What are \(E_{\text {cell }}^{\circ}\) and \(\Delta G^{\circ}\) of a redox reaction at \(25^{\circ} \mathrm{C}\) for which \(n=2\) and \(K=0.065 ?\)

Short Answer

Expert verified
For the given redox reaction, \( E_{\text {cell }}^{\text{°}} = -0.0355 \text{ V } \) and \( \triangle G^{\text{o}} ≈ 6.847 \text{ kJ/mol } \).

Step by step solution

01

Identify Given Values

Identify the given values in the problem: - Temperature: 25°C (which is 298 K)- Number of moles of electrons transferred () = 2- Equilibrium constant (K) = 0.065
02

Calculate Standard Electrode Potential (E_{\text {cell }}^{\text{o}})

Use the Nernst equation at standard conditions to determine the standard electrode potential: \[ E_{\text {cell }}^{\text {o}} = \frac{RT}{nF} \times \text{ln}(K) \] Where: - R = 8.314 J/(mol·K) (universal gas constant) - T = 298 K (temperature in Kelvins) - F = 96485 C/mol (Faraday's constant) - K = 0.065 (equilibrium constant) Plug in the given values: \[ E_{\text {cell }}^{\text {o}} = \frac{(8.314)(298)}{(2)(96485)} \times \text{ln}(0.065) \] Calculate the natural logarithm of 0.065 and then solve: \[ E_{\text {cell }}^{\text {o}} = \frac{(8.314)(298)}{(2)(96485)} \times (-2.733) \] Finally: \[ E_{\text {cell }}^{\text {o}} ≈ -0.0355 \text{ V} \]
03

Calculate Standard Gibbs Free Energy Change (ΔG°)

Use the relationship between the Gibbs free energy change and the standard electrode potential: \[ \triangle G^{\text{o}} = -nFE_{\text {cell }}^{\text{o}} \] Where: - n = 2 - F = 96485 C/mol - E_{\text {cell }}^{\text{o}} = -0.0355 V Plug in the values: \[ \triangle G^{\text{o}} = -2 \times 96485 \times (-0.0355) \] Calculate: \[ \triangle G^{\text{o}} ≈ 6847 \text{ J/mol or 6.847 kJ/mol} \]

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Key Concepts

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

Nernst equation
The Nernst equation is crucial when it comes to understanding electrochemical cells. In simple terms, it relates the reduction potential of a half-cell in an electrochemical cell to the standard electrode potential, temperature, activity, and the reaction quotient. At standard conditions, the Nernst equation for calculating the cell potential is:

\[ E_{\text {cell }}^{\text {o}} = \frac{RT}{nF} \times \text{ln}(K) \]

Here’s a quick guide to the important components:
  • \text{R} = 8.314 J/(mol·K) is the universal gas constant.
  • \text{T} is the temperature in Kelvins; for standard conditions, it's 298 K (25°C).
  • \text{n} is the number of moles of electrons transferred in the reaction.
  • \text{F} is Faraday's constant (96485 C/mol).
  • \text{K} is the equilibrium constant.

This equation helps us find \( E_{\text {cell }}^{\text {o}} \) when we're given \( K \), translating the equilibrium state information into the cell potential. This relationship shows us how the spontaneity of a reaction changes with different species concentrations.
standard Gibbs free energy change
The standard Gibbs free energy change (ΔG°) indicates the spontaneity of a reaction under standard conditions. The formula linking ΔG° to the standard electrode potential is:

\[ \triangle G^{\text{o}} = -nFE_{\text {cell }}^{\text{o}} \]

Let’s break down what this means:
  • \text{n} is the number of moles of electrons transferred.
  • \text{F} is Faraday's constant (96485 C/mol).
  • \text{E_{\text {cell }}^{\text{o}}} is the standard electrode potential found using the Nernst equation.

