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Redox reactions consist of oxidation and reduction processes occurring simultaneously. What does this actually mean? Redox reaction involves the transfer of electrons between species (Easy way to remember: increase in oxidation number = oxidation, hence decrease in oxidation number = reduction). These processes occur together at the same time at different ...


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Pourbaix Diagram In Environmental Chemistry Pourbaix diagrams are very useful to identify the nature chemical species in different pH environment. The figure contains additional information on the stability of water. The lines at 1.23 V and 0.00 V represent the redox chemistry of water. At 1.23 V water is being oxidised to oxygen. At 0.00 V the water is ...


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what actually happens inside the battery once it is fully discharged? Alkaline batteries use the exchange of electrons from zinc to manganese dioxide to produce electricity $$\begin{array}{rcl}\\ \ce{Zn (s) +2 OH- (aq)}& \ce{->}& \ce{ZnO (s) + H2O (l) + 2 e-}\qquad \text{(anode)}\\ \text{(high potential anode)}\quad \ce{2e-}&\ce{->[load][\...


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For the acidic electrolysis, use the reactions where $\ce{H+}$ occurs. As $\ce{OH-}$ is not available in considerable amount there as a reagent, neither it is created as a product. Generally, for a reaction choice, apply the principle of availability and stability, allowing for a reagent to exist in (relative) abundance. $\ce{OH-}$ or anions of ...


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To extend @M.Farooq's answer, let's look at this a slightly different way. From the Nernst equation: $$\mathscr{E} = \mathscr{E}^{\circ} - \frac{RT}{nF}\ln Q$$ As the reaction progresses, the value of $Q$, the reaction quotient, increases because product concentrations are higher and reactant concentrations are lower. As a result, the cell potential $\...


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A discharged battery means that cell reaction has achieved equilibrium. When this system is at equilibrium no (useful) work can be done by the battery. It cannot drive or pump electrons in the circuit. In other words, people say that its Gibbs free energy is zero.


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The anode is the electrode, where substances are losing electrons and are oxidated. The cathode is the electrode, where substances are gaining electrons and are reduced. The tricky part for the memorising is, anodes and cathodes flip the position, when the current is reversed, depending on if the cell is in the mode of electrolysis the cell is in the mode ...


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There is no completed electronic circuit in an electrochemical cell In an electrochemical cell, the anode is the source of electrons to the external circuit and the cathode is the sink. The circuit of charge transport gets completed by ions traveling inside the cell. A solar cell is different from an electrochemical cell in that their is no net chemical ...


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The above post answers your questions, but let me add that $$\ce{2 H+ + 2 e- <=> H2}\label{rxn:1}\tag{1}$$ and $$\ce{2 H2O + 2 e- <=> H2 + 2 OH-}\label{rxn:2}\tag{2}$$ are two different experimental procedures. Hence these are two distinct half cell reactions. It happens that you get $\ce{H2}$ as a product in $\eqref{rxn:1}$ and $\eqref{rxn:...


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Note that the actual potential for a particular redox reaction is not a fixed value, but depends on concentrations ( more exactly activities ) of reagents. The standard redox potentials are potentials with activities equal to 1, If we consider reactions $$\begin{align}\ce{ 2 H+ + 2e- &<=> H2 \\ 2 H2O + 2 e- &<=> 2 OH- + H2 \\ }\...


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I still teach switching the sign. I find it easier to remember adding up reduction potential and oxidation potential. The half reactions are written as reduction in a table of reduction potentials, so it makes sense that you have to treat the oxidation half reaction differently. If the cell potential is calculated from reduction potential of the cathode ...


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I have done a significant amount of research over the past ten years to trace to origins of these electrochemical conventions and luckily got a chance to discuss these with some top electrochemists. I have been planning to write an article on this issue since it is a perpetual confusion. Basically, the origin of these "signs" issues originated in Germany and ...


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The sign of the electrode reduction potential is invariant. If reflects the sign of the electrostatic charge of the electrode with respect to the hydrogen electrode.* Also remember that all electrode potentials are written as reduction these days. This is a convention set by all electrochemists all over the world along with the other conventions. The ...


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As noted in your question, the standard reduction potentials are given as such in tables. $$ \begin{align} \ce{O2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l)} &\quad E^\circ &= \pu{+1.23 V} \\ \ce{Ag+ + e- &→ Ag(s)} &\quad E^\circ &= \pu{+0.799 V} \\ \ce{Br2(l) + 2 e- &→ 2 Br-(aq)} &\quad E^\circ &= \pu{+1.065 V} \end{align} $$ ...


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You do need the electromotive series or reactivity series to answer that question. If you will not have access to it during such exams, you'll need to memorize the order of some common elements, and to understand the arrangement of the Periodic Table. The most reactive metals are at the bottom left of the table, low in Group I. The "noble" or coinage metals ...


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There are many half-reactions where the electrode does not participate in the redox reaction (and no metal is deposited on the electrode, which arguably would change the electrode such that it now does participate). The prime example is the standard hydrogen electrode, where hydrogen gas is bubbled into a solution in a way that the bubbles come in contact ...


