Couple more GC Destroyer Questions

[Muggs]

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#262
Consider the equation Cr2O7-2 + H+ + I- --> Cr+3 + H2O + I3-
Which coefficient would be needed to balance I-?

It's to my understanding that you don't have to use H2O and H+ to balance out the two half reactions.
So for Cr2O7-2 --> 2Cr+3
The oxidation state of Cr2 would be (+12) and that for 2Cr+3 would be (+6).

Now my question for this part is, is this the correct way to look at the oxidation states? So for example, you wouldn't look at the oxidation of one Cr of Cr2, which would be +6, nor would you look at one Cr of the 2Cr+3, which would be +3, because that gives a change of 3 electrons for the half reaction, when you are looking for 6. In other words, when the half reaction is balanced, you would look at the oxidation state for the 2 moles of Cr and not just the single Cr, right?

#266
Organize in decreasing order with respect to atomic radii.
Br- > I- > F-
K+ > Ca++ > Al+3

Aren't both of these answer choices correct? The solutions states that only the second one is.

#256
Which statement is false?
Strontium cation is larger than sulfur anion since negative ions gain electrons.

I know that this statement is false given that it is stating that cations are larger than anions (which is false), but I thought that this comparison was valid for an isoelectronic series (elements with same number of electrons). Even though this question is asking about a strontium cation, it would still have 37 electrons vs. a Sulfur anion, which would have 17 electrons. Does it then not matter what shell the atoms are as long as one is a cation and the other is an anion?

 

[Dgeorg6]

on #266 Atomic Radius goes down and to the left if they don't have the same amount of electrons or whatever so it would be I > Br > F i think..

in the 2nd case they all have same electrons so when they all have the same electrons you always pick the one with the least amount of protons

 

[Muggs]

on #266 Atomic Radius goes down and to the left if they don't have the same amount of electrons or whatever so it would be I > Br > F i think..

in the 2nd case they all have same electrons so when they all have the same electrons you always pick the one with the least amount of protons

wow. i thought I was looking at Br- > Cl- > F-
the I- looked like a Cl- haha. thanks

 

[Muggs]

anyone have an explanation for 262 and 256?

 

[aqz]

#256:

Atomic radius is determined by how closely an atom's nucleus can pull in its electrons. The reason it increases down a column is because you're adding more electron shells (principal quantum number increases...or "n" increases) and electron shells are able to shield the proton attraction force. Similarly, the reason why atomic radius decreases going left to right across a period is because there are more protons = more positive attraction. Even though you gain more electrons, you are not adding new electron shells (e.g. n does not increase) and cannot shield the increasing proton attraction.

Therefore, it is true that Sr+2 and S-2 have the same amount of electrons, meaning they share the same amount of electron shells. However, strontium has more protons, which generates more force on the electrons than sulfur does, making its atomic radius smaller.

To answer your question, the only way a cation would have a larger radius than an anion is if the cation had more electron shells than the anion (e.g. cation was in a lower period than anion).

#262

I got lost trying to figure out what your question is so this might not answer your question but....

You don't need to consider oxidation states when balancing redox reactions. The coefficient DOES affect the charge, but it does NOT affect oxidation state/oxidation number afaik.

 

[Muggs]

#256:


Therefore, it is true that Sr+2 and S-2 have the same amount of electrons, meaning they share the same amount of electron shells. However, strontium has more protons, which generates more force on the electrons than sulfur does, making its atomic radius smaller.

To answer your question, the only way a cation would have a larger radius than an anion is if the cation had more electron shells than the anion (e.g. cation was in a lower period than anion).

#262

I got lost trying to figure out what your question is so this might not answer your question but....

You don't need to consider oxidation states when balancing redox reactions. The coefficient DOES affect the charge, but it does NOT affect oxidation state/oxidation number afaik.

But isn't that what is pictured? Sr+2 is equivalent to Kr in electrons (4th shell), whereas S-2 is equivalent to Ar in electrons (3rd shell).

 

[Muggs]

#256:

#262

I got lost trying to figure out what your question is so this might not answer your question but....

You don't need to consider oxidation states when balancing redox reactions. The coefficient DOES affect the charge, but it does NOT affect oxidation state/oxidation number afaik.

Then how would you balance MnO4- + I- --> Mn2+ + I2
Chad uses the oxidation states to find the number of electrons required.

 

[aqz]

Okay...I tried looking up past threads on this and I think the only explanation is that either the answer in Destroyer is wrong and Sr++ is bigger than S--, or the answer is right because that one electron shell difference between Sr++ and S-- is not enough to make Sr++ bigger. Personally, I'd just go with cations are smaller than anions and leave it at that, there's more important things to worry about.

For MnO4- + I- --> Mn2+ +I2, depending on if the problem stated in basic or acidic solution, you would balance it using H2O and H+

My fingers hurt from typing, hope you don't mind if I refer you to some links~

Acidic solution:

http://answers.yahoo.com/question/index?qid=20110425174610AAtHZTc

Basic solution:

http://answers.yahoo.com/question/index?qid=20080423161705AAH3IV9

I see what you mean by oxidation state method, but in my opinion this method is more accurate, and not really all that time-consuming once you get it. Takes me about 2 minutes max.

 

[Muggs]

thanks for the links and help!