E-Cell System version 4 — E-Cell4 latest documentation
E-Cell4
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A software platform for modeling, simulation and analysis of a cell
E-Cell System version 4
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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... →
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電極電勢到底指的是什么？ - 知乎首頁知乎知學(xué)堂發(fā)現(xiàn)等你來答?切換模式登錄/注冊物理學(xué)化學(xué)電極電勢到底指的是什么？根據(jù)《無機化學(xué)》(中國科學(xué)技術(shù)大學(xué)出版社，張祖德編著，2008年)的說法： 金屬的電極電勢= 金屬的表面電勢 - 金屬與溶液界面處的相間電勢 那這個"…顯示全部 ?關(guān)注者96被瀏覽196,798關(guān)注問題?寫回答?邀請回答?好問題 13?添加評論?分享?10 個回答默認排序?qū)m非?臺灣科技大學(xué) 工學(xué)博士? 關(guān)注2015-12-21電極電勢就是由金屬的「表面電勢」和金屬與溶液界面處的「相間電勢」所組成。把電池作為一個整體，籠統(tǒng)地說電池做電功的本領(lǐng)，是由于電池內(nèi)物質(zhì)的「化學(xué)能」轉(zhuǎn)化來的。也就是說，電池之所以有電動勢是因為電池內(nèi)化學(xué)反應(yīng)有自發(fā)趨勢所致，現(xiàn)在我們試著把注意力集中在電池的兩個電極上，具體地研究電極和溶液的相界面上，電勢差究竟是如何產(chǎn)生的。我相信你要問的就是這個部分。以金屬電極為例，先來討論電極與溶液界面電勢差的產(chǎn)生。根據(jù)現(xiàn)代金屬理論，金屬晶格中有金屬離子和能夠自由移動的電子存在。當(dāng)把一金屬電極浸入含有該種金屬離子的溶液時，如果金屬離子在電極相中與溶液相中的「化學(xué)勢不相等」，則金屬離子會從化學(xué)勢較高的相轉(zhuǎn)移到化學(xué)勢較低的相中。這可能發(fā)生兩種情況：是金屬離子由電極相進入溶液相，而將電子留在電極上，導(dǎo)致電極相荷負電而溶液相荷正電，如 Zn | ZnS0_{4} 界面；或者是金屬離子由溶液相進入電極相，使電極相荷正電而溶液相荷負電，如 Cu | CuS0_{4} 界面。無論那種情況，都破壞了電極和溶液各相的電中性，使相間出現(xiàn)「電勢差」。由于靜電的作用，這種金屬離子的相間轉(zhuǎn)移很快會停止，達到平衡狀態(tài)，于是相間電勢差亦趨于穩(wěn)定。在靜電作用下，電極相所帶的電荷是集中在電極表面的，而溶液中的帶異號電荷的離子，一方面受到電極「表面電荷」的吸引，趨向于排列在緊靠「電極表面」附近；另一方面，由于離子的熱運動使這種集中于電極表面附近，離子又會向遠離電極的方向分散，當(dāng)靜電吸引與熱運動分散平衡時，在電極與溶液界面處就形成了一個雙電層。圖 1 是以電極荷負電為例示意出雙電層的結(jié)構(gòu)。雙電層是由電極表面上的電荷層與溶液中過剩的反號離子層構(gòu)成，而溶液中又分為「緊密層」和「分散層」兩部分。與金屬靠得較緊密的緊密層厚度 d 約為 10^{-10}m ，而較遠離金屬的分散層的厚度與溶液的濃度、金屬的電荷以及溫度等有關(guān)。溶液濃度越大，分散層厚度越小，其變動范圍通常從 10^{-10}m-10^{-6} m 。如果規(guī)定溶液本體中的電位為零，電極相的電位為 ε ，則電極與溶液界面電勢差就是 ε 。 ε 在雙電層中的分布情況如圖 2 所示，即 ε是緊密層電位 ψ_1 和分散層電位 ψ_2 之加和值： ε = ψ_1 + ψ_2而 ε 的數(shù)值與電極的種類，溶液中相應(yīng)離子的「活度」以及「溫度」等因素有關(guān)。所以，教科書的內(nèi)容無誤。分類：科普 >>化學(xué) >>電化學(xué)編輯于 2020-10-09 14:49?贊同 227??13 條評論?分享?收藏?喜歡收起?果核剝核? 關(guān)注如果將金屬棒浸泡在其鹽的水溶液中，一段時間后，金屬棒相對于溶液會帶正電荷或帶負電荷。這樣，金屬棒和溶液之間就產(chǎn)生了電荷差(電位差)。這種電位差稱為相應(yīng)電極的電極電位。例如，如果一根鋅棒浸泡在ZnSO4的水溶液中，一段時間后，該棒相對于溶液帶負電，即在鋅電極上形成負電極電位。 同樣，如果銅棒浸泡在CuSO4的水溶液中，一段時間后，銅棒相對于溶液帶正電，即在銅電極上形成正極電位。事實上，當(dāng)金屬棒浸泡在其鹽的水溶液中時，溶液中的金屬離子不斷地撞擊金屬棒，可能會出現(xiàn)以下兩種重要情況：(1)金屬棒的金屬原子在棒上失去電子，以金屬離子的形式進入溶液。這個過程是可逆的，在平衡狀態(tài)下，由于存在過剩的電子，金屬棒帶負電。例如，鋅電極。 (2) 溶液中的金屬離子從棒上奪取電子，還原成金屬原子。這些金屬原子要么沉積在金屬棒上，要么沉淀在底部。這個過程也是可逆的，在平衡時金屬棒由于電子不足而帶正電。因此，金屬棒上產(chǎn)生了正極電位。例如，銅電極。電極電位的形成是由于金屬原子給電子的傾向（金屬的溶液壓力）和金屬離子取電子的傾向不同（金屬離子的滲透壓）。