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Beginning with the equations that relate \(E^{\circ}, \Delta G^{\circ}\), and \(K\), show that \(\Delta G^{\circ}\) is negative and \(K>1\) for a reaction that has a positive value of \(E^{\circ}\)

Short Answer

Expert verified
If \(E^{\circ} > 0\), then \(\Delta G^{\circ} < 0\) and \(K > 1\).

Step by step solution

01

Introduction to Relevant Equations

First, recall the equations that relate the standard cell potential \(E^{\circ}\), the standard Gibbs free energy change \(\Delta G^{\circ}\), and the equilibrium constant \(K\). The equations are: \(\Delta G^{\circ} = -nFE^{\circ}\) and \(\Delta G^{\circ} = -RT \ln K\), where \(n\) is the number of moles of electrons transferred, \(F\) is the Faraday constant, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin.
02

Determining Sign of \(\Delta G^{\circ}\)

Given that \(E^{\circ}\) is positive, substitute into the equation \(\Delta G^{\circ} = -nFE^{\circ}\). Since \(\Delta G^{\circ}\) is the product of a negative sign and a positive number (positive \(E^{\circ}\)), \(\Delta G^{\circ}\) becomes negative.
03

Relation Between \(\Delta G^{\circ}\) and \(K\)

From the equation \(\Delta G^{\circ} = -RT \ln K\), when \(\Delta G^{\circ}\) is negative, the term \(\ln K\) must be positive because \(R\) and \(T\) are always positive. This implies that \(K > 1\) as the natural logarithm of a number greater than 1 is positive.
04

Conclusion

From the derivations, if \(E^{\circ} > 0\), then \(\Delta G^{\circ} < 0\), leading to \(K > 1\). This demonstrates the initial assertion using the relevant equations and relationships.

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Key Concepts

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

Standard Cell Potential
The standard cell potential, represented by the symbol \(E^{\circ}\), is a key concept in electrochemistry. It is the measure of the electromotive force (emf) of a galvanic cell under standard conditions, which typically include a temperature of 298 K, a pressure of 1 atm, and concentrations of 1 M for all solutions involved. \(E^{\circ}\) is expressed in volts (V). This potential provides insight into the capability of a chemical reaction to produce electric current.

The standard cell potential is calculated as the difference between the standard reduction potentials of the cathode and anode in a cell. The formula used is:
\[ E^{\circ} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \]
When \(E^{\circ} > 0\), it indicates a spontaneous reaction under standard conditions. This positive potential signifies that the reaction can do work and drive electrons from the anode to the cathode in the electrochemical cell.
Gibbs Free Energy
Gibbs free energy, denoted as \(\Delta G^{\circ}\), is a thermodynamic quantity that helps predict whether a chemical reaction will proceed spontaneously. Expressed in joules per mole (J/mol), \(\Delta G^{\circ}\) is defined under standard conditions, which are 298 K for temperature, 1 atm for pressure, and 1 M concentration for all substances.

The relationship between Gibbs free energy and standard cell potential is given by:
\[ \Delta G^{\circ} = -nFE^{\circ} \]
Here, \(n\) is the number of moles of electrons exchanged in the reaction, \(F\) is the Faraday constant (approximately 96485 C/mol), and \(E^{\circ}\) is the standard cell potential. When \(E^{\circ} > 0\), the term \(-nFE^{\circ}\) results in a negative \(\Delta G^{\circ}\), indicating a spontaneous process.

Spontaneity in this context means that the reaction can occur without any input of energy. Therefore, a negative \(\Delta G^{\circ}\) corresponds to a reaction that can naturally proceed forward.
Equilibrium Constant
The equilibrium constant, symbolized by \(K\), is a fundamental aspect in the study of chemical equilibrium. It provides the ratio of the concentrations of products to reactants at equilibrium, each raised to the power of their stoichiometric coefficients. For a general reaction, \( K \) is expressed using the equation:
\[ K = \frac{[\text{products}]}{[\text{reactants}]} \]

The standard Gibbs free energy change is directly related to \(K\) via:
\[ \Delta G^{\circ} = -RT \ln K \]
Where \(R\) is the universal gas constant (8.314 J/mol·K) and \(T\) is the temperature in Kelvin. When \(\Delta G^{\circ} < 0\), as derived from a positive \(E^{\circ}\), the natural logarithm \(\ln K\) must be positive, implying \(K > 1\).

