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A test for completeness of electrodeposition of \(\mathrm{Cu}\) from a solution of \(\mathrm{Cu}^{2+}(\mathrm{aq})\) is to add \(\mathrm{NH}_{3}(\mathrm{aq}) .\) A blue color signifies the formation of the complex ion \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\left(K_{\mathrm{f}}=1.1 \times 10^{13}\right) .\) Let \(250.0 \mathrm{mL}\) of \(0.1000 \mathrm{M} \mathrm{CuSO}_{4}(\text { aq })\) be electrolyzed with a \(3.512 \mathrm{A}\) current for 1368 s. At this time, add a sufficient quantity of \(\mathrm{NH}_{3}(\text { aq })\) to complex any remaining \(\mathrm{Cu}^{2+}\) and to maintain a free \(\left[\mathrm{NH}_{3}\right]=0.10 \mathrm{M} .\) If \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) is detectable at concentrations as low as \(1 \times 10^{-5} \mathrm{M}\) should the blue color appear?

Short Answer

Expert verified
After performing the steps and calculations, if the calculated concentration of \([Cu(NH_3)_4]^{2+}\) is greater than \(1 \times 10^{-5} M\), then the blue color will appear. Otherwise, it will not be observed.

Step by step solution

01

Calculate Moles of Copper Electrolyzed

Use Faraday's law of electrolysis to find the amount of Copper (Cu) that gets electrolyzed. Faraday's law states that the amount of substance discharged at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the solution. This can be written as: \( n = \dfrac{I \cdot t}{F \cdot z} \) where, \( n \) is the number of moles, \( I \) is current in amperes, \( t \) is time in seconds, \( F \) is Faraday's constant (96485 C/mol) and \( z \) is the number of electrons transferred per molecule during electrolysis. Given that the current (I) is 3.512 A and time (t) is 1368 s, the number of moles of Cu that gets electrolyzed can be calculated.
02

Calculate Remaining Concentration of Copper

After calculating the moles of Copper that have been electrolyzed in step 1, we can find the remaining concentration of Copper in the solution. This can be found by subtracting the moles of Copper electrolyzed from the initial moles of Copper (initial concentration times volume), and then dividing by the volume of the solution in Liters.
03

Equilibrium of Complex ion formation

In this step, we address the reaction between the remaining Copper in the solution and the ammonia that is added. This forms a complex ion according to the reaction: \( Cu^{2+} + 4NH_3 \rightarrow [Cu(NH_3)_4]^{2+} \). The ammonia is added until its concentration is 0.1M. The equilibrium constant (\(K_f\)) of this reaction is given as \(1.1 \times 10^{13}\). Use the Reaction Quotient expression \(Q = \dfrac{[[Cu(NH_3)_4]^{2+}]}{[NH_3]^4[Cu^{2+}]}\) to calculate the reaction quotient. If Q is less than \(K_f\), the reaction will proceed to the right and more complex will form. If Q is greater than \(K_f\), the reaction will move to the left and less complex will form. If Q is equal to \(K_f\), then the reaction is at equilibrium.
04

Determine the Observable Color Change

Compare the concentration of the complex that is formed (\([Cu(NH_3)_4]^{2+}\)) with the detectable concentration level (1X10^-5 M). If the concentration of the complex formed is greater than the detectable concentration, then the blue color will appear.

