Problem 19
Question
Write the equations relating \(\Delta G^{\circ}\) and \(K\) to the standard emf of a cell. Define all the terms.
Step-by-Step Solution
Verified Answer
ΔG° = -nFE° and ΔG° = -RT ln K relate ΔG° to E° and K. E° = (RT/nF) ln K relates E° to K.
1Step 1: Understand the Relationship Between ΔG° and E°
The relationship between the standard free energy change (ΔG°) and the standard cell potential (E°) is given by the equation:\[ΔG° = -nFE°\]where ΔG° is the standard free energy change, n is the number of moles of electrons transferred in the reaction, F is the Faraday constant (approximately 96,485 C/mol), and E° is the standard cell potential in volts (V).
2Step 2: Relate ΔG° to the Equilibrium Constant (K)
The relationship between the standard free energy change (ΔG°) and the equilibrium constant (K) is expressed as:\[ΔG° = -RT \, ext{ln} \, K\]where R is the universal gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and K is the equilibrium constant of the reaction.
3Step 3: Relate E° to the Equilibrium Constant (K)
By combining the relationship between ΔG° and E° with that of ΔG° and K, we can derive a direct relationship between E° and K:\[E° = \frac{RT}{nF} \ln K\]This equation links the standard cell potential with the equilibrium constant. To convert this to base 10 logarithms, the equation often appears as:\[E° = \frac{0.0592}{n} \log K \quad (at \ 298 \ K)\]
Key Concepts
Equation RelationsStandard EmfEquilibrium ConstantStandard Cell Potential
Equation Relations
Understanding the equations that connect Gibbs Free Energy and other factors is key in chemistry and thermodynamics. The relationships form a fundamental basis for predicting and explaining chemical reactions.
\(\Delta G^{\circ}\) (standard free energy change) is linked to two important quantities through equations. First, it relates to the standard emf (electromotive force, E°) of a cell via the equation:
Another important equation is:
\(\Delta G^{\circ}\) (standard free energy change) is linked to two important quantities through equations. First, it relates to the standard emf (electromotive force, E°) of a cell via the equation:
- \(\Delta G^{\circ} = -nFE^{\circ}\)
- \(n\) stands for the number of moles of electrons in the redox reaction.
- \(F\) is Faraday's constant, approximately 96,485 C/mol.
Another important equation is:
- \(\Delta G^{\circ} = -RT \ln K\)
- \(R\) is the universal gas constant (8.314 J/mol·K).
- \(T\) represents the temperature in Kelvin.
- \(K\) is the equilibrium constant of the reaction.
Standard Emf
The standard emf, or standard cell potential (E°), measures the voltage or electrical potential difference of a cell under standard conditions. This is a way to gauge the tendency of a redox reaction to occur.
When calculating E°, remember it reflects the potential energy difference between the cathode and anode in an electrochemical cell. The sign of E° offers insight:
When calculating E°, remember it reflects the potential energy difference between the cathode and anode in an electrochemical cell. The sign of E° offers insight:
- Positive E° indicates a spontaneous reaction under standard conditions.
- Negative E° suggests a non-spontaneous reaction unless additional energy is provided.
- \(\Delta G^{\circ} = -nFE^{\circ}\)
Equilibrium Constant
The equilibrium constant (K) plays a vital role in chemistry, reflecting the balance of a reaction at equilibrium. It gives a quantitative idea of the position of equilibrium and tells how far a reaction progresses.
In the relationship \(\Delta G^{\circ} = -RT \ln K\), we can see:
In the relationship \(\Delta G^{\circ} = -RT \ln K\), we can see:
- A large K value (>1) corresponds to a negative \(\Delta G^{\circ}\), suggesting the reaction favors products and is spontaneous.
- A small K value (<1) means positive \(\Delta G^{\circ}\), where the reactants are favored.
- \(E^{\circ} = \frac{RT}{nF} \ln K\)
Standard Cell Potential
The standard cell potential (E°) is a significant measure in electrochemistry since it indicates the voltage a cell can supply when operating reversibly under standard conditions (298 K, 1 atm pressure, and 1 M concentration solutions).
It is calculated using the standard reduction potentials of the cathode and anode:
Understanding E° helps predict the energetics and directionality of electrochemical reactions. Its integration with \(\Delta G^{\circ}\) and K makes it a central concept in understanding both thermodynamic stability and chemical equilibrium of reactions.
It is calculated using the standard reduction potentials of the cathode and anode:
- \(E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}\)
Understanding E° helps predict the energetics and directionality of electrochemical reactions. Its integration with \(\Delta G^{\circ}\) and K makes it a central concept in understanding both thermodynamic stability and chemical equilibrium of reactions.
Other exercises in this chapter
Problem 10
Calculate the standard emf of a cell that uses the \(\mathrm{Mg} / \mathrm{Mg}^{2+}\) and \(\mathrm{Cu} / \mathrm{Cu}^{2+}\) half-cell reactions at \(25^{\circ}
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Predict whether the following reactions would occur spontaneously in aqueous solution at \(25^{\circ} \mathrm{C}\). Assume that the initial concentrations of di
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The equilibrium constant for the reaction: $$\operatorname{Sr}(s)+\mathrm{Mg}^{2+}(a q) \rightleftharpoons \mathrm{Sr}^{2+}(a q)+\mathrm{Mg}(s)$$ is \(2.69 \tim
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Given that \(E^{\circ}=0.52 \mathrm{~V}\) for the reduction \(\mathrm{Cu}^{+}(a q)+e^{-}\) \(\longrightarrow \mathrm{Cu}(s),\) calculate \(E^{\circ}, \Delta G^{
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