2016年1月24日 · If you move the element around, you will get $\ce{MnO(OH)_2}$ is actually $\ce{H_2MnO_3}$. $\ce{H_2MnO_3}$ is dihydroxy(oxo)manganese which is an acidic substance. For the question regarding the structure for both substances, I'm sorry that I don't haven't the image but I can describe it (Hope that would do).

2021年7月16日 · As for the question, the answer is evident if you apply Le Chatelier's principle (hint: temperature aside, $\ce{MnO(OH)}$ is precipitate). $\endgroup$ – andselisk ♦ Commented Jul 16, 2021 at 22:11

2014年7月24日 · An alkaline battery (or more precise alkaline manganese cell) consists of a zinc anode, a $\gamma\!-\!\ce{MnO2}$ cathode and $\ce{KOH}$ (in paste form) electrolyte.

I want to ask about character of manganese(II) compounds. I read that $\ce{MnO}$ is basic, just like $\ce{Mn(OH)2}$. I started doing research and I read ions like $\ce{MnO2^{2-}}$ and $\ce{HMnO2^-}$ exist too, but how? Are they related to reactions in solid state?

2017年1月20日 · The $\ce{I3-}$ forms the indication complex with the starch, which can then be titrated with thiosulfate solution. When the colour is gone you have successfully converted all of the prepared $\ce{MnO(OH)2}$ back to manganese(II) ions. $$\ce{I3- (aq) + 2 S2O3^2- (aq) -> 3 I- (aq) + S4O6^2- (aq)}\tag6\label{triiodide-reduction}$$

2015年4月25日 · $\begingroup$ This formular $\ce{[Zr4(OH)8 *16 H2O] ^{8+} }$ is a "white gel called zirconium hydroxide but generally considered to be a hydrous oxide" from Nielsen 2005 $\endgroup$ – zs11 Commented May 13, 2022 at 13:17

The data I have been given is Mn(OH)2's pKs=12.7, as well as the standard potential for MnO2(s)/Mn(OH)2(s), which is 1.23V, and the one of O2/H2O, which is also 1.23V; nothing more. Using Gibbs free energy for each half reaction, I have managed to calculate the quilibrium constant for the reaction at 298K, which I found to be 10^-25.4, making ...

2021年5月23日 · $\begingroup$ Oh, and you'd like a real mechanism, or a reaction scheme? The first is most likely unknown. The first is most likely unknown. The second you should try to write yourself. $\endgroup$

2019年1月6日 · The sum of the oxidation states of all of the atoms needs to equal the total charge on the ion. For example, for $\ce{NH4+}$, if each hydrogen atom has an oxidation state of +1, and the overall charge is +1, then we can solve for the oxidation state of nitrogen:

2018年12月27日 · Generally, I add $\ce{OH-}$ ions on the oxygen deficient side and then add water to compensate the hydrogen deficiency on the other side. Note: your first half-reaction balancing the conversion of permanaganate to $\ce{MnO2}$ must be regarded incorrect, as $\ce{MnO2}$ , being soluble in acidic conditions (presence of $\ce{H+}$ , as the equation ...