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A Cu electrode is immersed in a solution that is \(1.00 \mathrm{M}\) in \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) and \(1.00 \mathrm{M}\) in \(\mathrm{NH}_{3}\). When the cathode is a standard hydrogen electrode, the emf of the cell is found to be \(+0.08 \mathrm{~V}\). What is the formation constant for \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) ?

Short Answer

Expert verified
The formation constant for the \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) complex ion is 1.

Step by step solution

01

Write down the half-cell reactions and overall cell reaction

First, let us write down the half-cell reactions. The standard hydrogen electrode will act as the reference half-cell with the following reaction, which is considered the reduction half-cell: \[\mathrm{2H}^{+}(aq) + 2e^{-} \rightarrow \mathrm{H}_{2}(g)\] Now, let's consider the Cu electrode as the oxidation half-cell, in which copper ion forms complex with ammonia: \[\mathrm{Cu}^{2+}(aq) + 4\mathrm{NH}_{3}(aq) \rightarrow \left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}(aq)\] Therefore, the overall cell reaction will be: \[\mathrm{2H}^{+}(aq) + \mathrm{Cu}^{2+}(aq) + 4\mathrm{NH}_{3}(aq) \rightarrow \mathrm{H}_{2}(g) + \left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}(aq)\]
02

Calculate the standard cell potential of the given cell

The standard cell potential of the given cell (\(E^{0}_{cell}\)) is calculated from the given emf and the reduction half-cell potential of hydrogen electrode (\(E^{0}_{H^{+}/H_{2}}\)) as follows: \[E^{0}_{cell} = E^{0}_{Cu^{2+}/Cu(NH_{3})_{4}^{2+}} - E^{0}_{H^{+}/H_{2}}\] We know that the standard reduction potential of hydrogen electrode, \(\mathrm{E}^{0}_{H^{+}/H_{2}} = 0\mathrm{~V}\), and the given emf of the cell is \(+0.08\mathrm{~V}\), which we will use to calculate \(\mathrm{E}^{0}_{Cu^{2+}/Cu(NH_{3})_{4}^{2+}}\): \[E^{0}_{Cu^{2+}/Cu(NH_{3})_{4}^{2+}} = E^{0}_{cell} = +0.08 \mathrm{~V}\]
03

Apply the Nernst equation to determine the formation constant of the complex

We can apply the Nernst equation to determine the relationship between cell potential and the concentration of chemical species: \[E_{cell} = E^{0}_{cell} - \frac{0.05916}{n} \log Q\] where \(E_{cell}\) is the cell potential at the given concentration, \(E^{0}_{cell}\) is the standard cell potential, \(n\) is the number of electrons transferred, and \(Q\) is the reaction quotient. For the given overall cell reaction, we have \(n = 2\) and the reaction quotient, \(Q = \frac{[\mathrm{Cu(NH_{3})_{4}]^{2+}]}{[\mathrm{Cu^{2+}][\mathrm{NH}_{3}]^{4}}\), which represents the formation constant of the complex (\(K_f\)). Let's plug the values into the Nernst equation: \[+0.08 \mathrm{~V} = +0.08 \mathrm{~V} - \frac{0.05916}{2} \log K_f\] Here, the cell potential under standard conditions and at the given concentration are equal. Solving for \(K_f\), we get: \[\log K_f = 0\] Taking the antilog (or 10 to the power of both sides): \[K_f = 10^{0} = 1\] Thus, the formation constant for the \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) complex ion is 1.

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Key Concepts

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

Electrode Potential
Understanding the concept of electrode potential is essential for studying electrochemistry. This potential is a measure of a chemical species' ability to be reduced or oxidized at an electrode. It is crucial because it affects how electrons transfer between chemical species and electrodes, which is fundamental to the operation of electrochemical cells and batteries.

When we immerse a metal electrode into a solution containing its ions, there is a tendency for the metal atoms to lose electrons and become ions (oxidation), or for the metal ions in the solution to gain electrons and become metal atoms (reduction). The direction in which the reaction proceeds depends on both the electrode and the ion's standard reduction potentials.

The potential difference that develops between the electrode and the solution is what we refer to as the 'electrode potential'. When we measure this potential against a standard reference electrode, such as the standard hydrogen electrode (SHE), we obtain a value that tells us how readily the electrode's metal will gain or lose electrons.
Nernst Equation
The Nernst equation is a fundamental equation in electrochemistry that links the electrode potential of an electrochemical cell to the concentrations of the reacting species. The equation provides a quantitative way to incorporate the effect of concentration (or activities for more precise work) on cell potential.

The Nernst equation is given by: \[E_{cell} = E^{0}_{cell} - \frac{0.05916}{n} \log Q\]Here, \(E_{cell}\) is the cell potential under non-standard conditions, \(E^{0}_{cell}\) is the standard cell potential (when all concentrations are 1 M), \(n\) is the number of moles of electrons transferred, and \(Q\) is the reaction quotient, a measure of the ratio of the concentrations of products to reactants at any point in time.

