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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 ...


2

The little problem with the standard addition in potentiometry is, that the response to the addition is not linear, but logarithmic. If we consider the simplified Nernst equation in context of parameters $A, B$, and if we consider $c, c_\mathrm{0}$ as unknown concentration and concentration increment, respectively, we can follow this derivation: $$\...


1

As you can see from figures in Zhe's answer that it is possible by adding waves to make a wavepacket. The characteristic of a wave is that superpositions are possible and these lead to interference/diffraction phenomena. By analysing the wavepacket's spread in wavelength and position, the expression $\Delta k.\Delta x \ge 1$ is found where $k=2\pi/\lambda$. ...


4

Many recommended procedures such as this one call for connecting (+) to (+), then (-) on the "good" battery to a metal component in the "dead" car engine. The negative terminal of the battery is grounded to the metal components, so we may think of the last connection as (-) on the dead car. If a spark were to form when the circuit is established on the ...


3

Done that dozens of times, have never seens sparks fly. The current flowing is not exactly huge, you are connecting (first) + to +, and (then) - to -, after all. The voltage difference is perhaps 3V, not 12. I don't think there is an electrochemical reason to do it in this order. Afaik it's just so you dont short-circuit one battery if you slip with the ...


1

One problem is there is not enough voltage for this thing to sell. Say you have molar acid and 1 molar base, both strong and both functioning as ideal solutions. Thermodynamically you have one electron transfer associated with the net reaction $\ce{H^+}_{aq}+\ce{OH^-}_{aq}\to\ce{H2O}$ and the equilibrium constant is $10^{14}$ at room temperature. Put ...


3

I can provide some physical intuition, but it's still going to require math and previous work to make it happen (aside on this at the end). Let's consider a particle in just one dimension. There are two quantities of interest: (1) the position of the particle along this coordinate, (2) the scalar momentum (related to the velocity) along this coordinate. I'...


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There is a simple way to understand this if you understand the relationship between frequency and wavelength in a vibrating string But you can't avoid at least some mathematics. They key to understanding where the position momentum uncertainly relationship comes from is to think about the relationship between the frequencies present in a musical note and ...


1

It's simply a consequence of wave mechanics and operators. Any observable, i.e., any observable which can be measured in a physical experiment should be associated with a quantum-mechanical operator (like + or - operators, just more complicated). Furthermore, if these operators are intrinsically non-communicative, or if the order in which they are applied ...


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 ...


0

A reliable method to obtain metallic nanoparticles in general is to reduce a metallic ion in presence of a capping agent. For a example, a traditional method to obtain gold nanoparticles (AuNPs) is to reduce a gold salt in presence of dodecanothiol. For example, here is a method to obtain Pd nanoparticles by thermal decomposition, still the precursor is a ...


1

Can the Volta Battery use a different ionic solution? Yes you can use any soluble salt as long as it does not react with electrodes, and certainly such a battery would be able to produce current. My chemistry high school teacher had a clock which ran on salt water. See for example, https://science.howstuffworks.com/environmental/green-tech/sustainable/water-...


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Here is a figure of the famed Daniell cell: Note the spelling of the name: it was invented in 1836 by John Frederic Daniell. At 25°C, and with n = 2 equivalents/mole, as correct here, (ln10)RT/nF = 0.02958 V = 29.58 mV.


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....


0

It is a very good question, and I have wondered about it for years, talking to highly seasoned electrochemists, only to find that nobody knows the answer! The sole reason is that the problem is overwhelmingly complex despite its deceptive simplicity. Nernst offered him theory similar to osmosis, but it did not work- 100 years ago. However there are couples ...


0

Those $\pu{-0.74V}$ are measured against a standard hydrogen electrode. That's the universally accepted but totally arbitrary standard. You don't have a second electrode, hydrogen or other, so the absolute number is totally meaningless. If someone had decided to use the zinc electrode instead, as zero point for the electrochemical series, then the voltage ...


2

The potassium ions in the reduction half-cell are inert, merely serving as part of the electrolyte. The are no problem at all. Indeed, you actually need to allow for the passage of more potassium ions through your separator, to balance the potassium metal that's dissolving and make up for the cations being plated out at the cathode.


0

The derivation of equation for cyclic voltammetry requires a lot of advanced calculus and differential equations based on the Fick's law of diffusion. This is not to discourage you but to show that the teacher is oversimplifying it. The derivation are provided in Allan J. Bard's Electroanalytical Chemistry. Why are they teaching cyclic voltammetry in ...


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


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For a quick general answer, this equation represents the charge balance. In a solution, the charges of all ions should add up to zero. You charge balance reads: $$\ce{[H+] = [Cl-] + [OH-]}$$ If you had divalent ions, you would have to correct for that, e.g. if the solution also contained magnesium ions: $$\ce{[H+] + 2 [Mg^2+] = [Cl-] + [OH-]}$$ How ...


1

Let's take the degree of dissociation of water be $\alpha$ and the $[\ce{H2O}] = c$, then at equilibrium the $[\ce{H^+}] = [\ce{OH^-}] = c\alpha$. And since $\ce{HCl}$ is a strong acid it will completely dissociate. So $\ce{HCl}$ is a then the $[\ce{H^+}] = [\ce{Cl^-}] = a$. So $[\ce{H^+}]_{\textrm{net}} = a+ c\alpha$, which is equal to $[\ce{OH^-}] + [\ce{...


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