New answers tagged redox
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When we say that a metal falls below hydrogen in the electrochemical series, we typically refer to the oxidizing power of hydrogen ion at 1 molar concentration in water solution. The oxidizing power of concentrated sulfuric acid, forming sulfur dioxide and water when it reacts, could be greater than that of the aqueous hydrogen ions; if so, then ...
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One of the best known examples is nitroglycerin, $\ce{C3H5N3O9}$, which has $18$ oxygen atoms in two molecules whereas only $17$ are required to oxidize all the carbon and hydrogen in those molecules. Not surprisingly:
In its undiluted form, nitroglycerin is a contact explosive, with physical shock causing it to explode. If it has not been adequately ...
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As shown in the question, to fully oxidize one carbon and two hydrogen atoms you need three oxygens. Such a molecule, $\ce{HCOOOH}$, exists and it's called performic acid. It is used as a bleach and disinfectant and is explosive in high concentrations.
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Ammonium dichromate $\ce{(NH4)2Cr2O7}$ is well known for being able to burn with its own oxygen. If you dip a match into a crucible containing about $10\text{–}\pu{15 g}$ of ammonium dichromate, it will start to burn softly and throw out sparks like an active volcano, according to the equation $$\ce{(NH4)2Cr2O7 -> N2 + Cr2O3 + 4 H2O}$$ This operation is ...
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Disclaimer: None of this constitutes proof; it's just an informed guess. Any corrections with reputable sources are welcome.
The simplest proposal seems to be single-electron transfer (SET) from $\ce{Cp-}$ to $\ce{Fe^3+}$. This forms the neutral cyclopentadienyl radical as well as $\ce{Fe^2+}$. This $\ce{Fe^2+}$ then reacts with any remaining unreduced $\ce{...
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As suggested by Poutnik in the comments I considered $\ce{O2-}$ in the reduction half reaction which then became:
$\ce{2OH- + 2e- -> H2 + 2O^2-}$
After multiplying each half-reaction by the number of electrons exchanged by the other one and adding back the spectator ions the balanced equation becomes:
$\ce{14NaOH + 2B -> 2NaBO2 + 3H2 + 6Na2O + 4H2O}$
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Wet-oxidation of pyrite lead to iron(II) sulfate.
$$\ce{2FeS2 + 2H2O + 7O2 ->[100 °C] 2FeSO4 + 2H2SO4}$$
iron(II) sulfate formed is eventually further oxidized to iron(III) sulfate, iron(III) oxide and iron(III) hydroxide. Note that the reaction is not that simple as it proceeds through many Fe(II) and Fe(III) intermediates:
There are different proposed ...
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Typically, roasting sulfides leads to the generation of sulfur dioxide and metal oxide. In the case of pyrite, this may be described by
$$\ce{4 FeS2 + 10 O2 -> 2 Fe2O3 + 8 SO2 }$$
The fire is needed to initiate the reaction. If well maintained, the heat generated by roasting may sustain the reaction.
However, and especially for marcasite as an other ...
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Those are not Coupled reactions; the burning of carbon simply supplies the energy to decompose the carbonate. Another way of looking at it would be if you were the fireman on an old steam locomotive would you prefer to shovel pure coal or a mix of coal and limestone into the firebox for several hours trying to keep the train on time. The reactions are ...
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In the event that this question is actually more than just a theoretical exercise, the general electrolysis half-reactions, as cited by Maurice in my opinion, may not actually be in accord with this particular's complex reaction system per a review of the possible underlying mechanics.
More precisely, here is a suggested overview of reaction mechanics that ...
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I think to understand heterogenious hydrogenation you need to understand homogenious hydrogenation of metals by Wilkinson's cataylst and hydroformylation of alkenes by a mixture of hydrogen and carbon monoxide using a cobalt carbonyl cataylst. If we consider for a moment the reaction of an acetylene with a complex such as RuHCl(CO)(PPh3)2. Then what will ...
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Forget about the equivalent weights. Work with moles, and only with moles. Look how it goes.
$102$ g $\ce{H2O2}$ is $\ce{\frac{102 g}{34 g/mol} = 3.00 mol H2O2}$. The production of $1$ mol $\ce{H2O2}$ requires $1$ mol $\ce{(NH4)2S2O8}$. Now we will show that the production of $1$ mol $\ce{(NH4)2S2O8}$ requires $2$ moles electrons. Before doing this, we will ...
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You can find examples where your thoughts fit what is actually happening. For example, a reaction of magnesium oxide with barium (I don't know if this reaction would actually happen):
$$\ce{MgO(s) + Ba(s) -> Mg(s) + BaO(s)}$$
Here, it is correct to speak of the negative and positive charges because we are looking at ionic compounds. You could show the ...
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An easier approach would be to think about redox-reactions as an exchange of electrons, rather than of oxygen. This is more general and equally allows oxygen to be oxidized (e.g., fluorine is even more electronegative than oxygen, as in oxygen fluorides).
Retain that an oxidation is the removal of electrons, and reduction is the addition of electrons. The «...
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Your explanation reminds me of the historical definition of the words "oxidation" and "reduction". In the early 19th century, oxidation was the reaction an element with oxygen. The only thing that the chemist could measure with precision was the weight, the mass, before and after the reaction. When iron or copper gets oxidized, it ...
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