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E-Cell System version 4 — E-Cell4 latest documentation
E-Cell System version 4 — E-Cell4 latest documentation
E-Cell4
ecell4
A software platform for modeling, simulation and analysis of a cell
E-Cell System version 4
Installation
Features
Tutorials
Models
For Developers
Citation
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E-Cell System version 4¶
The E-Cell System is a software platform for modeling, simulation and analysis of complex, heterogeneous and multi-scale systems like the cell.
Its latest version, E-Cell4, accepts multi-algorithms, multi-timescales and multi-spatial-representations as its central feature.
E-Cell4 is a free and open-source software licensed under the GNU General Public License version 3. The source code is available on GitHub (ecell4 and ecell4_base).
This document is generated from ecell4_docs.
A Slack workspace is open for questions here.
Installation¶
Installation of the E-Cell System Version 4
Features¶
Single particle simulations, i.e. The enhanced Green’s Function Reaction Dynamics (eGFRD) method, Spatiocyte (a lattice-based method), and the Reaction Brownian Dynamics (RBD) method
Ordinary differential equations, Gillespie algorithm (the direct method), and spatial Gillespie algorithm (the next subvolume method)
Rule-based modeling
Python programmable
Tutorials¶
Welcome to E-Cell4 in Jupyter Notebook!
1. Brief Tour of E-Cell4 Simulations
2. How to Build a Model
3. How to Setup the Initial Condition
4. How to Run a Simulation
5. How to Log and Visualize Simulations
6. How to Solve ODEs with Rate Law Functions
7. Introduction of Rule-based Modeling
8. More about 1. Brief Tour of E-Cell4 Simulations
9. Spatial Gillespie Method
10. Spatiocyte Simulations at Single-Molecule Resolution
Models¶
Examples
Attractors
Drosophila Circadian Clock
Dual Phosphorylation Cycle
Simple EGFR model
Glycolysis Model and Metabolic Control Analysis
Action Potentials in Neurons
Lotka-Volterra 2D
MinDE System with Mesoscopic Simulator
MinDE System with Spatiocyte Simulator
Simple Equilibrium
Tyson1991
How to Use the Unit System
Multiary Complex Model of GPCR Signaling Activations
The sGFRD Method
Tests
Birth-Death
Homodimerization and Annihilation
Reversible
Reversible (Diffusion-limited)
Mean Square Displacement (MSD)
For Developers¶
For Developers
Citation¶
If this package contributes to a project which leads to a scientific publication, I would appreciate a citation.
Licensing Terms¶
This product is licensed under the terms of the GNU General Public License v3.
See also LICENSE for the software included in this product.
Copyright (c) 2010-, RIKEN
All rights reserved.
API¶
ecell4 module
ecell4_base module
Indices and tables
Links¶
https://www.e-cell.org
https://github.com/ecell/
https://spatiocyte.org/
https://gfrd.org/
Installation of the E-Cell... →
© Copyright 2015-, E-Cell project.
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designed by Hsiaoming Yang.
电极电势到底指的是什么？ - 知乎
电极电势到底指的是什么？ - 知乎首页知乎知学堂发现等你来答切换模式登录/注册物理学化学电极电势到底指的是什么？根据《无机化学》(中国科学技术大学出版社，张祖德编著，2008年)的说法： 金属的电极电势= 金属的表面电势 - 金属与溶液界面处的相间电势 那这个"…显示全部 关注者96被浏览196,798关注问题写回答邀请回答好问题 13添加评论分享10 个回答默认排序宫非台湾科技大学 工学博士 关注2015-12-21电极电势就是由金属的「表面电势」和金属与溶液界面处的「相间电势」所组成。把电池作为一个整体，笼统地说电池做电功的本领，是由于电池内物质的「化学能」转化来的。也就是说，电池之所以有电动势是因为电池内化学反应有自发趋势所致，现在我们试着把注意力集中在电池的两个电极上，具体地研究电极和溶液的相界面上，电势差究竟是如何产生的。我相信你要问的就是这个部分。以金属电极为例，先来讨论电极与溶液界面电势差的产生。根据现代金属理论，金属晶格中有金属离子和能够自由移动的电子存在。当把一金属电极浸入含有该种金属离子的溶液时，如果金属离子在电极相中与溶液相中的「化学势不相等」，则金属离子会从化学势较高的相转移到化学势较低的相中。这可能发生两种情况：是金属离子由电极相进入溶液相，而将电子留在电极上，导致电极相荷负电而溶液相荷正电，如 Zn | ZnS0_{4} 界面；或者是金属离子由溶液相进入电极相，使电极相荷正电而溶液相荷负电，如 Cu | CuS0_{4} 界面。无论那种情况，都破坏了电极和溶液各相的电中性，使相间出现「电势差」。由于静电的作用，这种金属离子的相间转移很快会停止，达到平衡状态，于是相间电势差亦趋于稳定。在静电作用下，电极相所带的电荷是集中在电极表面的，而溶液中的带异号电荷的离子，一方面受到电极「表面电荷」的吸引，趋向于排列在紧靠「电极表面」附近；另一方面，由于离子的热运动使这种集中于电极表面附近，离子又会向远离电极的方向分散，当静电吸引与热运动分散平衡时，在电极与溶液界面处就形成了一个双电层。图 1 是以电极荷负电为例示意出双电层的结构。双电层是由电极表面上的电荷层与溶液中过剩的反号离子层构成，而溶液中又分为「紧密层」和「分散层」两部分。与金属靠得较紧密的紧密层厚度 d 约为 10^{-10}m ，而较远离金属的分散层的厚度与溶液的浓度、金属的电荷以及温度等有关。溶液浓度越大，分散层厚度越小，其变动范围通常从 10^{-10}m-10^{-6} m 。如果规定溶液本体中的电位为零，电极相的电位为 ε ，则电极与溶液界面电势差就是 ε 。 ε 在双电层中的分布情况如图 2 所示，即 ε是紧密层电位 ψ_1 和分散层电位 ψ_2 之加和值： ε = ψ_1 + ψ_2而 ε 的数值与电极的种类，溶液中相应离子的「活度」以及「温度」等因素有关。所以，教科书的内容无误。分类：科普 >>化学 >>电化学编辑于 2020-10-09 14:49赞同 22713 条评论分享收藏喜欢收起果核剥核 关注如果将金属棒浸泡在其盐的水溶液中，一段时间后，金属棒相对于溶液会带正电荷或带负电荷。