completely confused gen chem question

[eravin99]

During a titration it was determined that 30.00
mL of a 0.100 M Ce4+ solution was required to
react completely with 20.00 mL of a 0.150 M
Fe2+ solution. Which one of the following reactions
occurred?
A. Ce4+ + 3Fe2+ + H2O  3Fe3+ + CeO- + 2H+
B. 2Ce4+ + Fe2+  Fe4+ + 2Ce3+
C. Ce4+ + Fe2+  Fe3+ + Ce3+
D. Ce4+ + 2Fe2+  2Fe3+ + Ce2+
E. Ce4+ + 2Fe2+  2Fe4+ + Ce2+ + 2e-

I dont understand it at all. how do you determine from this info which is reduced or oxidized?? answer is C

Thanks!!!

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[thefifthbeatle]

My first answer was C.......I picked it because nowhere in the question does it say the # of moles (coefficients in front of the elements)..C is the only one free of this

 

[thefifthbeatle]

Also how I got to this:

A has impossible/not expected products..can't be A

You dont have to worry which is oxidized..there is no mystery, each one is doing the same thing...@ different ox. numbers. Like in B/E, Fe +4 doesn't exist...so wrong

and as for D, like my post above, there doesn't need to be any coefficients because it was not asked.

 

[levyusupov]

big hint: CONVERT to MOLES and see the ratio between them.

recall that (M)(liter)= moles.

that is 100% correct !!!

 

[Danny289]

During a titration it was determined that 30.00
mL of a 0.100 M Ce4+ solution was required to
react completely with 20.00 mL of a 0.150 M
Fe2+ solution. Which one of the following reactions
occurred?
A. Ce4+ + 3Fe2+ + H2O  3Fe3+ + CeO- + 2H+
B. 2Ce4+ + Fe2+  Fe4+ + 2Ce3+
C. Ce4+ + Fe2+  Fe3+ + Ce3+
D. Ce4+ + 2Fe2+  2Fe3+ + Ce2+
E. Ce4+ + 2Fe2+  2Fe4+ + Ce2+ + 2e-

I dont understand it at all. how do you determine from this info which is reduced or oxidized?? answer is C

Thanks!!!

again concept! if you know the concept you wish all questions being like this,
look at this words: "30 mL of a 0.100 M Ce4+
and 20.00 mL of a 0.150 M Fe2+ then:

30 X 0.100 = 3 for Ce4+
20 X 0.150 = 3 for Fe2+
result : "n" just can be "1" what is "n"?
recall from Normalite : N= M x n and for solutions we have N1V1=N2V2


"n" is number "OH" in bases or number of "H" in acids or the number of oxidation and reduction changes in oxidation reduction reactions

we must look to the reaction that having n=1 for both reacant( in this problem we don't have acid or base, then we must look for reduction and oxidation)
notice : we must find one reduction and one oxidation agent, it is not possible both being oxide or both being redox( because we have transfer elctron from one substance to another, it is not possible both loosing elctron(oxide) or both gain electron(redox) happens ).

disscution:
choise A :
Ce+4 + 3e =====> Ce+1 (CeO- Ce has oxidation number +1 then n=3 redox can not be answer , we are looking for n=1
choise B :
Fe2+ =====> Fe+4 +2e n=2 oxide same reason can not be answer
Choise D: n=2 for Ce4+ same reason already is disscused can not be the answer
choise E :
can not be the answer already we talked about that plus we have 2 free electrons? the whole porpose of redox & oxdize reaction is transfering electron from one agent to another agent and in the end not having free electrons! ( review the concepts)
hopefully it helped.

 

[AaBbZz]

Just adding to what Danny298 said you can also just look at what cation Fe and Ce prefer to be
Fe is prefers to be 3+ and Ce prefers to be Ce3+

 

[levyusupov]

big hint: CONVERT to MOLES and see the ratio between them.

recall that (M)(liter)= moles.

that is 100% correct !!!

this solution also was approved by my instructor who happen to have PhD in chemistry.

 

[IsRaEl21]

I was also checking the answers and Lev's answer was correct and most accurate.

