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Chapter 18: Problem 135

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 \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\)

Short Answer

Expert verified
The emf of the cell at 25 degrees Celsius is calculated as 1.10 V. The emf would decrease when NH3 is added to the CuSO4 solution, whereas the emf would remain unaffected when NH3 is added to the ZnSO4 solution. The formation constant (Kf) for the complex Cu(NH3)42+ is obtained from Nernst equation.

Step by step solution

01

Calculation of emf

The cell notation of our galvanic cell is Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s). From this, we find that the anode is Zn and cathode is Cu. Using standard reduction potentials, we get E(cell)= E(cathode) – E(anode) = [Cu2+/Cu] - [Zn2+/Zn] = +0.34 V - (-0.76) = 1.10 V.
02

Prediction of cell behaviour with addition of NH3

Addition of NH3 to CuSO4 solution will result in the formation of a complex Cu(NH3)42+ ion. This withdrawal of Cu2+ ions from the solution will make the Cu2+/Cu half-cell reduction potential less positive or more negative, resulting in a decrease in cell potential. But if we add NH3 to ZnSO4 solution, it doesn't significantly affect the Zn2+/Zn couple because Zn2+ doesn't form complex with NH3.
03

Calculation of Formation Constant (Kf)

For a complex Cu(NH3)42+, The reaction involved is, Cu2+ + 4NH3 ⇌ Cu(NH3)42+. Now the new half reaction becomes Cu(NH3)42+ + 2e- ⇌ Cu + 4NH3. The Nernst equation can be written as E = E⁰ - (0.0591/n) log ([Cu] / [Cu(NH3)42+]). Given that, E = 0.68 V, and E⁰ = 0.34 V, we can calculate for [Cu(NH3)42+] / [Cu] which will give us the value of Kf.

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Most popular questions from this chapter

What is the emf of a cell consisting of a \(\mathrm{Pb}^{2+} / \mathrm{Pb}\) half-cell and a \(\mathrm{Pt} / \mathrm{H}^{+} / \mathrm{H}_{2}\) half-cell if \(\left[\mathrm{Pb}^{2+}\right]=0.10 \mathrm{M}\), \(\left[\mathrm{H}^{+}\right]=0.050 \mathrm{M},\) and \(P_{\mathrm{H}}=2.0 \mathrm{~atm} ?\)

Tarnished silver contains \(\mathrm{Ag}_{2} \mathrm{~S}\). The tarnish can be removed by placing silverware in an aluminum pan containing an inert electrolyte solution, such as \(\mathrm{NaCl} .\) Explain the electrochemical principle for this procedure. [The standard reduction potential for the half- cell reaction \(\mathrm{Ag}_{2} \mathrm{~S}(s)+2 e^{-} \longrightarrow 2 \mathrm{Ag}(s)+\) \(\mathrm{S}^{2-}(a q)\) is \(-0.71 \mathrm{~V} .1\)

Industrially, copper metal can be purified electrolytically according to the following arrangement. The anode is made of the impure Cu electrode and the cathode is the pure Cu electrode. The electrodes are immersed in a \(\mathrm{CuSO}_{4}\) solution. (a) Write the half-cell reactions at the electrodes. (b) Calculate the mass (in grams) of Cu purified after passing a current of 20 A for \(10 \mathrm{~h}\). (c) Explain why impurities such as \(\mathrm{Zn}, \mathrm{Fe}, \mathrm{Au},\) and \(\mathrm{Ag}\) are not deposited at the electrodes.

Which of the following reagents can oxidize \(\mathrm{H}_{2} \mathrm{O}\) to \(\mathrm{O}_{2}(g)\) under standard-state conditions: \(\mathrm{H}^{+}(a q), \mathrm{Cl}^{-}(a q)\) \(\mathrm{Cl}_{2}(g), \mathrm{Cu}^{2+}(a q), \mathrm{Pb}^{2+}(a q), \mathrm{MnO}_{4}^{-}(a q)\) (in acid)?

Write the Nernst equation and explain all the terms.
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