这样，金属棒和溶液之间就产生了电荷差(电位差)。这种电位差称为相应电极的电极电位。例如，如果一根锌棒浸泡在ZnSO4的水溶液中，一段时间后，该棒相对于溶液带负电，即在锌电极上形成负电极电位。 同样，如果铜棒浸泡在CuSO4的水溶液中，一段时间后，铜棒相对于溶液带正电，即在铜电极上形成正极电位。事实上，当金属棒浸泡在其盐的水溶液中时，溶液中的金属离子不断地撞击金属棒，可能会出现以下两种重要情况：(1)金属棒的金属原子在棒上失去电子，以金属离子的形式进入溶液。这个过程是可逆的，在平衡状态下，由于存在过剩的电子，金属棒带负电。例如，锌电极。 (2) 溶液中的金属离子从棒上夺取电子，还原成金属原子。这些金属原子要么沉积在金属棒上，要么沉淀在底部。这个过程也是可逆的，在平衡时金属棒由于电子不足而带正电。因此，金属棒上产生了正极电位。例如，铜电极。电极电位的形成是由于金属原子给电子的倾向（金属的溶液压力）和金属离子取电子的倾向不同（金属离子的渗透压）。如果金属原子给出电子（金属的溶液压力）大于金属离子获取电子的趋势（金属离子的渗透压），然后金属棒由于存在过量电子而在平衡时带负电，即负电极电位在金属棒上形成。例如锌电极。 另一方面，如果金属原子给电子的倾向（金属的溶液压力）小于金属离子取电子的倾向（金属离子的渗透压），则金属棒由于缺乏而处于平衡状态时带正电电子，即正电极电位在金属棒上产生。例如，铜电极。实际上，金属的电极电位等于该金属的溶液压力与其存在于溶液中的离子的渗透压之差。电极电位的种类电极电位可以有以下两种：（1）氧化电位电极被氧化即产生电子的趋势称为电极的氧化电极电位。例如， 氧化电极电位是衡量电极被氧化能力的指标。氧化电极电位值越大，电极被氧化的能力就越大。 (2) 还原电位电极被还原的趋势，即取电子的电极称为还原电极电位。例如，还原电极电位是电极被还原的能力的量度。还原电极电位值越大，电极被还原的能力越强。所有电极都具有氧化和还原电极电位。这两个电极电位的数值对于一个电极也相同，但符号相反。因此，例如，锌电极的标准氧化电极电位为+0.76V，则其标准还原电极电位为-0.76V。同理，铜电极的标准氧化电极电位为-0.34V，因此其标准还原电极电位为+0.34V。按照欧洲惯例，在印度，标准还原电位作为电极的标准电极电位。电池电位（Ecell）在电化学电池中，电极具有不同的电极电位值（还原电位）。由于电极电位的这种差异，电子从低还原电极移动电位（高氧化电位）到具有高还原电位的电位（低氧化电位）。电化学电池的两个电极的电极电位（还原电位）的差异称为电池的电位。电池电位电池电位： 电化学电池在标准条件下（298 K 温度、1atm 压力和 1M 电解液浓度）下的电池电位称为电池的标准电池电位。电池的电动势——电化学电池在开路时的电池电势，当没有电流或从电池汲取的电流很小时，称为电池的电动势（EMF）。补充视频：知乎用户编辑于 2022-01-29 12:47赞同 403 条评论分享收藏喜欢
20.3: Ecell, ΔG, and K - Chemistry LibreTexts
20.3: Ecell, ΔG, and K - Chemistry LibreTexts
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20: ElectrochemistryMap: General Chemistry (Petrucci et al.){ }{ "20.1:_Electrode_Potentials_and_their_Measurement" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.
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Learning ObjectivesThe Relationship between Cell Potential & Gibbs EnergyMichael Faraday (1791–1867)Example \(\PageIndex{1}\)Strategy:SolutionABExercise \(\PageIndex{1}\)Potentials for the Sums of Half-ReactionsThe Relationship between Cell Potential & the Equilibrium ConstantExample \(\PageIndex{2}\)Strategy:SolutionExercise \(\PageIndex{2}\)Summary
↵
Learning Objectives
To understand the relationship between cell potential and the equilibrium constant.
To use cell potentials to calculate solution concentrations.
Changes in reaction conditions can have a tremendous effect on the course of a redox reaction. For example, under standard conditions, the reaction of \(\ce{Co(s)}\) with \(\ce{Ni^{2+}(aq)}\) to form \(\ce{Ni(s)}\) and \(\ce{Co^{2+}(aq)}\) occurs spontaneously, but if we reduce the concentration of \(\ce{Ni^{2+}}\) by a factor of 100, so that \(\ce{[Ni^{2+}]}\) is 0.01 M, then the reverse reaction occurs spontaneously instead. The relationship between voltage and concentration is one of the factors that must be understood to predict whether a reaction will be spontaneous.
The Relationship between Cell Potential & Gibbs Energy
Electrochemical cells convert chemical energy to electrical energy and vice versa. The total amount of energy produced by an electrochemical cell, and thus the amount of energy available to do electrical work, depends on both the cell potential and the total number of electrons that are transferred from the reductant to the oxidant during the course of a reaction. The resulting electric current is measured in coulombs (C), an SI unit that measures the number of electrons passing a given point in 1 s. A coulomb relates energy (in joules) to electrical potential (in volts). Electric current is measured in amperes (A); 1 A is defined as the flow of 1 C/s past a given point (1 C = 1 A·s):
\[\dfrac{\textrm{1 J}}{\textrm{1 V}}=\textrm{1 C}=\mathrm{A\cdot s} \label{20.5.1} \]
In chemical reactions, however, we need to relate the coulomb to the charge on a mole of electrons. Multiplying the charge on the electron by Avogadro’s number gives us the charge on 1 mol of electrons, which is called the faraday (F), named after the English physicist and chemist Michael Faraday (1791–1867):
\[\begin{align}F &=(1.60218\times10^{-19}\textrm{ C})\left(\dfrac{6.02214 \times 10^{23}\, J}{\textrm{1 mol e}^-}\right)
\\[4pt] &=9.64833212\times10^4\textrm{ C/mol e}^- \\[4pt] &\simeq 96,485\, J/(\mathrm{V\cdot mol\;e^-})\end{align} \label{20.5.2} \]
The total charge transferred from the reductant to the oxidant is therefore \(nF\), where \(n\) is the number of moles of electrons.
