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If the equilibrium constant for a two-electron redox reaction at 298 \(\mathrm{K}\) is \(1.5 \times 10^{-4}\) , calculate the corresponding \(\Delta G^{\circ}\) and \(E^{\circ} .\)

Short Answer

Expert verified
The standard Gibbs free energy change \(\Delta G^{\circ}\) can be calculated using the equation \(\Delta G^{\circ} = -RT \ln{K}\), where R is the Universal gas constant, T is the temperature, and K is the equilibrium constant. Plugging in the values, we get \(\Delta G^{\circ} = -(8.314 \frac{J}{K.mol})(298 K)\ln{(1.5 \times 10^{-4})}\), which equals approximately \(5.77 \times 10^3 J/mol\). The standard cell potential \(E^{\circ}\) can be calculated using the equation \(E^{\circ} = -\frac{\Delta G^{\circ}}{nF}\), where n is the number of electrons transferred and F is the Faraday constant. Plugging in the values, we get \(E^{\circ} = -\frac{5.77 \times 10^3 J/mol}{(2)(96485 \frac{C}{mol})}\), which equals approximately \(0.0299 V\).

Step by step solution

01

Calculate the standard Gibbs free energy change

We can calculate the standard Gibbs free energy change for the given reaction using the first equation: $$\Delta G^{\circ} = -RT \ln{K}$$ Here, R is the universal gas constant which equals \( 8.314 \frac{J}{K.mol} \), T is the temperature in Kelvins (298 K), and K is the equilibrium constant which is given as \( 1.5 \times 10^{-4} \). Plugging these values into the equation, we can calculate the standard Gibbs free energy change: $$\Delta G^{\circ} = -(8.314 \frac{J}{K.mol})(298 K)\ln{(1.5 \times 10^{-4})}$$ Calculate the value of \(\Delta G^{\circ}\) to find the standard Gibbs free energy change.
02

Calculate the standard cell potential

Now, we can calculate the standard cell potential for the redox reaction using the second equation: $$\Delta G^{\circ} = -nFE^{\circ}$$ Here, n is the number of electrons transferred in the redox reaction (given as 2 electrons), F is the Faraday constant which equals \( 96485 \frac{C}{mol} \), and \(\Delta G^{\circ}\) is the standard Gibbs free energy change calculated in Step 1. We can rearrange the equation to solve for the standard cell potential: $$E^{\circ} = -\frac{\Delta G^{\circ}}{nF}$$ Plugging the values for \(\Delta G^{\circ}\), n, and F into the equation, we can calculate the standard cell potential: $$E^{\circ} = -\frac{\Delta G^{\circ}}{(2)(96485 \frac{C}{mol})}$$ Calculate the value of \(E^{\circ}\) to find the standard cell potential for the redox reaction. By following these steps, you can find the standard Gibbs free energy change and the standard cell potential for the given redox reaction with the provided equilibrium constant and temperature.

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Key Concepts

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

Gibbs Free Energy Change
The Gibbs free energy change, denoted as \( \Delta G^{\box} \), is a pivotal concept in thermodynamics and chemistry, which indicates the spontaneity of a chemical reaction at constant temperature and pressure. In simple terms, it helps predict whether a reaction will occur without additional energy. If \( \Delta G^{\box} < 0 \), the reaction is spontaneous; if \( \Delta G^{\box} > 0 \), the reaction is non-spontaneous; and if \( \Delta G^{\box} = 0 \), the system is at equilibrium.

To calculate \( \Delta G^{\box} \) for a reaction, you can use the relationship between free energy change and the equilibrium constant \( K \) at a particular temperature \( T \) provided by the equation \( \Delta G^{\box} = -RT \ln{K} \). The negative sign indicates that as the equilibrium constant increases, indicating a greater ratio of products to reactants, the free energy decreases, making the reaction more likely to proceed. The gas constant \( R = 8.314 \frac{J}{K\cdot mol} \) links the reaction to absolute temperature in Kelvins.

In the context of the exercise, one can deduce that the small value of \( K = 1.5 \times 10^{-4} \) suggests the reaction favors the reactants at equilibrium, leading to a positive \( \Delta G^{\box} \) value, meaning the reaction is non-spontaneous under standard conditions.
Standard Cell Potential
Standard cell potential \( E^{\box} \) represents the voltage or electromotive force of a cell when all components are in their standard states (which usually means concentrations of 1 molar and gases at 1 atmosphere pressure) at a temperature of 298 K (25°C). It's a measure of how much voltage, or potential energy, a cell can produce, which is directly linked to the spontaneity of the redox reaction taking place within the cell.

The standard cell potential can be linked to the Gibbs free energy change via the equation \( \Delta G^{\box} = -nFE^{\box} \), where \( n \) is the number of moles of electrons transferred, and \( F \) is the Faraday constant. By calculating \( \Delta G^{\box} \) first, as done in the exercise, one can then find \( E^{\box} \) to determine if the cell will release or require energy to operate. A positive value of \( E^{\box} \) signifies a spontaneous redox reaction, while a negative value signifies a non-spontaneous reaction.

