oxalic acid can be oxidised into carbon dioxide by acidified potassium permanganate 6.3 grams of oxalic acid cannot be oxidised by
a)3.16g of kmno4
b)200ml of 0.1M KMnO4
c)0.1 mole of kmno4
d)0.02 moles of kmno4
PLS NO SPAM ,ANSWER IF U KNOW
Answers
Answer:
19.1(c)
This statement is about the oxidation of methanoic and ethanedioic (oxalic) acids.
Before you go on, you should find and read the statement in your copy of the syllabus.
Statement 19.1(c)(i): Oxidising methanoic acid
Methanoic acid, HCOOH, has this structure:

If you look at the top half of this, you will see an aldehyde group, with a hydrogen attached to a carbon-oxygen double bond.
Aldehyde groups can be oxidised using things like Fehling's solution or Tollens' reagent. The reactions of methanoic acid look just the same as those with aldehydes. You will find these on the page oxidation of aldehydes and ketones.
The methanoic acid is oxidised to carbon dioxide and water.
With Fehling's solution you get a red precipitate of copper(I) oxide

and with Tollens' reagent you get a silver mirror.

This is new for the 2016 syllabus, and there is no clue as to whether these equations are needed or not.
I suspect they won't be asked and that most people would get them wrong in an exam anyway. An alternative might be to use the version that you would use if you were talking about oxidation with acidified potassium dichromate(VI) or acidified potassium manganate(VII).

Statement 19.1(c)(ii): Oxidising ethanedioic (oxalic) acid
Ethanedioic acid is:

It is oxidised to carbon dioxide by warm potassium manganate(VII) solution acidified with dilute sulphuric acid.
The reaction is often used to standardise potassium manganate(VII) solution. You can't make up a solution of potassium manganate(VII) with an exactly accurate concentration, but you can with ethanedioic acid.
A standard solution of ethanedioic acid is acidified with dilute sulphuric acid, warmed and titrated with potassium manganate(VII) solution from a burette.
The potassium manganate(VII) solution is decolourised until you get to the end point when a trace of pale pink remains.
You can then do a calculation to find the actual concentration of the potassium manganate(VII) solution.
It is possible that you might need to write an equation for this in a practical context, and you will have to build it up from half-equations.
The half-equation for the ethanedioic acid is

The half-equation for the manganate(VII) ion is

If you combine those it gives:

Check that you can do this! If you can't get it right, you need to spend some time looking at the page about writing equations for redox reactions. This example is similar to the hydrogen peroxide and manganate(VII) ion case.