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0

When heated in air, a piece of copper metal gets oxidized into black copper oxide $\ce{CuO}$ according to $$\ce{2 Cu + O2 -> 2 CuO}$$ When hot copper oxide $\ce{CuO}$ is in contact with some ethanol vapor $\ce{C2H5OH}$, it reacts according to : $$\ce{CuO + C2H5OH -> Cu + CH3CHO + H2O}$$ And the copper metal is regenerated. So if a copper plate is ...


-1

The reaction with Cu metal is a catalytic dehydrogenation not a chemical reduction of trace copper oxide: R-CH2-OH = R-CH=O + H2


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You likely had either or both of the following: Copper had some oxide on it, and this oxidized the alcohol leaving the metal behind. If there was air in your system, especially on heating copper could have reacted with it to generate the oxide (a relatively fast reaction with heating) and you're back to (1). The metal would then, of course, be available ...


1

For lighter elements, the shells fill in order. Starting at the transition metals, an outer s orbital may fill before an inner d orbital, so the electron configuration of unioninzed cobalt is written $\ce{[Ar]}4\mathrm s^1\,3\mathrm d^7$, rather than $\ce{[Ar]}3\mathrm d^7\,4\mathrm s^1$. There is a video diagramming the electron configuration of $\ce{Co}$, $...


1

For the first example, you are given that there are $6\, \ce{CN}$ ligands which are -1. That means that there is a total of 6- from the ligands. We also know the overall charge of the complex is -3. That means the charge of the central metal must be +3. In the second example, we can look at the charge again. Because the $\ce{CH3-}$ has a charge of -1, the ...


4

Actually it is diamagnetic but due to a phenomenon called charge transfer spectrum (CTS), oxygen transfers one electron to manganese and $\ce{KMnO4}$ as a whole becomes paramagnetic. This is also the reason why it shows a characteristic colour while diamagnetic complexes are colourless due to absence of d-d transitions.


3

The oxidation state of $\ce{Zn}$ as calculated by you is incorrect. The correct oxidation state of $\ce{Zn}$ is $+2$ as it is associated with two mono negatively charged $\ce{NO3-}$ ions. Thus the unbalanced reaction with correct oxidation states would be: $$\ce{\overset{0}{Zn} + \overset{+1}{H}\overset{+5}{N}\overset{-2}{O_3}\longrightarrow \overset{+2}{Zn}(...


4

Besides steric factors related to the small size of the transition metal core, we could be seeing an electronic effect described in this answer. Iron is fairly early in the transition metal series, so when pushed beyond the $+3$ oxidation state it has few $d$ electrons in the central core. As explained in the referenced answer, this makes the iron strongly ...


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Assuming you meant $\ce{MnO^-4}$ and not $\ce{MnO^{4-}}$, your reduction half reaction is incorrect. Because according to your equation $\ce{Mn^{2+}}$ is being oxidised to $\ce{MnO^-4}$ rather than being reduced. The correct half cell reactions would be: $$ \ce{Zn -> Zn^{2+} + 2e-}\\ \ce{MnO^-4 + 8H+ + 5e- -> Mn^{2+} + 4H2O}\\ $$ And so the overall ...


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