> there is no evidence for $\ce{F_2^{2–}}$ in the gas phase
> We are not surprised as we recall the general observation that in the absence of solvation or counterions, multiply charged molecular ions are rarely stable relative to the loss of an electron and/or bond cleavage
> The presence of F2^2– ion in the condensed phase has been suggested as part of discussions of terms of the NMR spectra of concentrated fluoride ion solu- tions and explicit quantum chemical calculations of F2^2– hydrates.
> Although 28 is overall a dication, the bonding about the central carbon is 10-C-5. Various types of evidence support our claim that it is in fact a TBP ten-electron species rather than the Td 8-C-4 species, 29.
So you can stop talking about how pentavalent carbon is a mistake
Sugden
Above all, his name is Sugden, not Sudgen.
What he said
He did not say that the octet rule is never violated.
On the contrary, he suggested that "the maximum number of electrons in the valency orbit can not [sic] exceed eight". (source)
This means, while he denies that the octet rule ...
Sugden
Above all, his name is Sugden, not Sudgen.
What he said
He did not say that the octet rule is never violated.
On the contrary, he suggested that "the maximum number of electrons in the valency orbit can not [sic] exceed eight". (source)
This means, while he denies that the octet rule ...
When I was new on the stack exchange( now i consider myself an old member(1 month)),the problem i faced was most of my questions were marked as duplicate of other questions.
I can't understand that why people mark some questions as duplicate,there are several possibilities,,maybe the person who h...
@pentavalentcarbon it's a transition state for inversion :)
My only experience with comp chem so far was to investigate this energy barrier for inversion. And we found one imaginary vibrational frequency for planar ammonia. That was actually all we did haha
According to most sources, $\ce{CrO3}$ has the following Lewis structure (courtesy ChemSpider):
However, it is also not planar (courtesy Wikipedia):
Implying that there is a lone pair somewhere.
What is the electron configuration of the $\ce{Cr}$ atom?
Which orbital is responsible for the...
From what I was taught in middle school, cations are those ions that move towards the cathode, likewise anions are those ions which move towards the anode.
I didn't have issues with this back then, since all we studied were electrolytic cells. But now that we've crossed over to electrochemical c...
An electrochemical cell is a device capable of either generating electrical energy from chemical reactions or facilitating chemical reactions through the introduction of electrical energy. A common example of an electrochemical cell is a standard 1.5-volt cell meant for consumer use. This type of device is known as a single Galvanic cell. A battery consists of two or more cells, connected in either parallel or series pattern.
== Half-cells ==
An electrochemical cell consists of two half-cells. Each half-cell consists of an electrode and an electrolyte. The two half-cells may use the same...
An electrolytic cell is an electrochemical cell that undergoes a redox reaction when electrical energy is applied. It is most often used to decompose chemical compounds, in a process called electrolysis—the Greek word lysis means to break up. When electrical energy is added to the system, the chemical energy is increased. Similarly to a galvanic cell, electrolytic cells usually consist of two half cells.
Important examples of electrolysis are the decomposition of water into hydrogen and oxygen, and bauxite into aluminium and other chemicals. Electroplating (e.g. of copper, silver, nickel or chromium...
@Ramanujan C-H bond enathalpies. I guess the bonds in CH4 are sp3 hybrids, in CH3 they would be sp2 hybrids, in CH2 they'd be sp hybrids and in CH then the bond would be between a p orbital on the C and the H s orbital so no hybridization at all. Therefore a weaker bond and lower enthalpy
@DHMO Keeping this in mind first: If a species has a high reduction potential it has a high tendency to get reduced, that is, it is a very good oxidizer. The reduction potential is inversely proportional to the oxidation potential. Going by the same logic...if a species has high oxidation potential it has a high tendency to get oxidized, that is, it is a good reducing agent. Reduction and oxidation potentials are inversely proportional to each other (Continuing...)
(Continued) If a species has a high reduction potential (high tendency to act as an oxidizer) then it is also said to have a low oxidation potential (low tendency to act as a reducer). Clear with this bit?
The values of standard electrode potentials are given in the table below in volts relative to the standard hydrogen electrode and are for the following conditions:
A temperature of 298.15 K (25 °C);
An effective concentration of 1 mol/L for each aqueous species or a species in a mercury amalgam;
A partial pressure of 101.325 kPa (absolute) (1 atm, 1.01325 bar) for each gaseous reagent. This pressure is used because most literature data are still given for this value rather than for the current standard of 100 kPa.
