# Tag Info

## Hot answers tagged redox

4

Boron mostly favours the +3 oxidation state. However, boron does form some compounds in +1 O.S. but they are unstable and rearranges to stable molecule (such as boron monofluoride in @Oscar's answer which polymerizes by itself to compounds containing 10-12 boron atoms). Apart from the monofluoride, I found one example of boron(I) complex which is a radical ...

3

We may not have all boron atoms equivalently bonded to carbon in $\ce{B4C}$, so we might not be able to assign all boron atoms an equal oxidation state of +1. For that matter, assuming a carbon oxidation state of -4 might also be a reach. A better candidate for boron(I) is boron monofluoride, $\ce{BF}$, which is unstable but isolable: Boron monofluoride can ...

3

I use dilute hydrochloric acid to keep my galinstan looking like mercury. The oxide doesn't just increase wetting it its the sole cause of it. You need at least 0.1 M HCl to get maximum surface tension and to keep the surface shiny. Paper on it 0.1 M is about 0.37% concentration. Any concentration of HCl under 5% is "safe" to handle and only over ...

3

Why does mercury in dilute HNO3 give mercurous nitrate while hot concentrated HNO3 produces mercuric nitrate? “Mercury dissolves in oxidizing acids, producing either Hg${^{2+}}$ or Hg$_2$$^{2+}$, depending on which reagent (mercury or e.g., nitric acid) is in excess.” Ref 1 Wikipedia says “Mercuric nitrate can be reacted with elemental mercury to form ...

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Add coin into nitric acid and sulfuric acid. Nitric should be 60% weight of the coin and sulfuric should be 1.6 times. It will dissolve in minutes. Heat it up to boil unused nitric. Add some urea to neutralize acids. Add iron fillings to remove copper. Filter and melt the copper powder. You will have pure copper. Take rest of the solution, dilute it a bit, ...

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You separate the reduction and oxidation half-equations, and then there is no issue with the oxidation state of oxygen on either side. For the oxidation half-equation you start with (in acid solution) $\ce{H2O2 -> O2}$ Clearly the hydrogen on the left is given off as $\ce{H^+(aq)}$ and the two oxygen atoms, being oxidized from -1 to 0 each, lose two ...

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Essentially, it's because nitric acid is a strong acid. At concentrations where we generally perform aqueous chemistry, it is essentially completely dissociated to give $\ce{H^+(aq)}$ and $\ce{NO3^-}$. So the nitrogen-bearing species that does the oxidizing is primarily the dissociated ion $\ce{NO3^-}$. As you balance the reaction you will also see the \$\ce{...

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Balancing a reaction equation is a sequential process. One approach is that you identify the starting materials by their formulae, and the reaction products by theirs. you balance the equation such the same number of atoms on the left hand side equates the number of atoms on the left hand side. Depending on the reaction, there are some groups of atoms, or ...

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