crayons U4 exam questions ~

[crayolé]

Alright, it's that time of year again ;]
So before i start going kenhung ( :smitten: )on your asses by spamming up this board with individual question threads I thought I might stick it all in here for everyones convenience ;]

Much appreciation and many karmas coming your way

First up
How would one go about finding the enthalpy change of H1?

[vexx]

Hi, hopefully you have the answers, but what i would probably do (someone correct me if i'm wrong/let me know if this is right with answers) is considering you are forming C + 4H, and atomisation of Hydrogen is H2--> 2H, is times that figure by 2 = 717*2, and then add the atomisation of carbon because there is only 1, and thennn since you are breaking down methane add that too, to get 1726kj/mol
And the rule of thumb is if bonds break it is endothermic, and so it would be =+1726kj/mol

[crayolé]

Cheers ;]

Okay, For this question, i have all the concentrations and stuff right its just my equilibrium constant equation that I can't get right.

So in this closed system, the products are [Cl2] and [CO] right? So they should be the numerator with the [CoCl2] being the denominator.
The answers do it the other way around though?

Also, would the reduction reaction during the corrosion of Fe(s) use the
O₂ +4H⁺ +4e- ----> 2H₂O
or
O₂ +2H₂O +4e- ----> 4OH-

[Edmund]

Cheers ;]

Okay, For this question, i have all the concentrations and stuff right its just my equilibrium constant equation that I can't get right.

So in this closed system, the products are [Cl2] and [CO] right? So they should be the numerator with the [CoCl2] being the denominator.
The answers do it the other way around though?

Also, would the reduction reaction during the corrosion of Fe(s) use the
O₂ +4H⁺ +4e- ----> 2H₂O
or
O₂ +2H₂O +4e- ----> 4OH-

The equilibrium constant should be derived from the given reaction. The question does say that phosgene gas is the product of the reaction:



The reduction reaction would be since oxygen gas and water is required for corrosion, to provide the electrons.

[crayolé]

Cheers guys, heres a couple more to bug you about.

Just got semi-fucked over with NEAP 09, need these questions answered ;3

Okay, so for the first one.
For the cathode reaction, is O₂ + 4H⁺ + 4e^- -----> 2H₂O correct also?

Second one, Can someone please run through the method on how to derive individual half reactions from a full one?

Cheers

[fady_22]

For the cathode reaction, is O₂ + 4H⁺ + 4e^- -----> 2H₂O correct also?

No, that isn't right.
Where is the source of H+? You have to look at what reactants are available to you.

Second one, Can someone please run through the method on how to derive individual half reactions from a full one?

I usually look at the oxidation number changes in the reactants. From there, you can see if reduction/oxidation has occurred, and you can write half equations easily (just in the normal way).

So, for the anode, look for oxidation. Oxidation is an increase in oxidation number, which occurs with zinc.

[crayolé]

Thanks fady ;]

For this one, there's the question and I've also included the answer

So let me get this right,  if an iron electrode is used as the anode, we would see the reaction
instead.

but this reaction is beneath the in the electrochemical series

Is it possible for an electrolytic cell to go from a non-spontaneous reaction to a spontaneous one just like that?

EDIT: thanks Studyinghard

[Studyinghard]

If there is a power source present it doesnt matter if its a spontaneous or non spontaneous reaction it will be still classified as a electrolytic cell so you cant call it a galvanic reaction/cell.

[stonecold]

According to my teacher, you balance oxygen with water for an acidic cell, then if it is an alkaline cell, you neutralize the H+ with OH-.  If it is neutral, then just O2 + 4e- > 2O2-

[Studyinghard]

According to my teacher, you balance oxygen with water for an acidic cell, then if it is an alkaline cell, you neutralize the H+ with OH-.  If it is neutral, then just O2 + 4e- > 2O2-

my tutor explained a bit differently. He followed the original steps for balancing redox reactions and then told us to add OH- ions to which ever side that didnt have the OH- ions and then combine them with H+ to form water. (if there were any H+ present)

[stonecold]

Cheers guys, heres a couple more to bug you about.

Just got semi-fucked over with NEAP 09, need these questions answered ;3

Okay, so for the first one.
For the cathode reaction, is O₂ + 4H⁺ + 4e^- -----> 2H₂O correct also?

Second one, Can someone please run through the method on how to derive individual half reactions from a full one?

Cheers

But would that work for this question?

[crayolé]

Thanks guys.
Just for clarification,  can spontaneous reactions still occur in an electrolytic cell?


And theres another question that I'm stumped on, not sure if its because of the wording or what but I just can't seem to get my head around it

Water and the iron(II) ion are both able to act as either an oxidant or a reductant during a chemical reaction. Select the equation representing a reaction that would be expected to proceed substantially to the right spontaneously.
A.   Fe (s) + 2 Fe3+ (aq) -----------------> 3 Fe2+ (aq)
B.   2 H2O (l) -----------------> O2 (g) + 2 H2 (g)
C.   2 H2O (l) + 2 Fe2+ (aq) -----------------> O2 (g) + 4 H+ (aq) + 2 Fe (s)
D.   2 H2O (l) + 4 Fe3+ (aq) -----------------> O2 (g) + 4 H+ (aq) + 4 Fe2+ (aq)

[vexx]

^ It's asking pretty much to refer to the electrochemical series and find a reaction between two reactants that can have a spontaneous reaction
ie: negative gradient when connecting the two reactants

look at A)
Fe(s) is a relatively strong reductant that can react with the relatively strong oxidant Fe3+(aq). They have a negative gradiant when connected, and so are spontaneous. So balance charges and check to see if that is the correct answer.
'to the right' is basically just in the forward direction so it doesn't really mean anything being there..

B-D If you look at reactions with H20, you can see that water is a weak reductant or a weak oxidant when it is on it's own (ie not with O2) and so will not be likely to react spontaneous as it will form a positive gradient with most reactants (that being it is not spontaneous), but check each of the above using the electrochemical series, balancing out charges and such to see if they are there.

Then you should get the correct answer, which though i haven't done the entire question seems to be somewhat obvious.

[crayolé]

Cheers vexxyboy :smitten:

How would one go about separating this equation into two half-reactions?

 

[vexx]

Cheers vexxyboy :smitten:

How would one go about separating this equation into two half-reactions?

 

^^

Well, you can see that Zn ---> Zn0
And Ag2O ----> 2Ag
So just balance them by adding in the waters, H+ ions and electrons in the correct way.

Pretty sure that's it.