The importance of \( E^{\box} \) in electrochemistry cannot be overstated, as it underpins the operation of batteries, fuel cells, and electrolysis processes, all of which convert chemical energy into electrical energy or vice-versa.
Faraday Constant
The Faraday constant \( F \) is a fundamental physical constant used in calculations involving electrochemistry, named after the English scientist Michael Faraday. It represents the total electric charge carried by one mole of electrons, with a magnitude of approximately \( 96485 \frac{C}{mol} \) (coulombs per mole).

This constant is crucial when translating between the chemical scale of moles and the physical scale of electric charge in coulombs. It allows for the quantification of the electrical work—which is directly related to Gibbs free energy change—involving redox reactions, as showcased in our exercise.

In the context of the standard cell potential equation, \( F \) is used to highlight the relationship between energy change and electric charge: \( \Delta G^{\box} = -nFE^{\box} \). It is fundamental in calculating how much electrical energy is produced or required in a reaction involving electron transfer, such as the operation of galvanic or electrolytic cells.
Nernst Equation
The Nernst equation is another cornerstone of electrochemistry, enabling the determination of cell potential under non-standard conditions, particularly when concentrations are not at standard conditions. It states that the voltage of an electrochemical cell depends logarithmically on the ratio of the concentrations of the products to the reactants.

The equation is given by: \( E = E^{\box} - \frac{RT}{nF} \ln{\frac{[products]}{[reactants]}} \) where \( E \) is the cell potential, \( E^{\box} \) is the standard cell potential, \( R \) is the ideal gas constant, \( T \) is the temperature in Kelvins, \( n \) is the number of moles of electrons transferred, \( F \) is the Faraday constant, and \( [products]/[reactants] \) represents the reaction quotient.

The Nernst equation elegantly combines thermodynamics with equilibrium to describe how electrochemical potentials shift as concentrations change, crucial for understanding batteries, corrosion, sensors, and any application where the chemical composition affects electrical properties. While the Nernst equation isn't directly applied in our exercise with given standard conditions, in practical scenarios, it is omnipresent in the adaptation of standard potentials to real-world conditions.

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Most popular questions from this chapter

A voltaic cell is based on \(\mathrm{Ag}^{+}(a q) / \mathrm{Ag}(s)\) and \(\mathrm{Fe}^{3+}(a q) /\) \(\mathrm{Fe}^{2+}(a q)\) half-cells. (a) What is the standard emf of the cell? (b) Which reaction occurs at the cathode and which at the anode of the cell? (c) Use \(S^{\circ}\) values in Appendix \(\mathrm{C}\) and the relationship between cell potential and free-energy change to predict whether the standard cell potential increases or decreases when the temperature is raised above \(25^{\circ} \mathrm{C}\) .

Is each of the following substances likely to serve as an oxidant or a reductant: (a) \(\mathrm{Ce}^{3+}(a q),\)(b) \(\mathrm{Ca}(s),\) (c) \(\mathrm{ClO}_{3}(a q)\) (d) \(\mathrm{N}_{2} \mathrm{O}_{5}(g) ?\)

Magnesium is obtained by electrolysis of molten \(\mathrm{MgCl}_{2}\) . (a) Why is an aqueous solution of MgCl_ not used in the electrolysis? (b) Several cells are connected in parallel by very large copper bars that convey current to the cells. Assuming that the cells are 96\(\%\) efficient in producing the desired products in electrolysis, what mass of Mg is formed by passing a current of \(97,000\) A for a period of 24 \(\mathrm{h} ?\)

For each of the following balanced oxidation-reduction reactions, (i) identify the oxidation numbers for all the elements in the reactants and products and (ii) state the total number of electrons transferred in each reaction. $$ \begin{array}{l}{\text { (a) } \mathrm{I}_{2} \mathrm{O}_{5}(s)+5 \mathrm{CO}(g) \longrightarrow \mathrm{I}_{2}(s)+5 \mathrm{CO}_{2}(g)} \\\ {\text { (b) } 2 \mathrm{Hg}^{2+}(a q)+\mathrm{N}_{2} \mathrm{H}_{4}(a q) \longrightarrow 2 \mathrm{Hg}(l)+\mathrm{N}_{2}(g)+4 \mathrm{H}^{+}(a q)} \\\ {\text { (c) } 3 \mathrm{H}_{2} \mathrm{S}(a q)+2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q) \longrightarrow 3 \mathrm{S}(s)+} \\\\{\quad\quad 2 \mathrm{NO}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)}\end{array} $$

(a) Write the half-reaction that occurs at a hydrogen electrode in acidic aqueous solution when it serves as the cathode of a voltaic cell.(b) Write the half-reaction that occurs at a hydrogen electrode in acidic aqueous solution when it serves as the anode of a voltaic cell. (c) What is standard about the standard hydrogen electrode?

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