[bacondude]

Hello, our Chemistry class is doing a lab on Oxidation-Reduction Titrations, and I really need to get these 3 questions done and I'd really appreciate the help..

I apologize if I sound like a person who just wants answers without having put effort but I really am having trouble. I am a 8th grader taking HS/AP chemistry class because I was told to go there by my biology teacher in the 4th quarter of this year! Anyways...

1. Write the balanced net ionic equation for the reaction between MnO4- (4 a subscript, the negative sign a superscript) and H2C2O4 in acid solution.(#s are subscripts)

2. How many moles of Fe+2 ions will be oxidized by 0.043 moles of MnO4- ions?

3. Iron can be analyzed by dissolving it in acid to convert Fe(0) to Fe+2 ions and then titratiing the Fe+2 ions with KMnO4. 1.630 g of iron ore are dissolved in an acidic solution. The solution is titrated to a pink endpoint with 27.15 mL of 0.020M KMnO4 solution.
a) How many moles of MnO4- were consumed?
b) How many moles of Fe+2 were in the iron ore sample?
c) What is the % of iron in the iron ore sample?

Any help is highly appreciated!! Any! thank you all! I am happy to be a new member of this forum!!

[Astrokel]

1. Do you know how to write a half equation or an ionic equation? It seems you've just started chemistry.

These are the steps to write a half equation.

1) Write the substance and its oxidized or reduced states.
2) Balance the oxygen atoms by adding H2O
3) Balance the hydrogen atoms by adding H+ (only in acidic solution)
4) Balance charges by adding e-

Example: 

MnO4-   --->  Mn2+  (step 1)
MnO4-   --->  Mn2+ + 4H20 (step 2)
MnO4- + 8H+  ---> Mn2+ + 4H20 (step 3)
MnO4- + 8H+ + 5e- ---> Mn2+ + 4H20 (step 4)

You try doing the same steps for C2042- and combine the final equations by eliminating e- to get the net ionic equations.

2. You have to write a net ionic equation between Fe2+ and MnO4-, and to find the stoichiometric coefficients.

3a. You have the volume and amounts of KMnO4 solution, you can get the moles of MnO4- consume.

3b. Same as question 2. Same stoichiometric coefficients

3c. Once you get the amounts of Fe2+, you can get the mass.

and welcome! If you still don't understand, let me know 

 
 

[bacondude]

Astro, thanks for the great explanation, but I seem to be a bit confused  You said..."Example:

MnO4-   --->  Mn2+  (step 1)
MnO4-   --->  Mn2+ + 4H20 (step 2)
MnO4- + 8H+  ---> Mn2+ + 4H20 (step 3)
MnO4- + 8H+ + 5e- ---> Mn2+ + 4H20 (step 4)"

But why doesn't anything happen to the Oxygen? Shouldnt it be something like... e- + MnO4- --> Mn+2 + 2O2-2   to start with? please get back to me on this one, thanks

[Astrokel]

Astro, thanks for the great explanation, but I seem to be a bit confused  You said..."Example:

MnO4-   --->  Mn2+  (step 1)
MnO4-   --->  Mn2+ + 4H20 (step 2)
MnO4- + 8H+  ---> Mn2+ + 4H20 (step 3)
MnO4- + 8H+ + 5e- ---> Mn2+ + 4H20 (step 4)"

But why doesn't anything happen to the Oxygen? Shouldnt it be something like... e- + MnO4- --> Mn+2 + 2O2-2   to start with? please get back to me on this one, thanks

Hey,
  because oxygen is not being oxidised nor reduced. It's the Mn ions thats being reduced.

The equation you wrote is incorrect, it's Mn²⁺ and not Mn⁺². We only use +2 when we say Oxidation state of maganese ion is being reduced from +7 in MnO₄^- to +2 in Mn²⁺ See the difference? We only use the +/- before digit when describing the oxidation state, however when we write chemical symbol, digit must comes before -/+

and there's no such thing as O₂^-2, O₂ is a molecule, if you write a -2 there, you are implying it is an ion which is contradictory. Check oxidation state of oxygen, from -2 in MnO₄^- to -2 in H₂O, the oxidation state of oxygen doesn't change so it doesn't really participate in the redox reaction.

When some thing is reduced, there must be other thing that is being oxidized and it is the C₂O₄^2- that undergoes oxidation.

 

[Borek]

it's Mn²⁺ and not Mn⁺². We only use +2 when we say Oxidation state of maganese ion is being reduced from +7 in MnO₄^- to +2 in Mn²⁺

Generally yes, but it is not enforced too rigorously; don't read too much from what you see. Especially when the formulas are not correctly formatted - SO₄^2- is obvious, SO42- is ambiguous, so some will write it as SO4-2 Note also that oxidation states do not exist in reality, they are just used as a convenient way to count electrons.

and there's no such thing as O₂^-2

It is not present in this reaction, but some peroxides are ionic and such ion can be detected in their crystallic structure.