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Key Concepts

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

Faraday's Law of Electrolysis
Faraday's Law of Electrolysis is crucial for understanding how substances change during electrolysis. It states that the amount of a substance transformed at an electrode is proportional to the electricity passed through the electrolyte. This is mathematically expressed as \( n = \frac{I \cdot t}{F \cdot z} \) where \( n \) is the number of moles, \( I \) is the current in amperes, \( t \) is the time in seconds, \( F \) is Faraday's constant (96,485 C/mol), and \( z \) is the number of electrons exchanged per ion.
  • This equation allows us to calculate the number of moles of a compound deposited or dissolved through the electrolysis process.
  • For instance, in the electrolysis of copper, knowing the charge needed to deposit one mole of copper can help predict how much will be deposited given a current and a time period.
  • In our problem, the copper has a valency of 2, meaning \( z = 2 \), which becomes a critical component in determining the electrolysis outcome.
By using this law, we can conclude how much copper is removed from or added to a solution during experimentation, ultimately helping students grasp the quantitative nature of electrochemical processes.
Complex Ion Formation
When transition metals like copper react with ligands (molecules or ions that donate pairs of electrons), they can form complex ions. This aspect of electrochemistry deals with how metal ions bond with ligands in solutions, forming structurally distinct and often colorful complexes.
  • In the problem provided, copper(II) ions ( Cu^{2+}) react with ammonia ( NH_3) to form a complex ion. The reaction is \( Cu^{2+} + 4NH_3 \rightarrow [Cu(NH_3)_4]^{2+} \).
  • The formation of this blue-colored complex is used as an indication of remaining copper ions in the solution post-electrolysis.
  • The equilibrium constant ( K_f) for this complex formation is very high, suggesting that the reaction proceeds almost completely in the forward direction under standard conditions.
Once the complex ion is formed, it typically alters the solution's properties, including its color, which can be observed visually, aiding qualitative analyses in a laboratory setting. The complexation principle not only plays a role in scientific experiments but also in everyday phenomena, such as the relations of metals within biological systems.
Copper Electrodeposition
Electrodeposition is a form of electrolysis used to deposit a scale of metal onto an electrode. In the context of copper, electrodeposition involves placing copper onto a conductive surface by applying an electrical current.
  • When a copper solution undergoes electrolysis, electrons are transferred from the cathode, reducing copper ions to solid copper that deposits onto the electrode.
  • This process is affected by factors such as the concentration of copper ions, the surface area of electrodes, the current density, and time.
  • In the practical exercise, accurately timing and controlling this procedure ensures that the desired amount of copper is deposited or tested for removal.
After electrodeposition, understanding if the process is complete or sufficient often involves chemical tests, like the formation of a complex ion, to see if any copper remains. Students explore electrodeposition in electrochemistry to learn how metals purify, plate, or recover in various industrial and research applications. Such knowledge effectively bridges principles of chemistry with real-world technological advances.

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Most popular questions from this chapter

The following voltaic cell registers an \(E_{\text {cell }}=0.108 \mathrm{V}\) What is the pH of the unknown solution? $$\operatorname{Pt}\left|\mathrm{H}_{2}(\mathrm{g}, 1 \mathrm{atm})\right| \mathrm{H}^{+}(x \mathrm{M}) \| \mathrm{H}^{+}(1.00 \mathrm{M}) |$$ $$\mathrm{H}_{2}(\mathrm{g}, 1 \mathrm{atm}) | \mathrm{Pt}$$

Natural gas transmission pipes are sometimes protected against corrosion by the maintenance of a small potential difference between the pipe and an inert electrode buried in the ground. Describe how the method works.

Predict whether, to any significant extent, (a) \(\mathrm{Fe}(\mathrm{s})\) will displace \(\mathrm{Zn}^{2+}(\mathrm{aq})\) (b) \(\mathrm{MnO}_{4}^{-}(\mathrm{aq})\) will oxidize \(\mathrm{Cl}^{-}(\mathrm{aq})\) to \(\mathrm{Cl}_{2}(\mathrm{g})\) in acidic solution; (c) \(\mathrm{Ag}(\mathrm{s})\) will react with \(1 \mathrm{M} \mathrm{HCl}(\mathrm{aq})\) (d) \(\mathrm{O}_{2}(\mathrm{g})\) will oxidize \(\mathrm{Cl}^{-}(\mathrm{aq})\) to \(\mathrm{Cl}_{2}(\mathrm{g})\) in acidic solution.

Use the Nernst equation and data from Appendix D to calculate \(E_{\text {rell for each of the following cells. }}\) (a) \(\operatorname{Mn}(\mathrm{s}) | \mathrm{Mn}^{2+}(0.40 \mathrm{M}) \| \mathrm{Cr}^{3+}(0.35 \mathrm{M})\) \(\mathrm{Cr}^{2+}(0.25 \mathrm{M}) | \mathrm{Pt}(\mathrm{s})\) (b) \(\operatorname{Mg}\left(\text { s) } | \operatorname{Mg}^{2+}(0.016 \mathrm{M}) \|\left[\mathrm{Al}(\mathrm{OH})_{4}\right]^{-}(0.25 \mathrm{M})\right.\) \(\mathrm{OH}^{-}(0.042 \mathrm{M}) | \mathrm{Al}(\mathrm{s})\)

From the observations listed, estimate the value of \(E^{\circ}\) for the half- reaction \(\mathrm{M}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{M}(\mathrm{s})\) (a) The metal M reacts with HNO \(_{3}(\text { aq })\), but not with \(\mathrm{HCl}(\mathrm{aq}) ; \mathrm{M}\) displaces \(\mathrm{Ag}^{+}(\mathrm{aq}),\) but not \(\mathrm{Cu}^{2+}(\mathrm{aq})\) (b) The metal \(M\) reacts with \(\mathrm{HCl}(\mathrm{aq}),\) producing \(\mathrm{H}_{2}(\mathrm{g}),\) but displaces neither \(\mathrm{Zn}^{2+}(\text { aq })\) nor \(\mathrm{Fe}^{2+}(\mathrm{aq})\).

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