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1

I'm going to be a literalist, and then an extremist, and then a speculator. If you put a silver wire into silver nitrate (which is literally a solid, mp = 210 C), no visible reaction will occur. However, silver nitrate decomposes when heated: 2 AgNO3(l) → 2 Ag(s) + O2(g) + 2 NO2(g) Qualitatively, decomposition is negligible below the 210 C melting point, ...


3

Between the solution of silver ions and the silver atoms on the electrode there is ongoing exchange of silver ions for silver atoms with a zero net change. Since Ag mined from the earth has no radioactive isotopes, it would possible to detect the change by using a radioisotope of silver say $\ce{^{108\mathrm{m}}Ag}$ in the solution. After immersing the Ag ...


2

Short answer: No Long answer: Honestly, there's really no good way to tell. Think about it this way: the only plausible redox reaction between the Silver(I) cations and the Silver metal is the exchange of electrons on the surface of the metal. And if this happened, the $\ce{Ag(s)}$ would donate one of it's valence electrons to $\ce{Ag+(aq)}$, resulting in ...


3

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 ...


0

A is Zinc or Iron. C is Copper. B is Silver. I don't know what is D. The reason is that when dipped in Cu2+ or Ag+ solution, a piece of iron or zinc reacts with both and produces a metallic deposit of Cu or Ag. A metallic strip of silver never reacts with Zn2+ or Cu2+. A metallic strip of Copper reacts with a Ag+ solution, producing a nice silver deposit. ...


3

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): $$\ce{A(NO3)2(aq)}$$ ...


0

In addition to price advantage, CuSO4 also provides a source of sulfate ions, which are relatively stable. The sulfate anion is also a known scavenger of radicals that may be present in the solution. For example, a reaction between sulfate and the hydroxyl radical (a possible electrolysis product): $\ce{SO4^{2-} + .OH -> .SO4^- + OH^-}$ And, the ...


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 ...


1

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 ...


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 ...


0

Another way of describing the electric behavior of the electrode is to state that when dipped in water, one Zn atom quickly looses two electrons and gets dissolved in the solution. The electron remains in the metal and prevents any second Zn atom from bringing more electrons in the metal. Simultaneously the solution gets positively charged and the first $\ce{...


0

If you dip a Copper plate and a Zinc plate into a salt solution, you create a galvanic cell, where Zinc is oxidized and is the negative pole. If you want to apply an external tension from another cell, the result depends on the choice of the positive and the negative poles. If you connect positive poles to the negative poles, both cells produce courant, and ...


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 ...


0

I found a 2019 source work on Chromium peroxide. To quote the key preparation equation: Chromium peroxide (Cr(O2)2.H2O or briefly CrO5) is an extremely potent oxidant. This compound is a product of the following reaction: $\ce{(NH4)2Cr2O7 + 4 H2O2 + 2 H+ -> 2 Cr(O2)2.H2O + 4 H2O + ammonium salt}$ (1) Here are associated corresponding comments ...


0

Without taking oxidation numbers into account, I have first balanced the Cr atoms, then the H atoms, and finally the Oxygen atoms. And I have found :$$\ce{4CrO_5 + 12 H^+ -> 4 Cr^{3+} + 7 O_2 + 6 H_2O}$$ With the sulfate ions it gives : $$\ce{6 H_2SO_4 + 4 CrO_5 -> 2 Cr_2(SO_4)_3 + 7 O_2 + 6 H_2O}$$


0

IUPAC defines the electrode potential of an electrode to be the EMF of a cell in which the electrode in question is on the right hand side of the line representation of the cell and a SHE is on the left hand side. As such, the electrode potential of the $M^+/M$ couple is $E^o_{M^+/M} = \phi_{M} - \phi_{R}$ where $R$ refers to the reference electrode, in ...


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 ...


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 ...


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 ...


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 ...


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 ...


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 ...


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