H2 Chem TYS MCQ qns :D

Charmy

21 Oct 09, 17:17

hi, would like to ask about these few questions:

1) The protein albumin (eggwhite) is precipitated from its aqueous solution when it is poured into X. What could X be?
1. a saturated aqueous solution of (NH4)2SO4
2. 0.1 mol dm-3 sulphuric acid
3. 0.1 mol dm-3 Pb(NO3)2
(Ans: all three)

i have no idea what this question is asking. i'm guessing the denaturisation of proteins? if so, why would 1. be an answer? i know 2. messes with the pH and 3. is a metal ion if i'm not wrong.

2)A domestic carbon monoxide detector is based on an electrochemical cell consisting of the electrodes:
CO2(g) + 2H+(aq) + 2e- --> CO(g) + H2O(l) E= -0.10V
H2O(l) --> 1/2O2(g) + 2H+(aq) +2e- E= +1.23V

Which statements about the cell are true?
1. In the cell, carbon is oxidised
2. Under standard conditions the cell potential is +1.33V
3. As the concentration of carbon monoxide in the detector increases, the cell potential becomes more positive.
(Ans: all three)

for one i know carbon is oxidised because it changes from CO to CO2.
and 2. is right cause E(red) - E(oxi) = +1.23 - (-0.10) = +1.33V
but i can't find anything concerning 3. in my notes. o.o

on a side note though, why wouldn't i need to flip the equations since its going from CO to CO2? and if i really flip, 2. would be wrong. [E(red) - E(oxi) = -1.23 - (0.10) = -1.33V]

and for some questions, the E(red) - E(oxi) equation doesn't apply. like you have to add up the E values instead. i'm confused.

3) Which reagent can be used to distinguish between seperate samples of ethanol and propanone?
A [Ag(NH3)2]+
B 2,4-DNPH
C Iodine and aq NaOH
D Sodium borohydride
(ans: A)

my answer is B and i'm pretty sure it's B because A tests for aldehydes and there isn't any aldehyde here? C would work for both ethanol and propanone so no. and i'm not quite sure what D is, a reducing reagent if i'm not wrong? i think it doesn't show any colour changes/bubbles for it's reaction with propanone so no. so am i overlooking anything?

yes, thats all. please help! and thanks lots! (:

Eyelessz

21 Oct 09, 18:38

Hi,

1) Addition of (NH4)SO4 is an ionic salt, NH4+ and SO42-, it disrupts the ionic bonds in protein

2) As conc. of CO increases, By Le chatelier's principle, what will happen? Favour the ____ reaction, hence ur E' for oxidation becomes ______ and sub in into the E cell eqn of Red - Oxi, what will ur E'Cell be? Higher? Lower?

3) 2,4-DNPH seems to be the right answer, as it differntiates between carbonyl compound and non- cabonyl compounds. I'm guessing that the answer is wrong.

Charmy

21 Oct 09, 20:03

thanks eyeless!

for 2), it will favour the reaction CO -> CO2. but how does that affect the E' value? i'm so sorry this seems so basic. ><

CF18

21 Oct 09, 20:21

Simple...drop chem...easy as ABC, 123, yee er san, satu dua tiga, onru, irandu, munru. Racial Harmony at work

