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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 \(\Delta G^{\text {t }}\) and \(E_{\text {red }}\)

Short Answer

Expert verified
The corresponding change in Gibbs free energy (∆G) for this two-electron redox reaction at 298 K is 16386.6 J/mol, and the reduction potential (Ered) is -0.085 V.

Step by step solution

01

Identify the known variables and required formula

We know the following variables: - Equilibrium constant (K) = \(1.5 \times 10^{-4}\) - Temperature (T) = 298 K - The number of electrons transferred in the redox reaction (n) = 2 We need to find: 1. Change in Gibbs free energy (∆G) 2. Reduction potential (Ered) For this, we'll use the following formulae: 1. Relationship between K and ΔG: \[\Delta G = -RT\ln K\] Where R is the universal gas constant, 8.314 J/molK 2. Relationship between ΔG and E: \[\Delta G = -nFE_{\text{red}}\] Where F is the Faraday's constant, 96485 C/mol
02

Calculate the change in Gibbs free energy (∆G)

Now, we'll use the equation to find the change in Gibbs free energy (∆G): \[\Delta G = -RT\ln K\] \[\Delta G = - (8.314\, \text{J/molK})(298\, \text{K})\ln(1.5 \times 10^{-4})\] Now, calculate the value: \[\Delta G = 16386.6\, \text{J/mol}\]
03

Calculate the reduction potential (Ered)

Next, we'll find the reduction potential (Ered) using the relationship between ΔG and E: \[\Delta G = -nFE_{\text{red}}\] Rearrange the equation for Ered: \[E_{\text{red}} = -\frac{\Delta G}{nF}\] Now, plug in the values \[E_{\text{red}} = -\frac{16386.6 \, \text{J/mol}}{(2)(96485\, \text{C/mol})}\] Solve for Ered: \[E_{\text{red}} = -0.085\, \text{V}\] So, the corresponding change in Gibbs free energy (∆G) for this two-electron redox reaction at 298 K is 16386.6 J/mol, and the reduction potential (Ered) is -0.085 V.

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Key Concepts

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

Gibbs Free Energy
Gibbs free energy (G) is a thermodynamic quantity that serves as a measure of the maximum amount of work a system can perform at constant temperature and pressure. It is an important concept when studying chemical reactions, such as redox reactions, as it helps predict the spontaneity of the reaction.

In the context of redox reactions, the change in Gibbs free energy, denoted as \( \Delta G \), is particularly significant as it determines the feasibility of the reaction. A negative value of \( \Delta G \) generally indicates that the reaction is spontaneous, while a positive value suggests non-spontaneity.

The relationship between the equilibrium constant (K) and \( \Delta G \) can be identified through the formula:
\[\Delta G = -RT\ln K\]
Where:\
  • \( R \) is the universal gas constant
  • \( T \) is the temperature in Kelvin
  • \( K \) is the equilibrium constant
By knowing the equilibrium constant and the temperature, one can compute the change in Gibbs free energy for the reaction, providing insight into the driving force behind the process.
Reduction Potential
Reduction potential, expressed as \( E_{\text{red}} \), is a measure of the tendency of a chemical species to acquire electrons and be reduced. In electrochemistry, it provides a quantitative means of understanding how different substances donate or accept electrons. This concept is pivotal in determining the direction and flow of electrons in redox reactions.

\

Significance in Redox Reactions\

In any given redox reaction, the species with higher reduction potential will act as the oxidizing agent and gain electrons, while the one with lower reduction potential will lose electrons and act as the reducing agent.

