1. Redox Calculations

2. Equations Mass = Moles x Mr Moles = Concentration x Volume
To convert volume from
cm3 → dm3, divide by 1000

3. Redox Calculations
Example 1: 3.00 g of lawn sand containing an iron(II) salt was shaken with dilute sulphuric acid
The resulting solution needed cm3 of mol dm-3 KMnO4 to oxidise the Fe2+ ions in the solution to Fe3+ ions
Use this information to calculate the percentage by mass of Fe2+ ions in this sample of lawn sand

4. MnO4- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
Redox Calculations
1) Equation:
MnO4- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
2) Moles of MnO4- = Conc x Vol
= x (25 ÷ 1000)
= 5 x 10-3 mol
3) Moles of Fe2+ = Moles of MnO4- x 5 (Ratio 1:5)
= 5 x 10-3 x 5
= mol

5. Redox Calculations 4) Mass of Fe2+ = Moles x Mr = 0.025 x 55.8
= 1.40 g
5) % by mass of Fe2+ = (1.40 ÷ 3.00) x 100
= 46.7 %
The most important part is to write the equation for reaction as it allows you to work out the ratio (step 3)

6. Redox Calculations
Example 2: In a titration, g of a moss killer reacted with cm3 of acidified mol dm-3 K2Cr2O7 solution
Calculate the percentage by mass of iron in the moss killer
Assume that all of the iron in the moss killer is in the form of iron(II)

7. Cr2O72- + 14H+ + 6Fe2+ → 2Cr3+ + 7H2O + 6Fe3+
Redox Calculations
1) Equation:
Cr2O H+ + 6Fe2+ → 2Cr3+ + 7H2O + 6Fe3+
2) Moles of Cr2O72- = Conc x Vol
= x (23.60 ÷ 1000)
= x 10-3 mol
3) Moles of Fe2+ = Moles of Cr2O72- x 6 (Ratio 1:6)
= x 10-3 x 6
= x 10-3 mol

8. Redox Calculations 4) Mass of Fe2+ = Moles x Mr = 3.087 x 10-3 x 55.8
= g
5) % by mass of Fe2+ = (0.172 ÷ 0.321) x 100
= 53.6 %
The most important part is to write the equation for reaction as it allows you to work out the ratio (step 3)

9. Redox Calculations
Example 3: A 1.27g sample of impure iron was reacted with an excess of dilute sulphuric acid
The solution formed was made up to 250 cm3
A 25.0 cm3 sample of this solution reacted completely with exactly 19.6 cm3 of a mol dm-3 solution of potassium manganate(VII)
Calculate the percentage by mass of iron in the sample

10. MnO4- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
Redox Calculations
1) Equation:
MnO4- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
2) Moles of MnO4- = Conc x Vol
= x (19.6 ÷ 1000)
= 4.31 x 10-4 mol
3) Moles of Fe2+ = Moles of MnO4- x 5 (Ratio 1:5)
= 4.31 x 10-4 x 5
= 2.16 x 10-3 mol in 25 cm3

11. Redox Calculations 4) Moles of Fe2+ in 250 cm3 = 2.16 x 10-3 x 10
= 2.16 x 10-2 mol
5) Mass of Fe2+ = Moles x Mr
= 2.16 x 10-2 x 55.8
= 1.21 g
6) % by mass of Fe2+ = (1.21 ÷ 1.27) x 100
= 95.3 %

12. Redox Calculations
Example 4: A solution of iron(II) sulfate was prepared by dissolving g of FeSO4.7H2O (Mr = 277.9) in water and making up to 250 cm3 of solution
The solution was left to stand, exposed to air, and some of the iron(II) ions became oxidised to iron(III) ions
A 25.0 cm3 sample of the partially oxidised solution required cm3 of mol dm-3 potassium dichromate(VI) solution for complete reaction in the presence of an excess of dilute sulfuric acid
Calculate the percentage of iron(II) ions that had been oxidised by the air

13. Cr2O72- + 14H+ + 6Fe2+ → 2Cr3+ + 7H2O + 6Fe3+
Redox Calculations
1) Equation:
Cr2O H+ + 6Fe2+ → 2Cr3+ + 7H2O + 6Fe3+
2) Moles of Cr2O72- = Conc x Vol
= x (23.70 ÷ 1000)
= 2.37 x 10-4 mol
3) Moles of Fe2+ = Moles of Cr2O72- x 6 (Ratio 1:6)
= 2.37 x 10-4 x 6
= 1.42 x 10-3 mol in 25 cm3

14. Redox Calculations 4) Moles of Fe2+ in 250 cm3 = 1.42 x 10-3 x 10
= 1.42 x 10-2 mol
5) Original moles of Fe2+ = Mass ÷ Mr
= ÷ 277.9
= mol
6) Moles of Fe2+ oxidised = Original - Fe2+ remaining
= x 10-2
= mol
7) % by mass of Fe2+ oxidised = ( ÷ 0.036) x 100
= 60.6 %