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I know I'm not supposed to give my opinion on here, but you asked..I would go with option 2. The HCl won't do anything (not dissolve copper anyway).This is tough to answer because the way the options are phrased is confusing...


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The blue color was caused by copper being stripped from the wire. btw, when your hair turns orange, it is because it is being stripped of color, among other other things (chemical bonds), which will cause it to be brittle and also absorb any hair dye to the extreme that it feels waxy...I'd suggest you cut off a piece of hair for a test before dunking your ...


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According to my notes and many sources on the internet, electrons and cations both travel from the anode (A in the image) to the cathode (B in the image). The idea of the salt bridge is to prevent electrolytes mixing while providing ion flow. When you have a high concentration of inert ions in the salt bridge, cations in the salt bridge will flow into B, ...


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Equilibrium constant When a reaction is at equilibrium, the species will have concentrations called the equilibrium concentrations. They do not depend on whether the reaction is written in one direction or in the reverse direction, and whether all the stoichiometric coefficients are multiplied by the same value. However, your equilibrium expression will be ...


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It is a bit of a tricky question. The trick is in the wording "flow through". The positive ions on the A side do not actually "flow through" the salt bridge from A to B. Rather positive ions exit the salt bridge on the B side to compensate for the cations depositing on the cathode (B). Likewise negative ions exit the salt bridge on the A side to balance the ...


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It is a bad question because "Metal A is more reactive than Metal B" has no real meaning. It is a vague choice of word, A is more reactive, but with respect to what? Your logic is correct Since A is more reactive, it undergoes oxidation to form A2+ ions, while B2+ ions undergo reduction and form B. From this we can deduce that A is the anode and B is ...


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Energies are additive. Concentrations, partial pressures, and activities which are components of equilibrium constants are strongly related to probabilities. The probability of two things happening at the same time is computed by multiplying the probabilities, and likewise, the equilibrium constant is computed from products of concentrations/activities. ...


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In my Opinion, 1) Conductance= It is reciprocal of the electrical resistance. Or you can say it is the ease with which a conductor or an electrolyte allows the flow of electrons through it. It does not take dimensions into account. It's SI unit is Siemens. 2) Conductivity = It is the Conductance of a solution of 1cm length and having 1 Sq. cm as the area ...


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A cell with two electrodes separated from each other and an aluminum divider plate in the middle is really two cells: one has an aluminum plate as a cathode, and the other cell has the same aluminum plate as the anode. In this situation, the solution does not directly connect the external anode and cathode; the aluminum divider separates the solution into ...


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There are 2 competing principles. $\ce{[Co(H2O)6]^3+}$ has strong oxidation effect, but... $\ce{[Co(NH3)6]^3+}$ is much more stable than $\ce{[Co(NH3)6]^2+}$. As the consequence, the former from the next 2 equations is significantly shifted to the right. $$\begin{align} \ce{ [Co(H2O)6]^3+ + 6 NH3 &<=>> [Co(NH3)]^3+ + 6 H2O} \\ \ce{[Co(...


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Note that at molecular/atomic level, there is no such a thing like being idle. If a particular system ( like the particular lead acid half-cell ) is in equilibrium, so macroscopically "nothing happens", there is ongoing dynamic equilibrium. The rate of the oxidation reaction, providing electrons to the electrode wire, increases exponentially with raising ...


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I was introduced to this concept using an equilibrium between the metal and its ions. I was then told that those that have a higher tendency to give up electrons will have equilibria lying to the left, and for the others equilibria will be to the right. Then I was told that this results in a charges on the metal. Then it was that the metals with ...


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The main thing to point out is that your question doesn't actually make sense the way you want it to. What does it mean if a reduction potential is zero? Or positive? Or negative? Is that favorable or not? It turns out we don't specify this at all! As an example, consider the standard hydrogen electrode and it's reduction potential: $$\ce{2H+ + 2e- -> ...


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Quoting your comment A better way to say it maybe that the two form ions just as readily. Then why should electrode B itself not be charged comparatively more negative? Eo is not a direct measure of the tendency but rather Gibbs free energy. They are related as Delta Go= -nFEo; Write it the other way, you get Eo = Delta Go/(-nF) did you notice ...


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Let's write the half-cell equations in standard form as half-cell reductions. In a table of standard electrode potentials it would silly to write both the reduction and oxidation reactions since that would needlessly double the size of the table. $$ \begin{align} \ce{A+ + e- &-> A} &\quad V_\ce{A} \\ \ce{B^{2+} + 2e- &-> B} &\quad ...


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Here is a plot of the cell potential for your example: Lucky for us, it does not change much until the very end. This means numerical integration would work quite well. Notice that the x-axis is the progress of the reaction (related in a linear way to which charge has been transferred from oxidation to reduction half reaction. To integrate, you would have ...


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My question is that suppose we have two metals, A and B. A produces +1 ions. B produces +2 ions. Now even if the two have the same tendency to lose or gain electrons, the B plate would be 'more negative'. That is not a problem because the tendency of the metal to lose or gain electrons has no direct bearing on the reactivity of the metal or the reduction ...


