[777888]

When you write chemical equation / ionic equations, do you put the acid or base (use of manipulating the pH of the solution) in the chemical equation?

H2SO4 (corrosive acid) is added to KMnO4 solution. And then, NaHSO3 is added! Mn 2+ are present after the reaction. WRITE a balanced equation (ignore the Na+ spectator ions) IDENTIFY the substance reduced and oxidized.
Notes: HSO3 - is always oxidized to SO4 2-

How can I write the reactants and products? Should I wirte an ionic equation or non-ionic?

Is this right?
MnO4 - + HSO3 - -> Mn 2+ + SO42-
This seems like an ionic equation! Is this ok?

Besides, is the ion colour of MnO4 2- green?

Thanks!

[AWK]

All reactants are present. Balance them!
NaHSO3 + KMnO4 + H2SO4 = Na2SO4 + K2SO4 + MnSO4 + H2O
HSO4(-) + MnO4(-) + H(+) = SO4(2-) + Mn(2+) + H2O

Besides, it is green.

[kevins]

Please read the following first

Oxidants
Potassium permanganate (acid): MnO4 - +8H+ +5e -> Mn2+ + 4H2O
Potassium permanganate (neutral):MnO4 - +2H2O +3e -> MnO2 + 4OH-
Potassium permanganate (strongly alkaline):MnO4 - + e -> MnO4 2-
Cerium (IV) sulfate: Ce 4+  + e -> Ce 3+
Potassium dichromate: Cr2O7 2- + 14H + 6e -> 2Cr 3+ + 7H2O
Chlorine: Cl2 + 2e -> 2Cl-
Bromine: Br2 + 2e -> 2Br-
Iron (III) : Fe3+ +e -> Fe2+
Potassium Bromate: BrO3- +6H+ +6e -> Br- + 3H2O
Potassium Iodate:IO3- + 6H+ + 6e -> I- + 3H2O
sodium hyprochlorite: ClO- + H2O + 2e -> Cl- + 2OH-
Hydrogen peroxide: H2O2 + 2H+  +2e -> 2H2O
Nitric acid (conc): NO3- +2H+ +e -> NO2 + H2O
Nitric acid (dilute): NO3- + 4H+ +3e -> NO + 2H2O

Reductants
Hydrogen:  H2 -> 2H+ + 2e
Zinc : Zn -> Zn2+ + 2e
Hydrogen iodide: 2HI -> I2 + 2H+ + 2e
Oxalic acid: C2O4 2- -> 2CO2 +2e
Iron (II) sulfate: Fe2+ -> Fe3+ + e
Sulfurous acid: H2SO3 + H2O -> SO4 2- + 4H+ +2e
sodium thiosulfate: 2S2O3 2- -> S4O6 2- +2e
Hydrogen peroxide: H2O2 -> 2H+ + O2 + 2e


As per your question, it is very easy to find that the oxidant is acid potassium permanganate and the reductant is HSO3- ( this will generate a sulfurous acid).

Can you got it!
 

[kevins]

the MnO4- is purple in colour and the Mn2+ is colourless.

[777888]

Thanks for helping!

For this:
NaHSO3 + KMnO4 + H2SO4 = Na2SO4 + K2SO4 + MnSO4 + H2O

How do you know the products and what type of reaction is this? (double displacement?)

The question says ingnore spectator ions!
So should I write an skeleton equation (ionic quation)? (excluding K, Na?)

What's the different between the first and second equation?
NaHSO3 + KMnO4 + H2SO4 = Na2SO4 + K2SO4 + MnSO4 + H2O
HSO4(-) + MnO4(-) + H(+) = SO4(2-) + Mn(2+) + H2O

[777888]

When the question asks for SUBSTANCE REDUCED and SUBSTANCE OXIDIZED, does it mean the whole compound or only the ion?

For example, in my first question, will the substance reduced be Mn in MnO4(-) or KMnO4?

Thank you!

[AWK]

To Kevin:
The question was: Besides, is the ion colour of MnO4 2- green?
Your answer is on other question
the MnO4- is purple in colour and the Mn2+ is colourless.

To 777888
NaHSO3 + KMnO4 + H2SO4 = Na2SO4 + K2SO4 + MnSO4 + H2O
This is unbalanced redox reaction. It can be write down in molecular form or ionic form, as I showed previously
NaHSO3 + KMnO4 + H2SO4 = Na2SO4 + K2SO4 + MnSO4 + H2O
HSO4(-) + MnO4(-) + H(+) = SO4(2-) + Mn(2+) + H2O
Kevin pointed out that KMnO4 or MnO4(-) is an oxidiser and NaHSO3 or HSO3(-) is a reducer
During reaction Mn(+7) is reduced to Mn(+2) and S(+4) is oxidised to S(+6)