\mathrm{B}\). At \(25^{\circ} \mathrm{C}\), the osmotic pressures of the two solutions are 48.9 atm and 4.89 atm, respectively. Calculate the \(E_{\text {cell }},\) assuming no ion-pair formation.

Short Answer

Expert verified
The cell potential \(E_{cell}\) of the constructed concentration cell is -0.0146 V

Step by step solution

01

Calculate the concentrations

The first step involves calculating the molar concentrations of the two solutions using the formula for osmotic pressure \(Π = CRT\). The molar concentrations of solution A (\(C_A\)) and solution B (\(C_B\)) are: \(C_A = Π_A/RT\) and \(C_B = Π_B/RT\). Here, \(Π_A=48.9\) atm, \(Π_B=4.89\) atm, R (the ideal gas constant) equals \(0.0821 L∙atm/K∙mol\) and T (the absolute temperature) equals \(25°C + 273.15 = 298.15 K\)
02

Apply the Nernst equation

The next step is to apply the Nernst equation to calculate the potential of the cell. A concentration cell at equilibrium always has a potential of zero so \(E^0 = 0\). Then we have \(E_{cell} = -\frac{RT}{2F}ln(C_A/C_B)\). The value of \(F\) (Faraday's constant) is \(96485 C/mol\), and \(n=2\) because the balanced chemical reaction occurring in the cell is \(Cu^{2+}(B) + 2e^{-} → Cu(A)\).
03

Perform the calculation

Lastly, substitute the known values into the equation to calculate the cell potential. The natural logarithm \(\ln(C_A/C_B)\) can be written as \(\ln((Π_A/RT)/(Π_B/RT))\), which simplifies to \(\ln(Π_A/Π_B)\). So we get: \(E_{cell} = -\frac{RT}{2F}\ln(Π_A/Π_B)=-\frac{0.0821 L∙atm/K∙mol × 298.15 K}{2 × 96485 C/mol}×ln(48.9/4.89)=-0.0146 V\) Here, the osmotic pressure values were used instead of concentrations, because the ratio of pressures equals the ratio of concentrations in ideal solutions.

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Most popular questions from this chapter

Calculate the standard emf of a cell that uses the \(\mathrm{Mg} / \mathrm{Mg}^{2+}\) and \(\mathrm{Cu} / \mathrm{Cu}^{2+}\) half-cell reactions at \(25^{\circ} \mathrm{C}\) Write the equation for the cell reaction that occurs under standard-state conditions.

Describe the electrolysis of an aqueous solution of \(\mathrm{KNO}_{3}\)

A galvanic cell is constructed by immersing a piece of copper wire in \(25.0 \mathrm{~mL}\) of a \(0.20 \mathrm{M} \mathrm{CuSO}_{4}\) solution and a zinc strip in \(25.0 \mathrm{~mL}\) of a \(0.20 \mathrm{M} \mathrm{ZnSO}_{4}\) solution. (a) Calculate the emf of the cell at \(25^{\circ} \mathrm{C}\) and predict what would happen if a small amount of concentrated \(\mathrm{NH}_{3}\) solution were added to (i) the \(\mathrm{CuSO}_{4}\) solution and (ii) the \(\mathrm{ZnSO}_{4}\) solution. Assume that the volume in each compartment remains constant at \(25.0 \mathrm{~mL}\). (b) In a separate experiment, \(25.0 \mathrm{~mL}\) of \(3.00 \mathrm{M} \mathrm{NH}_{3}\) are added to the \(\mathrm{CuSO}_{4}\) so- lution. If the emf of the cell is \(0.68 \mathrm{~V},\) calculate the formation constant \(\left(K_{\mathrm{f}}\right)\) of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\)

Explain why most useful galvanic cells give voltages of no more than 1.5 to \(2.5 \mathrm{~V}\). What are the prospects for developing practical galvanic cells with voltages of \(5 \mathrm{~V}\) or more?

An aqueous solution of a platinum salt is electrolyzed at a current of 2.50 A for \(2.00 \mathrm{~h}\). As a result, \(9.09 \mathrm{~g}\) of metallic \(\mathrm{Pt}\) are formed at the cathode. Calculate the charge on the Pt ions in this solution.
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