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If the equilibrium constant for a two-electron redox reaction at \(298 \mathrm{~K}\) is \(2.2 \times 10^{5},\) calculate the corresponding \(\Delta G^{\circ}\) and \(E^{\circ}\).

Short Answer

Expert verified
The standard Gibbs free energy change (\(\Delta G^{\circ}\)) for the two-electron redox reaction is approximately \(-33922.3\mathrm{~J~mol^{-1}}\) and the standard electrode potential (\(E^{\circ}\)) is approximately \(0.176\mathrm{~V}\).

Step by step solution

01

Calculate \(\Delta G^{\circ}\) using the given equilibrium constant

We can use the equation \(\Delta G^{\circ} = -RT \ln K\) to calculate the standard Gibbs free energy change: $$\Delta G^{\circ} = - (8.314\mathrm{~J~mol^{-1}K^{-1}}) (298\mathrm{~K}) \ln (2.2 \times 10^{5})$$ Computing this gives: $$\Delta G^{\circ} \approx -33922.3\mathrm{~J~mol^{-1}}$$
02

Calculate the standard electrode potential (\(E^{\circ}\)) using the Nernst equation

Now that we have the \(\Delta G^{\circ}\) value, we can use the Nernst equation to calculate \(E^{\circ}\): $$\Delta G^{\circ} = -nFE^{\circ}$$ Rearrange the equation to solve for \(E^{\circ}\): $$E^{\circ} = -\dfrac{\Delta G^{\circ}}{nF}$$ Plug in the values for \(\Delta G^{\circ}\), \(n\) (number of electrons transferred, two in this case), and \(F\): $$E^{\circ} = -\dfrac{-33922.3\mathrm{~J~mol^{-1}}}{(2)(96,485\mathrm{~C~mol^{-1})}}$$ Computing this gives: $$E^{\circ} \approx 0.176\mathrm{~V}$$ So, the standard Gibbs free energy change (\(\Delta G^{\circ}\)) for the reaction is approximately \(-33922.3\mathrm{~J~mol^{-1}}\) and the standard electrode potential (\(E^{\circ}\)) is approximately \(0.176\mathrm{~V}\).

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Key Concepts

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

Equilibrium Constant
In chemistry, the equilibrium constant is a measure of the extent to which a chemical reaction proceeds to completion under standard conditions. When a chemical reaction reaches equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. The equilibrium constant, denoted as \( K \), provides insights into the reaction's favorability and position.
  • If \( K > 1 \), the reaction favors the formation of products.
  • If \( K < 1 \), the reaction favors the formation of reactants.
  • If \( K \approx 1 \), both reactants and products are present in significant amounts.
For a redox reaction, the equilibrium constant is crucial because it helps determine the Gibbs free energy change \( \Delta G^{\circ} \), which indicates if a process is spontaneous under standard conditions. Using the equation \( \Delta G^{\circ} = -RT \ln K \), at 298 K, the relationship between the equilibrium constant and Gibbs energy becomes straightforward and widely used in thermodynamic calculations.
Gibbs Free Energy
Gibbs free energy is a thermodynamic potential that reflects the capacity of a system to perform non-mechanical work and is a crucial indicator of the spontaneity of a reaction. When we talk about the standard Gibbs free energy change, denoted as \( \Delta G^{\circ} \), it refers to the change in energy between reactants and products under standard conditions (pressure, concentration, and a reference temperature, usually 298 K).
  • If \( \Delta G^{\circ} < 0 \), the reaction is spontaneous and will proceed without any external energy input.
  • If \( \Delta G^{\circ} > 0 \), the reaction is non-spontaneous, meaning it requires energy to proceed.
  • If \( \Delta G^{\circ} = 0 \), the system is at equilibrium, and no net change occurs.
In the context of redox reactions, Gibbs free energy change can link to the equilibrium constant, providing insight into the electrochemical potential energy available to drive the reaction. This is particularly important in calculating the electrochemical cell potential (\( E^{\circ} \) ), employing the relationships shown in the exercise.
Nernst Equation
The Nernst equation is a fundamental tool in electrochemistry, providing a link between the Gibbs free energy change of a reaction and its electric potential difference.For any electrochemical cell, the Nernst equation calculates the cell potential under non-standard conditions by considering the concentration of reactants and products. Under standard conditions, it simplifies to calculate the standard electrode potential \( E^{\circ} \), as illustrated in the exercise.The general form of the equation is:\[ E = E^{\circ} - \frac{RT}{nF} \ln \frac{[\text{products}]}{[\text{reactants}]}\]Where:
  • \( E \) is the cell potential at non-standard conditions.
  • \( E^{\circ} \) is the standard cell potential.
  • \( R \) is the universal gas constant (8.314 J/mol·K).
  • \( T \) is the temperature in Kelvin.
  • \( n \) is the number of electrons transferred in the reaction.
  • \( F \) is the Faraday constant, approximately 96485 C/mol.
Utilizing the Nernst equation allows chemists to determine real-time reaction conditions, making it an indispensable tool in the analysis and prediction of redox reactions and electrochemical cells.

