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1

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 ...


1

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 ...


1

Let's subtract the two reactions from each other to simplify things a bit: $$\ce{2ClF + 2KBr -> 2KF + Cl2 + Br2}\ \ \ \ \ \ \ \ \ \ [1]$$ $$\ce{ClF + 2KBr -> KCl + KF + Br2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [2]$$ $$\ce{ClF + KCl -> KF + Cl2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ [3] = [1]-[2]]$$ Now we can discuss [2] and [3] separately by splitting them up ...


0

If you actually passed current thru the cell, you would have been cathodically protecting the cathode - that is, preventing the dissolution of iron. On top of this "cathodic protection", however, you have copper deposition. And the copper could have been loosely deposited as black nano particles. At the anode, you are discharging SO4-- dianions, which then ...


4

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, ...


1

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 ...


2

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 ...


0

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 ...


0

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 ...


3

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- -> ...


0

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 ...


6

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 ...


2

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 ...


1

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? ...


-1

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.


3

You have rightfully assumed the reaction takes place in either basic or neutral medium, otherwise $\ce{Cr(OH)3}$ wouldn't be the product. This automatically suggests $\ce{H+}$ are not used for balancing half reactions. Your solution is fine for the provided reaction given that chromium(III) hydroxide stays precipitated. Here I've basically formatted your ...


2

The standard cell potential is temperature-dependent itself, just like the Gibbs free energy. The easiest way to see this is to write the Gibbs free energy in terms of enthalpy, entropy and temperature: $$\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$$ This shows you that $\Delta G^\circ$ is strongly temperature dependent (even if $ \Delta H^\circ$ ...


1

I'm wondering why is it not necessary to mix a metal like Fe into its own solution $\ce{Fe(NO3)2}$ (like you would do when setting up an electrochemical cell)? First remember what the purpose of your experiment was: To determine the relative activity of several metals we observed reactions between the metals and aqueous solutions of dissolved metals. In ...


2

Yes, the reaction must be taking place for the voltmeter to measure the difference in potential. Remember $$V = iR$$ thus a voltmeter works by knowing its internal resistance, $R$, and measuring $i$. Also notice that if $i=0$ then $V = 0$, so a voltmeter must have at least some tiny current to work. The problem with a voltmeter is that you can imagine ...


2

An ideal voltmeter has an infinite resistance, so does not let any current flow in the system. However nothing is ideal in real life, it consumes a very small amount of current to move the magnet and needle arrangement or electronics. An ideal ammeter has zero resistance. Assume that you have a water tank on your ceiling which supplies water to your kitchen ...


2

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 ...


2

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. ...


2

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 ...


1

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 ...


1

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.


0

You can rust metal quickly but not rapidly the least amount of time is 10 minutes using 2 cups hydrogen peroxide 1 half teaspoon of salt🎓


0

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+(...


1

As previous comments said, this isn't a redox reaction. The patina or corrosion is made of metal salts -- for copper, it's copper oxide, sulfide, carbonate, and perhaps other things. Vinegar readily converts all these to soluble copper acetate. Most coins are made of metal alloys, and those alloys sometimes contain metals that are attacked by acid. Nickel ...


2

Rust is a hydrated iron oxide that has a volume greater than the iron it contains. When iron rusts, the oxide grows in layers and curls away from the iron, exposing it to more corrosion. Phosphoric acid is a moderately strong acid, but it forms an iron phosphate that is insoluble but adherent to iron (and steel). In that sense, phosphoric acid deactivates ...


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