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Gold pentafluoride has been reacted with elemental fluorine to yield $\ce{AuF_7$}$, which is actually a complex of the pentafluoride with a molecule of $\ce{F2}$. This difluorine ligand takes on a positive partial charge as discussed here.


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Try to remember that, in the $\ce{Zn/Cu}$ cell, like in all other electrochemical devices, whatever the working type (cell or electrolysis), the following sentences may be memorized : the anode is always where oxidation happens (transforming $\ce{M}$ into $\ce{M^{z+}}$ ion), so that the cathode is always where reduction happens. the zinc electrode is ...


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I frequently get confused by the terms cathode and anode when they are used without specifying where they are being used. Electrochemists have to juggle words that are very similar. In an active cell, the electrode dissolves and positive CAT-ions leave the AN-ode and leave it negative so it can push an electronic current (electrons go from - to +) thru an ...


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Here you have two devices producing or absorbing electrons. First there is a Daniell cell, made of two compartments, one made of zinc and one made of copper. If there was no outer voltage source, the cell would work as usual. Zinc would be the negative pole, work as an anode and produce electrons that would be used by copper ions at the cathode to produce a ...


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Your analysis about $\ce{PbCl2}$ in four points is correct. Then ($\ce{A}$ and ($\ce{B}$) are correct, if they are in aqueous solution. The following equation shows that $\ce{SnCl2}$ is a reducing agent that is easily oxidized by air : $$\ce{6 SnCl2 + O2 + H2O -> SnCl4 + 4 Sn(OH)Cl}$$ and, in ($\ce{B}$), $\ce{SnO2}$ is soluble in $\ce{KOH}$ solutions ...


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I already made a comment about some of what I am about to say but I will provide a partial answer. I say partially because I could not find any mechanism for the second product. However, from literature, what I found was that in acidic conditions, $\ce{KMnO4}$ will oxidize naphthalene into the first product, phthalic acid1. In basic/alkaline conditions, $\ce{...


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First iron(II) hydroxide $\ce{Fe(OH)2}$ is a precipitate made in aqueous solution, and it does not react with ammonia. Second, the iron(II) hydroxide $\ce{Fe(OH)2}$ is a green substance which is extremely sensitive to the oxygen of the air. In a couple of minutes, it gets brown, due to the formation of iron(III) hydroxide $\ce{Fe(OH)3}$ according to $$\ce{4 ...


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As @Poutnik has already very well described the intuitive way to solve such problem so I will try to describe a standard method for such problems. First analyse the reaction and find out the species which are being oxidised and reduced by finding the change in their oxidation state. Initially ignore all other ions which are not included in the redox reaction....


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The final arbiter of formal correctness of chemical reaction enumeration are laws of mass and charge conservation. If total counts of charge and atoms of every element are not the same on each side, the equation is wrong. Reaction enumeration by following these laws may be troublesome, as general solution leads to resolving a set of linear equations. That ...


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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}(...


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Oxidation number is a deliberate approximation to make keeping track of electrons in molecules easy You are quite right to say that oxidation number isn't "true". It is clearly a simplified idea and not a perfect representation of "reality" whatever that is. But there is a point to it. That point is to enable chemists to keep track of the ...


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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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$\ce{Zn + 2AgCN -> 2Ag + Zn(CN)2}$ Zinc changes its oxidation state from $0$ to $+2$, and silver changes its oxidation state from $+1$ to $0$. So, zinc is reducing agent and silver is oxidising agent. $\ce{Zn^0 ->[-2e^-] Zn^{+2}}$ (oxidation); $\ce{2Ag^{+1} ->[+2e^-] 2Ag^0}$ (reduction). Other reactions are just equations of ion exchange without ...


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Apparently the discussion does not go much ahead. So I will try to speed it up and show how it goes. First try to establish the half-reaction with ions, not with molecules and formula units. We will go back to neutral compounds later on and get rid of the positive and negative charges at the end. Here, in the first half-reaction, the ion $\ce{MnO4-}$ is ...


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I have no expertise, but Pourbaix does. Let's take a look at some of his diagrams which offer a hint to what is going on, with both silver and gold. Or more accurately, why gold is indeed less vulnerable to attack by hydrogen peroxide than silver. Begin with just the diagram for hydrogen and oxygen in aqueous solution. Ross[1] gives a diagram that ...


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If you are going to do anything differently than is normally done, you should understand what you are going to do and why it should work.(*) The essential (but often not sufficient) conditions for successful electroplating by any metal are: The used metal salts/compounds are soluble, so ions can migrate within the solution in applied electrostatic gradient. ...


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I think that the following method would look parallel to the method mentioned in the above post: $\ce{4H2O + Cu2S + H2SO4 -> 2CuSO4 + 10H+ + 10e^- }$ (oxidation) $5(\ce{Cu2S + 3O2 + H2SO4 + 2H+ + 2e^- ->2CuSO4 + 2H2O})$ (reduction) $\ce{6Cu2S + 15O2 + 6H2SO4 -> 12CuSO4 + 6H2O}$ $\ce{2Cu2S + 5O2 + 2H2SO4 -> 4CuSO4 + 2H2O}$


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