New answers tagged redox
2
Copper(III) nitrate cannot be obtained from aqueous nitric acid, and likely doesn't exist.
Reaction carried under strongly oxidative conditions between $\ce{Cu(NO3)2}$ and fuming $\ce{HNO3}$ yields in nitrosyl copper(II) trinitrate $\ce{[NO+][Cu(NO3)3−]},$ [1, 2] sometimes written as adduct $\ce{Cu(NO3)2 · N2O4},$ which is contradictory to the crystal ...
1
Say we have a given redox reaction:
$$\ce{MnO4- + Fe^2+ -> Fe^3+ + Products}$$
This is an unbalanced redox reaction, just depicting how $\ce{Fe^2+}$ is oxidized to $\ce{Fe^3+}$ by $\ce{KMnO4}$. Now, depending on the medium, the following reactions may take place:
In an acidic medium, $\ce{MnO4-}$ is reduced to $\ce{Mn^2+}$ by accepting 5 electrons
$$\ce{...
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In acidic medium the equation must mention somewhere that some $\ce{H+}$ ions appear somewhere in the equation. In basic medium, the equation must mention somewhere that some $\ce{OH-}$ ions appear somewhere in the equation,
For example, you may say that permanganate ion reacts in acidic conditions to produce $\ce{Mn^2+}$. In basic conditions, it could not ...
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If you consider the oxidation states of chlorine and oxygen, oxygen is REDUCING the chlorine as it goes from +5 to -1 (oxygen in turn is OXIDISED from -2 to 0). This makes no sense as O is more electronegative than Cl and I can only assume that since the reaction requires heat, this energy is driving a non-spontaneous reaction.
-1
The H atom is neither oxidized nor reduced in the redox semi-equation. It remains at +1. Only Manganese reacts, as it is reduced from +7 to +2. The redox couple should be written : MnO4-/Mn2+, without mentioning Hydrogen
And MnO4 does not lose electrons, as you say. On the contrary, MnO4 fixes or consumes 5 electrons in its reduction to Mn2+.
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The author has written it wrong because you cannot have random redox pairs such as $\ce{H+/MnO4^{-}}$. The copper, zinc notation is fine. He should have written $\ce{MnO4^{-}/Mn^{2+}}$ rather for this pair. He is simply following the electrochemical conventions of half cells which shows every redox process as a reduction equation. You can have a quick look ...
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