Redox Titrations Revision (OCR)

0.151 g of steel wool was dissolved 25 cm³ of 1mol dm^-3 sulphuric acid (a large excess), forming a solution of Fe²⁺. This was then titrated against a solution of 0.0200 mol dm^-3 KMnO₄, of which 25.45 cm³ was required to reach the end point. The equation for the reaction between Fe²⁺ and MnO₄^- is:

MnO₄^- + 5Fe²⁺ + 8H⁺ --> Mn²⁺ + 5Fe³⁺ + 4H₂O

i) Calculate the moles of MnO₄^- used in the titration.

0.151 g of steel wool was dissolved 25 cm³ of 1mol dm^-3 sulphuric acid (a large excess), forming a solution of Fe²⁺. This was then titrated against a solution of 0.0200 mol dm^-3 KMnO₄, of which 25.45 cm³ was required to reach the end point. The equation for the reaction between Fe²⁺ and MnO₄^- is:

MnO₄^- + 5Fe²⁺ + 8H⁺ --> Mn²⁺ + 5Fe³⁺ + 4H₂O

ii) Calculate the moles of Fe²⁺ in the solution.

0.151 g of steel wool was dissolved 25 cm³ of 1mol dm^-3 sulphuric acid (a large excess), forming a solution of Fe²⁺. This was then titrated against a solution of 0.0200 mol dm^-3 KMnO₄, of which 25.45 cm³ was required to reach the end point. The equation for the reaction between Fe²⁺ and MnO₄^- is:

MnO₄^- + 5Fe²⁺ + 8H⁺ --> Mn²⁺ + 5Fe³⁺ + 4H₂O

iii) Calculate the mass of Fe in the steel wool

0.151 g of steel wool was dissolved 25 cm³ of 1mol dm^-3 sulphuric acid (a large excess), forming a solution of Fe²⁺. This was then titrated against a solution of 0.0200 mol dm^-3 KMnO₄, of which 25.45 cm³ was required to reach the end point. The equation for the reaction between Fe²⁺ and MnO₄^- is:

MnO₄^- + 5Fe²⁺ + 8H⁺ --> Mn²⁺ + 5Fe³⁺ + 4H₂O

iv) Calculate the % by mass of Fe in the steel wool.

A metal ore contains Fe(II). 6.46 g of the ore is dissolved in sulfuric acid and the resulting solution is made up to 250cm3. 25cm3 of this solution is titrated against 21.40cm3 of 0.0200 mol dm-3 potassium manganate (VII) solution. Calculate the percentage by mass of iron(II) in the sample of ore.

30 cm³ of bleach was added to an excess of acidified iodide ions. Bleach contains chlorate (I) ions, ClO^- which will oxidise I^- to I2 (ClO^- is reduced to Cl^-).

The I₂ formed was titrated against 0.200 mol dm^-3 of sodium thiosulphate 29.45cm³ was required.

2ClO^-(aq) + 4H⁺(aq) + 4I^-(aq) → 2Cl^-(aq) + 2I₂(aq) + 2H₂O(l)

I₂ + 2S₂O₃^2- --> 2I^- + S₄O₆^2-

Calculate the moles of S₂O₃^2-

30 cm³ of bleach was added to an excess of acidified iodide ions. Bleach contains chlorate (I) ions, ClO^- which will oxidise I^- to I2 (ClO^- is reduced to Cl^-).

The I₂ formed was titrated against 0.200 mol dm^-3 of sodium thiosulphate 29.45cm³ was required.

2ClO^-(aq) + 4H⁺(aq) + 4I^-(aq) → 2Cl^-(aq) + 2I₂(aq) + 2H₂O(l)

I₂ + 2S₂O₃^2- --> 2I^- + S₄O₆^2-

Calculate the moles of I₂

30 cm³ of bleach was added to an excess of acidified iodide ions. Bleach contains chlorate (I) ions, ClO^- which will oxidise I^- to I2 (ClO^- is reduced to Cl^-).

The I₂ formed was titrated against 0.200 mol dm^-3 of sodium thiosulphate 29.45cm³ was required.

Calculate the concentration of chlorate ions in the bleach.

2ClO^-(aq) + 4H⁺(aq) + 4I^-(aq) → 2Cl^-(aq) + 2I₂(aq) + 2H₂O(l)

I₂ + 2S₂O₃^2- --> 2I^- + S₄O₆^2-

Calculate the concentration of ClO^-

0.500g of bronze was reacted with nitric acid giving a Cu²⁺ solution. The solution was reacted with iodide ions, I^- forming a solution of iodine, I₂.

This iodine solution was then titrated with 0.200 mol dm^-3 solution of sodium thiosulphate, 22.40cm³ of this was required.

Calculate the % by mass of copper in bronze

2S₂O₃^2– (aq) + I₂(aq) ==> S₄O₆^2– (aq) + 2I^– (aq)

25.0 cm³ of a solution of iodine in potassium iodide solution required 26.5 cm³ of 0.0950 mol dm^–3 sodium thiosulphate solution to titrate the iodine.

What is the concentration of iodine to 3s.f?

1.51 g of the wire was dissolved in excess of dilute sulphuric acid and the solution made up to 250 cm³ in a standard graduated flask.

A 25.0cm³ sample of this solution was pipetted into a conical flask and needed 25.45 cm³ of 0.0200 mol dm^–3 KMnO₄ for complete oxidation.

Calculate the percentage of iron in a sample of steel wire to 3s.f.

3. 8.25g of an iron(II) salt was dissolved in 250 cm³ of pure water. 25.0 cm³ samples were pipetted from this stock solution and titrated with 0.0200 mol dm^–3 potassium manganate(VII) solution. The titration value obtained was 23.87 cm³ . Remember MnO₄^- : 5Fe²⁺

Guided steps:

Calculate the moles of manganate(VII) used in the titration.

Calculate the moles of iron(II) ion titrated

Calculate the mass of iron(II) titrated

Calculate the total mass of iron in the original sample of the iron(II) salt.

Calculate the % iron in the salt and enter the numbers to 3s.f below

3.00 g of a lawn sand containing an iron(II) salt was shaken with dilute H₂SO₄. The resulting solution required 25.00 cm³ of 0.0200 mol dm^-3 potassium manganate(VII) to oxidise the Fe²⁺ ions in the solution to Fe³⁺ ions. Use this to calculate the percentage by mass of Fe²⁺ ions in this sample of lawn sand.