chem redox

• AsimpleKid

  14 Mar 10, 23:20

  25.0cm3 of solution of iron (II) ethandioate FeC2O4 was titrated against a 0.0200moldm-3 solution of acidified potassium manganate (VII) solution. It was found that 33.75cm3 of potassium manganate (VII) solution was needed for complete reaction. What was the con. of iron(II) ethandioate solution? Reminder: both Fe2+ and C2O4(2-) are oxidised by MnO4-

  Thanks :)

• UltimaOnline

  14 Mar 10, 23:26

  Originally posted by AsimpleKid:

  25.0cm3 of solution of iron (II) ethandioate FeC2O4 was titrated against a 0.0200moldm-3 solution of acidified potassium manganate (VII) solution. It was found that 33.75cm3 of potassium manganate (VII) solution was needed for complete reaction. What was the con. of iron(II) ethandioate solution? Reminder: both Fe2+ and C2O4(2-) are oxidised by MnO4-

  Thanks :)

  
  When I give this question to my students, I'm not so generous as to give that "reminder". I expect my students to be able to figure that out for themselves.

   

  Write two separate balanced redox equations, one for MnO4- oxidizing the cation Fe2+, and another for MnO4- oxidizing the anion C2O4 2-.

   

  Let x be the molarity of FeC2O4. Hence find the amt of Fe2+ and C2O4 2- to be oxidized, in terms of x.

   

  Based on stoichiometry with both balanced redox equations, work out how much (in total) MnO4- is required.

   

  From experimental data, you're given how much MnO4- you actually used.

   

  Hence form an equation in x, and solve for x. 

   

• AsimpleKid

  14 Mar 10, 23:50

  i am sry but i am stuck here.
  MnO4- + 8H+ + 5Fe2+ --> Mn2+ + 4H2O + 5Fe3+

  2MnO4- + 16H+ +5C2O4 2- -->2Mn2+ +8H2O + 10 CO2

   

  1 mol of MnO4- react with 5 mol of Fe2+

  2 mol of MnO4- react with 5 mol of C2O4-

• AsimpleKid

  14 Mar 10, 23:58

  thank you, it's ok, i have got it.

  1mol of MnO4- react with x/5 mol of Fe2+

  !mol of Mn04- react with x/(5/2) mol of C2O4 2-

  3x/5= 0.0200 * 33.75/1000

  x = 0.001125

  n=cV

  c= 0.001125/ 25/1000

  =0.0450 moldm-3

• UltimaOnline

  15 Mar 10, 00:02

  Originally posted by AsimpleKid:

  i am sry but i am stuck here.
  MnO4- + 8H+ + 5Fe2+ --> Mn2+ + 4H2O + 5Fe3+

  2MnO4- + 16H+ +5C2O4 2- -->2Mn2+ +8H2O + 10 CO2

   

  1 mol of MnO4- react with 5 mol of Fe2+

  2 mol of MnO4- react with 5 mol of C2O4-

  
  Let y be the molarity of FeC2O4. Hence amt of Fe2+ and C2O4 2- to be oxidized = 0.025y Fe2+ and 0.025y C2O4 2-

   

  Hence no. of mol of MnO4- needed to oxidize 0.025y Fe2+ = (5x10^-3)y

  Hence no. of mol of MnO4- needed to oxidize 0.025y C2O4 2- = 0.01y

   

  Hence total amt of MnO4- needed = (5x10^-3)y + 0.01y

   

  Which we know from the data given in the question, is equals to (33.75/1000)*(0.02)

   

  Hence solve for y in the equation (5x10^-3)y + 0.01y = (33.75/1000)*(0.02).

   

   

  Edited :

  Ok great that you've solved it by yourself. Keep enjoying Chemistry!