10

There is important to know the actual potential depends on actual activities of reagents ( approximately concentrations for dilute solutions). Additionally, the electrode potential is the thermodynamic quantity. If some net reaction kinetic is involved, other potentials affects the actual potential at the solution border, like the diffusion potential and ...


7

Of course, we're starting by acknowledging a perpetual motion machine is impossible. The question, then, is how do we understand, through chemical thermodynamics, why your specific set of steps can't constitute a perpetual motion machine. The answer is that what you're not accounting for is that the electrical energy required for your step 2 is greater ...


6

After some searching, I found this 1956 Thesis(Reference), which describes the production of peroxydisulfuric by electrolyzing concentrated sulfuric acid. The yield varied according to the concentration of the acid, temperature, current, and the nature of the electrode. In the Thesis, the chemist noted that discharged $\ce{HSO4-}$ united together to form ...


5

This is pure scam! All he is doing is using an electrode which is corroding during electrolysis. The flocculated material you see in tap water is just a metallic hydroxide. If tap water had this much of metals, nobody will survive. Such a fraudulent salesman should be shooed away!


5

First, sodium burns very nicely in air. However, it also burns in chlorine, returning back to the $\ce{NaCl}$ with which you started. Most welders provide alternating current, meaning the the electrode becomes alternately positive and then negative (50 times per second [50 Hz] in much of the world, and 60 Hz in North America). Therefore, both $\ce{Cl2}$ gas ...


4

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 \\ }\...


4

Using up to 30 kilovolts is not uncommon with aqueous systems in a technique called capillary electrophoresis. Similarly, applying kilohertz current is also common in a technique called conductometry to measure the resistance of aqueous solutions. In general, when you pass high frequency current, electrolysis does not occur. You have to check the dielectric ...


4

In electrochemistry, the rate of electrolysis depends on rate of charge entering the cell. Second most important point is that one can either control potential or current but not both during electrolysis i.e. you cannot have both values set. If you are fixing potential at 2.5 V, the value of current is not in your hand. I don't think Ohm's law remains valid ...


3

Supposing there is a powerful DC source, then quite hot, unless small enough electrode current density is managed. 6 V is quite high voltage for electrolysis. But it is possible the source will not be able to provide the current for the voltage 6 V, unless the electrode area is small. Consider also the graphite electrodes may get deteriorated by the ...


3

As someone mentioned in his/her answer last year, you cannot get sodium metal from electrolysis of aqueous sodium chloride. But, contrary to his/her explanation, you do actually get metallic sodium at the cathode in this reaction. Even if the half-cells are separated with Nafion (chloralkali process), the elemental sodium produced at the cathode instantly ...


3

The standard electrode potential, $E^\circ$, in volts, does not depend on the surface area of the electrodes. However the standard electrode potential is measured with an infinitesimal current flow. In the simplest model you can imagine a cell, galvanic or electrolytic, as having an internal resistance. So as the current of the cell increases the voltage ...


3

Consider the molecular mass, and the mass of chlorine in 1 g of $\ce{PdCl2}$. Then consider the volume of the work space, and how quickly the $\ce{Cl2}$ disperses to fill that space. Until it disperses, the concentration at the source might be 1E6 ppm (i.e. 100%), but is that meaningful? If it disperses to fill a flask, what is that volume? Are you doing ...


3

For a general electrolytic reaction, $$\ce{M^{n+} + $n$ e^- -> M}$$ where $n$ moles of electrons are consumed in depositing 1 mole of substance ($n$ is also known as $n$-factor). So, for $x$ moles of substance, $nx$ moles of electrons would be consumed. Therefore, $$Q = nxF,$$ dividing by $nF$ on both sides, we have $$\frac{Q}{nF} = x,$$ which is ...


3

The conductivity of molten lithium carbonate has been determined for vast range of temperatures (Ref.1). The abstract of which states that: The present communication reports the results of an investigation of the properties of surface tension, density, and electrical conductance for molten $\ce{Li2CO3}$, $\ce{Na2CO3}$, and $\ce{K2CO3}$ and some mixtures ...


2

You've described the classic preparation of sodium hypochlorite, better known as "bleach". If you electrolyze sodium chloride solution, you form hydrogen at one electrode and chlorine at the other, and sodium hydroxide in solution. But chlorine tends to dissolve in alkaline solutions to form chloride and hypochlorite ions. So you get "sodium hypochlorite" ...