An activity of unity for each pure solid, pure liquid, or for water (solvent)....
Side note: The reduction of 2H+ is arbitrarily assigned a neutral value- Zero volts, and I'll be referring to reactions (oxidation-reduction) as Upper-scale (having negative/low reduction potential) and Lower-scale (having positive/high reduction potential) reactions...
In your second case, you replace Fe^3+ with K^+ (None taken....and continuing......I'm trying to deal with the space limitations of the chatbox...so bear with me)
> let's say I have a chemical cell consisting of two beakers connected by a KNO3(aq) salt bridge the left beaker has KI(aq) and the right has Fe2(SO4)3(aq) the electrode is inert platinum connected to a voltmeter which has reading so iodide ion is oxidized to iodine and ferric ion to ferrous ion problem: would my cell still work if I replace ferric sulfate solution by acidified potassium dichromate solution?
(Playing my bit here: @DHMO 's argument hardly has any punctuation, he writes in clauses, doesn't format his symbols correctly, believes that the ferric is oxidised to ferrous, lays no emphasis on the voltmeter reading, nor does he mention how the voltmeter's connected. Gentlemen, clearly DHMO is asking a question not worthy of your consideration. It is a poorly framed question, peppered with errors of all sorts. I rest my case. Advocatus Diaboli out......
Sorry, my booboo, Fe3+ --> Fe2+ in the original cell - that's a reductive process and that will drive the 2I- --> I2 in the LHS cell. So acidified dichromate will also work as a reductive process causing the oxidation of I-
The dichromate system is electrochemically analogous to the Fe2+/Fe3+ system. It shouldn't matter that the dichromate ion is negatively charged to start off with. It should still suck electrons off the cathode and get reduced to Cr3+ creating an electrochemical potential. So, you'd get a reading of 0.79V. Now, if any current were to actually start flowing, i.e. if there would actually be a reaction - I doubt it.
Well, with a perfect voltmeter - i.e. infinite resistance - no electrons will actually flow between the electrodes so no reaction will actually occur and the voltmeter will measure the full (theoretical) cell potential of 0.79V. If any electrons were to flow, then the cell would run into the problem you've identified which is that as electrons
build up on the electrode (I'm not going to commit as to whether it's the anode or the cathode) on the RHS then the dichromate ions are going to be repelled.
@DHMO As far as calculating the cell potential is concerned, no, it doesn't matter. As far as the reaction actually proceeding, then yes, it does matter.
@DHMO Well, it does, which is why I think the reaction would not actually proceed. But in terms of the cell potentials, which is basically the energetics involved, then the reaction is spontaneous and therefore you'd get a potential.
@DHMO No, in fact, experimentally, you get values that are closer to the theoretical cell potentials when no reaction is occurring i.e. when the resistance of the voltmeter is infinite and no current is flowing. As soon as current starts flowing, then you get resistive effects occuring and the measure energy per coulomb drops.
Because the chemical energetics work out okay. So, the chemical potential can be worked out as 0.79V. However, as you point out, if the reaction does proceed then the cathode becomes negatively charged with electrons coming from the iodide at the anode and that builds up an electrostatic barrier to the reaction of the dichromate, also negatively charged, at the cathode.
@GavinMachell why does permanganate reduce to +4 in alkaline medium but +2 in acidic medium?
(MnO2 and Mn^2+ respectively)
It amazes me, how most catabolic reactions are exothermic yet bond-breaking is endothermic; and how most anabolic reactions are endothermic yet bond-forming is exothermic.
@DHMO That's a very good question. The answer probably has a lot to do with the reduction potentials of H+/H2O/OH- Allow me to consult the table a moment.
@DHMO What happens during the electolysis of water? Hydrogen and oxygen are produced. Hydrogen at the cathode (where reduction of H+ occurs) and oxygen at the anode (where oxidation of OH- is happening)
But to answer your first question, I think that it is mostly to do with the availability of H+. I've done plenty of oxidations using acidified manganate when I have ended up with a lot of brown MnO2 - usaully because I ran out of acid. It takes 8H+ to mop up all the oxygen on MNO7- to take the Mn all the way to Mn2+ If there isn't enough H+ (at high pH for example) then you get MnO2
That's almost certainly not the complete answer, but it fits.