Redox Question!

• Jetihyun

  15 Mar 12, 17:43

  25.0 cm3 of a solution of M2O5 of concentration 0.100moldm-3 is reduced by sulphur dioxide to a lower oxidation state. To re-oxidise M to its original oxidation number, this required 50.00cm3 of 0.0200moldm-3 potassium manganate (VII) in acidified medium. To what oxidation number was M reduced to by sulfur dioxide?

  I am really confused by the question. How do I use the mol ratio to find the oxidation number. I can't solve similar questions due to this problem too....

  Thanks in advance

• UltimaOnline

  15 Mar 12, 18:55

  Originally posted by Jetihyun:

  25.0 cm3 of a solution of M2O5 of concentration 0.100moldm-3 is reduced by sulphur dioxide to a lower oxidation state. To re-oxidise M to its original oxidation number, this required 50.00cm3 of 0.0200moldm-3 potassium manganate (VII) in acidified medium. To what oxidation number was M reduced to by sulfur dioxide?

  I am really confused by the question. How do I use the mol ratio to find the oxidation number. I can't solve similar questions due to this problem too....

  Thanks in advance

  
  Write reduction half-equation of MnO4- to Mn2+.

   MnO4– + 8H+ + 5e– � Mn2+ + 4H2O

   

  Find moles of MnO4- required.

  50.00cm3 of 0.0200moldm-3 = 1 x 10^-3 mol

   

  Hence find moles of electrons transferred.

  1 x 10^-3 x 5 = 5 x 10^-3 mol

   

  Find moles of M2O5 present.

  25.0 cm3 of 0.100moldm-3 = 2.5 x 10^-3 mol

   

  Find moles of M atoms present.

  2.5 x 10^-3 x 2 = 5 x 10^-3

   

  Hence find moles of electrons transferred per mole of M atoms.

  (5 x 10^-3) / (5 x 10^-3) = 1

   

  Find OS of M in M2O5.

  +5

   

  Hence find reduced OS of M.

  (+5) + (-1) = +4