2

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:...


2

In a chemical sense, "anode" of an electrolytic cell is the most powerful oxidizing reagent known, so much so that it oxidize F(-) to elemental fluorine. Don't forget that the electron balance in an electrolytic cell is always maintained so the gram-equivalents of x reduced at the cathode must equal gram equivalents of y oxidized at the anode. As a result, ...


2

Unfortunately, the little guy would be a bit disappointed down the line. The energy of the cell comes from the oxidation of the aluminum (the soda pop can) and that Al anode is eventually consumed. Aluminum is produced at relatively high cost, using electrolysis, so wet soil (or seawater) is not the issue. Same deal with the standard Zn and Cu electrodes ...


2

A good mnemonic to remember which is the anode or the cathode is to remember that Right side represents Reduction in a cell diagram. This is fixed by those who invented cell notation.. Also, by definition, reduction occurs at the cathode. Whether it is a spontaneous cell or not, is another question. Now as EdV said, calculate the cell potential as written i....


2

Catalysts generally decrease the electrode overpotential, which leads to decreasing the needed voltage to perform electrolysis at given current densities toward its minimal theoretical value. If higher current is possible at the same voltage due the catalyst, the lower voltage is needed for the same current, both leading to better efficiency. As the energy ...


2

The primary problem is too low permeability = too high resistence of the gloves, causing high voltage drop where it should be minimal one. The gloves are not hydrophilic enough for electrolyte soaking to get low resistance. The remaining voltage - after subtracting the glove voltage drop - is not high enough to cause electrolysis by significant current. ...


1

I can answer your first two questions. I am confused by 'require', do ions upon discharging gain energy or lose energy? Ions lose free energy during discharging, as you are referring to the spontaneous process of transferring charge at the electrodes with conversion of some of the ions into neutral species (in general you are speaking about the process ...


1

Absolutely a good experiment! Diffusion of ions, propelled by the electric field (i.e., current) will occur at a given, calculable rate. Deposition of those ions (and conversion to $\ce{O2 + H2}$) will occur and the $\mathrm{pH}$ will rise at the cathode (negative electrode) and fall at the anode (positive). Then back diffusion will occur as the $\ce{OH-}$ ...


1

A current is simply the rate of flow charge (and hence the rate of flow of electrons); the more electrons flowing, the higher the current. Anions migrate across the potential gradient between electrodes, carrying the electrons to the anode and the anions are then oxidised. The oxidised species (often uncharged) must then diffuse across a concentration ...


1

As you noticed there is a small amount of current which keeps on flowing when there is no electrolytic decomposition. This "problem" is aptly named as non-Faradaic current. Faradaic current refers to current due to oxidation and reduction. Here is a detailed review: The difference between Faradaic and Nonfaradaic processes in Electrochemistry


1

For hydrogen to burn or explode in air, its concentration needs to be at least 4%. Unless you're using very high power levels, you're unlikely to reach that concentration in the room's air. You could ignite bubbles of hydrogen as soon as they pop at the surface, but if you don't, the hydrogen (with about 1/8 the density of air) will rise quickly, diffusing ...


1

Do I use the total surface area of both Cathode and Anode or just Anode? You use the cross-section, just like it were a wire. Cross-section of both Anode and Cathode or just Anode? Just one. Like in a wire, you would measure the cross-sectional area once, not cut the wire into many piece and add up all the areas.


1

The galvanic cells from the same material are possible. They are called concentration cells and they use two electrolytes of the different concentration of the same electrolyte. But their voltage is low. In the case of $\ce{CuSO4}$ $$U = 0.0295 \log\frac{c_1}{c_2}\ \mathrm V$$ Note that the electrode naming is based on the direction of the electron flow, ...


1

It does, as an intermediate product. The peroxide formed is on the surface of the electrode which changes it's chemical behavior. Electrolysis is a four-electron process. In the first step, an electron is removed from a water molecule, producing a neutral OH adsorbed to the surface of the electrode. The second step removes another electron from the adsorbed ...


1

What is the "correct" answer You are asking why the answer key has water and not saltwater as an answer. Decomposition is defined as: A reaction where a single compound breaks down into simpler compounds. For salt water, the salt reacts at one electrode, and the water at the other, and you are making NaOH. For pure water, you are splitting water into ...


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