Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: plaito_999 on May 29, 2010, 04:58:13 AM
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I have been doing an assignment, and I finished most of the questions, but there are a couple I just don't know how to do. I don't even know where to start. I don't want anyone to give me just the answers, but if someone could work through these problems with me, I would really appreciate it. That way I'll know how to answer questions like these come exam time.
The first question:
Write a balanced chemical equation for the oxidation of Ag(s) by concentrated nitric acid. Two products of the reaction are NO2 (g) and Ag+ (aq). What is the oxidising agent and what is the reducing agent?
I tried to take a stab at the equation, but I'm not sure if it's right. I got this: Ag(s) + 2HNO3(aq) :rarrow: Ag+(aq) + 2NO2 +H2O(l). I think a big part of my problem is that I have no idea how to identify oxidising and reducing agents.
The second question:
Use the standard reduction potentials below determine which compound or ion is the best oxidizing agent?
Cl2(g) + 2 e- :rarrow: 2 Cl-(aq) Eº = +1.36 V
Ag2+(aq) + e- :rarrow: Ag(s) Eº = +0.80 V
Fe+(aq) + 2 e- :rarrow: Fe(s) Eº = -0.44 V
This one I just don't have a clue about. If someone could work through this and show me how to solve it, I would be really grateful.
Third question:
Write a balanced chemical equation for the following reaction in a basic solution. H2O2(aq) + Cr(OH)3(s) H2O + CrO42-(aq).
I think I can figure out how to write the actual equation, but according to my textbook, to do this I need to know the 2 half reactions, and I don't know how to identify those.
Fourth question:
Consider the following half-reactions:
F2(g) + 2 e- :rarrow: 2 F-(aq) Eº = +2.87 V
Cu2+(aq) + 2 e- :rarrow: Cu(s) Eº = +0.337 V
Sn2+(aq) + 2 e- :rarrow: Sn(s) Eº = -0.14 V
Al3+(aq) + 3 e- :rarrow: Al(s) Eº = -1.66 V
Na+(aq) + e- :rarrow: Na(s) Eº = -2.714 V
Which of the above compounds or ions are able to reduce Al3+?
Once again, I just have no idea.
Last question:
Calculate the cell potential, at 25 ºC, based upon the overall reaction 3 Fe3+(aq) + Al(s) -> 3 Fe2+(aq) + Al3+(aq) if [Fe3+] = 0.300 M, [Fe2+] = 0.150 M, and [Al3+] = 0.300 M. The standard reduction potentials are as follows:
Fe3+(aq) + e- :rarrow: Fe2+(aq) Eº = +0.771 V
Al3+(aq) + 3 e- :rarrow: Al(s) Eº = -1.66 V
I think this one involves the use of this equation E = Eº - 0.0592/n log Q, but I don't know for sure, and if it does I'm not sure what the Q is supposed to stand for.
Any help that anyone could give me would be greatly appreciated, I'm running against the clock here. I really did try, but by the time I was ready to pull my hair out, I figured I should get some help.
These are due Monday the 1st of June, so any help before then would be great.
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To identify half reaction find out what is being reduced and what is being oxidized - you can do it finding oxidation numbers of all elements before reaction and after reaction. Change in ON means either reduction or oxidation. OIL RIG.
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1) First, since you are working with an aqueous solution, it would help if you converted your molecular equation into an ionic equation:
Ag(s) + 2H^+(aq) + NO3^-(aq) --> NO2^- + Ag^+(aq) + H2O(l)
This way is much easier to balance the reaction as well.
Next, assign oxidation numbers. Remember, ON numbers must add up to the compound's actual charge. Neutral elemental atoms have an oxidation number of zero. H usually has ON of + if bonded to a more electronegative atom and - if bonded to a less electronegative atom (usually a metal). Oxygen likes to have the charge of 2- . so try assigning the oxidation numbers for each atom. :) I can help if you get stuck
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Oxygen likes to have the charge of 2-
Beware: while you are right in general, there is H2O2 present in one of the questions.
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Oxygen likes to have the charge of 2-
Beware: while you are right in general, there is H2O2 present in one of the questions.
Right. These aren't hard and fast rules. Oxygen prefers to have a charge of 2-, but that is not always the case. Hydrogen peroxide has an O-O bond, which is an exception to the general rule
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Thanks for all the help, I tried to do the first question again, and came up with these as the half reactions:
oxidation 1/2 : Ag(s) :rarrow: Ag+(aq) + e-
reduction 1/2 : 2H+(aq) + e- :rarrow: H2O(l)
or is it this:
oxidation 1/2 : Ag(s) :rarrow: Ag+(aq) + e-
reduction 1/2 : NO3- + 2 e- :rarrow: NO2-
I'm really confused, about all of this.
