Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 21: Problem 64

What are \(E_{\text {cell }}^{\circ}\) and \(\Delta G^{\circ}\) of a redox reaction at \(25^{\circ} \mathrm{C}\) for which \(n=1\) and \(K=5.0 \times 10^{4} ?\)

Short Answer

Expert verified
Standard cell potential \( E_{\text{cell}}^{\text{o}} \) is 0.278 V and the standard Gibbs free energy change \( \text{Δ}G^{\text{o}} \) is -26.8 kJ/mol.

Step by step solution

01

- Determine the relationship between \( E_{\text{cell}}^{\text{o}} \) and K

The standard cell potential \( E_{\text{cell}}^{\text{o}} \) can be related to the equilibrium constant \( K \) using the Nernst equation simplified for standard conditions. This relationship is given by \[ E_{\text{cell}}^{\text{o}} = \frac{0.0592}{n} \times \text{log}(K) \] where \( n \) is the number of electrons transferred.
02

- Plug in the values

Given that \( n = 1 \) and \( K = 5.0 \times 10^{4} \), substitute these values into the equation: \[ E_{\text{cell}}^{\text{o}} = \frac{0.0592}{1} \times \text{log}(5.0 \times 10^{4}) \]
03

- Calculate \( \text{log}(5.0 \times 10^{4}) \)

First, calculate the logarithm: \[ \text{log}(5.0 \times 10^{4}) = \text{log}(5.0) + \text{log}(10^{4}) = 0.699 + 4 = 4.699 \]
04

- Compute \( E_{\text{cell}}^{\text{o}} \)

Now, substitute \( \text{log}(5.0 \times 10^{4}) = 4.699 \) into the equation: \[ E_{\text{cell}}^{\text{o}} = 0.0592 \times 4.699 = 0.278 \text{ V} \]
05

- Determine the relationship between \( \text{Δ}G^{\text{o}} \) and \( E_{\text{cell}}^{\text{o}} \)

The standard Gibbs free energy change \( \text{Δ}G^{\text{o}} \) is related to the standard cell potential by the equation: \[ \text{Δ}G^{\text{o}} = -nFE_{\text{cell}}^{\text{o}} \] where \( F \) is the Faraday constant, \( F = 96485 \) C/mol.
06

- Calculate \( \text{Δ}G^{\text{o}} \)

Substitute the values \( n = 1, F = 96485 \) C/mol, and \( E_{\text{cell}}^{\text{o}} = 0.278 \) V into the equation: \[ \text{Δ}G^{\text{o}} = -1 \times 96485 \times 0.278 = -26836.93 \text{ J/mol} = -26.8 \text{ kJ/mol} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Nernst equation
The Nernst equation is a fundamental formula in electrochemistry. It connects the standard cell potential (E⁰_cell) with the equilibrium constant (K). The simplified form for standard conditions is given by: \[ E_{\text{cell}}^{\text{∘}} = \frac{0.0592}{n} \times \text{log}(K) \] In this equation:
  • **E⁰_cell**: Standard cell potential

  • **n**: Number of moles of electrons transferred in the reaction

  • **log**: Base-10 logarithm

  • **K**: Equilibrium constant

The equation shows that higher equilibrium constants result in higher standard cell potentials. Plugging in the given values, the calculation of E⁰_cell becomes straightforward.
Gibbs free energy
Gibbs free energy (ΔG⁰) is an essential quantity that indicates the spontaneity of a reaction. It's related to the standard cell potential through the equation: \[ \text{Δ}G^{\text{∘}} = -nF E_{\text{cell}}^{\text{∘}} \] Where:
  • **ΔG⁰**: Standard Gibbs free energy change

  • **n**: Moles of electrons transferred

  • **F**: Faraday constant (96485 C/mol)

