New answers tagged

-1

$$\ce{4 KClO3 -> 3 KClO4 + KCl}$$


1

Sometimes the best chemistry, is to simply PREVENT the chemistry of rust in the first place. So in addition to the already well covered chemistry issues above, I think it's very important, that before coating the underlying surface, that that surface is both smooth and solid. Old guys refer to it being 'sound'. So I've found that one important reason to ...


-2

Let's discuss the problem in a qualitative way. In such a cell, the metal will get dissolved in the right-hand side of the cell, where the concentration is low, so that $[M^{z+}]$ increases in this compartment, which becomes the anode. In the left-hand side, the ions $[M^{z+}]$ are discharged and deposited as a metal layer on the electrode, which is the ...


0

For us to understand the redox nature of the above examples, we have to use the definition of redox as gain or loss of electron density, not just electrons. For instance, in the reaction between sodium hydroxide and hydrochloric acid to give sodium chloride and water, chlorine is reduced by gaining electron density in its bond with sodium compared to its ...


2

It is a very good questions from a student who seems to be thinking deeply. I have been thinking to write on this topic for the last 7-8 years. In short the poles of the batteries are electrostatically charged and one can actually "sense" this charge with the help of sensitive electroscopes which were known in the time of Volta. The label anode or cathode ...


-1

An anode is the place where oxidation occurs. In a galvanic cell, like the Daniell cell ($Zn - Cu$), the $Zn$ is oxidized at the Minus pole. $Zn$ is oxidized into $Zn^{2+}$ ion and the electrons are sent in the electric connexion going to the $Cu$ electrode. $Zn$ is the anode (negative sign) and $Cu$ is the cathode (positive sign) If the same setup is used ...


-1

Actually, some have voiced an opinion that FeI3 can apparently be created with difficulty, and is reputedly very unstable, decomposing into FeI2 and I2. Similarly, with CuI2, which is likewise unstable. To quote a source reference. $$ FeCl3 + 3KI = FeI3 + 3KCl ; FeI3 = FeI2 + I2 $$ Observation- Large anions diminish the lattice energy of the ...


0

When an atom $M$ looses one or more electrons, it becomes positively charged and becomes the ion ${M^{z+}}$. In aqueous solution this positive charge attracts the unused doublets of the Oxygen atom from $H_2O$, and it repells the proton of $H^+$. If $Z$ = $1$, the ion $M^+$ attracts several molecules water around it, and it makes a Debye layer of oriented $...


2

If an enzyme is called $\mathrm{NAD(P)H}$-dependent, that means it is using both $\mathrm{NAD^+/NADH}$ and $\mathrm{NADP^+/NADPH}$ as its cofactors. For example, most oxidoreductases are $\mathrm{NAD(P)H}$-dependent, which are able to oxidize a relevant substrate by transferring a hydride ($\ce{H−}$) group to a nicotinamide adenine dinucleotide cofactor (...


1

If you want to understand the reaction of $NaOH$ on Aluminium, you may describe it this way. First, the aluminium reacts spontaneously with water, like sodium or calcium, producing gaseous Hydrogen $H_2$ and an metallic oxide or hydroxide. With sodium or calcium, the hydroxide produced by this reaction quits the metal surface, and get dissolved, leaving a ...


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,...


0

One path I accept is as proceeding via the created presence of HOI from the interaction of iodine and SO2 with water in a series of reactions: $\ce{I2 + H2O = H+ + I- + HOI}$ $\ce{SO2 + H2O = H+ + HSO3-}$ $\ce{HOI + HSO3- -> H+ + I- + HSO4-}$ However, the reaction could also proceed to some extent via the following reactions following a ...


2

Firstly, I would replace the word "feasible" with "favorable". I'm also going to replace $E_{cell}$ with $E_{OCV}$, where the open circuit voltage (OCV) is the potential of the cell without any applied electric potential. So you're right in that if the $E_{OCV}$ is positive, the net reaction is thermodynamically favorable: it will occur spontaneously if ...


0

Acidity as $\mathrm{pH}$ does affect many redox potentials. But the compound acidic character itself has no direct relation to reducing power. E.g. superacids like $\ce{HSO3F}$ or $\ce{H2F+ + SbF6-}$ have practically no reducing power. The former probably just against free fluorine or fluorinating compounds like $\ce{ClF3}$, the latter none at all.


0

No. Acidic character is not related to reducing power. For example, nitric acid $\ce{HNO3}$ has no reducing properties. It is a strong acid and a strong oxidant, able to carry out the oxidation of copper metal $\ce{Cu}$ to $\ce{Cu^{2+}}$ ion.


0

An Acidic compound donates protons (Lowry Brönsted definition) $$\ce{HI -> H+ + I-}$$ By releasing $\ce{H+}$, you need to note that the compound as a whole releases the proton. A reducing/oxidizing agent donates/accepts electrons into it's own shells, differentiating it from an acid. $$\ce{2HI + 2OH- -> 2H2O + I2 + 2e-}$$ Here, $\ce{I-}$ donates two ...


0

Ionic solids like NaCl, KOH, KBr, PbCl2 and KBrO3 have ratios of individual elements, but the "molecule", if you could use that term to describe it, would be the whole crystal, because all the ions are attracted to nearby ions, ad infinitum. The white solid could be written as "K23.4% Br47.8% O28.8%", and every time this experiment is run, the same product ...


3

The main mistake is rather algebraic and forgetting the electrochemical conventions. When we write this equation, we follow the convention that all electrode potential values will be inserted right from the table with their signs intact, whether it is an oxidation or reduction. Don't change the sign of the electrode potential value. The sign of the electrode ...


3

It has been a while since I took electrochemistry so I'm not sure about the formal procedure that you're using. You wrote the correct overall reaction. $$\ce{2H3O+ + Fe -> H2 + Fe^{2+} +2H2O}\tag{1}$$ and you almost wrote the correct half cells reactions (should be $\ce{Fe^{2+}}$ in equation 3): $$reduction: \ \ \ce{2H3O+ + 2e− ->H2}\tag{2}$$ $$...


-4

Your mistake is that $E^\circ_\ce{Fe/Fe^{2+}} = \pu{-0.44 V}.$ So that $-E^\circ_\ce{Fe/Fe^{2+}} = \pu{+0.44 V}.$


1

In electrolysis, $Cu^{2+}$ is reduced in two steps, first to $Cu^+$ and later on to $ Cu$. Of course $Cu^+$ is automatically disproportioned into $Cu$ and $Cu^{2+}$ in acidic conditions. But in neutral solutions, $Cu^+$ may react with $H_2O$, forming an unwanted red precipitate of $Cu_2O$. This will affect electroplating of copper.


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