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The n-factor of a compound in a reaction gives the number of moles of electron lost or gained by it (during oxidation and reduction respectively) per mole of the given compound. In a disproportionation, the compound undergoing both the oxidation and reduction is the same. It is important to understand that the n-factor depends not just on the change in ...


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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 $\...


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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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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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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 ...


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Can you please elaborate on that part? I know that the E value of Mno4- is +1.51, and the one for BrO3- is +1.48. What does that suggest?? that Mno4- will more likely to be reduced by Fe+ than Bro3-?? please explain The first and foremost hint, even before you calculate the electrode potentials is the solubility rule taught in general chemistry. ...


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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 ...


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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 ...


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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 $\ce{Fe3O4}$ ...


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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} &...


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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 ...


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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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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.


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Do you know that the American Chemical Society has listed zinc as an "endangered" element? According to them, within hundred years the supply of zinc will be scarce. The best reducing agent in the world is the cathode of an electrolytic cell. Copper(II) is quite easy to reduce by electricity. Why waste an endangered element?


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We can. But I see few reasons why it is not used: Iron is much cheaper than zinc. There can be remaining residue of iron/zinc, coated by copper, or just being excessive. While copper can be melted away and iron stays, zinc would melt together with copper, causing unwanted impurity (unless wanted for making brass alloys) If we remove copper for ...


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Answer: Both reactions involving Cu+ have the lowest and highest Standard Electrode Potentials respectively. Thus, Cu+ has to be the strongest oxidizing/reducing agents, and therefore it undergoes disproportionation. Because the formation of Cu has a higher SEP than Cu+, it is more likely to get reduced and therefore is a better oxidizing agent - NOT a ...


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I think the answer would be a black hole made up of antimatter (specifically positrons) . Here you have the annihilation energy along with immense gravitational energy and immense electromagnetic energy. The gravitational tidal forces are strong coupled with the fact that the electromagnetic force repels the nucleus and attracts the electrons it is even more....


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The disproportionation happens to elements, not to compounds. In this particular case, to the part of oxygen, $$ \overset{\ce{2 KO2}}{ 4 \times -0.5}= \overset{\ce{H2O2}} {2\times -1} +\overset{\ce{O2}}{ 2\times 0}$$ while the other part keeps the oxidation state constant: $$ \overset{\ce{2 H2O}} {2 \times -2}= \overset{\ce{2 KOH}} {2\times -2}$$ $\...


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You can solve by using the Faradays second law: use the following expression: $$\frac{\text{mass of}~\ce{Ag}}{\text{equivalent mass of}~\ce{Ag}} = \frac{\text{mass of}~\ce{O2}} {\text{equivalent mass of}~\ce{O2}}$$ $$\frac{x}{108} = \frac{\text{mass of}~\ce{O2}} {8 }$$ so :mass of $\ce{O2} =\frac{x}{108}\times{\pu{8 g}}$ and the amount of $\ce{O2} = \frac{x}...


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