如果金屬原子給出電子（金屬的溶液壓力）大于金屬離子獲取電子的趨勢（金屬離子的滲透壓），然后金屬棒由于存在過量電子而在平衡時帶負電，即負電極電位在金屬棒上形成。例如鋅電極。 另一方面，如果金屬原子給電子的傾向（金屬的溶液壓力）小于金屬離子取電子的傾向（金屬離子的滲透壓），則金屬棒由于缺乏而處于平衡狀態(tài)時帶正電電子，即正電極電位在金屬棒上產(chǎn)生。例如，銅電極。實際上，金屬的電極電位等于該金屬的溶液壓力與其存在于溶液中的離子的滲透壓之差。電極電位的種類電極電位可以有以下兩種：（1）氧化電位電極被氧化即產(chǎn)生電子的趨勢稱為電極的氧化電極電位。例如， 氧化電極電位是衡量電極被氧化能力的指標(biāo)。氧化電極電位值越大，電極被氧化的能力就越大。 (2) 還原電位電極被還原的趨勢，即取電子的電極稱為還原電極電位。例如，還原電極電位是電極被還原的能力的量度。還原電極電位值越大，電極被還原的能力越強。所有電極都具有氧化和還原電極電位。這兩個電極電位的數(shù)值對于一個電極也相同，但符號相反。因此，例如，鋅電極的標(biāo)準(zhǔn)氧化電極電位為+0.76V，則其標(biāo)準(zhǔn)還原電極電位為-0.76V。同理，銅電極的標(biāo)準(zhǔn)氧化電極電位為-0.34V，因此其標(biāo)準(zhǔn)還原電極電位為+0.34V。按照歐洲慣例，在印度，標(biāo)準(zhǔn)還原電位作為電極的標(biāo)準(zhǔn)電極電位。電池電位（Ecell）在電化學(xué)電池中，電極具有不同的電極電位值（還原電位）。由于電極電位的這種差異，電子從低還原電極移動電位（高氧化電位）到具有高還原電位的電位（低氧化電位）。電化學(xué)電池的兩個電極的電極電位（還原電位）的差異稱為電池的電位。電池電位電池電位： 電化學(xué)電池在標(biāo)準(zhǔn)條件下（298 K 溫度、1atm 壓力和 1M 電解液濃度）下的電池電位稱為電池的標(biāo)準(zhǔn)電池電位。電池的電動勢——電化學(xué)電池在開路時的電池電勢，當(dāng)沒有電流或從電池汲取的電流很小時，稱為電池的電動勢（EMF）。補充視頻：知乎用戶編輯于 2022-01-29 12:47?贊同 40??3 條評論?分享?收藏?喜歡
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
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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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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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用于無細胞蛋白質(zhì)合成、生物傳感和修復(fù)的 eCell 技術(shù)。,Biotechnology of Isoprenoids - X-MOL
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用于無細胞蛋白質(zhì)合成、生物傳感和修復(fù)的 eCell 技術(shù)。
Biotechnology of Isoprenoids
Pub Date?:?2023-01-01
, DOI:
10.1007/10_2023_225
Damian Van Raad
1
,
Thomas Huber
1
Affiliation ?
Research School of Chemistry, Australian National University, Canberra, ACT, Australia.
eCell 技術(shù)是最近推出的一種專門的蛋白質(zhì)生產(chǎn)平臺，可用于多種生物技術(shù)應(yīng)用。本章總結(jié)了 eCell 技術(shù)在四個選定應(yīng)用領(lǐng)域的使用。首先，用于檢測體外蛋白質(zhì)表達系統(tǒng)中的重金屬離子，特別是汞。結(jié)果顯示，與同類體內(nèi)系統(tǒng)相比，靈敏度更高，檢測限更低。其次，eCell 具有半滲透性、穩(wěn)定且可以長期保存，使其成為一種便攜式且易于使用的技術(shù)，用于極端環(huán)境下有毒物質(zhì)的生物修復(fù)。第三和第四，eCell技術(shù)的應(yīng)用被證明可以促進正確折疊的富含二硫鍵的蛋白質(zhì)的表達，并將化學(xué)上感興趣的氨基酸衍生物摻入對體內(nèi)蛋白質(zhì)表達有毒的蛋白質(zhì)中。總體而言，eCell 技術(shù)為生物傳感、生物修復(fù)和蛋白質(zhì)生產(chǎn)提供了一種經(jīng)濟高效的方法。
"點擊查看英文標(biāo)題和摘要"
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
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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
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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
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