  • **E⁰_cell**: Standard cell potential

This equation implies that a positive standard cell potential corresponds to a negative ΔG⁰, signifying a spontaneous reaction. Conversely, a negative E⁰_cell indicates a non-spontaneous reaction. For the redox reaction with E⁰_cell = 0.278 V, we calculate ΔG⁰ as -26.8 kJ/mol, confirming the process is spontaneous due to the negative ΔG⁰.
Equilibrium constant
The equilibrium constant (K) represents the ratio of the concentrations of products to reactants at equilibrium for a chemical reaction. For electrochemical cells, K can be linked to the standard cell potential using the Nernst equation. Specifically: \[ E_{\text{cell}}^{\text{∘}} = \frac{0.0592}{n} \times \text{log}(K) \] The equilibrium constant is a measure of the extent to which a chemical reaction proceeds to completion. A higher value of K indicates a reaction that favors product formation. In this exercise, K = 5.0 × 10⁴:
  • **High K value**: Signifies that at equilibrium, the concentration of products is much higher than that of reactants.

  • **Calculating E⁰_cell**: Plugging K into the Nernst equation helps determine the E⁰_cell, showing the direct quantitative relationship between K and standard cell potential.

Electrochemistry
Electrochemistry is the branch of chemistry that deals with the relationship between electrical energy and chemical reactions. It involves studying redox (reduction-oxidation) reactions where electrons are transferred between chemical species. Key concepts in electrochemistry include:
  • **Redox reactions**: Reactions involving electron transfer. One species is oxidized (loses electrons), and another is reduced (gains electrons).

  • **Standard cell potential (E⁰_cell)**: The potential difference between two half-cells in an electrochemical cell under standard conditions.

  • **Nernst equation**: Used to calculate the cell potential at any conditions, relating E⁰_cell to the equilibrium constant, temperature, and reactant/product concentrations.

  • **Gibbs free energy (ΔG)**: Indicates whether a reaction is spontaneous (-ΔG) or non-spontaneous (+ΔG).

Overall, electrochemistry provides vital tools and equations to understand how chemical energy can be converted into electrical energy and vice versa, which is crucial in applications like batteries, fuel cells, and electrolysis.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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) \(\mathrm{Fe}(s)\left|\mathrm{Fe}^{2+}(a q) \| \mathrm{NO}_{3}^{-}(a q)\right| \mathrm{NO}(g) \mid \operatorname{Pt}(s)\)

Use the following half-reactions to write three spontaneous reactions, calculate \(E_{\text {cell }}^{\circ}\) for each reaction, and rank the strengths of the oxidizing and reducing agents: (1) \(2 \mathrm{HClO}(a q)+2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(I)\) \(E^{\circ}=1.63 \mathrm{~V}\) (2) \(\mathrm{Pt}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}(s) \quad E^{\circ}=1.20 \mathrm{~V}\) (3) \(\mathrm{PbSO}_{4}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pb}(s)+\mathrm{SO}_{4}^{2-}(a q) \quad E^{\circ}=-0.31 \mathrm{~V}\)

Commercial electrolytic cells for producing aluminum operate at \(5.0 \mathrm{~V}\) and \(100,000 \mathrm{~A}\). (a) How long does it take to produce exactly 1 metric ton ( \(1000 \mathrm{~kg}\) ) of aluminum? (b) How much electrical power (in kilowatt-hours, \(\mathrm{kW} \cdot \mathrm{h}\) ) is used \(\left(1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s} ; 1 \mathrm{~kW} \cdot \mathrm{h}=3.6 \times 10^{3} \mathrm{~kJ}\right) ?\) (c) If electricity costs \(\$ 0.123\) per \(\mathrm{kW} \cdot \mathrm{h}\) and cell efficiency is \(90 . \%\) what is the cost of electricity to produce exactly 1 lb of aluminum?

Balance the following skeleton reactions, and identify the oxidizing and reducing agents: (a) \(\mathrm{As}_{4} \mathrm{O}_{6}(s)+\mathrm{MnO}_{4}^{-}(a q) \longrightarrow \mathrm{AsO}_{4}^{3-}(a q)+\mathrm{Mn}^{2+}(a q)\) [acidic] (b) \(\mathrm{P}_{4}(s) \longrightarrow \mathrm{HPO}_{3}^{2-}(a q)+\mathrm{PH}_{3}(g)\) [acidic] (c) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{CN}^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)+\mathrm{CNO}^{-}(a q)\) [basic]

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

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Chemical Reactions

Read Explanation

The Earths Atmosphere

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free