 

[Danny289]

😳

this solution also was approved by my instructor who happen to have PhD in chemistry.

Ok Dear Levy,
Now I am going to change this problem a little bit, try to solve it with your instructor method, and explain how you get the answer. this is the problem:
(attention: this problem has been changed from its orginal (30 ml===> 10 ml) )
During a titration it was determined that 10.00
mL of a 0.100 M Ce4+ solution was required to
react completely with 20.00 mL of a 0.150 M
Fe2+ solution. Which one of the following reactions
occurred?( reactions are not balanced)
A. Ce4+ + Fe2+ + H2O  Fe3+ + CeO- + H+
B. Ce4+ + Fe2+  Fe4+ + Ce3+
C. Ce4+ + Fe2+  Fe3+ + Ce3+
D. Ce4+ + Fe2+  Fe3+ + Ce2+
E. Ce4+ + Fe2+  Fe4+ + Ce2+ + 2e-
??
answer is : A

 

[levyusupov]

again concept! if you know the concept you wish all questions being like this,
look at this words: "30 mL of a 0.100 M Ce4+
and 20.00 mL of a 0.150 M Fe2+ then:

30 X 0.100 = 3 for Ce4+
20 X 0.150 = 3 for Fe2+
result : "n" just can be "1" what is "n"?
recall from Normalite : N= M x n and for solutions we have N1V1=N2V2


"n" is number "OH" in bases or number of "H" in acids or the number of oxidation and reduction changes in oxidation reduction reactions

we must look to the reaction that having n=1 for both reacant( in this problem we don't have acid or base, then we must look for reduction and oxidation)
notice : we must find one reduction and one oxidation agent, it is not possible both being oxide or both being redox( because we have transfer elctron from one substance to another, it is not possible both loosing elctron(oxide) or both gain electron(redox) happens ).

disscution:
choise A :
Ce+4 + 3e =====> Ce+1 (CeO- Ce has oxidation number +1 then n=3 redox can not be answer , we are looking for n=1
choise B :
Fe2+ =====> Fe+4 +2e n=2 oxide same reason can not be answer
Choise D: n=2 for Ce4+ same reason already is disscused can not be the answer
choise E :
can not be the answer already we talked about that plus we have 2 free electrons? the whole porpose of redox & oxdize reaction is transfering electron from one agent to another agent and in the end not having free electrons! ( review the concepts)
hopefully it helped.

first, you have problem with your solution !!

you give me 10 mL with 0.100M
and 20mL with 0.15 M (moles ratio of 3:1)


but you solve the problem with:
30 ml and 0.100M
and 20mL and 0.15 M (moles ration of 1:1)

of course i will not get the right answer, none of the choices has my solution, and ur solution is based on the wrong concentrations.

second, this problem got nothing to do with normality, dont comlicate the poor kid, it is a simple problem.

 

[IsRaEl21]

I think u need to rewrite ur quest.

 

[Danny289]

I think u need to rewrite ur quest.

My question is totally new question and it is not related to orginal question
why I changed the question in new form? not because testing you, or saying your instructor is wrong. actually your professor method is based on equivalants ( you can ask from him/her) and equivalant is base for Normality
the whole point is : in complete reactions the substances react in same equivalant.

lets give you a simple example look at this simple reaction:
H2SO4 + 2 NaOH =======> Na2SO4 + 2 H2O
you can translate this reaction in words like this:
one mole Sulforic acid reacts with two mole Sodium Hydroxide and give .......
nothing wrong about this translation but the better way to say is:
one mole Sulforic acid in complete reactions is equivalant to two mole Sodium Hydroxide.
Now lets approch for solving the problem:
we always have N1V1=N2V2 and recall of Normality N=M x n and we know whats n (I explained in orginal problem)
now plugs the number you have
10 X 0.1 X n1 = 20 X 0.15 X n2
and we we come to this ratio (n1/n2)=3/1
you know what is "n". in this reaction ''n'' is the changes of oxidation Number and look at your answers which one satsified your ratio? !!!!