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My question is that suppose we have to metals, A and B. A produces +1 ions. B produces +2 ions. Now even if the two have the same tendency to lose or gain electrons, the B plate would be 'more negative'( As in this plate would have more negative charge). How has the voltage helped us determine the difference in the tendencies of these two metals? ...


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Which plate would be negative depends on concentration of ions, as the plate with 1 electron transfer would have double rate of the potential change with the ion concentration at the electrode. But if at particular concentrations both electrodes have the same potential, the cell voltage will be zero and neither of plates would be negative wrt the other.


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We should not extend the idea of electrode potentials for very high concentrations such as concentrated sulfuric acid which is on the order of 18 M. The interesting property of these acids (HNO3 or H2SO4) is that they behave as oxidizing agents or even as dehydrating agents (H2SO4) especially at high temperature and at high concentrations. This is not to be ...


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Given that the standard potential for a fuel cell is 1.229 V what is causing the difference in our measurement? Multiple sources (e.g. here, here and there) estimate the voltage to be much less than the standard potential, especially when there is a load. Individual fuel cells produce relatively small electrical potentials, about 0.7 volts, so cells are "...


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It is a very good thought experiment as to why we don't cap the salt bridge ends with gold plugs. By capping, you are essentially providing an interface which can source and sink for electrons in the cells, thus each end will act as an electrode in the cell. One end of the salt bridge will act as a cathode and one end will act as an anode (strange as it may ...


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Energy storage depends on the electromotive potential (i.e. difference between species in the electromotive series) and on the number of electrons available. Li, for example, has an oxidation potential of ~3.04 V relative to hydrogen, but Al has one of 1.66 V, so Li has the greater potential. On the other hand, Li has only one freely available outer ...


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My book tells me to keep the E∘half-cells as they are written in the tables and simply put them in Your book is then one of the few books which teaches electrochemistry properly. The sign of the electrode reduction potential is invariant. If reflects the sign of the electrostatic charge of the electrode with respect to hydrogen electrode. I don;t know if ...


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To the extent that it's appropriate to treat it as a thermodynamic quantity, given that it's generally a description of a specific, macroscopic system (and thus violates the statistical-mechanical large-numbers assumption of thermo), resistance is extensive, always. Resistance of a wire: The resistance of a wire depends on area and length. This implies ...


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Take a look at the two half reactions: $$ \begin{align} \ce{Ag+(aq) + e- &→ Ag(s)} &\qquad E^\circ &= \pu{0.80 V} \\ \ce{Sn^2+(aq) + 2 e- &→ Sn(s)} &\qquad E^\circ &= \pu{-0.14 V} \end{align} $$ If there is an electron for grabs (like the ones in the wire of a voltaic cell), $\ce{Ag+(aq)}$ and $\ce{Sn^2+(aq)}$ are competing for it. ...


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Free energy is a state function. No matter how you run the reaction, if you start with a certain state (set of concentration) and end with a certain state (set of concentrations at equilibrium), the change will be the same. The free energy of reaction is the maximal amount of non-pV work (electrical work in this case) the reaction can do. If it does not do ...


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All depends on the cell chemistry and geometry. It is usually a combination of keeping the energy and conversion of the free energy to thermal energy, when reagents diffuse and happen to meet each other. Or, some side reaction with solvent or auxiliary components may occur, like for $\ce{Li-ion}$ cells. Some primary lithium cells last many years, while some ...


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The Nernst equation and electrochemical potentials relate to redox systems, not to reagents and products. The forward and reversed reactions are the same redox system. Imagine you would want to make a galvanical cell with the same electrodes. Flipping the sign would grant you a Nobel price for inventing a perpetuum mobile.


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You can ask yourself: if the universe is expanding (that is, if it is in a dynamic state), is thermodynamics truly ever applicable? The answer is that in science we need to take approximations, during measurement or when extracting information from data. So it is in thermodynamics, which works well enough even though on long enough scales the approximations ...


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If the Zn2+ ions aren't involved in the redox reaction, then why does the extra concentration affect the cell potential? The overall reaction is $$\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)}$$ If this reaction were happening in a common reaction mixture (maybe with Zn and Cu powder on the bottom of a solution) we would write it as $$\ce{Zn(s) + Cu^2+(...


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First off, read carefully: Different metals will increase the likelihood of corrosion. It doesn't say: Different metals will increase the likelihood of corrosion of gold With that in mind, it's called galvanic corrosion and occurs when contact between two (or more) dissimilar metals causes the more reactive one to corrode when it normally would not ...


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You're questioning the intuitive disconnect caused by most galvanic cell drawings which seem to assume the electrolyte solution in the salt bridge does not conduct electricity, so let's investigate. Imagine a Zn/Cu$^{2+}$ cell with electrodes 5 cm apart in a 3.5% NaCl solution with a tube (1 cm$^2$ cross-section) of solution as the salt bridge for balancing ...


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Just keep in mind that Eo is determined under highly controlled conditions and concentrations. I sense two key problems here. The major problem is already stated above that the resistance of a wet filter paper must be high, and this is giving you lower voltage. Evaporation of water from the filter paper would make the situation worse. Since you are not ...


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