The following question i solve for hours but i still cant solve it hope some people can help me :)
Q: 25cm3 of a solution X2O5 of concentration 0.1 mol dm-3 is reduced by sulfur dioxide to a lower oxidation state. Re oxidation of X to its original oxidation state required 50 cm3 of 0.2 mol dm-3 of acifified potassium manganate (VII) solution. Assuming that the manganate (VII) solution is reduced to Mn2+ to what oxidation number was X reduced by sulfur dioxide ?
The answer is +4 but how to get this????
Originally posted by Tanbrenda:The following question i solve for hours but i still cant solve it hope some people can help me :)
Q: 25cm3 of a solution X2O5 of concentration 0.1 mol dm-3 is reduced by sulfur dioxide to a lower oxidation state. Re oxidation of X to its original oxidation state required 50 cm3 of 0.2 mol dm-3 of acifified potassium manganate (VII) solution. Assuming that the manganate (VII) solution is reduced to Mn2+ to what oxidation number was X reduced by sulfur dioxide ?
The answer is +4 but how to get this????
Calculate no. of mols of X2O5 present, hence no. of mols of X 5+ cations. (5 X 10^-3 mol)
Write a statement indicating that this is also the no. of mols of X n+ cations present after reduction by SO2. (5 x 10^-3 mol)
Write reduction half equation for MnO4-. Hence state no. of mols of electrons accepted per mol of MnO4-. (5 electrons)
Calculate mols of MnO4- present. (0.01 mol)
Calculate total no. of electrons accepted by all the MnO4- present. (0.05 mol)
Hence calculate no. of mols of electrons removed from each mol of X n+ cations. (10 electrons)
This is obviously improbable. Hence, the question must have contained a typo error. Either 50 cm3 of 0.02 mol dm-3 of acidified potassium manganate (VII) solution was used, or 5 cm3 of 0.2 mol dm-3 of acidified potassium manganate (VII) solution was used, or 25cm3 of a solution X2O5 of concentration 1.0 mol dm-3 was used.
Then, the no. of mols of electrons removed from each mol of X n+ cations would be (1 electron).
Hence calculate the lower O.S. of X :
n + 1 = +5
n = +4