B:
Ce+4 + 1e =========> Ce+3
Fe+2 ==============>Fe+4 + 2e
ratio 1/2

C:
Ce+4 + 1e =========> Ce+3
Fe+2 ==============>Fe+3 + 1e
ratio 1/1



D:
Ce+4 + 2e =========> Ce+2
Fe+2 ==============>Fe+3 + 1e
ratio 2/1

E: same plus we can see 2e can not be answer

A ( correct answer)
Ce+4 + 3e =========> Ce+1
Fe+2 ==============>Fe+3 + 1e
ratio 3/1

how we got oxidation number +1 for Ce in A

CeO-
Ce + (-2 X1)= -1
Ce= +1
tough question, just I wanted to show: knowing the concepts helps for solving for hard question like this, and for sure we will not see this kind of tough questions in DAT

good luck for your DAT

 

[levyusupov]

your solution is correct but complicated. the question did not ask for oxidatin/reduction or electron, etc. it only asks find the correct equation. why r u complicate it so much.

your concept that being used is too long (like u said) and will not appear on the DAT (according to you). But you are wrong. The original problem that eravin99 posted came straight from the ADA sample exam - so u need to know it. thus, in the DAT u will not gonna have a lot of time doing what u did, but u will have time to solve the problem by looking at the mole ratios, which is much simpler !!!!

GOOD LUCK !!!!

 

[levyusupov]

Just adding to what Danny298 said you can also just look at what cation Fe and Ce prefer to be
Fe is prefers to be 3+ and Ce prefers to be Ce3+

you can not look at what the cations preffer to be because these are transition metals, the charge on those can vary !!!!!

 

[Shunwei]

During a titration it was determined that 30.00
mL of a 0.100 M Ce4+ solution was required to
react completely with 20.00 mL of a 0.150 M
Fe2+ solution. Which one of the following reactions
occurred?
A. Ce4+ + 3Fe2+ + H2O ? 3Fe3+ + CeO- + 2H+
B. 2Ce4+ + Fe2+ ? Fe4+ + 2Ce3+
C. Ce4+ + Fe2+ ? Fe3+ + Ce3+
D. Ce4+ + 2Fe2+ ? 2Fe3+ + Ce2+
E. Ce4+ + 2Fe2+ ? 2Fe4+ + Ce2+ + 2e-

I dont understand it at all. how do you determine from this info which is reduced or oxidized?? answer is C

Thanks!!!

Actually this is not that hard. You know from the question and answser choices that this is a redox reaction, where Ce is functioning as an oxidizing agent to oxidize Fe. Then, if you look at the numbers, you can see that the actual # of moles being exchanged is the same from Ce to Fe. The only answer choice that fits this description is C, where a 1:1 ration exists of electron transfer/electron acceptance between Ce and Fe.

 

[Danny289]

your solution is correct but complicated. the question did not ask for oxidatin/reduction or electron, etc. it only asks find the correct equation. why r u complicate it so much.

your concept that being used is too long (like u said) and will not appear on the DAT (according to you). But you are wrong. The original problem that eravin99 posted came straight from the ADA sample exam - so u need to know it. thus, in the DAT u will not gonna have a lot of time doing what u did, but u will have time to solve the problem by looking at the mole ratios, which is much simpler !!!!

GOOD LUCK !!!!

Dear levy,
Just I wanted to say we must know the concept in general chemistry, thats it, and of course we are not going to see question like what i designed here in real DAT and you are absulatly right. for eravin orginal question your method works very well. and plus I am not smart like you guys, I must know the subjects before approaching for problems.
and thank you for your wishes in my DAT 👍

 

[Danny289]

Actually this is not that hard. You know from the question and answser choices that this is a redox reaction, where Ce is functioning as an oxidizing agent to oxidize Fe. Then, if you look at the numbers, you can see that the actual # of moles being exchanged is the same from Ce to Fe. The only answer choice that fits this description is C, where a 1:1 ration exists of electron transfer/electron acceptance between Ce and Fe.

very nice Quote from you, good and clear explanation 👍