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13

I will be using an approach which has been enlisted in the following book for answering this question: Arrow Pushing in Inorganic Chemistry ;A Logical Approach to the Chemistry of the Main-Group Elements The preface of the book says: The approach: These reactions represent important facets of the elements involved but are typically presented as no more than ...


12

Gold(III) chloride does not exist as $\ce{AuCl3}$. According to Wikipedia, the name gold trichloride is a simplification of the name, which is referring to the empirical formula, $\ce{AuCl3}$. The X-ray defraction studies revealed that the chemical exists as a chloride-bridged dimer, $\ce{Au2Cl6}$ (Ref.1): In $\ce{Au2Cl6}$, each gold center is square planar,...


11

I'm guessing your teacher is looking for sulfur dioxide as the answer, but I don't see how or why you're supposed to be able to arrive to this answer logically. Either you'd need to read about it specifically, or maybe you're supposed to stare at a table of standard reduction potentials and notice that $\ce{SO2}$ appears as both a reagent and as a product ...


11

Virtually every analytical chemistry textbook's chapter on anion determination mentions that Devarda's alloy is very brittle and can be easily pulverized in a mortar, unlike $\ce{Al}$ and $\ce{Zn}$ metals. This is quite handy property as for qualitative analysis of nitrates in a heterogeneous media the reducing phase should be ground to fine powder in order ...


11

Not even zinc would react with neutral nitrate + chloride, why should gold ? Aqua regia must be strongly acidic for nitrates to have oxidative properties for oxidation of chlorides to chlorine and to oxidize gold to trace amounts of $\ce{Au^3+}$(what redox potential allows) Mixing of acids leads to these reactions: $$\ce{HNO3 + 3 HCl -> NOCl + Cl2 + 2 ...


10

What you have written is right, but it is just the first step. The release and solvation of an electron is fast. Reaction of an electron with ammonia to release hydrogen $$\ce{2 e-(solv) + 2 NH3 -> H2 + 2 NH2-(solv)}$$ is slow. Deep blue (diluted) or bronze (concentrated) "sodium electride" solution in liquid ammonia slowly converts itself to ...


8

Very good question and thought provoking too. It appears that the nail somehow knows or is it is able to decide which part will be a cathode and which part will be an anode! Why is that so? Why one end was not a cathode and the other end an anode, just the one shown in the cartoon taken from here Imaging metal corrosion So the fact you should keep in mind ...


8

Gold dissolves in aqua regia because there is a redox reaction: The nitrate ion oxidizes the gold in the presence of $\ce{Cl-}$ ion into $\ce{Au^3+}$, which combine with the chloride ions to form $\ce{Au(Cl)4-}$ ion. In this reaction, the presence of $\ce{Cl-}$ ion is important because it reduces the $\ce{Au/Au^3+}$ couple's potential as follows: $$\ce{Au &...


8

The answer appears to be "it depends" (perhaps this should not surprise an experienced organic chemist). According to the abstract of the paper by Mendelovici and Glotter [1] in steroid chemistry: Treatment of 3β- and 3α-hydroxy-(acetoxy-)cholest-4-en-6-one and of 6β-and 6α-hydroxy-(acetoxy-)cholest-4-en-3-one with MCPBA gives two types of product, ...


8

As per the OP's request, here is a solution. The equation is: $$\ce{a XeF4 + b H2O -> c XeO3 + d Xe + e O2 + f HF \tag 1}$$ So atom balance yields: $$4a=1f \tag 2$$ $$2b=1f \tag 3$$ $$1a=1c + 1d \tag 4$$ $$1b = 3c + 2e \tag 5$$ The equations relate the six unknown coefficients $a$ through $f$. But each coefficient is an integer ...


7

The general textbook formula for the Li ion battery is $\ce{Li_{1-a}Ni_{1-x-y}Mn_{x}Co_{y}O2}$ , and the simplest/earliest type has $y=1$, i.e. cobalt-only. These "mixed (lithium transition metal) oxides" have a layered structure (see e.g. wikipedia), where you can relatively easily electrochemically remove (and later reinsert) a part $a$ of the Li ...


7

Unfortunately, the OP's starting reaction expression is simply wrong: the left hand side has only a neutral compound while the right hand side has anions among the product species. From the wikipedia article on fluorine nitrate, the unbalanced reaction of fluorine nitrate and water is $$\ce{a FNO3 + b H2O -> c O2 + d OF2 + e HF + f HNO3 \tag 1}$$ For ...


6

Copper(III) nitrate cannot be obtained from aqueous nitric acid, and likely doesn't exist. Reaction carried under strongly oxidative conditions between $\ce{Cu(NO3)2}$ and fuming $\ce{HNO3}$ yields in nitrosyl copper(II) trinitrate $\ce{[NO+][Cu(NO3)3−]},$ [1, 2] sometimes written as adduct $\ce{Cu(NO3)2 · N2O4},$ which is contradictory to the crystal ...


6

Interesting question. First of all understand what is on the x-axis and on the y-axis. You y-axis is Gibbs free energy/Faraday constant. Choose three consecutive oxidation states, the middle oxidation state, if it lies above the two, is likely to disproportionate.Read more here on Frost and Latimer digram In your case look at the example of $Mn^{3+}$, if ...