Michael Faraday (1791–1867)
Faraday was a British physicist and chemist who was arguably one of the greatest experimental scientists in history. The son of a blacksmith, Faraday was self-educated and became an apprentice bookbinder at age 14 before turning to science. His experiments in electricity and magnetism made electricity a routine tool in science and led to both the electric motor and the electric generator. He discovered the phenomenon of electrolysis and laid the foundations of electrochemistry. In fact, most of the specialized terms introduced in this chapter (electrode, anode, cathode, and so forth) are due to Faraday. In addition, he discovered benzene and invented the system of oxidation state numbers that we use today. Faraday is probably best known for “The Chemical History of a Candle,” a series of public lectures on the chemistry and physics of flames.
The maximum amount of work that can be produced by an electrochemical cell (\(w_{max}\)) is equal to the product of the cell potential (\(E^°_{cell}\)) and the total charge transferred during the reaction (\(nF\)):
\[ w_{max} = nFE_{cell} \label{20.5.3} \]
Work is expressed as a negative number because work is being done by a system (an electrochemical cell with a positive potential) on its surroundings.
The change in free energy (\(\Delta{G}\)) is also a measure of the maximum amount of work that can be performed during a chemical process (\(ΔG = w_{max}\)). Consequently, there must be a relationship between the potential of an electrochemical cell and \(\Delta{G}\); this relationship is as follows:
\[\Delta{G} = −nFE_{cell} \label{20.5.4} \]
A spontaneous redox reaction is therefore characterized by a negative value of \(\Delta{G}\) and a positive value of \(E^°_{cell}\), consistent with our earlier discussions. When both reactants and products are in their standard states, the relationship between ΔG° and \(E^°_{cell}\) is as follows:
\[\Delta{G^°} = −nFE^°_{cell} \label{20.5.5} \]
A spontaneous redox reaction is characterized by a negative value of ΔG°, which corresponds to a positive value of E°cell.
Example \(\PageIndex{1}\)
Suppose you want to prepare elemental bromine from bromide using the dichromate ion as an oxidant. Using the data in Table P2, calculate the free-energy change (ΔG°) for this redox reaction under standard conditions. Is the reaction spontaneous?
Given: redox reaction
Asked for: \(ΔG^o\) for the reaction and spontaneity
Strategy:
From the relevant half-reactions and the corresponding values of \(E^o\), write the overall reaction and calculate \(E^°_{cell}\).
Determine the number of electrons transferred in the overall reaction. Then use Equation \ref{20.5.5} to calculate \(ΔG^o\). If \(ΔG^o\) is negative, then the reaction is spontaneous.
Solution
A
As always, the first step is to write the relevant half-reactions and use them to obtain the overall reaction and the magnitude of \(E^o\). From Table P2, we can find the reduction and oxidation half-reactions and corresponding \(E^o\) values:
\[\begin{align*}
& \textrm{cathode:} &\quad & \mathrm{Cr_2O_7^{2-}(aq)} + \mathrm{14H^+(aq)}+\mathrm{6e^-}\rightarrow \mathrm{2Cr^{3+}(aq)} +\mathrm{7H_2O(l)}
&\quad & E^\circ_{\textrm{cathode}} =\textrm{1.36 V} \\
& \textrm{anode:} &\quad & \mathrm{2Br^{-}(aq)} \rightarrow \mathrm{Br_2(aq)} +\mathrm{2e^-}
&\quad & E^\circ_{\textrm{anode}} =\textrm{1.09 V}
\end{align*} \nonumber \]
To obtain the overall balanced chemical equation, we must multiply both sides of the oxidation half-reaction by 3 to obtain the same number of electrons as in the reduction half-reaction, remembering that the magnitude of \(E^o\) is not affected:
\[\begin{align*}
& \textrm{cathode:} &\quad & \mathrm{Cr_2O_7^{2-}(aq)} + \mathrm{14H^+(aq)}+\mathrm{6e^-}\rightarrow \mathrm{2Cr^{3+}(aq)} +\mathrm{7H_2O(l)}
&\quad & E^\circ_{\textrm{cathode}} =\textrm{1.36 V} \\[4pt]
& \textrm{anode:} &\quad & \mathrm{6Br^{-}(aq)} \rightarrow \mathrm{3Br_2(aq)} +\mathrm{6e^-}
&\quad & E^\circ_{\textrm{anode}} =\textrm{1.09 V} \\[4pt] \hline
& \textrm{overall:} &\quad & \mathrm{Cr_2O_7^{2-}(aq)} + \mathrm{6Br^{-}(aq)} + \mathrm{14H^+(aq)} \rightarrow \mathrm{2Cr^{3+}(aq)} + \mathrm{3Br_2(aq)} +\mathrm{7H_2O(l)}
&\quad & E^\circ_{\textrm{cell}} =\textrm{0.27 V}
\end{align*} \nonumber \]
B
We can now calculate ΔG° using Equation \(\ref{20.5.5}\). Because six electrons are transferred in the overall reaction, the value of \(n\) is 6:
\[\begin{align*}\Delta G^\circ &=-(n)(F)(E^\circ_{\textrm{cell}}) \\[4pt] & =-(\textrm{6 mole})[96,485\;\mathrm{J/(V\cdot mol})(\textrm{0.27 V})] \\& =-15.6 \times 10^4\textrm{ J} \\ & =-156\;\mathrm{kJ/mol\;Cr_2O_7^{2-}} \end{align*} \nonumber \]
Thus \(ΔG^o\) is −168 kJ/mol for the reaction as written, and the reaction is spontaneous.
