Post-Lab Questions 1. Re-write the iron-permanganate reaction and indicate: a. o
ID: 975811 • Letter: P
Question
Post-Lab Questions 1. Re-write the iron-permanganate reaction and indicate: a. oxidizing agent, b. reducing agent, c. the species that gets oxidized, and d. the species that gets reduced. 2. An acidic solution of permanganate ion reacts with oxalate (C,0 ion to form carbon dioxide and Mn(ID. Write a balanced equation for the reaction. 3. Referring to the reaction in Question 38.4 mL of a 0.150 M KMno, solution is required to titrate 2 mL of a sodium oxalate solution. What is the concentration of oxalate ion in the solution? 25.2 mL of a sodium oxalate solution. What is the concentrati 4. Suggest an alternative way to determine Fe quantitatively.Explanation / Answer
Iron-Permanganate reaction:
MnO4- + 5Fe2+ + 8H+ ---------> Mn2+ + 5Fe3+ + 4H2O
i) Mn in MnO4- is at Oxidation State (+7) and it changes to (+2) in Mn2+. i.e. electrons are added to the valence shell of Mn and hence Mn said to be Reduced and called as Oxidizing agent (Oxidant).
ii) Fe2+ changes to Fe3+ it means 1 electron is removed from valence shell of Fe2+ and hence Fe2+ is said to be oxidized and called as Reducing agent.
Hence we can write,
MnO4: Is Oxidizing agent or Oxidant.
And, Mn2+: Is the Reduced species.
Fe2+: Is Reducing agent or Reductant.
And Fe3+: Is Oxidized species.
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2) and 3)
Reaction between permanganate ion and oxalate ion in acidic medium.
2MnO4- + 5C2O42- + 16H+ --------> 2Mn2+ + 10CO2 + 8H2O
From above redox reaction stoichiometry it clear that at equivalence point,
5[MnO4-] = 2[C2O42-]
i.e.
5 x V(MnO4-) x M(MnO4-) = 2 x V (C2O42-) x M(C2O42-)
Let us put given values,
5 x 38.4 x 0.150 = 2 x 25.2 x M(C2O42-)
28.8 = 50.4 x M(C2O42-)
M(C2O42-) = 28.8/50.4
M(C2O42-) = 0.571 M
I.e. [C2O42-] = 0.571 M
Concentration of oxalate ion in the solution is 0.571 M.
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