Question #9cee6

1 Answer
Oct 10, 2016

Answer:

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.