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7

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 ...


7

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 ...


6

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 ...


4

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....


3

The usual way of solving such a problem is to write two half-equations, and make the sum of them, after multiplying each one by a factor so as to get the same number of electrons. Each equation is about one of the atoms that changes its oxidation number. Let's start from the first such atom, iodine in the ion iodate $\ce{IO3-}$. We will build its own half-...


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}(...


1

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 ...


1

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 ...


1

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 ...


1

The anion $\ce{AuCl4−}$ is the most known complex of gold(III). It is very stable complex and have very high formation constant ($K_f$). That is the reason the metal chloride complex have a lower reduction potential than metal aqua ion. As pointed out in the other answer, this is a consequence of Nernst's law. Let's look at the two redox equations we ...


1

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. ...


1

I think that the half-reaction method for this reaction would look something like this: $\ce{Cu2S + O2 + H2SO4 -> CuSO4 + H2O}$ $\ce{2Cu^+ -2e^- -> 2Cu^2+}$ (oxidation) $\ce{S^2- + 4H2O - 8e^- -> SO4^- + 8H^+}$ (oxidation) $\ce{O2 + 4H^+ +4e^- -> 2H2O}$ (reduction) $\ce{2Cu2S + 5O2 + 2H2SO4 -> 4CuSO4 + 2H2O}$


1

No, adding acid to a ferrous sulfate solution will not prevent its being oxidized by bromine water. Fresh FeSO$_4$ solutions are light blue-green, almost colorless, but pick up O$_2$ readily: Fe$^3$$^+$ + e$^-$ --> Fe$^2$$^+$ +0.770 V and O$_2$ + 2 H$_2$O + 4e$^-$ --> 4 OH$^-$ +0.401 V (The half cell voltages are only indicative of what's ...


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