Exercise \(\PageIndex{1}\)
Use the data in Table P2 to calculate \(ΔG^o\) for the reduction of ferric ion by iodide:
\[\ce{2Fe^{3+}(aq) + 2I^{−}(aq) → 2Fe^{2+}(aq) + I2(s)}\nonumber \]
Is the reaction spontaneous?
Answer
−44 kJ/mol I2; yes
Relating G and Ecell: Relating G and Ecell(opens in new window) [youtu.be]
Potentials for the Sums of Half-Reactions
Although Table P2 list several half-reactions, many more are known. When the standard potential for a half-reaction is not available, we can use relationships between standard potentials and free energy to obtain the potential of any other half-reaction that can be written as the sum of two or more half-reactions whose standard potentials are available. For example, the potential for the reduction of \(\ce{Fe^{3+}(aq)}\) to \(\ce{Fe(s)}\) is not listed in the table, but two related reductions are given:
\[\ce{Fe^{3+}(aq) + e^{−} -> Fe^{2+}(aq)} \;\;\;E^° = +0.77 V \label{20.5.6} \]
\[\ce{Fe^{2+}(aq) + 2e^{−} -> Fe(s)} \;\;\;E^° = −0.45 V \label{20.5.7} \]
Although the sum of these two half-reactions gives the desired half-reaction, we cannot simply add the potentials of two reductive half-reactions to obtain the potential of a third reductive half-reaction because \(E^o\) is not a state function. However, because \(ΔG^o\) is a state function, the sum of the \(ΔG^o\) values for the individual reactions gives us \(ΔG^o\) for the overall reaction, which is proportional to both the potential and the number of electrons (\(n\)) transferred. To obtain the value of \(E^o\) for the overall half-reaction, we first must add the values of \(ΔG^o (= −nFE^o)\) for each individual half-reaction to obtain \(ΔG^o\) for the overall half-reaction:
\[\begin{align*}\ce{Fe^{3+}(aq)} + \ce{e^-} &\rightarrow \ce\mathrm{Fe^{2+}(aq)} &\quad \Delta G^\circ &=-(1)(F)(\textrm{0.77 V})\\[4pt]
\ce{Fe^{2+}(aq)}+\ce{2e^-} &\rightarrow\ce{Fe(s)} &\quad\Delta G^\circ &=-(2)(F)(-\textrm{0.45 V})\\[4pt]
\ce{Fe^{3+}(aq)}+\ce{3e^-} &\rightarrow \ce{Fe(s)} &\quad\Delta G^\circ & =[-(1)(F)(\textrm{0.77 V})]+[-(2)(F)(-\textrm{0.45 V})] \end{align*} \nonumber \]
Solving the last expression for ΔG° for the overall half-reaction,
\[\Delta{G^°} = F[(−0.77 V) + (−2)(−0.45 V)] = F(0.13 V) \label{20.5.9} \]
Three electrons (\(n = 3\)) are transferred in the overall reaction, so substituting into Equation \(\ref{20.5.5}\) and solving for \(E^o\) gives the following:
\[\begin{align*}\Delta G^\circ & =-nFE^\circ_{\textrm{cell}} \\[4pt]
F(\textrm{0.13 V}) & =-(3)(F)(E^\circ_{\textrm{cell}}) \\[4pt]
E^\circ & =-\dfrac{0.13\textrm{ V}}{3}=-0.043\textrm{ V}\end{align*} \nonumber \]
This value of \(E^o\) is very different from the value that is obtained by simply adding the potentials for the two half-reactions (0.32 V) and even has the opposite sign.
Values of \(E^o\) for half-reactions cannot be added to give \(E^o\) for the sum of the half-reactions; only values of \(ΔG^o = −nFE^°_{cell}\) for half-reactions can be added.
The Relationship between Cell Potential & the Equilibrium Constant
We can use the relationship between \(\Delta{G^°}\) and the equilibrium constant \(K\), to obtain a relationship between \(E^°_{cell}\) and \(K\). Recall that for a general reaction of the type \(aA + bB \rightarrow cC + dD\), the standard free-energy change and the equilibrium constant are related by the following equation:
\[\Delta{G°} = −RT \ln K \label{20.5.10} \]
Given the relationship between the standard free-energy change and the standard cell potential (Equation \(\ref{20.5.5}\)), we can write
\[−nFE^°_{cell} = −RT \ln K \label{20.5.12} \]
Rearranging this equation,
\[E^\circ_{\textrm{cell}}= \left( \dfrac{RT}{nF} \right) \ln K \label{20.5.12B} \]
For \(T = 298\, K\), Equation \(\ref{20.5.12B}\) can be simplified as follows:
\[ \begin{align} E^\circ_{\textrm{cell}} &=\left(\dfrac{RT}{nF}\right)\ln K \\[4pt] &=\left[ \dfrac{[8.314\;\mathrm{J/(mol\cdot K})(\textrm{298 K})]}{n[96,485\;\mathrm{J/(V\cdot mol)}]}\right]2.303 \log K \\[4pt] &=\left(\dfrac{\textrm{0.0592 V}}{n}\right)\log K \label{20.5.13} \end{align} \]
Thus \(E^°_{cell}\) is directly proportional to the logarithm of the equilibrium constant. This means that large equilibrium constants correspond to large positive values of \(E^°_{cell}\) and vice versa.
Example \(\PageIndex{2}\)
Use the data in Table P2 to calculate the equilibrium constant for the reaction of metallic lead with PbO2 in the presence of sulfate ions to give PbSO4 under standard conditions. (This reaction occurs when a car battery is discharged.) Report your answer to two significant figures.
Given: redox reaction
Asked for: \(K\)
Strategy:
Write the relevant half-reactions and potentials. From these, obtain the overall reaction and \(E^o_{cell}\).
Determine the number of electrons transferred in the overall reaction. Use Equation \(\ref{20.5.13}\) to solve for \(\log K\) and then \(K\).