ΔG° tells us if a reaction is spontaneous (ΔG° < 0) or non-spontaneous (ΔG° > 0) under standard conditions. If ΔG° is zero, the system is at equilibrium. In essence, this relationship connects electrochemistry with thermodynamics.
In our exercise, the calculated ΔG° is approximately +6.847 kJ/mol, indicating that the reaction is non-spontaneous at standard conditions.
equilibrium constant
The equilibrium constant (K) is a reflection of the ratio of products to reactants at equilibrium for a particular reaction. A larger K (>1) generally means a reaction favors the formation of products, while a smaller K (<1) suggests that reactants are preferred.

In electrochemical cells, the equilibrium constant impacts the standard electrode potential (\text{E_{\text {cell }}^{\text{o}}}). Remember, the Nernst equation:

\[ E_{\text {cell }}^{\text{o}} = \frac{RT}{nF} \times \text{ln}(K) \]

This equation helps to link the equilibrium constant with the measurable electrode potential of the cell. By calculating the natural logarithm of K and incorporating temperature, number of electrons, and constants like \text{R} and \text{F}, we can find \text{E_{\text {cell }}^{\text{o}}}.
In our given problem, K was 0.065, indicating the reaction heavily favors reactants over products at equilibrium. This is reflected in a small and slightly negative \text{E_{\text {cell }}^{\text{o}}}.

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Most popular questions from this chapter

Comparing the standard electrode potentials \(\left(E^{\circ}\right)\) of the Group \(1 \mathrm{~A}(1)\) metals \(\mathrm{Li}, \mathrm{Na},\) and \(\mathrm{K}\) with the negative of their first ionization energies reveals a discrepancy: Ionization process reversed: \(\mathrm{M}^{+}(g)+\mathrm{e}^{-} \rightleftharpoons \mathrm{M}(g)\) Electrode reaction: $$\mathrm{M}^{+}(a q)+\mathrm{e}^{-} \rightleftharpoons \mathrm{M}(s)$$ $$\begin{array}{lcc}\text { Metal } & -\text { IE (kJ/mol) } & E^{\circ} \text { (V) } \\\\\hline \text { Li } & -520 & -3.05 \\\\\text { Na } & -496 & -2.71 \\\ \text { K } & -419 & -2.93\end{array}$$ Note that the electrode potentials do not decrease smoothly down the group, whereas the ionization energies do. You might expect that, if it is more difficult to remove an electron from an atom to form a gaseous ion (larger IE), then it would be less difficult to add an electron to an aqueous ion to form an atom (smaller \(E^{\circ}\) ), yet \(\mathrm{Li}^{+}(a q)\) is more difficult to reduce than \(\mathrm{Na}^{+}(a q) .\) Applying Hess's law, use an approach similar to a Born-Haber cycle to break down the process occurring at the electrode into three steps, and label the energy involved in each step. How can you account for the discrepancy?

What does a negative \(E_{\mathrm{cell}}^{\circ}\) indicate about a redox reaction? What does it indicate about the reverse reaction?

When metal \(\mathrm{A}\) is placed in a solution of a salt of metal \(\mathrm{B}\), the surface of metal A changes color. When metal \(\mathrm{B}\) is placed in acid solution, gas bubbles form on the surface of the metal. When metal \(\mathrm{A}\) is placed in a solution of a salt of metal \(\mathrm{C},\) no change is observed in the solution or on the surface of metal A. (a) Will metal C cause formation of \(\mathrm{H}_{2}\) when placed in acid solution? (b) Rank metals \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) in order of decreasing reducing strength.

How many grams of aluminum can form by passing 305 C through an electrolytic cell containing a molten aluminum salt?

Write a balanced equation from each cell notation: (a) \(\operatorname{Mn}(s)\left|\mathrm{Mn}^{2+}(a q) \| \mathrm{Cd}^{2+}(a q)\right| \mathrm{Cd}(s)\) (b) \(\mathrm{Fe}(s)\left|\mathrm{Fe}^{2+}(a q) \| \mathrm{NO}_{3}^{-}(a q)\right| \mathrm{NO}(g) \mid \operatorname{Pt}(s)\)

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