Balance this redox rxn

[rocketbooster]

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Fe2+ + Na2Cr2O7 ---> Fe3+ + Cr2O3

I don't really care if you go as far as to add the OH-/H+ and H2O. Just go far enough to get the coefficients of the actual reacting species above.

I keep getting 3:1---->1:3

Supposedly, it's 6:1---->1:6 because each chromium gets 3 electrons, so since there are 2 chromiums then they get a total of 6 electrons. I don't get it, though, because it goes from (Cr2O7)-2 to Cr2O3. Simply figure out the oxidation # of each chromium in each compound and you get +6 and +3. That's an addition of 3 electrons total and those oxidation #s are based on taking into account the 2 chromiums to begin with.

Anyone?

 

[MSUSpartan642]

This is not so much useful advice, but I just dont bother with balancing redox equations. To much work for one point on the MCAT. Though I know this is of no help to you. Mark and move on, if you have time go back. Haha, sorry for being worthless.

 

[rocketbooster]

This is not so much useful advice, but I just dont bother with balancing redox equations. To much work for one point on the MCAT. Though I know this is of no help to you. Mark and move on, if you have time go back. Haha, sorry for being worthless.

LOL, thanks for the reply. 😛

balancing redox questions is REALLY easy. it seriously takes 30 seconds as long as you have practiced doing it a few times before your test. it's an easy thing to do and worth the point as long as you can do it fast!

I know how to balance them fine, but this one makes no sense to me. based on the way I was taught (BR), it should be 3:1--->1:3, not 6:1!

Anyone?

 

[Bernoull]

Fe2+ + Na2Cr2O7 ---> Fe3+ + Cr2O3

I don't really care if you go as far as to add the OH-/H+ and H2O. Just go far enough to get the coefficients of the actual reacting species above.

I keep getting 3:1---->1:3

Supposedly, it's 6:1---->1:6 because each chromium gets 3 electrons, so since there are 2 chromiums then they get a total of 6 electrons. I don't get it, though, because it goes from (Cr2O7)-2 to Cr2O3. Simply figure out the oxidation # of each chromium in each compound and you get +6 and +3. That's an addition of 3 electrons total and those oxidation #s are based on taking into account the 2 chromiums to begin with.

Anyone?

I got a unbalanced redox eq of

6Fe(2+) + 6H(+) + Na2Cr2O7 ---> Cr2o3 +3H2O +6Fe (3+)

It's unbalanced "atomically" wrt oxygen atoms because I think there's a sodium oxide product that you left out (Na2O). Nonetheless, its balanced electronically.

I think you mistake is in assigning the ox. state for Cr. You say "I don't get it, though, because it goes from (Cr2O7)-2 to Cr2O3. Simply figure out the oxidation # of each chromium in each compound and you get +6 and +3."

however this is incorrect bcos since Na2Cr2O7 is neutral,
the ox # of Cr2 = sum of -14(O) +2(Na) = +12
therefore each Cr = +6,
Alternative you correctly stated that (Cr2O7) have -2 charge and since O7 has -14, this leaves Cr2 with +12. Either way once you figure this out the rest is easy.

 

[rocketbooster]

I got a unbalanced redox eq of

6Fe(2+) + 6H(+) + Na2Cr2O7 ---> Cr2o3 +3H2O +6Fe (3+)

It's unbalanced "atomically" wrt oxygen atoms because I think there's a sodium oxide product that you left out (Na2O). Nonetheless, its balanced electronically.

I think you mistake is in assigning the ox. state for Cr. You say "I don't get it, though, because it goes from (Cr2O7)-2 to Cr2O3. Simply figure out the oxidation # of each chromium in each compound and you get +6 and +3."

however this is incorrect bcos since Na2Cr2O7 is neutral,
the ox # of Cr2 = sum of -14(O) +2(Na) = +12
therefore each Cr = +6,
Alternative you correctly stated that (Cr2O7) have -2 charge and since O7 has -14, this leaves Cr2 with +12. Either way once you figure this out the rest is easy.

oh, I get it now. What I meant with +6 and +3 is the chromium in the reactant Cr is +6 and the the chromium in the product Cr is +3. that's correct.

the thing is the reaction is dealing with Cr2, though, not Cr. so while there are 3 e- between +6 and +3 of Cr, you have to multiply the 3e- by 2 since the reaction is using Cr2, not Cr.

I don't know how your explanation helped me exactly, but it did because I get it now! Thanks

ALSO, you just made me figure out a redox problem I couldn't balance from 2 months ago! I never realized you had to multiply the # of e- if the atom you are looking at has a subscript.


HOWEVER, let's say based on that same reaction, the question asked "How does the formal charge of chromium change from reactant to product?"

The answer would be +6 to +3, NOT +12 to +6, right? If it asked how Cr2 changed or something, then it would be +12 to +6, correct?

 

[Bernoull]

oh, I get it now. What I meant with +6 and +3 is the chromium in the reactant Cr is +6 and the the chromium in the product Cr is +3. that's correct.

the thing is the reaction is dealing with Cr2, though, not Cr. so while there are 3 e- between +6 and +3 of Cr, you have to multiply the 3e- by 2 since the reaction is using Cr2, not Cr.

