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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 ...


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 ...


3

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 ...


3

As noted in the comments, $\ce{Cu+}$ is not stable in aqueous solution. It tends to disproportionate to $\ce{Cu^0}$ and $\ce{Cu^{2+}}$. You can therefore assume that all but a negligible amount turned into $\ce{Cu^{2+}}$. Source: Holleman/Wiberg: Lehrbuch der Anorganischen Chemie, de Gruyter, 101. edition, 1995.


3

The final solution contains $Zn^{2+}$ ions, which reacts with $H_2S$ to produce white $ZnS$. But you may not avoid that maybe some of the initial $Cu^{2+}$ ions still remains in the solution. And these rare $Cu^{2+}$ ions do react with $H_2S$ to produce Copper sulphide $CuS$ which is dark black. $$\ce{Cu^{2+} + H_2S -> CuS + 2 H^+}$$So the obtained ...


3

Oxalic acid is a relatively strong acid for a carboxylic acid, and according to my sources below, can auto-catalyze the reaction with Potassium Permanganate. So, the reaction you performed was likely just the same mechanism that you have seen everywhere else. My source is as follows: Kovacs K.A.; Grof P.; Burai L.; Riedel M. (2004). "Revising the mechanism ...


2

No, electrodes do not know about each other. When an electrode is inserted to an electrolyte, the electrochemical reaction is ongoing in both directions. If reduction direction overruns oxidation, the potential of the electrode is increasing ( or vice versa ) until the rate of both reaction gets equal, the net reaction rate is zero and the electrode reaches ...


2

This is a rather unusual case of what is discussed in answers like this one, where we circumvent problems with multiple atoms being oxidized or reduced by considering whole compounds as oxidizing or reducing agents. Here, the whole-compound redox-active material is $\ce{CrO5}$, and as in peroxide disproportionations generally this is both an oxidizing agent ...


1

Can I write $\ce{X + 2 e- -> X^2-}?$ Why can't I start with $\ce{X}$ instead of $\ce{X^2+}?$ The question says that the metals are $\ce{A(s), B(s), C(s) and D(s)}$. That means the letters stand for the uncharged element, not for something more arbitrary. In the table, the letters are used again, in the following way (shown for A): $$\ce{A(NO3)2(aq)}$$ ...


1

If I were not a chemist, and wanted blue, I would suspect cobalt first, but after 4 years, I might suspect copper, but not just Cu or CuO or Cu2O. Copper chlorides can be blue, but maybe not in a hot (or cold) glaze. I'm not a potter, but I wonder if you can use the idea behind the borax bead test (https://en.wikipedia.org/wiki/Bead_test) to get your ...


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One candidate is manganese dioxide, $\ce{MnO2}$. This decomposes at 535°C, releasing oxygen. As the phase diagram from Ref. [1] shows below, the reduced manganese fuses with excess silica between 1250 and 1300°C, forming rhodonite ($\ce{MnSiO3}$). You would see this as a pink color. But, of course, I'm a chemist rather than a potter, so I make no ...


1

I don't know why your textbook author is confusing and still teaching American & European conventions. It is obsolete now. I will show you how oxidation potentials were quoted by American electrochemists in the 1950s-60s. Have a look at the table Latimer's book: Oxidation States of the Elements and their Potentials in Aqueous Solutions pg 340. This was a ...


1

You are seeing the difference between a metal that can be precipitated by hydrogen sulfide in acidic solution and one that requires the sulfide bearing solution to be alkaline or at least not as acidic. In my student days (1970s-1980s) they were called Group II and Group III metals in qualitative analysis, although that may have changed. Both copper and ...


1

The two organic compounds are ultimately being converted to carbon dioxide, $\ce{CO2}$, and water, $\ce{H2O}$, just as they would be in complete combustion with pure oxygen gas, $\ce{O2}$, in a bomb calorimeter. Thus the two balanced combustion equations would be as follows: $$\ce{C29H24O12 + 29 O2 -> 29 CO2 + 12 H2O} \tag{1}$$ $$\ce{C22H18O11 + 21 O2 -&...


1

It is really surprising that your molecules are transformed to $\ce{CO2}$. But if it is the case, if all your theaflavin is oxidized into $\ce{CO2}$, the half-equation is : $$\ce{C29H24O12 + 46 H2O -> 29 CO2 + 116 H+ + 116 e-}$$ The other half-equation depends on the pH of the solution. In highly acidic solution, it produces $\ce{Mn^2+}$. In nearly ...


1

I think Bard's is more general than Newman's, Newman's assumes Cx=Cx* where Bard does not. Bard says the current is proportional to the forward rate, and if there are metal ions near the electrode, the forward reaction removes electrons from the electrode to reduce the ions. That means the current flows into the electrode. Bard then says the applied ...


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