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Most popular questions from this chapter

Given the following reduction half-reactions: $$ \begin{aligned} \mathrm{Fe}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}(a q) & E_{\mathrm{red}}^{\circ}=+0.77 \mathrm{~V} \\ \mathrm{~S}_{2} \mathrm{O}_{6}^{2-}(a q)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{SO}_{3}(a q) & E_{\mathrm{red}}^{\circ}=+0.60 \mathrm{~V} \\ \mathrm{~N}_{2} \mathrm{O}(g)+2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & E_{\mathrm{red}}^{\circ}=-1.77 \mathrm{~V} \\ \mathrm{VO}_{2}^{+}(a q)+2 \mathrm{H}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{VO}^{2+}+\mathrm{H}_{2} \mathrm{O}(l) & E_{\mathrm{red}}^{\circ}=+1.00 \mathrm{~V} \end{aligned} $$ (a) Write balanced chemical equations for the oxidation of \(\mathrm{Fe}^{2+}(a q)\) by \(\mathrm{S}_{2} \mathrm{O}_{6}^{2-}(a q),\) by \(\mathrm{N}_{2} \mathrm{O}(a q),\) and by \(\mathrm{VO}_{2}^{+}(a q)\) (b) Calculate \(\Delta G^{\circ}\) for each reaction at \(298 \mathrm{~K}\). (c) Calculate the equilibrium constant \(K\) for each reaction at \(298 \mathrm{~K}\).

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 \(\mathrm{CO}_{2}\).

Consider a redox reaction for which \(E^{\circ}\) is a negative number. (a) What is the sign of \(\Delta G^{\circ}\) for the reaction? (b) Will the equilibrium constant for the reaction be larger or smaller than \(1 ?\) (c) Can an electrochemical cell based on this reaction accomplish work on its surroundings?

Aqueous solutions of ammonia \(\left(\mathrm{NH}_{3}\right)\) and bleach (active ingredient \(\mathrm{NaOCl}\) ) are sold as cleaning fluids, but bottles of both of them warn: "Never mix ammonia and bleach, as toxic gases may be produced." One of the toxic gases that can be produced is chloroamine, \(\mathrm{NH}_{2} \mathrm{Cl}\). (a) What is the oxidation number of chlorine in bleach? (b) What is the oxidation number of chlorine in chloramine? (c) Is Cl oxidized, reduced, or neither, upon the conversion of bleach to chloramine? (d) Another toxic gas that can be produced is nitrogen trichloride, \(\mathrm{NCl}_{3}\). What is the oxidation number of \(\mathrm{N}\) in nitrogen trichloride? \((\mathbf{e})\) Is \(\mathrm{N}\) oxidized, reduced, or neither, upon the conversion of ammonia to nitrogen trichloride?

(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: $$ \begin{aligned} \mathrm{Cd}(\mathrm{OH})_{2}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cd}(s)+2 \mathrm{OH}^{-}(a q) & \\ E_{\mathrm{red}}^{\circ} &=-0.76 \mathrm{~V} \\ \mathrm{NiO}(\mathrm{OH})(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{e}^{-} \longrightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{OH}^{-}(a q) \\ E_{\mathrm{red}}^{\circ} &=+0.49 \mathrm{~V} \end{aligned} $$ (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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