Solution
A The relevant half-reactions and potentials from Table P2 are as follows:
\[\begin{align*} & \textrm {cathode:}
& & \mathrm{PbO_2(s)}+\mathrm{SO_4^{2-}(aq)}+\mathrm{4H^+(aq)}+\mathrm{2e^-}\rightarrow\mathrm{PbSO_4(s)}+\mathrm{2H_2O(l)}
& & E^\circ_\textrm{cathode}=\textrm{1.69 V} \\[4pt]
& \textrm{anode:}
& & \mathrm{Pb(s)}+\mathrm{SO_4^{2-}(aq)}\rightarrow\mathrm{PbSO_4(s)}+\mathrm{2e^-}
& & E^\circ_\textrm{anode}=-\textrm{0.36 V} \\[4pt] \hline
& \textrm {overall:}
& & \mathrm{Pb(s)}+\mathrm{PbO_2(s)}+\mathrm{2SO_4^{2-}(aq)}+\mathrm{4H^+(aq)}\rightarrow\mathrm{2PbSO_4(s)}+\mathrm{2H_2O(l)}
& & E^\circ_\textrm{cell}=\textrm{2.05 V} \end{align*} \nonumber \]
B Two electrons are transferred in the overall reaction, so \(n = 2\). Solving Equation \(\ref{20.5.13}\) for log K and inserting the values of \(n\) and \(E^o\),
\[\begin{align*}\log K & =\dfrac{nE^\circ}{\textrm{0.0591 V}}=\dfrac{2(\textrm{2.05 V})}{\textrm{0.0591 V}}=69.37 \\[4pt]
K & =2.3\times10^{69}\end{align*} \nonumber \]
Thus the equilibrium lies far to the right, favoring a discharged battery (as anyone who has ever tried unsuccessfully to start a car after letting it sit for a long time will know).
Exercise \(\PageIndex{2}\)
Use the data in Table P2 to calculate the equilibrium constant for the reaction of \(\ce{Sn^{2+}(aq)}\) with oxygen to produce \(\ce{Sn^{4+}(aq)}\) and water under standard conditions. Report your answer to two significant figures. The reaction is as follows:
\[\ce{2Sn^{2+}(aq) + O2(g) + 4H^{+}(aq) <=> 2Sn^{4+}(aq) + 2H2O(l)} \nonumber \]
Answer
\(5.7 \times 10^{72}\)
Figure \(\PageIndex{1}\) summarizes the relationships that we have developed based on properties of the system—that is, based on the equilibrium constant, standard free-energy change, and standard cell potential—and the criteria for spontaneity (ΔG° < 0). Unfortunately, these criteria apply only to systems in which all reactants and products are present in their standard states, a situation that is seldom encountered in the real world. A more generally useful relationship between cell potential and reactant and product concentrations, as we are about to see, uses the relationship between \(\Delta{G}\) and the reaction quotient \(Q\).
Figure \(\PageIndex{1}\): The Relationships among Criteria for Thermodynamic Spontaneity. The three properties of a system that can be used to predict the spontaneity of a redox reaction under standard conditions are K, ΔG°, and E°cell. If we know the value of one of these quantities, then these relationships enable us to calculate the value of the other two. The signs of ΔG° and E°cell and the magnitude of K determine the direction of spontaneous reaction under standard conditions. (CC BY-NC-SA; Anonymous by request)
If delta G is less than zero, E is greater than zero and K is greater than 1 then the direction of the reaction is spontaneous in forward direction. If delta G is greater than zero, E is less than zero and K is less than one then the direction of reaction is spontaneous in reverse direction. If delta G is zero, E is zero and k is one that there is no net reaction and the system is at equilibrium .
Electrode Potentials and ECell: Electrode Potentials and Ecell(opens in new window) [youtu.be]
Summary
A coulomb (C) relates electrical potential, expressed in volts, and energy, expressed in joules. The current generated from a redox reaction is measured in amperes (A), where 1 A is defined as the flow of 1 C/s past a given point. The faraday (F) is Avogadro’s number multiplied by the charge on an electron and corresponds to the charge on 1 mol of electrons. The product of the cell potential and the total charge is the maximum amount of energy available to do work, which is related to the change in free energy that occurs during the chemical process. Adding together the ΔG values for the half-reactions gives ΔG for the overall reaction, which is proportional to both the potential and the number of electrons (n) transferred. Spontaneous redox reactions have a negative ΔG and therefore a positive Ecell. Because the equilibrium constant K is related to ΔG, E°cell and K are also related. Large equilibrium constants correspond to large positive values of E°.
20.3: Ecell, ΔG, and K is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.
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20.2: Standard Electrode Potentials
20.4: Cell Potential as a Function of Concentrations
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The Cell Potential - Chemistry LibreTexts
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IntroductionElectrochemical CellHow does this relate to the cell potential?NoteElectrochemical cellHow to measure the cell potential?Cell DiagramStandard Cell PotentialStandard Cell Potential ExampleImportant Standard Electrode (Reduction) PotentialsProblemsAnswersReferencesContributors and Attributions
The batteries in your remote and the engine in your car are only a couple of examples of how chemical reactions create power through the flow of electrons. The cell potential is the way in which we can measure how much voltage exists between the two half cells of a battery. We will explain how this is done and what components allow us to find the voltage that exists in an electrochemical cell.
Introduction
The cell potential, \(E_{cell}\), is the measure of the potential difference between two half cells in an electrochemical cell. The potential difference is caused by the ability of electrons to flow from one half cell to the other. Electrons are able to move between electrodes because the chemical reaction is a redox reaction. A redox reaction occurs when a certain substance is oxidized, while another is reduced. During oxidation, the substance loses one or more electrons, and thus becomes positively charged. Conversely, during reduction, the substance gains electrons and becomes negatively charged. This relates to the measurement of the cell potential because the difference between the potential for the reducing agent to become oxidized and the oxidizing agent to become reduced will determine the cell potential. The cell potential (Ecell) is measured in voltage (V), which allows us to give a certain value to the cell potential.
