Chapter 18: Problem 154
For the following half-reaction, \(E^{\circ}=1.103 \mathrm{~V}\) : $$ \mathrm{Cu}^{2+}(a q)+2 \mathrm{CN}^{-}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}(\mathrm{CN})_{2}^{-}(a q) $$ Calculate the formation constant \(K_{\mathrm{f}}\) for \(\mathrm{Cu}(\mathrm{CN})_{2}^{-}\).
Short Answer
Expert verified
The formation constant \(K_f\) for \(\mathrm{Cu(CN)_2^{-}}\) is approximately \(1.47 \times 10^{18}.\)
Step by step solution
01
Understand Relationship between E° and Kf
The standard electrode potential \(E^{\circ}\) can be related to the equilibrium constant \(K\) using the Nernst equation for equilibrium: \[ \Delta G^{\circ} = -RT \ln K = -nFE^{\circ} \]. Here, \(n\) is the number of moles of electrons transferred, \(F\) is Faraday's constant \(96485\,\text{C/mol}\), and \(R\) is the ideal gas constant \(8.314\,\text{J/(mol K)}\). \(T\) is the temperature in Kelvin, usually taken as standard room temperature \(298\,\text{K}\).
02
Identify Necessary Variables
For the given half-reaction, \(n = 1\) electron is transferred as we observe from the half-reaction equation in the question. \(\Delta G^{\circ}\) and \(K_f\) are related through the formula we will calculate in the next steps.
03
Calculate ΔG° Using E°
Use the formula \(\Delta G^{\circ} = -nFE^{\circ} \) to calculate \(\Delta G^{\circ}\). Substituting for the given values: \[\Delta G^{\circ} = -(1)(96485)(1.103)\, \text{J/mol} = -106375.355\, \text{J/mol}.\]
04
Convert ΔG° from Joules to Kilojoules
\(\Delta G^{\circ}\) typically is more conveniently expressed in kJ/mol. Convert \(-106375.355\, \text{J/mol}\) to kJ/mol: \(\Delta G^{\circ} = -106.375\, \text{kJ/mol}\).
05
Relate ΔG° to K_f
Use the relation \(\Delta G^{\circ} = -RT \ln K_f\). Rearrange to find \(K_f\): \[ \ln K_f = -\frac{\Delta G^{\circ}}{RT} \]. Substituting \(\Delta G^{\circ} = -106.375\, \text{kJ/mol} = -106375\, \text{J/mol} \), \(R = 8.314\, \text{J/(mol K)} \), and \(T = 298\, \text{K}\): \[ \ln K_f = \frac{106375}{(8.314)(298)} \approx 42.92.\]
06
Calculate K_f from ln K_f
Now solve for \( K_f \) by exponentiating: \[ K_f = e^{42.92} \approx 1.47 \times 10^{18}.\]
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Nernst Equation
The Nernst Equation is a fundamental formula in electrochemistry. It connects the properties of electrochemical cells with the concentration of the chemical species involved. This equation is expressed as:\[ E = E^{\circ} - \frac{RT}{nF} \ln Q \]Where:
- \( E \) is the electrode potential at non-standard conditions.
- \( E^{\circ} \) is the standard electrode potential.
- \( R \) is the universal gas constant \(8.314\,\text{J/(mol K)}\).
- \( T \) is the temperature in Kelvin.
- \( n \) is the number of moles of electrons transferred.
- \( F \) is Faraday's constant \(96485\,\text{C/mol}\).
- \( Q \) is the reaction quotient, which represents the concentration of reactants and products.
Electrode Potential
Electrode Potential is a measure of the ability of an electrochemical cell to drive an electric current from the electrode to the electrolyte or vice versa. It is denoted as \( E \) and is measured in volts. The standard electrode potential \( E^{\circ} \) refers to the potential measured under standard conditions: 1 M concentrations, 1 atm pressure, and 298 K temperature.Electrode potential is crucial in determining how easily a chemical species is reduced or oxidized in a reaction:
- A positive \( E^{\circ} \) indicates a species is predisposed to accept electrons and undergo reduction.
- A negative \( E^{\circ} \) suggests a species will likely give up electrons and undergo oxidation.
- The greater the \( E^{\circ} \), the stronger its oxidizing power.
Gibbs Free Energy
Gibbs Free Energy, denoted as \( \Delta G \), is a thermodynamic term that measures the amount of "useful" work obtainable from a process at constant temperature and pressure. It helps determine the spontaneity of a process:\[ \Delta G = \Delta H - T\Delta S \]Where:
- \( \Delta G \) is the change in free energy.
- \( \Delta H \) is the change in enthalpy.
- \( T \) is the temperature in Kelvin.
- \( \Delta S \) is the change in entropy.