Titration UNIT 5 (CHEM)

Q. An experiment is carried out to check the oxidation number of chromium in
chromium(II) ethanoate Cr2(CH3CO2)4(H2O)2.

1.00 g (2.66 × 10−3 mol) of chromium(II) ethanoate is dissolved in 25 cm3
of 1.00 mol dm−3 sulfuric acid.
The solution is diluted with distilled water until the volume is 250 cm3.

25.0 cm3 portions of the diluted solution are titrated with 0.00750 mol dm−3 potassium manganate(VII).

Calculate the volume of potassium manganate(VII) needed to oxidize the chromium(II) ions present in each 25.0 cm3 portion to the +6 oxidation state.
The manganese is reduced to the +2 oxidation state.

Reply 1

Aleksander Krol

My working:
i found the mole of cr2+ ions in diluted 25 cm^3 solution by dividing the original moles of chromium (||) ethanoate with 10 = 2.66x10^-4 mol

then i took 1 equation from the booklet for Mn where it changes its oxidation no. from 7+ to 2+
and wrote a balanced equation of Cr2+ getting oxidised to 6+.[cr2+ -----> Cr6+ + 4e-]
after both half equations: (final redox equation)
4MnO4- + 32H+ + 5Cr2+
-----> 4Mn2+ + 16H2O + 5Cr6+

the molar ratio of Mn7+ and Cr2+ in titration is 4:5
so, to get the vol. of KMnO4 required in titration:
[ (4×2.66×10^-4)/5 ] ÷ 0.0075 = 0.02837 dm^3

so it's giving me, 28.37 cm^3 as the final answer.

but the ms says the answer is: 56.75cm^3

HOW??

i'd be grateful if you can point where i went wrong or what i missed out.
thank you in advance!!

(edited 1 year ago)