Are my Redox Calculations correct?

[PBimages]

Homework Statement

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...

Producing...

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

Homework 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...

The Attempt at a Solution

To Balance the equation...

Reduction

MnO4- (aq) = Mn2+

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

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

Oxidation

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

 

[Borek]

Calculations seem OK. Think about significant figures.

 

[PBimages]

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...