By using this equation, you can calculate the actual potential of an electrochemical cell under any conditions, not just the standard conditions. It gives us a powerful tool to predict how a cell's potential will change as the reaction proceeds or as the concentrations of reactants or products change.
Chemical Equilibrium
At the heart of understanding reactions in chemistry is the concept of chemical equilibrium. This is the point in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentrations of reactants and products.

In the context of electrochemistry and electrode potential, the formation constant (\(K_f\)) is a specific type of equilibrium constant that tells us about the stability of a complex ion in solution. It is mathematically expressed in a similar manner to other equilibrium constants, but it applies to the formation of a complex ion from its constituent ions and molecules.

Understanding chemical equilibrium is crucial when predicting how reactions will respond to changes in conditions, such as concentration and temperature. The principle of Le Chatelier's states that a system at equilibrium will counteract any changes imposed upon it. If a reaction at equilibrium is subjected to changed conditions, the equilibrium will adjust itself to partially oppose the effect of the change.
Complex Ion Formation
Complex ions are species where a central metal ion is bonded to one or more ligands, which are molecules or ions that can donate a pair of electrons. The process by which ligands attach to the central ion is called complex ion formation, and it is a reversible reaction.

In the context of the exercise, the central metal ion is copper (\(Cu^{2+}\)), and the ligand is ammonia (\(NH_3\)). The formation of \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) from \(Cu^{2+}\) ions and \(NH_3\) molecules can be represented by an equilibrium expression with its formation constant \(K_f\). This constant is a measure of the strength of the interaction between the metal ion and the ligands. A high \(K_f\) value indicates a strong attachment and a stable complex ion.

It's important to note that the formation constant is actually a ratio of the concentration of the complex ion to the product of the concentrations of the metal ion and the ligands to the power of their stoichiometry in the complex formation. This value can be determined through experiments, such as measuring the electromotive force (emf) of an electrochemical cell under specific conditions, as illustrated in the exercise.

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Most popular questions from this chapter

Give the number of (valence) \(d\) electrons associated with the central metal ion in each of the following complexes: (a) \(\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\), (b) \(\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]\left(\mathrm{NO}_{3}\right)_{2}\) (c) \(\mathrm{Na}\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]\), (d) \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right] \mathrm{ClO}_{4}\), (c) \([\mathrm{Sr}(\mathrm{EDTA})]^{2-}\) -

Write names for the following coordination compounds: (a) \(\left[\mathrm{Cd}(\mathrm{en}) \mathrm{Cl}_{2}\right]\) (b) \(\mathrm{K}_{4}\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]\) (c) \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{s}\left(\mathrm{CO}_{3}\right)\right] \mathrm{Cl}\) (d) \(\left[\operatorname{lr}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]\left(\mathrm{NO}_{3}\right)_{3}\)

For each of the following metals, write the electronic configuration of the atom and its \(2+\) ion: (a) \(\mathrm{Mn}\), (b) \(\mathrm{Ru}\), (c) \(\mathrm{Rh}\). Draw the crystal-field energy-level diagram for the \(d\) orbitals of an octahedral complex, and show the placement of the \(d\) electrons for each \(2+\) ion, assuming a strong-field complex. How many unpaired electrons are there in each case?

Indicate the likely coordination number of the metal in each of the following complexes (a) \(\left[\mathrm{Rh}(\text { bipy })_{3}\right]\left(\mathrm{NO}_{3}\right)_{3}\) (b) \(\mathrm{Na}_{4}\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2} \mathrm{Cl}_{2}\right]\) (c) \(\left[\mathrm{Cr}(0 \text {-phen })_{3}\right]\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{3}\) (d) \(\mathrm{Na}_{2}[\mathrm{Co}(\mathrm{EDTA}) \mathrm{Br}]\)

The complex \(\left[\mathrm{Ru}(\mathrm{EDTA})\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{-}\)undergoes substitution reactions with several ligands, replacing the water molecule with the ligand. In all cases, the ruthenium stays in the \(+3\) oxidation state and the ligands use a nitrogen donor atom to bind to the metal. $$ \left[\operatorname{Ru}(\mathrm{EDTA})\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{-}+\mathrm{L} \longrightarrow[\operatorname{Ru}(\mathrm{EDTA}) \mathrm{L}]^{-}+\mathrm{H}_{2} \mathrm{O} $$ The rate constants for several ligands are as follows: (a) One possible mechanism for this substitution reaction is that the water molecule dissociates from the Ru(III) in the rate-determining step, and then the ligand L binds to Ru(III) in a rapid second step. A second possible mechanism is that L approaches the complex, begins to form a new bond to the Ru(III), and displaces the water molecule, all in a single concerted step. Which of these two mechanisms is more consistent with the data? Explain. (b) What do the results suggest about the relative donor ability of the nitrogens of the three ligands toward Ru(TII))? (c) Assuming that the complexes are all low spin, how many unpaired electrons are in each?

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