Electrochemical Cell
An electrochemical cell is comprised of two half cells. In one half cell, the oxidation of a metal electrode occurs, and in the other half cell, the reduction of metal ions in solution occurs. The half cell essentially consists of a metal electrode of a certain metal submerged in an aqueous solution of the same metal ions. The electrode is connected to the other half cell, which contains an electrode of some metal submerged in an aqueous solution of subsequent metal ions. The first half cell, in this case, will be marked as the anode. In this half cell, the metal in atoms in the electrode become oxidized and join the other metal ions in the aqueous solution. An example of this would be a copper electrode, in which the Cu atoms in the electrode loses two electrons and becomes Cu2+ .
The Cu2+ ions would then join the aqueous solution that already has a certain molarity of Cu2+ ions. The electrons lost by the Cu atoms in the electrode are then transferred to the second half cell, which will be the cathode. In this example, we will assume that the second half cell consists of a silver electrode in an aqueous solution of silver ions. As the electrons are passed to the Ag electrode, the Ag+ ions in solution will become reduced and become an Ag atom on the Ag electrode. In order to balance the charge on both sides of the cell, the half cells are connected by a salt bridge. As the anode half cell becomes overwhelmed with Cu2+ ions, the negative anion of the salt will enter the solution and stabilized the charge. Similarly, in the cathode half cell, as the solution becomes more negatively charged, cations from the salt bridge will stabilize the charge.
How does this relate to the cell potential?
For electrons to be transferred from the anode to the cathode, there must be some sort of energy potential that makes this phenomenon favorable. The potential energy that drives the redox reactions involved in electrochemical cells is the potential for the anode to become oxidized and the potential for the cathode to become reduced. The electrons involved in these cells will fall from the anode, which has a higher potential to become oxidized to the cathode, which has a lower potential to become oxidized. This is analogous to a rock falling from a cliff in which the rock will fall from a higher potential energy to a lower potential energy.
Note
The difference between the anode's potential to become reduced and the cathode's potential to become reduced is the cell potential.
\[E^o_{Cell}= E^o_{Red,Cathode} - E^o_{Red,Anode} \nonumber \]
Note:
Both potentials used in this equation are standard reduction potentials, which are typically what you find in tables (e.g., Table P1 and Table P2). However, the reaction at the anode is actually an oxidation reaction -- the reverse of a reduction reaction. This explains the minus sign. We would have used a plus sign had we been given an oxidation potential \(E^o_{Ox,Anode}\) instead, since \(E^o_{Red}=E^o_{Ox}\).
The superscript "o" in E^o indicates that these potentials are correct only when concentrations are 1 M and pressures are 1 bar. A correction called the "Nernst Equation" must be applied if conditions are different.
Electrochemical cell
Here is the list of the all the components:
Two half cells
Two metal electrodes
One voltmeter
One salt bridge
Two aqueous solutions for each half cell
All of these components create the Electrochemical Cell.
How to measure the cell potential?
The image above is an electrochemical cell. The voltmeter at the very top in the gold color is what measures the cell voltage, or the amount of energy being produced by the electrodes. This reading from the voltmeter is called the voltage of the electrochemical cell. This can also be called the potential difference between the half cells, Ecell. Volts are the amount of energy for each electrical charge; 1V=1J/C: V= voltage, J=joules, C=coulomb. The voltage is basically what propels the electrons to move. If there is a high voltage, that means there is high movement of electrons. The voltmeter reads the transfer of electrons from the anode to the cathode in Joules per Coulomb.
Cell Diagram
The image above is called the cell diagram. The cell diagram is a representation of the overall reaction in the electrochemical cell. The chemicals involved are what are actually reacting during the reduction and oxidation reactions. (The spectator ions are left out). In the cell diagram, the anode half cell is always written on the left side of the diagram, and in the cathode half cell is always written on the right side of the diagram. Both the anode and cathode are seperated by two vertical lines (ll) as seen in the blue cloud above. The electrodes (yellow circles) of both the anode and cathode solutions are seperated by a single vertical line (l). When there are more chemicals involved in the aqueous solution, they are added to the diagram by adding a comma and then the chemical. For example, in the image above, if copper wasn't being oxidized alone, and another chemical like K was involved, you would denote it as (Cu, K) in the diagram. The cell diagram makes it easier to see what is being oxidized and what is being reduced. These are the reactions that create the cell potential.
Standard Cell Potential
The standard cell potential (\(E^o_{cell}\)) is the difference of the two electrodes, which forms the voltage of that cell. To find the difference of the two half cells, the following equation is used:
\[E^o_{Cell}= E^o_{Red,Cathode} - E^o_{Red,Anode} \tag{1a} \]
with
\(E^o_{Cell}\) is the standard cell potential (under 1M, 1 Barr and 298 K).
\(E^o_{Red,Cathode}\) is the standard reduction potential for the reduction half reaction occurring at the cathode
\(E^o_{Red,Anode}\) is the standard reduction potential for the oxidation half reaction occurring at the anode
The units of the potentials are typically measured in volts (V). Note that this equation can also be written as a sum rather than a difference
\[E^o_{Cell}= E^o_{Red,Cathode} + E^o_{Ox,Anode} \tag{1b} \]
where we have switched our strategy from taking the difference between two reduction potentials (which are traditionally what one finds in reference tables) to taking the sum of the oxidation potential and the reduction potential (which are the reactions that actually occur). Since E^o_{Red}=-E^o_{Ox}, the two approaches are equivalent.
Standard Cell Potential Example
The example will be using the picture of the Copper and Silver cell diagram. The oxidation half cell of the redox equation is:
Cu(s) → Cu2+(aq) + 2e- EoOx= -0.340 V
where we have negated the reduction potential EoRed= 0.340 V, which is the quantity we found from a list of standard reduction potentials, to find the oxidation potential EoOx. The reduction half cell is:
( Ag+ + e- → Ag(s) ) x2 EoRed= 0.800 V
where we have multiplied the reduction chemical equation by two in order to balance the electron count but we have not doubled EoRed since Eo values are given in units of voltage. Voltage is energy per charge, not energy per reaction, so it does not need to account for the number of reactions required to produce or consume the quantity of charge you are using to balance the equation. The chemical equations can be summed to find:
Cu(s) + 2Ag+ + 2e- → Cu2+(aq) + 2Ag(s) + 2e-
and simplified to find the overall reaction:
Cu(s) + 2Ag+ → Cu2+(aq) + 2Ag(s)
where the potentials of the half-cell reactions can be summed
EoCell= EoRed,Cathode+EoOx,Anode
EoCell = 0.800 V + (-0.340 V)
EoCell = 0.460V
to find that the standard cell potential of this cell is 0.460 V. We are done.
