Why does sulfur dioxide decolourise acidified potassium manganate (VII) solution?

When SO2 gas is passed through it acts as a reducing agent & it decolorizes purple colored to brown MnO2

2KMnO4→→K2O+ 2MnO2+3[O]

SO2+[O]+H2O→H2SO4

thus

2KMnO4+3SO2+ 3H2O→→K2O+3MnO2+3H2O.

‘Acidified potassium tetraoxo manganate (VII) is more commonly known as potassium permanganate (KMnO4). Under acidic or basic conditions, this compound oxidizes primary alcohols (like ethanol, CH3-CH2-OH) to carboxylic acids. So, for the oxidation of ethanol with potassium permanganate, the product is ethanoic acid (aka, acetic acid, CH3-COOH) and the chemical reaction is:

CH3-CH2-OH + KMnO4 → CH3-COOH + MnO2

You will watch iron reacting with hydrochloric acid and beautiful yellow sulfur floating on top of it! :-D

The only reaction possible is:

Fe + 2 HCl → H2 + FeCl2

The elemental sulfur will not react in aqueous environment!

The formula for Sulphur dioxide is:
SO2.
An easy way to do this for yourself, is by knowing the abbreviations of the periodic table , and the index numbers.
Periodic abbreviations are easily found on google, wiki, etc.
The index numbers are like this:
1. Mono
2. Di
3. Tri
4. Tetra
5. Penta
6. Hexa
7. Septa
8. Octo.
You can easily find these online.

I'll give you an example:

What is the formula for Titanium Dioxide?
Titanium=Ti
Oxygen= O
Di=2
Hence: Ti 2 O , so TiO2

Hope this helped.

Well, sulfur dioxide is LIKELY oxidized to $SO_{4}^{2-}$ by potassium permanganate, i.e. $K^{+}MnO_{4}^{-}$, the which is a $Mn(VII+)$ species…

Permanganate is REDUCED to COLOURLESS $Mn^{2+}$ ion … (because this is a $d^{5}$ ion, its electronic transitions are spin forbidden in the reduced metal ion…)

$underbrace{MnO_{4}^{-}+8H^{+} + 5e^{-} longrightarrow Mn^{2+} + 4H_{2}O(l)}_{ ext{purple to colourless}}$

And the oxidation half equation…

$SO_{2}(aq)+2H_{2}O(l) longrightarrow SO_{4}^{2-} + 4H^{+} + 2e^{-}$

And we add TWO of the reduction half-equation to FIVE of the oxidation half-equation…

$2MnO_{4}^{-}+16H^{+} + 10e^{-} +5SO_{2}(aq)+10H_{2}O(l) longrightarrow 2Mn^{2+} + 8H_{2}O(l)+5SO_{4}^{2-} + 20H^{+} + 10e^{-}$

And cancel away…

$underbrace{2MnO_{4}^{-} +5SO_{2}(aq)+2H_{2}O(l) longrightarrow 2Mn^{2+} +5SO_{4}^{2-} + 4H^{+}}_{ ext{purple to colourless}}$

It does not.

When sulfur dioxide gas (SO₂) is passed through aqueous potassium manganate (VII), the familiar KMnO₄, the deep purple colour of this solution turns into pale pink because a redox reaction goes on:

• KMnO₄, the oxidiser, oxidises SO₂ to SO₃;
• SO₂, the reducer, reduces the MnO₄¯ ion to Mn²⁺. This is faint pink, almost colourless in aqueous solution.

The complete reaction is:

2KMnO₄ + 5SO₂ → K₂SO₄ + 2MnSO₄ + 2SO₃

Since SO₃ is an acidic gas, as soon as it forms in water it reacts to yield dilute sulfuric acid, H₂SO₄.

KMnO4 is an oxidizing agent and an alkene like cyclohexene has an electron-rich pi cloud, which is to say it is a reducing agent. In the resulting redox reaction, the lovely violet KMnO4 is reduced to brown, insoluble MnO2. The cyclohexene will be oxidized to cis-1,2-cyclohexanediol (and sometimes further, depending on reaction conditions.)

When sulphur di-oxide is dissolved in water it produces sulphurous acid( H2SO3). So, the solution becomes acidic.

SO2 + H2O → H2SO3

Hope this helps.

This reaction is taken as an experimental verification for the presence of sulphur dioxide gas (SO2).

The equation for the reaction may be stated as follows:-

K2Cr2O7 + H2SO4 + 3SO2 ——— K2SO4 + Cr2(SO4)3 + H2O.

The orange-coloured dichromate solution will turn green due to the formation of chromium(III) sulphate, Cr2(SO4)3.

It is a redox reaction. Here:-

The oxidising agent is potassium dichromate and the reducing agent is sulphur dioxide.

Hope it helps!!!!

If paper impregnated with potassium dichromate solution turns green, sulfur dioxide may be present (it reduces chromium(VI) to chromium(III)).

Of course, this is a rather unsatisfactory test because any sufficiently powerful reducing agent will bring about the same reaction. A better one is to pass the gas sample through H2O2 (or add H2O2 to the solution, whatever). The following redox reaction takes place:

SO2 + H2O2 -----> H2SO4

causing a large and easily detectable drop in pH.

If the reaction goes to completion and all the reactants are used up, then it will be the same result as a strong acid/strong base reaction, producing a salt and water in its entirety. Because all of the strong acid will have neutralized the weak base, there will not be any of the weak base left in solution. All of it will be converted to salt. This is why titrations with weak base and strong acids are effective, because all of the weak base is reacted with the strong acid.

As with strong acid/strong base reactions, an excess of either will fully react the limiting reagent, producing a salt, water, and leaving some of the excess reagent left over. The only difference with weak acids and strong acids is that the pH of a weak acid will be closer to 7 than the pH of a strong acid of equal concentration (the same applies to strong/weak bases because they both do not fully dissociate).