Question

Acidified potassium manganate(VII) reacts with iron(II) ethanedioate, FeC2O4.
The reactions taking place are shown.
MnO4– + 8H+ + 5e– → Mn2+ + 4H2O
Fe2+ → Fe3+ + e–
C2O42– → 2CO2 + 2e–
How many moles of iron(II) ethanedioate react with one mole of potassium manganate(VII)?
(the answer is 1.67) but i want to know the method please.
(if you cant understand the question i typed then open the attachment and see question 10)

Answer 1

Answer:

1.67

Explanation:

As per the given equations, there is one reduction reaction and two oxidation reactions.

a) Reduction of  potassium manganate(VII)

b) oxidaiton of Ferrous ion

c) oxidation of oxalate ion

From given equations it is clear that reduction of each mole of manganate ion requires five moles of electrons and ferrous ion on oxidation gives one mole of electron and oxalate ion two moles of electrons.

Now in order to balance the oxidation and reduction reaction we will multiply the reduction reaction with three and oxidation reactions with five.

$MnO_{4}^{-}+ 8H^{+} + 5e ---> Mn^{+2}+ 4H_{2}O\\]X3Fe^{+2} ---> Fe^{+3} + e\\]X5C_{2}O_{4}^{-2} ---> 2CO_{2} + 2e]X5$

...........................................................................

$3MnO_{4}^{-}+ 24H^{+} + 15e ---> 3Mn^{+2}+ 12H_{2}O\\5Fe^{+2} ---> 5Fe^{+3} + 5e\\5C_{2}O_{4}^{-2} ---> 10CO_{2} + 10e$

Thus as per the equation from each three moles of manganate ion five moles of iron(II) ethanedioate will get oxidized.

Hence from each mole of manganate ion the moles of iron(II) ethanedioate oxidized will be =$\frac{5}{3}=1.67$