Note that since E^o_{Red}=-E^o_{Ox} we could have accomplished the same thing by taking the difference of the reduction potentials, where the absent or doubled negation accounts for the fact that the reverse of the reduction reaction is what actually occurs.
EoCell= EoRed,Cathode-EoRed,Anode
EoCell = 0.800V - 0.340V
EoCell = 0.460V
Important Standard Electrode (Reduction) Potentials
The table below is a list of important standard electrode potentials in the reduction state. To determine oxidation electrodes, the reduction equation can simply be flipped and its potential changed from positive to negative (and vice versa). When using the half cells below, instead of changing the potential the equation below can be used without changing any of the potentials from positive to negative (and vice versa):
EoCell= EoRed,Cathode - EoRed,Anode
Table: Reduction Half-Reaction Eo, V
Acidic Solution
F2(g) + 2e- → 2 F-(aq)
+2.866
O3(g) + 2H+(aq) + 2e- → O2(g) + H2O(l)
+2.075
S2O82-(aq) + 2e- → 2SO42-(aq)
+2.01
H2O2(aq) + 2H+(aq) +2e- → 2H2O(l)
+1.763
MnO4-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l)
+1.51
PbO2(s) + 4H+(aq) + 2e- → Pb2+(aq) + 4H2O(l)
+1.455
Cl2(g) + 2e- → 2Cl-(aq)
+1.358
Cr2O72-(aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H2O(l)
+1.33
MnO2(s) + 4H+(aq) +2e- -> Mn2+(aq) + 2H2O(l)
+1.23
O2(g) + 4H+(aq) + 4e- → 2H2O(l)
+1.229
2IO3-(aq) + 12H+(aq) + 10e- → I2(s) + 6H2O(l)
+1.20
Br2(l) + 2e- → 2Br-(aq)
+1.065
NO3-(aq) + 4H+(aq) + 3e- → NO(g) + 2 H2O(l)
+0.956
Ag+(aq) + e- → Ag(s)
+0.800
Fe3+(aq) + e- → Fe2+(aq)
+0.771
O2(g) + 2H+(ag) + 2e- → H2O2(aq)
+0.695
I2(s) + 2e- → 2I-(aq)
+0.535
Cu2+(aq) + 2e- → Cu(s)
+0.340
SO42-(aq) + 4H+(aq) + 2e- → 2H2O(l) + SO2(g)
+0.17
Sn4+(aq) + 2e- → Sn2+(aq)
+0.154
S(s) + 2H+(aq) + 2e- → H2S(g)
+0.14
2H+(aq) + 2e- → H2(g)
0
Pb2+(aq) + 2e- → Pb
-0.125
Sn2+(aq) + 2e- → Sn(s)
-0.137
Fe2+(aq) + 2e- → Fe(s)
-0.440
Zn2+ + 2e- → Zn(s)
-0.763
Al3+(aq) + 3e- → Al(s)
-1.676
Mg2+(aq) + 2e- → Mg(s)
-2.356
Na+(aq) + e- → Na(s)
-2.713
Ca2+(aq) + 2e- → Ca(s)
-2.84
K+(aq) + e- → K(s)
-2.924
Li+(aq) + e- → Li(s)
-3.040
Basic Solution
O3(aq) + H2O(l) + 2e- → O2(g) + 2OH-(aq)
+1.246
OCl-(aq) + H2O(l) + 2e- → Cl-(aq) + 2OH-(aq)
+0.890
O2(g) + 2H2O(l) +4e- → 4OH-(aq)
+0.401
2H2O(l) + + 2e- → H2(aq) + 2OH-(aq)
-0.0828
Problems
For this redox reaction \[Sn(s) + Pb^{2+}(aq) \rightarrow Sn^{2+}(aq) + Pb (s) \nonumber \] write out the oxidation and reduction half reactions. Create a cell diagram to match your equations.
From the image above, of the cell diagram, write the overall equation for the reaction.
If \(Cu^{2+}\) ions in solution around a \(Cu\) metal electrode is the cathode of a cell, and \(K^+\) ions in solution around a K metal electrode is the anode of a cell, which half cell has a higher potential to be reduced?
What type of reaction provides the basis for a cell potential?
How is the cell potential measured and with what device is it measured?
The \(E^o_{cell}\) for the equation \[ 4Al(s) + 3O_2(g) + 6H_2O(l) + 4OH^-(aq) \rightarrow 4[Al(OH)_4]^-(aq) \nonumber \] is +2.71 V. If the reduction of \(O_2\) in \(OH^-\) is +0.401 V. What is the reduction half-reaction for this reduction half reaction? \[[Al(OH)_4]^-(aq) + 3e^- \rightarrow Al(s) + 4OH^- \nonumber \]
Answers
oxidation: Sn(s) → Sn2+(aq) + 2e-(aq)
reduction: Pb2+(aq) + 2e-(aq) → Pb(s)
cell diagram: Sn(s) | Sn2+(aq) || Pb2+(aq) | Pb(s)
Cu(s) + 2Ag+(aq) → 2Ag(s) + Cu2+(aq)
Because the half cell containing the \(Cu\) electrode in \(Cu^{2+}\) solution is the cathode, this is the half cell where reduction is taking place. Therefore, this half cell has a higher potential to be reduced.
The redox reaction.
Cell potential is measured in Volts (=J/C). This can be measured with the use of a voltmeter.
We can divide the net cell equation into two half-equations.
Oxidation: {Al(s) + 4OH-(aq) → [Al(OH4)]-(aq) + 3e-} x4; -Eo= ? This is what we are solving for.
