In: Chemistry
As carried out in Section A, a 20
cm3 aliquot of a 0.0513 mol dm-3 vanadium(V)
solution wasreduced by excess zinc metal to an
unknown oxidation state (x). The resulting
lavender (lilac) coloured solution was then fully
reoxidised by 32.3 cm3 of a 0.0199 mol
dm-3 KMnO4solution.
Calculate the experimental value for the unknown oxidation state x
(to two decimal places)
Concentration of V5+ = 0.0513 mol dm-3
Volume of aliquot = 20 cm3 = 0.02 dm3
Moles of V5+ = 0.0513 * 0.02
= 0.001026
Concentration of KMnO4 solution = 0.0199 mol dm-3
Volume of solution = 32.3 cm3 = 0.0323 dm3
Moles of KMnO4 = 0.0199 * 0.0323
= 0.00064277
When V5+ is reduced to +x odidation state using Zn:
2V5+ + 2(5 - x)e- 2Vx+
(5 - x)Zn (5 -
x)Zn2+ + 2(5 - x)e-
---------------------------------------------------------------
2V5+ + (5 - x)Zn
2Vx+ + (5 - x)Zn2+
Now,
When Vx+ is odidized by KMnO4 solution.
3Vx+ 3V5+ + 3(5 - x)e-
(5 - x)MnO4- + 3(5
- x)e- (5 -
x)MnO2
---------------------------------------------------------------------------------------
3Vx+ + (5 -
x)MnO4-
3V5+ + (5 -
x)MnO2
As 3 moles Vx+ are oxidised by (5 - x) moles of KMnO4
So, 0.001026 moles Vx+ will be oxidised by = ((5 - x) / 3) * 0.001026 moles of KMnO4
But we have 0.00064277 moles of KMnO4
Therefore,
((5 - x) / 3) * 0.001026 = 0.00064277
(5 - x) * 0.001026 = 0.00192831
(5 - x) = 1.879
x = 5 - 1.879
x = 3.12