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Chapter 20: Problem 54

If the equilibrium censtant for a one-electron redox reaction at \(298 \mathrm{~K}\) is \(8.7 \times 10^{4}\), calculate the corresponding \(\Delta G^{\circ}\) and \(E_{\text {red }}\)

Short Answer

Expert verified
The standard Gibbs free energy change (螖G掳) for the given redox reaction is -171,410.3 J/mol, and the reduction potential (E_red) is 1.776 V.

Step by step solution

01

Calculate the standard Gibbs free energy change (螖G掳)

Given the equilibrium constant, K = 8.7 脳 10^4, temperature T = 298 K, and the gas constant R = 8.314 J/K/mol. We can now calculate the standard Gibbs free energy change, 螖G掳 using the formula: \( 螖G掳 = -RT \ln{K} \) Substitute the given values into the equation: \( 螖G掳 = -(8.314 \frac{J}{K \cdot mol})(298 K) \ln{(8.7 \times 10^4)} \) After calculation, we find that: \( 螖G掳 = -171,410.3 \frac{J}{mol} \)
02

Calculate the reduction potential (E_red)

We will use the Nernst equation to find the reduction potential E_red, considering the fact that it's a one-electron redox reaction (n = 1): \( E_red = E掳 - \frac{RT}{nF} \ln{Q} \) Since we need E_red, let's reorganize the equation to find it: \( E_red = \frac{螖G掳}{-nF} \) In this case, n = 1 (one-electron), and the Faraday's constant F = 96,485 C/mol. The 螖G掳 value we found in the first step is -171,410.3 J/mol. Plug all these values into the equation: \( E_red = \frac{-171,410.3 \frac{J}{mol}}{-1 \times 96,485 \frac{C}{mol}} \) After calculation, we find that: \( E_red = 1.776 V (rounded to three decimal places) \) In conclusion, the standard Gibbs free energy change, 螖G掳, is -171,410.3 J/mol, and the reduction potential, E_red, is 1.776 V.

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Key Concepts

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

Equilibrium Constant
In the world of chemistry, the equilibrium constant, often denoted as \( K \), is a significant value that indicates the extent of a reaction under a given set of conditions. It expresses the ratio of the concentrations of products to reactants at equilibrium. In essence, a high equilibrium constant, like \( 8.7 \times 10^4 \), suggests that at equilibrium, the products are overwhelmingly more abundant than the reactants. This value helps chemists understand the favorability of the reaction towards the products or reactants, thereby guiding various experimental and industrial processes.
  • When \( K \) is greater than one, the reaction favors products.
  • If \( K \) is less than one, it favors reactants.
  • \( K \) is dimensionless, as it's simply a ratio.
Gibbs Free Energy
Gibbs free energy, denoted as \( \Delta G \), is a thermodynamic property that can help predict the feasibility of a process. If \( \Delta G \) is negative, the reaction tends to be spontaneous, meaning it can proceed without any input of energy. The measure tells whether a chemical change is energetically favorable. The formula to find the standard Gibbs free energy change \( (\Delta G^{\circ}) \) involves the equilibrium constant:\[ \Delta G^{\circ} = -RT \ln{K} \].
Here:
  • \( R \) is the universal gas constant, which is \(8.314 \, \text{J/mol K}\).
  • \( T \) is the temperature in Kelvin.
By inserting the known values into this equation, you can find \( \Delta G^{\circ} \), which represents the change in Gibbs free energy under standard conditions.
Reduction Potential
Reduction potential, often represented as \( E_{\text{red}} \), is a measure of the propensity of a chemical species to acquire electrons and thus be reduced. Higher values of reduction potential indicate a greater tendency for reduction, signaling the ease with which a species can gain electrons. In electrochemistry, it is crucial as it gives insight into the relative strengths of different reducers.
To find the reduction potential, we can use the equation:\[ E_{\text{red}} = \frac{\Delta G^{\circ}}{-nF} \]where:
  • \( \Delta G^{\circ} \) is the standard Gibbs free energy change.
  • \( n \) is the number of moles of electrons transferred in the reaction (in this case, 1).
  • \( F \) is Faraday's constant \( (96,485 \, \text{C/mol}) \).
This formula helps calculate \( E_{\text{red}} \) using the Gibbs free energy value obtained earlier.
Nernst Equation
The Nernst equation is a pivotal equation in electrochemistry, used to determine the cell potential of an electrochemical cell under non-standard conditions. The equation provides a relationship between the reduction potential at any conditions and the reduction potential at standard conditions. It's expressed as:\[ E = E^{\circ} - \frac{RT}{nF} \ln{Q} \].
In this expression:
  • \( E \) is the electrode potential.
  • \( E^{\circ} \) is the standard electrode potential.
  • \( R \) is the universal gas constant \( (8.314 \, \text{J/mol K}) \).
  • \( n \) is the number of electrons participating in the redox reaction.
  • \( F \) is Faraday's constant \( (96,485 \, \text{C/mol}) \).
  • \( Q \) is the reaction quotient, representing the ratio of the concentration of products to reactants at any point during the reaction.
The Nernst equation elegantly shows how the cell potential varies with changes in concentration, temperature, and reaction progress.

