Advanced Level Chemistry - Redox Volumetric Titration Calculation Question 1

Fill in all the gaps, then press "CHECK" at any time, as long as you are not bothered about the score! You can bookmark this page, print it out, try the Q and go back on-line to CHECK!!! Thoughts on style of question? * Worksheet of Redox Titration Questions * Updated Doc Brown's Chemistry April 29th 2010 *

Question 1: Given the following two half-reactions:

(1) MnO4-(aq) + 8H+(aq) + 5e- ==> Mn2+(aq) + 4H2O(l)
(2) Fe3+(aq) + e- ==> Fe2+(aq)

and data (3) Ar(Fe) = 55.9

For the solutions to Question 1 work to x sf or x dp as indicated in [] (the 'fill' is unforgiving!) and for decimals <1 put 0.xxxx

Q1(a) Construct the fully balanced redox ionic equation for the manganate(VII) ion oxidising the iron(II) ion. Remember the basic half-cell ratio is deduced from e's lost = e's gained, ie oxidation state balance total up = total down.

ans (a) MnO4-(aq) + H+(aq) + Fe(aq) ==>
Mn2+(aq) + Fe(aq) + H2O(l)

Q1(b)(i) 24.3 cm3 of 0.02 mol dm-3 KMnO4 reacted with 20.0 cm3 of an iron(II) solution. Calculate the molarity of the iron(II) ion.

ans(b)(i) moles = molarity (moldm-3) x volume (dm3)
(i) mol MnO4- = [2dp] x [1dp] / [4sf]
= [6dp]

mol Fe2+ = x [6dp]
(from MnO4- : Fe2+ in equation)
mol Fe2+ = [5dp] in cm3[2sf]

so scaling up to 1 dm3, molarity of Fe2+
= [5dp] x [4sf] / [2sf]
molarity Fe2+ = mol dm-3 [4dp]

(b) (ii) How do you recognise the end-point in the titration?

ans (b) (ii) We do not use phenolphthalein for this titration! The end point is the first faint permanent due to a trace of excess KMnO4

Q1(c) Calculate the percentage of iron in a sample of steel wire if 1.51g of the wire was dissolved in excess of dilute sulphuric acid and the solution made up to 250cm3 in a standard flask. 25.0 cm3 of this solution was pipetted into a conical flask and needed 25.45 cm3 of O.02 mol dm-3 KMnO4 for complete oxidation.

ans (c) mol MnO4- = [2dp] x [4sf] /
= [6dp]

mol Fe = mol Fe2+ = x [6dp]
= [6dp]

mass Fe = x [1dp]
= [4dp] per cm3 aliqouot titration.

total Fe in wire = x [2sf]
= g (1/th of the made up solution used in titration)

so % Fe = [3dp] x /
= % [3sf]