# Tag Info

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 In aqueous solution the anode will oxidize the water instead, as this reaction can be done at much lower potential than oxidizing the sulfate ion. Sulfates can be oxidized if you take the water away. Potassium peroxydisulfate can be prepared by electrolyzing a solution of potassium bisulfate (\ce{KHSO4}) in sulfuric acid solvent. 3 Let's try to look at those electrons. C - cathode, A - anode. \begin{align} &\mathrm{C(-):} &\ce{Sn^2+ + 2e- &→ Sn^0} \\ &\mathrm{A(+):} &\ce{2Cl^- &→ Cl2^0 + 2e-} \end{align} $$So, your second reaction is right. But the first is wrong: \ce{Sn} is reduced, but there are no oxidation. You should use$$\ce{Sn + 2 Cl2 → SnCl4}$... 3 The standard electrode potential,$E^\circ$, in volts, does not depend on the surface area of the electrodes. However the standard electrode potential is measured with an infinitesimal current flow. In the simplest model you can imagine a cell, galvanic or electrolytic, as having an internal resistance. So as the current of the cell increases the voltage ... 3 Supposing there is a powerful DC source, then quite hot, unless small enough electrode current density is managed. 6 V is quite high voltage for electrolysis. But it is possible the source will not be able to provide the current for the voltage 6 V, unless the electrode area is small. Consider also the graphite electrodes may get deteriorated by the ... 3 Consider the molecular mass, and the mass of chlorine in 1 g of$\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 ... 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 In principle, a nitrate reduction in the presence or silver ions only would have to give silver oxide; but this decomposes between 200 and 300°C. So, perforce, you would ultimately get elemental silver from electrolyzing the nitrate melt. However, that does not mean it's best. With a little additional heating, to 440°C, you get decomposition to elemental ... 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 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

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....

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

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

The suppose total and half reaction are: \begin{align} \ce{2 MCl_n &-> 2 M(s) + n Cl2} \\ \ce{M^n+ + n e- &-> M(s)} \\ \ce{2 Cl- &-> Cl2 + 2 e-}\\ \end{align} The question is rather tricky. Note that I am addressing the supposed question in hand, not the electrochemical evaluation. In some sense, if we consider large scale ...

1

The redox potential is a measure of tendency of redox system to exist rather in oxidized form ( negative potential values) or reduced form ( positive values ). Therefore the higher the standard redox potential is, the stronger oxidation effect and weaker reduction effect have the respective oxidized/ reduced forms of the redox system. And vice versa. ...

1

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 ...

1

Any good general chemistry textbook has a chapter on electrolysis. If you wish to add more scholarly work check the Journal of Chemical Education. https://pubs.acs.org/action/doSearch?text1=Electrolysis&quickLinkYear=&quickLinkVolume=&field1=Title&type=within&publication=346464552 In the first part you electrolyze a solution of salt. ...

1

Add the information regarding the composition regarding the electrodes. However lead is very heavy it would not like to float on the top. Secondly, lead compounds are mostly white with an exception of oxides (yellow/ red) and iodide (yellowish). Most likely the black floating scum is carbon in iron. Wear gloves, and wash your object with a mild abrasive such ...

1

Pure water cannot conduct electricity fundamentally because of lack of rapid ion transport inside bulk solution. In fact, pure water electrolysis has been achieved, by using nanogap electrochemical cells, where the distance between two electrodes less than Debye-length of water (less than 100nm!). But the fundamental mechanism has been changed significantly....

1

Gosh! You would think that electrolyzing NaHCO3 with stainless electrodes would be a clean reaction to produce H2 and O2. Perhaps OH. is powerful enough to attach to some atoms at the surface and make it more insulating. So, significant degradation could occur at the positive electrode because of oxidation. Perhaps lower voltage and currents would be more ...

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