I don't know how your explanation helped me exactly, but it did because I get it now! Thanks

ALSO, you just made me figure out a redox problem I couldn't balance from 2 months ago! I never realized you had to multiply the # of e- if the atom you are looking at has a subscript.


HOWEVER, let's say based on that same reaction, the question asked "How does the formal charge of chromium change from reactant to product?"

The answer would be +6 to +3, NOT +12 to +6, right? If it asked how Cr2 changed or something, then it would be +12 to +6, correct?

"What I meant with +6 and +3 is the chromium in the reactant Cr is +6 and the the chromium in the product Cr is +3. that's correct." Ur absolutely right each Cr atom gains 3e'. Now you figured out ur error which is that stoichiometric coefficients are important. The reduction electrons MUST come from the oxidation half-reaction therefore the absolute number of e's lost in oxidation are gained in reduction and this is why stoichiometric coefficients are important.

As far as the difference b/t formal charge and oxidation states, they are usually different. FC tells u the actual charge of an atom within a compound whereas ox state is simply a model to explain electron transfer in redox reaction, the charges really don't exist, it's like ideal gas (they don't exist, it's simply a model) although sometimes the ox charges can coincide with FC. I can't calculate the FC for Cr, bcos I don't know the atoms in Na2Cr2O7 bond n i really don't want to figure that out. So I'll leave it to someone else to verify ur calculation, andif u have a way of finding FC w/o knowing how the atoms bond I'll love to know..

 

[rocketbooster]

"What I meant with +6 and +3 is the chromium in the reactant Cr is +6 and the the chromium in the product Cr is +3. that's correct." Ur absolutely right each Cr atom gains 3e'. Now you figured out ur error which is that stoichiometric coefficients are important. The reduction electrons MUST come from the oxidation half-reaction therefore the absolute number of e's lost in oxidation are gained in reduction and this is why stoichiometric coefficients are important.

As far as the difference b/t formal charge and oxidation states, they are usually different. FC tells u the actual charge of an atom within a compound whereas ox state is simply a model to explain electron transfer in redox reaction, the charges really don't exist, it's like ideal gas (they don't exist, it's simply a model) although sometimes the ox charges can coincide with FC. I can't calculate the FC for Cr, bcos I don't know the atoms in Na2Cr2O7 bond n i really don't want to figure that out. So I'll leave it to someone else to verify ur calculation, andif u have a way of finding FC w/o knowing how the atoms bond I'll love to know..

oh, MY FAULT. okay the question should be....How does the oxidation state of chromium change from reactant to product?

The answer is +6 to +3, not +12 to +6, right? In the actual redox question, you do have to use 6e-, not 3e-, correct?

 

[Bernoull]

oh, MY FAULT. okay the question should be....How does the oxidation state of chromium change from reactant to product?

The answer is +6 to +3, not +12 to +6, right? In the actual redox question, you do have to use 6e-, not 3e-, correct?

Pheww.. u got me there with the FC question!! Okay back to ox. states...
".How does the oxidation state of chromium change from reactant to product?"

Yes you are correct IF the question asks about a Cr atom, and if it's Cr2 (molecule), it's (ox. state/atom)* (#of atoms in molecule) = +6*2 = 12

It's purely a question of stoichiometric coeffs. and the same applies for electron transfer. Each Cr atom accepts 3e's but since there 2Cr atoms u multiply by 2. That's it. 😀😀😀

 

[rocketbooster]

Pheww.. u got me there with the FC question!! Okay back to ox. states...
".How does the oxidation state of chromium change from reactant to product?"

Yes you are correct IF the question asks about a Cr atom, and if it's Cr2 (molecule), it's (ox. state/atom)* (#of atoms in molecule) = +6*2 = 12

It's purely a question of stoichiometric coeffs. and the same applies for electron transfer. Each Cr atom accepts 3e's but since there 2Cr atoms u multiply by 2. That's it. 😀😀😀

YUP, and those questions always refer to single atoms for some reason. they never ask Cr2 or whatever. so, if the question asked how the oxidation # changes for chromium, you should assume they mean the chromium atom, not Cr2 or whatever Cr2 is called IDK.

I'm glad you caught my formal charge/oxidation state typo because in my head I was interchanging them, which is a big NO NO. I know how to calculate formal charge, but for some reason my brain kept referring to them as the samething. anyways, whatever, problem solved. 👍

 

[BerkReviewTeach]

YUP, and those questions always refer to single atoms for some reason. they never ask Cr2 or whatever. so, if the question asked how the oxidation # changes for chromium, you should assume they mean the chromium atom, not Cr2 or whatever Cr2 is called IDK.

I'm glad you caught my formal charge/oxidation state typo because in my head I was interchanging them, which is a big NO NO. I know how to calculate formal charge, but for some reason my brain kept referring to them as the samething. anyways, whatever, problem solved. 👍

Excellent processing and reasoning by both of you. You should note that besides Cr2O72- in terms of molecules where multiple atoms get reduced, there is also HOOH (peroxide).

I'm going to move this thread to the questions and answers section.