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Most popular questions from this chapter

During the discharge of an alkaline battery, \(4.50 \mathrm{~g}\) of \(\mathrm{Zn}\) is consumed at the anode of the battery. (a) What mass of \(\mathrm{MnO}_{2}\) is reduced at the cathode during this discharge? (b) How many coulombs of electrical charge are transferred from \(\mathrm{Zn}\) to \(\mathrm{MnO}_{2}\) ?

Cytochrome, a complicated molecule that we will represent as CyFe 2 , reacts with the air we breathe to supply energy required to synthesize adenosine triphosphate (ATP). The body uses ATP as an energy source to drive other reactions. (Section 19.7) At pH \(7.0\) the following reduction potentials pertain to this oxidation of \(\mathrm{CyFe}^{2+}\) : $$ \begin{aligned} \mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) & E_{\text {red }}^{*}=+0.82 \mathrm{~V} \\ \mathrm{CyFe}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{CyFe}^{2+}(a q) & E_{\text {red }}^{+}=+0.22 \mathrm{~V} \end{aligned} $$ (a) What is \(\Delta G\) for the oxidation of CyFe \({ }^{2+}\) by air? (b) If the synthesis of \(1.00 \mathrm{~mol}\) of ATP from adenosine diphosphate (ADP) requires a \(\Delta G\) of \(37.7 \mathrm{~kJ}\), how many moles of ATP are synthesized per mole of \(\mathrm{O}_{2}\) ?

Magnesium is obtained by electrolysis of molten \(\mathrm{MgCl}_{2}\). (a) Why is an aqueous solution of \(\mathrm{MgC}_{2}\) 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 \(\mathrm{Mg}\) is formed by passing a current of \(97,000 \mathrm{~A}\) for a period of \(24 \mathrm{~h}\) ?

Indicate whether each of the following statements is true or false: (a) If something is reduced, it is formally losing electrons. (b) A reducing agent gets oxidized as it reacts. (c) An oxidizing agent is needed to convert \(\mathrm{CO}\) into \(\mathrm{CO}_{2}\).

(a) Write the reactions for the discharge and charge of a nickel-cadmium (nicad) rechargeable battery. (b) Given the following reduction potentials, calculate the standard emf of the cell: $$ \begin{array}{r} \mathrm{Cd}(\mathrm{OH})_{2}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cd}(s)+2 \mathrm{OH}^{-}(a q) \\ E_{\mathrm{red}}^{\mathrm{e}}=-0.76 \mathrm{~V} \\ \mathrm{NiO}(\mathrm{OH})(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{e}^{-} \longrightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{OH}^{-}(a q) \\ E_{\text {red }}^{e}=+0.49 \mathrm{~V} \end{array} $$ (c) A typical nicad voltaic cell generates an emf of \(+1.30 \mathrm{~V}\). Why is there a difference between this value and the one you calculated in part (b)? (d) Calculate the equilibrium constant for the overall nicad reaction based on this typical emf value.
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