This relationship tells us that when \(K > 1\), the products are more favored than the reactants at equilibrium, confirming that the reaction proceeds in the forward direction. This aligns with the concept that a spontaneous reaction under standard conditions has a larger concentration of products compared to reactants at equilibrium.

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Most popular questions from this chapter

Consider a Daniell cell with \(1.0 \mathrm{M}\) ion concentrations: Does the cell voltage increase, decrease, or remain the same when each of the following changes is made? Explain. (a) \(5.0 \mathrm{M} \mathrm{CuSO}_{4}\) is added to the cathode compartment. (b) \(5.0 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) is added to the cathode compartment. (c) \(5.0 \mathrm{M} \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) is added to the anode compartment. (d) \(1.0 \mathrm{M} \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) is added to the anode compartment.

Given the following half-reactions, combine the two that give the cell reaction with the most positive \(E^{\circ} .\) Write a balanced equation for the cell reaction, and calculate \(E^{\circ}\) and \(\Delta G^{\circ}\). $$ \begin{array}{ll} \mathrm{Co}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Co}(s) & E^{\circ}=-0.28 \mathrm{~V} \\ \mathrm{I}_{2}(s)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{I}(a q) & E^{\circ}=0.54 \mathrm{~V} \\ \mathrm{Cu}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & E^{\circ}=0.34 \mathrm{~V} \end{array} $$

Copper reduces dilute nitric acid to nitric oxide (NO) but reduces concentrated nitric acid to nitrogen dioxide \(\left(\mathrm{NO}_{2}\right)\) : (1) \(3 \mathrm{Cu}(s)+2 \mathrm{NO}_{3}^{-}(a q)+8 \mathrm{H}^{+}(a q) \longrightarrow\) \(3 \mathrm{Cu}^{2+}(a q)+2 \mathrm{NO}(g)+4 \mathrm{H}_{2} \mathrm{O}(l) \quad E^{\circ}=0.62 \mathrm{~V}\) (2) \(\mathrm{Cu}(s)+2 \mathrm{NO}_{3}^{-}(a q)+4 \mathrm{H}^{+}(a q) \longrightarrow\) \(\mathrm{Cu}^{2+}(a q)+2 \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) \quad E^{\circ}=0.45 \mathrm{~V}\) Assuming that \(\left[\mathrm{Cu}^{2+}\right]=0.10 \mathrm{M}\) and that the partial pressures of \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) are \(1.0 \times 10^{-3} \mathrm{~atm}\), calculate the potential \((E)\) for reactions (1) and (2) at \(25^{\circ} \mathrm{C}\) and show which reaction has the greater thermodynamic tendency to occur when the concentration of \(\mathrm{HNO}_{3}\) is (a) \(1.0 \mathrm{M}\) (b) \(10.0 \mathrm{M}\) (c) At what \(\mathrm{HNO}_{3}\) concentration do reactions ( 1\()\) and (2) have the same value of \(E\) ?

The silver oxide-zinc battery used in watches delivers a voltage of \(1.60 \mathrm{~V}\). Calculate the free-energy change (in kilojoules) for the cell reaction $$ \mathrm{Zn}(s)+\mathrm{Ag}_{2} \mathrm{O}(s) \longrightarrow \mathrm{ZnO}(s)+2 \mathrm{Ag}(s) $$

(a) Sketch a cell with inert electrodes suitable for the electrolysis of an aqueous solution of sulfuric acid. Label the anode and cathode, and indicate the direction of electron and ion flow. Identify the positive and negative electrodes. (b) Write balanced equations for the anode, cathode, and overall cell reactions

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