Reduction: {O2(g) + 2H2O(l) + 4e- → 4OH-(aq)} x3 Eo= +0.401V
Net: 4Al(s) + 3O2(g) + 6H2O(l) + 4OH-(aq) → 4[Al(OH)4]-(aq) Eocell = 2.71V
Eocell= 2.71V= +0.401V - Eo{Al(OH)4]-(aq)/Al(s)}
Eo{[Al(OH)4]-(aq)/Al(s)} = 0.401V - 2.71V = -2.31V
Confirm this on the table of standard reduction potentials
References
Petrucci, Harwood, Herring, and Madura. General Chemistry: Principles and Modern Applications. 9th ed. Upper Saddle River, New Jersey: Pearson Education, 2007.
Contributors and Attributions
Katherine Barrett, Gianna Navarro, Joseph Koressel, Justin Kohn
The Cell Potential is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.
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用于无细胞蛋白质合成、生物传感和修复的 eCell 技术。
Biotechnology of Isoprenoids
Pub Date : 2023-01-01
, DOI:
10.1007/10_2023_225
Damian Van Raad
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Thomas Huber
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Affiliation
Research School of Chemistry, Australian National University, Canberra, ACT, Australia.
eCell 技术是最近推出的一种专门的蛋白质生产平台，可用于多种生物技术应用。本章总结了 eCell 技术在四个选定应用领域的使用。首先，用于检测体外蛋白质表达系统中的重金属离子，特别是汞。结果显示，与同类体内系统相比，灵敏度更高，检测限更低。其次，eCell 具有半渗透性、稳定且可以长期保存，使其成为一种便携式且易于使用的技术，用于极端环境下有毒物质的生物修复。第三和第四，eCell技术的应用被证明可以促进正确折叠的富含二硫键的蛋白质的表达，并将化学上感兴趣的氨基酸衍生物掺入对体内蛋白质表达有毒的蛋白质中。总体而言，eCell 技术为生物传感、生物修复和蛋白质生产提供了一种经济高效的方法。
"点击查看英文标题和摘要"
eCell Technology for Cell-Free Protein Synthesis, Biosensing, and Remediation.
The eCell technology is a recently introduced, specialized protein production platform with uses in a multitude of biotechnological applications. This chapter summarizes the use of eCell technology in four selected application areas. Firstly, for detecting heavy metal ions, specifically mercury, in an in vitro protein expression system. Results show improved sensitivity and lower limit of detection compared to comparable in vivo systems. Secondly, eCells are semipermeable, stable, and can be stored for extended periods of time, making them a portable and accessible technology for bioremediation of toxicants in extreme environments. Thirdly and fourthly, applications of eCell technology are shown to facilitate expression of correctly folded disulfide-rich proteins and incorporate chemically interesting derivatives of amino acids into proteins which are toxic to in vivo protein expression. Overall, eCell technology presents a cost-effective and efficient method for biosensing, bioremediation, and protein production.
更新日期：2023-06-13
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Cell-free Macromolecular Synthesis pp 129–146Cite as
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Cell-free Macromolecular Synthesis
Chapter
eCell Technology for Cell-Free Protein Synthesis, Biosensing, and Remediation
Damian Van Raad17 & Thomas Huber17
Chapter
First Online: 13 June 2023
177 Accesses
Part of the Advances in Biochemical Engineering/Biotechnology book series (ABE,volume 185)
AbstractThe eCell technology is a recently introduced, specialized protein production platform with uses in a multitude of biotechnological applications. This chapter summarizes the use of eCell technology in four selected application areas. Firstly, for detecting heavy metal ions, specifically mercury, in an in vitro protein expression system. Results show improved sensitivity and lower limit of detection compared to comparable in vivo systems. Secondly, eCells are semipermeable, stable, and can be stored for extended periods of time, making them a portable and accessible technology for bioremediation of toxicants in extreme environments. Thirdly and fourthly, applications of eCell technology are shown to facilitate expression of correctly folded disulfide-rich proteins and incorporate chemically interesting derivatives of amino acids into proteins which are toxic to in vivo protein expression. Overall, eCell technology presents a cost-effective and efficient method for biosensing, bioremediation, and protein production.Graphical Abstract
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AbbreviationsAA:
Amino acid
CFPS:
Cell-free protein synthesis
E. coli:
Escherichia coli
GSH:
Glutathione reduced
GSSG:
Glutathione oxidized
LbL:
Layer-by-layer assembly
ncAA:
Non-canonical amino acid
O-phospho-L-serine, SeP:
O-phospho-L-threonine, pThr
PylRS:
Pyrrolysl-tRNA synthetase
S30:
Supernatant 30,000 g
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Download references Author informationAuthors and AffiliationsResearch School of Chemistry, Australian National University, Canberra, ACT, AustraliaDamian Van Raad & Thomas HuberAuthorsDamian Van RaadView author publicationsYou can also search for this author in
PubMed Google ScholarThomas HuberView author publicationsYou can also search for this author in
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Thomas Huber . Editor informationEditors and AffiliationsTsinghua University, Beijing, ChinaYuan Lu Stanford University, Stanford, CA, USAMichael C. Jewett Rights and permissionsReprints and permissions Copyright information© 2023 The Author(s), under exclusive license to Springer Nature Switzerland AG About this chapterCite this chapterVan Raad, D., Huber, T. (2023). eCell Technology for Cell-Free Protein Synthesis, Biosensing, and Remediation.
In: Lu, Y., Jewett, M.C. (eds) Cell-free Macromolecular Synthesis. Advances in Biochemical Engineering/Biotechnology, vol 185. Springer, Cham. https://doi.org/10.1007/10_2023_225Download citation.RIS.ENW.BIBDOI: https://doi.org/10.1007/10_2023_225Published: 13 June 2023
Publisher Name: Springer, Cham
Print ISBN: 978-3-031-41286-8
Online ISBN: 978-3-031-41287-5eBook Packages: Chemistry and Materials ScienceChemistry and Material Science (R0)Share this chapterAnyone you share the following link with will be able to read this content:Get shareable linkSorry, a shareable link is not currently available for this article.Copy to clipboard
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