Also, I managed to solve the last one. I ended up with 2.46 V
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oxidation 1/2 : Ag(s) :rarrow: Ag+(aq) + e-
That's OK
reduction 1/2 : 2H+(aq) + e- :rarrow: H2O(l)
Where does the oxygen come from? What is oxidation number of hydrogen before and after - has it changed? Was hydrogen reduced?
reduction 1/2 : NO3- + 2 e- :rarrow: NO2-
Why NO2- when your original question asked for NO2?
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Thanks for all the help, I tried to do the first question again, and came up with these as the half reactions:
oxidation 1/2 : Ag(s) :rarrow: Ag+(aq) + e-
reduction 1/2 : 2H+(aq) + e- :rarrow: H2O(l)
or is it this:
oxidation 1/2 : Ag(s) :rarrow: Ag+(aq) + e-
reduction 1/2 : NO3- + 2 e- :rarrow: NO2-
I'm really confused, about all of this.
Also, I managed to solve the last one. I ended up with 2.46 V
You are close. Here is what I got:
Oxidation 1/2 rxn: Ag(s) :rarrow: Ag+(aq) + e-
Reduction 1/2 rxn: NO-3 (aq) + 2H+ + e- :rarrow: NO2 (g) + H2O(l)
Nitrogen is reduced because its oxidation number changes from 5+ to 3+, so it is the oxidizing agent and
Silver is being oxidized because its oxidation number changes from zero to + , so it is the reducing agent.
I know this can be very confusing at first, especially with all the new terminology. If you are ever lost, try working backwards. Look at a standard reduction table and find the two half reactions that are involved in the total chemical equation, then assign the oxidation numbers and convince yourself that silver is indeed losing an electron (pair) and Nitrogen has gained an electron (pair).
For the second part, the strongest oxidizing agent usually has the greatest reduction potential.
And, now that you've had practice writing half-reactions, take another stab and the third question :)
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Also, I managed to solve the last one. I ended up with 2.46 V
Yep. I got 0.771 + 1.66 = 2.431 V
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Reduction 1/2 rxn: NO-3 (aq) + 2H+ + e- :rarrow: NO2 (g) + H2O(l)
Nitrogen is reduced because its oxidation number changes from 5+ to 3+
3+?
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Reduction 1/2 rxn: NO-3 (aq) + 2H+ + e- :rarrow: NO2 (g) + H2O(l)
Nitrogen is reduced because its oxidation number changes from 5+ to 3+
3+?
just kidding. it's 4+ . I mistakenly wrote NO2- in my original notes. Thanks for correcting me
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For the third question I tried to follow the steps in my textbook and I ended up with this: 2CrO42- + 8H2O :rarrow: 2 Cr3+ + 3H2O2 + 10 OH-
Is this correct?
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For the third question I tried to follow the steps in my textbook and I ended up with this: 2CrO42- + 8H2O :rarrow: 2 Cr3+ + 3H2O2 + 10 OH-
Is this correct?
For H2O2 + Cr(OH)3 :rarrow: H2O + CrO42-
I got 3H2O2 + 2Cr(OH)3 :rarrow: 4H2O + 2CrO42- + 4H+
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I ended up with this: 2CrO42- + 8H2O :rarrow: 2 Cr3+ + 3H2O2 + 10 OH-
Not bad, but you were asked about chromium oxidation, this is reduction ;)
Just reverse the reaction.
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I got 3H2O2 + 2Cr(OH)3 :rarrow: 4H2O + 2CrO42- + 4H+
Question asked for the reaction to be balanced in basic solution.
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Borek, when you say reverse the reaction do you mean reverse it to this 2 Cr3+ + 3H2O2 + 10 OH- :rarrow: 2CrO42- + 8H2O
Also, thank you so much to everyone who has helped, I really appreciate it.
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2 Cr3+ + 3H2O2 + 10 OH- :rarrow: 2CrO42- + 8H2O
Cr(OH)3, but in general yes, that's what I meant - exchange lhs with rhs.
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Borek, so then the final answer is ... 3H2O2 + 2Cr(OH)3 + 4OH- :rarrow: 8H2O + 2CrO42- ?