Question

Using data from Appendix \(\mathrm{E}\), calculate the equilibrium constant for the disproportionation of the copper(I) ion at room temperature: \(2 \mathrm{Cu}^{+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{Cu}(s)\)

Short answer

The equilibrium constant for the disproportionation of the copper(I) ion at room temperature can be determined using the standard reduction potentials and the Nernst equation. The half-reactions are \(Cu^+(aq) \longrightarrow Cu^{2+}(aq) + e^-\) and \(Cu^+(aq) + e^- \longrightarrow Cu(s)\), with standard reduction potentials of +0.52V and +0.16V, respectively. The standard cell potential is \(E^0_{cell} = -0.36V\). Applying the Nernst equation, we find the equilibrium constant to be approximately \(K \approx 8.92 \cdot 10^{17}\).

Step-by-step solution

Determine the oxidation and reduction half-reactions

First, we need to identify the half-reactions involved in the disproportionation process. In this case, one copper(I) ion is being oxidized, and the other is being reduced. Oxidation half-reaction: \(Cu^{+}(aq) \longrightarrow Cu^{2+}(aq) + e^{-}\) Reduction half-reaction: \(Cu^{+}(aq) + e^{-} \longrightarrow Cu(s)\)

Look up standard reduction potentials

Consulting Appendix E, we can find the standard reduction potentials for both half-reactions: \(E^{0}_{ox}\) (Oxidation half-reaction): \[Cu^{+}(aq) + e^{-} \longrightarrow Cu(s),\, E^{0}_{ox} = +0.52V\] \(E^{0}_{red}\) (Reduction half-reaction): \[Cu^{2+}(aq) + e^{-} \longrightarrow Cu^{+}(aq),\, E^{0}_{red} = +0.16V\] Note that we need to reverse the standard reduction potentials for both half-reactions to match our desired disproportionation reaction.

Calculate the standard cell potential

Now, we can calculate the standard cell potential (\(E^{0}_{cell}\)) for the disproportionation reaction using the standard reduction potentials of the half-reactions: E^0_{cell} = E^0_{red} - E^0_{ox} \(E^{0}_{cell} = (+0.16V) - (+0.52V) = -0.36V\)

Apply the Nernst equation to determine the equilibrium constant

Next, we apply the Nernst equation to solve for the equilibrium constant (K): \[E^{0}_{cell} = -\frac{RT}{nF} \ln K\] At room temperature and for disproportionation n = 1, and using the values of the gas constant (R) and the Faraday's constant (F), we can rewrite the equation as: \(-0.36V = -\frac{8.314 J / (mol \cdot K) \cdot 298K}{1 \cdot 96,485 C / mol} \ln K\) Solving for K, we get: \(K \approx 8.92\cdot 10^{17}\) The equilibrium constant for the disproportionation of the copper(I) ion at room temperature is approximately \(8.92 \cdot 10^{17}\).

Key concepts

Disproportionation Reaction

A disproportionation reaction is a type of redox reaction wherein a single substance undergoes both oxidation and reduction, forming two different products. In these reactions, the same element is simultaneously oxidized and reduced.

In the given exercise, the disproportionation of copper(I) ions results in copper(II) ions and copper metal. This reaction is significant in electrochemistry because it showcases how an element can change oxidation states in a single reaction. It’s important to clearly identify which species is oxidized and which is reduced to proceed with calculations related to equilibrium and cell potential.

Standard Reduction Potential

The standard reduction potential (SRP), often denoted as E°, is the measure of the tendency of a chemical species to gain electrons and thereby be reduced. Each half-cell has its standard reduction potential, which can be found in electrochemical series tables. The more positive the standard reduction potential, the greater the species’ tendency to be reduced.

When dealing with a disproportionation reaction, we look up the standard reduction potentials for both half-reactions, bearing in mind that the oxidation process will have the sign of its SRP reversed. This helps in calculating the standard cell potential, an essential step towards finding the equilibrium constant for the reaction.

Nernst Equation

The Nernst Equation enables the calculation of cell potential under non-standard conditions. It relates the measurable cell potential to the reaction quotient and temperature. The equation is a fundamental principle in electrochemistry, expressed as:
\[E = E^{0}_{cell} - \frac{RT}{nF} \ln Q\]
E stands for the cell potential, E°_cell is the standard cell potential, R is the gas constant, T the temperature in Kelvin, n the number of moles of electrons transferred, F Faraday’s constant, and Q the reaction quotient. In the context of our exercise, the Nernst Equation is rearranged to solve for the equilibrium constant, K, which will give us an understanding of the position of equilibrium for the reaction.

Electrochemistry

Electrochemistry is the branch of chemistry that studies the movements of electrons in chemical reactions. It bridges the gap between electrical energy and chemical reactions through redox reactions. Understanding electrochemical processes involves exploring how chemical energy is converted to electrical energy in batteries, or vice versa in electrolysis.

In our example, electrochemistry principles are applied to calculate the equilibrium constant of the disproportionation reaction involving copper(I) ions. The reactions taking place at the electrodes, as well as the potential difference between them, represent core components of electrochemical investigation.

Oxidation and Reduction Half-Reactions

To fully comprehend electrochemical reactions, we must dissect them into oxidation and reduction half-reactions. Oxidation involves the loss of electrons, while reduction is the gain of electrons. For each redox reaction, there is an associated pair of half-reactions showing the separate processes occurring at the anode and cathode.

Oxidation and reduction half-reactions are written to reflect the actual electron transfer, with electrons included explicitly in the reactions. In practical applications like our textbook exercise, identifying and writing these half-reactions accurately is the precursor to finding cell potentials and equilibrium constants.