\

Calculating Reduction Potential from Gibbs Free Energy\

One can derive the reduction potential from the change in Gibbs free energy using the equation:\[\Delta G = -nFE_{\text{red}}\]
Where:\
  • \( n \) is the number of electrons transferred in the redox reaction
  • \( F \) is the Faraday constant
By rearranging this formula, \( E_{\text{red}} \) can be calculated if \( \Delta G \) is known, providing important insights about the redox couple involved in the reaction.
Nernst Equation
The Nernst equation is a fundamental equation in electrochemistry that relates the reduction potential of a redox reaction to the standard electrode potential, temperature, and the activities (or concentrations) of the species involved in the reaction. It can be expressed as:
\[E = E^0 - \frac{RT}{nF} \ln Q\]
Where:\
  • \( E \) is the reduction potential
  • \( E^0 \) is the standard electrode potential
  • \( R \) is the universal gas constant
  • \( T \) is the temperature in Kelvin
  • \( n \) is the number of electrons involved in the reaction
  • \( F \) is the Faraday constant
  • \( Q \) is the reaction quotient, which is a measure of the relative amounts of reactants and products
Through the Nernst equation, we are able to calculate the actual reduction potential at any point in the reaction, not just at standard conditions. This is critical for the understanding of how electrical potentials are affected by concentration changes, and it is widely used in the fields of batteries, corrosion, sensors, and more.

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Most popular questions from this chapter

Mercuric oxide dry-cell batteries are often used where a flat discharge voltage and long life are required, such as in watches and cameras. The two half-cell reactions that occur in the battery are $$ \begin{aligned} \mathrm{HgO}_{g}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} & \longrightarrow \mathrm{Hg}(l)+2 \mathrm{OH}^{-}(a q) \\ \mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q) & \longrightarrow \mathrm{ZnO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \end{aligned} $$ (a) Write the overall cell reaction. (b) The value of \(E_{\text {ied }}\) for the cathode reaction is \(+0.098 \mathrm{~V}\). The overall cell potential is \(+1.35 \mathrm{~V}\). Assuming that both half-cells operate under standard conditions, what is the standard reduction potential for the anode reaction? (c) Why is the potential of the anode reaction different than would be expected if the reaction occurred in an acidic medium?

The crude copper that is subjected to electrorefining contains tellurium as an impurity. The standard reduction potential between tellurium and its lowest common oxidation state, \(\mathrm{Te}^{4+}\), is $$ \mathrm{Te}^{4+}(a q)+4 e^{-} \longrightarrow \mathrm{Te}(s) \quad E_{\text {iod }}^{s}=0.57 \mathrm{~V} $$ Given this information, describe the probable fate of tellurium impurities during electrorefining. Do the impurities fall to the bottom of the refining bath, unchanged, as copper is oxidized, or do they go into solution as ions? If they go into solution, do they plate out on the cathode?

(a) In the Nernst equation what is the numerical value of the reaction quotient, \(Q\), under standard conditions? (b) Can the Nernst equation be used at temperatures other than room temperature?

Indicate whether the following balanced equations involve oxidation-reduction. If they do, identify the elements that undergo changes in oxidation number. (a) \(\mathrm{PBr}_{3}(l)+3 \mathrm{H}_{2} \mathrm{O}(I) \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{3}(a q)+3 \mathrm{HBr}(a q)\) (b) \(\mathrm{Nal}(a q)+3 \mathrm{HOCl}(a q) \longrightarrow \mathrm{NalO}_{3}(a q)+3 \mathrm{HCl}(a q)\) (c) \(3 \mathrm{SO}_{2}(g)+2 \mathrm{HNO}_{3}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) $$ 3 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NO}(g) $$

Indicate whether each of the following statements is true or false: (a) If something is axidized, it is formally losing electrons. (b) For the reaction \(\mathrm{Fe}^{3+}(a q)+\mathrm{Co}^{2+}(a q) \rightarrow \mathrm{Fe}^{2+}(a q)+\) \(\mathrm{Co}^{3+}(a q), \mathrm{Fe}^{3+}(a q)\) is the reducing agent and \(\mathrm{Co}^{2+}(a q)\) is the oxidizing agent. (c) If there are no changes in the oxidation state of the reactants or products of a particular reaction, that reaction is not a redox reaction.

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