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Homework Help: Are my Redox Calculations correct?

  1. Apr 25, 2009 #1
    1. The problem statement, all variables and given/known data

    0.9966g of Iron Wire was used...

    Iron Wire was placed in 200cm3 of 1m Sulphuric Acid, this was heated and the hydrogen gas vented off...


    Fe + H2SO4 = FeSO4 + H2

    Fe = Iron
    H2SO4 = Sulphuric Acid
    FeSO4 = Ferrous Sulphate
    H2 = Hydrogen

    This produced a Ferrous Sulphate solution, as shown above by the balanced equation...

    This was then filtered and made upto 250cm3 using 2M sulphuric Acid, to stop air oxidation effecting the solution...

    25cm3 aliquots of this solution were then titrated against 0.02M of Potassium Manganate (VII)

    This produced an average titre of 17.9cm3

    2. Relevant equations

    Work out the number of Moles of Manganate (VII) in your mean titre...

    Using the equations given, work out the moles of Fe2+ used in 25cm3 aliquots

    MnO4- (aq) + 8H+(aq) + 5e- = 4H2O(I) + Mn2+ (aq) Fe2+(aq) = Fe3+(aq) + e-

    Work out the number of moles present in 250cm3...

    Work out the mass in grams of iron in the sample...

    Work out the purity of the sample...

    3. The attempt at a solution

    To Balance the equation...


    MnO4- (aq) = Mn2+

    MnO4- (aq) + 8H+ (aq) = Mn2+ (aq) + 4H2O (I)

    MnO4- (aq) + 8H+ (aq) + 5e- = Mn2+ (aq) + 4H2O (I)


    Fe2+ (aq) = Fe3+ (aq) + e-

    To Balance the equation, the first half of the reaction needs to be multiplied by five, then the equations can be added together.

    MnO4- (aq) + 8H+ (aq) + 5e- = Mn2+ (aq) + 4H2O (I)

    Fe2+ (aq) = Fe3+ (aq) + 5e-

    MnO4- (aq) + 8H+ (aq) + 5e- + 5Fe2+ (aq) = Mn2+ (aq) + 4H2O (I) + 5Fe3+ (aq) + 5e-

    The overall redox balanced equation is therefore: -

    MnO4- (aq) + 8H+ (aq) + 5Fe2+(aq) = Mn2+ (aq) + 4H2O(I) + 5Fe3+(aq)

    The rest of the calculations follow...

    To calculate the number of moles of MnO4- used: -

    The volume is converted from cm3 into dm-3

    17.9cm3/1000 = 0.0179dm3

    The number of moles of MnO4- used was: -

    Moles of MnO4-= Molarity x Volume (dm3)

    Moles of MnO4-= 0.02 x 0.0179 = 3.58 x10-4

    So the total amount of manganate (VII) in the Titre is 3.58 x 10-4 mol

    MnO4- (aq) + 8H+ (aq) + 5Fe2+ (aq) = Mn2+ + 4H2O (I) + 5Fe3+ (aq)

    The balanced equation shows that a ratio of 1: 5 exists between the MnO4- and Fe2+.

    Therefore: -

    To work out the number of moles of Fe2+ in 25cm3:-

    Moles of Fe2+ = Moles of MnO4- x Ratio

    Moles of Fe2+ = 3.58x 10-4 x 5 =1.79x 10-3

    The number of moles of Fe2+ in 250cm3

    1.79 x 10-3 x 10 = 0.0179

    The total mass of iron in grams is: -

    Mass(g)= Moles of Fe2+ x RAM Fe2+

    Mass(g)= 0.0179 x 55.85 = 0.999715

    The percentage of purity is: -

    Purity= (Moles of Fe2+ x RAM Fe2+) / Weight of iron wire

    Purity=(0.0179 x 55.85)/0.9966 = 1.0031256

    The total percentage is: -

    1.0031256 x 100 = 100.31256

    Therefore the total percentage of purity is: 100.31
  2. jcsd
  3. Apr 25, 2009 #2


    User Avatar

    Staff: Mentor

    Calculations seem OK. Think about significant figures.
  4. Apr 25, 2009 #3
    Thanks Borek,

    I'll sort out the Significant Figures later...

    My main concern is the calculations...

    For some reason something is telling me that the calculations are wrong...
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