Question

Question #9cee6

Answer 1

I take it this is 2nd/3rd year university question?

Explanation:

Deep purple, potassium permanganate is reduced to colourless `Mn^(2+)`:

`MnO_4^(-)+8H^(+) +5e^(-) rarr Mn^(2+) +4H_2O` `(i)`

Charge and mass are balanced, so we know this is right.

And the `Pt^(2+)` is oxidized to `Pt^(4+)`:

`Pt^(2+) rarr Pt^(4+) + 2e^-` `(ii)`.

We wish to remove the electrons, so we take the cross product, `2xx(i)+5xx(ii)`:

`2MnO_4^(-)+16H^(+) +5Pt^(2+) rarr 2Mn^(2+) +5Pt^(4+) + 8H_2O`

The which I tink is balanced with respect to mass and charge. So this equation tells us that 2 equiv permanganate would oxidize 5 equiv platinum ion. The endpoint of the titration would be signalled by the persistence of the striking purple colour of `MnO_4^-` ion.

`26.87xx10^-3*Lxx0.0500*mol*L^-1=1.344xx10^-3*mol` `KMnO_4` were used. And thus `[Pt^(2+)]` `=` `(5/2xx1.344xxcancel(10^-3)*mol)/(250.0xxcancel(10^-3)*L)` `=` `1.344xx10^-2*mol*L^-1`.

If there were `250.0*mL` of the starting platinum solution, there were `250.0xx10^3Lxx1.344xx10^-2*mol*L^-1` `=` `??mol`, `??g` `Pt^(2+)`?

I do not particularly like this question, because they should not be asking you questions about experiments that have not been performed in the practical. I have never oxidized `Pt^(2+)` titrimetrically, and I doubt that the examiner has either.