6

My understanding is that a redox couple is an unordered pair of two conjugate species This is conceptually perfect and there is no problem when we talk about electrode potentials of half cells because as I had mentioned in your earlier queries, the electrode potential value and its associated sign do not know nor care how you write the half cell. What is ...


6

The chromate and dichromate ions exist in equilibrium as follows: $$\ce{2CrO4^2- + 2H+ <=> Cr2O7^2- + H2O}$$ In a basic medium, $\ce{[H+]}$ is less, which favours the formation of reactant $\ce{CrO4^2-}$ whereas in an acidic medium, $\ce{[H+]}$ is high, which favours the formation of product $\ce{Cr2O7^2-}$ Sources: Wikipedia


6

Actually there is no unique solution. You have four linearly independent equations and six compounds. In a typical redox reaction you need one less equation than the number of compounds. When the difference is greater than one, you have a combination of multiple reactions. We can get unique numbers only by assuming that just one oxidation product is ...


6

Starch is a long ribbon or a long filament made of a great number of glucose units attached to one another like the wagons in a train. But this filament is wound in a helicoidal way. And the inner part of this helix is a sort of long hole. And the dimension of this hole is just big enough to allow the iodine molecule $\ce{I2}$ to enter and stay there, ...


6

I get that this makes sense, but I have no idea how to derive it myself. Any help would be appreciated! Good question because such questions were pursued by prominent physical chemists and Nobel laureates (Ostwald/ Nernst and Hoffmann). The reduction of water can be tested visually and the apparatus to demonstrate it indeed named after Hoffman. Basically, ...


5

For OP: As I have realized from James Gaidis' answer, you were very minimalistic in describing context of your question, what is very bad practice, as much more burden is put on shoulders of responders via the question interpretation and clarification. As I have just now realized, you may, or may not speak about a solution. Yes, the reaction will occur, and ...


5

As I mentioned in the comments, I know of one case where a polyprotic acid has a second proton which is more acidic than the first: the aqueous pervanadyl complex, $\ce{[VO_2(H_2O)_4]^+}$. According to Wikipedia, $$ \begin{array}{rcl} \ce{[VO_2(H_2O)_4]^+ &<=>& H3VO4 + H+ + 2 H2O} & \mathrm{pK_{a_1}} = 4.2\\ \ce{H3VO4\ \ &<=>&...


5

The selective Baeyer-Villiger Oxidation of unsaturated ketones is definitely depends on the conditions used. I do not want to eloborate this fact any further because it was well documented in Waylander's answer. Howevr, so far into the literature suggest that use of peracids in presence of unsaturation in the substrate ketone is not a good idea. The quote "...


5

I like how Ed V has explained the mathematical solution logically. I'd like to show another way of solving the problem. The equation is as written by OP: $$\ce{a XeF4 + b H2O -> c XeO3 + d Xe + e O2 + f HF \tag 1}$$ So atom balance of equation $(1)$ yields: $$\text{Balancing } \ce{F}: \quad 4a = f \tag 2$$ $$\text{Balancing } \ce{H}: \quad 2b = ...


4

Because in the original question there is no elemental chlorine present. It says that you start from a solution of chloride and iodide, so both ions that can be oxidized to the corresponding halogen. And the rest of the answer is already given in your textbook example. Chlorine is a much stronger oxidizing agent. This means that it oxidizes others very well. ...


4

There is very well-written chapter "The chemistry of monovalent copper in aqueous solutions" in Advances in inorganic chemistry, Volume 64 [1, pp. 220–223, DOI: 10.1016/B978-0-12-396462-5.00007-6] which extensively covers as to why copper(I) is unlikely to exist in aqueous solution and why nitrate is a poor ligand for the purpose of preserving monovalent ...


4

Your question is a valid question, and ignore downvotes. They don't mean anything. Your understanding is very good and that you realized that the electrode potential is a property of the electrode and it really does not care how the reaction is written. However, a equation is $needed$ to keep track of the electrons lost or gained in the Nernst equation. So ...


4

You should read the Solution to nascent hydrogen challenge by Juris Meija & Alessandro D’Ulivo Link. The authors vehemently deny the existence of any nascent hydrogen. You will find that the concept of nascent hydrogen and nascent oxygen is still taught in some Indian chemistry textbooks. This term has literally vanished from US & European chemistry ...


4

I am afraid that as stated in another answer there is no simultaneous evolution of oxygen or chlorine during controlled electrolysis. It all depends on the concentration of ions in the solution. Keep in mind that the concentration of water is 55 M, so the ion concentration is nowhere near it no matter what you do! The key reason is that the electrode ...


4

I agreed with explanation given by M. Farooq. However, since OP is really a novice to the field, I'd like to clear some point for OP's benefit. The electrolysis cell has 3 species ($\ce{Na+}$, $\ce{Cl-}$, and $\ce{H2O}$), each of which can be either oxidized or or reduced. $\ce{Na+}$ can only be reduced while $\ce{Cl-}$ can be only oxidized. Water is a ...


4

This question can easily be addressed considering reduction potential of each compound. Concentrated nitric acid is a powerful oxidizing acid, which has a high reduction potential: $$ \begin{align} \ce{NO3- + 4H+ + 3e- &<=> NO + 2H2O} &\quad E^\circ &= \pu{0.957 V} \tag1\\ \ce{NO3- + 3H+ + 2e- &<=> HNO2 + H2O} &\quad E^\circ &...


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