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5

In your eq. (4), the copper/iodate ratio was adjusted to match the stoichiometry present in copper(II) iodate. 1 copper(II) ion to 2 iodate ions equals 2 copper(II) ions to 4 iodate ions.


4

The corrosion of metals like iron is essentially an electrochemical process. In corrosion, a metal is oxidised by loss of electrons to oxygen and formation of oxides. Corrosion of $\ce{Fe}$ (commonly known as rusting) occurs in presence of water and air. The chemistry of corrosion is quite complex, but it may be considered essentially as an electrochemical ...


3

Example 1 Ethylene glycol is often used to make cyclic acetals; its acetals are called ethylene acetals (or ethylene ketals).This interconversion makes acetals attractive as protecting groups to prevent ketones and aldehydes from reacting with strong bases and nucleophiles Example 2 Reference Page 861, Organic Chemistry , Eight edition , L.G.Wade


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The hint is to form the dioxolane (cyclic acetal) of the ketone with ethylene glycol then reduce the ester (LiAlH4 will do this, the dioxolane will not react) and finally deprotect the ketone. The reduction can also be accomplished by hydrolysing the ester and reducing the resulting acid with borane.


3

It can be prepared using hydrogen fluoride, in a double displacement reaction: $\ce{SbCl5 + 5 HF -> SbF5 + 5 HCl}$ However, hydrogen holds onto fluorine more tightly than does antimony in $\ce{SbF5}$. Consider the reverse of your proposed reaction, which is preferred energetically: $\ce{SbF5 + H2 -> SbF3 + 2HF}$


3

This is a nice example: the standard reduction potentials are close, so the Nernst equation ultimately reveals what is favored to happen spontaneously, ignoring all real world complications. Consider a galvanic cell having a tin $(\ce{Sn})$ metal electrode in $x~\pu{M}$ aqueous $\ce{Sn(NO3)2}$ and a lead $(\ce{Pb})$ metal electrode in $y~\pu{M}$ aqueous $\...


3

The reactions are ongoing this way: Relatively free electrons of potassium reduce water: $$\ce{2 e- + 2 H2O -> H2 + 2 OH-}\tag{1}$$ That leaves metal positively charged. Liquid ammonia, if exposed to alkali metal, reacts with electrons much slower than water, forming a dark blue solution of solvated electrons. As electrons progressively kick out ...


2

You guessed it right that hydrogen and oxygen mixtures will remain stable. Someone in Harvard waited for >30 years and found very little water if any. As you said, there is a energy barrier. What type of materials can lower this barrier? Catalysts. The electrode materials require some catalysts such as platinum or nickel. Of course, making cheaper and ...


2

The scheme starts with the production of OH radicals since Hydrogen dissociation is very endothermic (432 kJ/mole) compared to $\ce{H2 + O_2 \to 2OH\cdot}$ which is the initiation reaction with $\Delta H^{\mathrm{o}}_{298} = 72$ kJ/mole. There is a propagation reaction and two chain branching reactions and gas phase termination as well as wall termination ...


2

Reactions between solids are generally slow because diffusion only goes so fast. However, if you produced the two materials in nano size to increase surface area and contact, mixed thoroughly (in absence of air) and then perhaps under pressure (density of $\ce{FeO}$ is 5.745, density of $\ce{Fe2O3}$ is 5.25) with an increase in temperature ($\ce{FeO}$ is ...


2

Whether this qualifies as "without catalysis" is a matter of perspective, but it certainly does happen under ambient conditions. One mechanism of atmospheric rusting involves the $\ce{Fe(III)}$ bearing rust reacting with fresh iron to form $\ce{Fe3O4}$, which then oxidizes in air to form more rust. See Section 1.6 here. We can capture the $\ce{Fe3O4}$ ...


2

The balanced equation seems correct. Multiplying by $2$ we get the following: $$\ce{2 MnO4- + 16 H+ + 10 Cl- → 2 Mn^2+ + 5 Cl2 + 8 H2O}$$ which is also correct.


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The smoky black stuff is probably not tantalum oxide. The only such oxide that's well characterized is $\ce{Ta2O5}$, which is white. Tantalum itself is a rare element, so if the jewelry contains tantalum at all it's probably a thin coating or cladding on a base metal substrate.


1

You can see my detailed response on the sign of electrode potentials here: Still taught to reverse oxidation half cells in electrochemistry? To summarize, electrochemists all over the world have decided to quote all electrode potentials as reduction potentials. This tug of war of signs between the so-called European convention and American convention of ...


1

I'm unsure the level which this question is being asked, but hopefully this answers your question. As you pointed out "why did Oxygen take 2 electrons from magnesium" well if you would allow me to be rather 'hand-wavy' about the whole thing the reason is due to a activation energy vs ionization energy and bond-dissociation energy. If you stick some fresh ...


1

Actually, if you balance given redox reaction strictly following the half-reactions method where you have to equate the number of transferred electrons, you won't end up with fractional coefficients: $$ \begin{align} \ce{\overset{+7}{Mn}O4- + 8 H+ + 5 e- &→ \overset{+2}{Mn}^2+ + 4 H2O} &|\cdot 2 \tag{red}\\ \ce{2 \overset{-1}{Cl}^- &→ \overset{0}...


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Theoretically, yes. Just look at the following redox reaction: $$ \begin{align} \ce{Fe &<=> Fe^2+ + 2e-} &\quad \mathrm{E^\circ} &= \pu{0.447 V} \tag{Oxidation}\\ \ce{Fe^3+ + e- &<=> Fe^2+} &\quad \mathrm{E^\circ} &= \pu{0.771 V} \tag{Reduction}\\ \ce{2Fe^3+ + Fe &<=> 3Fe^2+ } &\quad \mathrm{E_{cell}^\circ} &...


1

Your initial comment exchange made me realize that you have some difficulties in balancing redox equations. Thus, I decided to include some clues for your benefit. An important part of writing half-reactions is making sure they're balanced by mass and charge. Since most redox reactions are done in aqueous medium, you can always balance $\ce{O}$ by $\ce{H2O}$...


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