UltimaOnline

21 Oct 09, 20:34

2)A domestic carbon monoxide detector is based on an electrochemical cell consisting of the electrodes: CO2(g) + 2H+(aq) + 2e- --> CO(g) + H2O(l) E= -0.10V H2O(l) --> 1/2O2(g) + 2H+(aq) +2e- E= +1.23V ~~~ Which statements about the cell are true? 1. In the cell, carbon is oxidised 2. Under standard conditions the cell potential is +1.33V 3. As the concentration of carbon monoxide in the detector increases, the cell potential becomes more positive. (Ans: all three) for one i know carbon is oxidised because it changes from CO to CO2. and 2. is right cause E(red) - E(oxi) = +1.23 - (-0.10) = +1.33V but i can't find anything concerning 3. in my notes. o.o ~~~ If based strictly upon the equations and information given by this question, then : When molarity of CO increases, position of equilibrium shifts left, ie. oxidation potential becomes more positive ( > +0.1V ), or the reduction potential becomes more negative ( < -0.1V ). Because the 2nd equation given is written as an oxidation half-equation, which has a positive value of +1.23V, it can be deduced that the 2nd equation (oxidation) occurs at the anode. Which means that the 1st equation (reduction) occurs at the cathode. Cell potential = oxidation potential at anode + reduction potential at cathode. Since the reduction potential becomes more negative, the cell potential becomes less positive. This is what Wikipedia has to say about "Electrochemical Carbon Monoxide Detector" : This is a type of fuel cell that instead of being designed to produce power, is designed to produce a current that is precisely related to the amount of the target gas (in this case carbon monoxide) in the atmosphere. Measurement of the current gives a measure of the concentration of carbon monoxide in the atmosphere. Essentially the electrochemical cell consists of a container, 2 electrodes, connection wires and an electrolyte - typically sulphuric acid. Carbon monoxide is oxidised at one electrode to carbon dioxide whilst oxygen is consumed at the other electrode. For carbon monoxide detection, the electrochemical cell has advantages over other technologies in that it has a highly accurate and linear output to carbon monoxide concentration, requires minimal power as it is operated at room temperature, and has a long lifetime (typically commercial available cells now have lifetimes of 5 years or greater). Until recently, the cost of these cells and concerns about their long term reliability had limited uptake of this technology in the marketplace, although these concerns are now largely overcome. Therefore, the correct half equations would be : Anode : CO(g) + H2O(l) --> CO2(g) + 2H+(aq) + 2e- E = +0.10V (assuming original question data given by Charmy is valid) Cathode : O2(g) + 4H+(aq) + 4eâ€“ --> 2H2O(l) E = +1.23V Therefore, when molarity of CO(g) increases, oxidation potential at anode increases (by Le Chatelier's principle). Since cell potential = oxidation potential at anode + reduction potential at cathode, hence overall cell potential increases. ~~~ on a side note though, why wouldn't i need to flip the equations since its going from CO to CO2? and if i really flip, 2. would be wrong. [E(red) - E(oxi) = -1.23 - (0.10) = -1.33V] and for some questions, the E(red) - E(oxi) equation doesn't apply. like you have to add up the E values instead. i'm confused. ~~~ To avoid confusion, choose either one formula and stick with it : EITHER : Cell potential = oxidation potential at anode + reduction potential at cathode. OR : Cell potential = reduction potential at cathode - reduction potential at anode.

dkcx

21 Oct 09, 20:37

Isn't it always preferred to just use reduction potential for both? Even at uni level we do not use the oxidation potential for any calculations.

Charmy

21 Oct 09, 20:38

right, i wish. haha. it's 3 weeks to exams. can't back out now ^^

Charmy

21 Oct 09, 20:50

so, it means that the answer given is wrong?? since the cell potential becomes less positive if strictly following the question?

and about the relation between the potential values and concentration, am i supposed to memorise that by hard??

and isn't the first equation supposed to be oxidation since carbon is oxidised?? ..

UltimaOnline

21 Oct 09, 23:02

~ Isn't it always preferred to just use reduction potential for both? Even at uni level we do not use the oxidation potential for any calculations. ~ Not true, actually. It's only a matter of convention (ie. standardization for the sake of convenience, not because it's more 'correct') that reduction potentials are more often used. Conceptually, oxidation and reduction potentials, are every bit as valid, correct, important and scientific, as oxidation and potential itself. In terms of calculations, regardless of whether oxidation or reduction potentials are used, there is no difference in the calculated result. There are actually professional chemists and university professors & lecturers, who even favour the use of oxidation potentials over reduction potentials. Personally for myself, I find it more meaningful to use both. The vast majority of students who use only reduction potentials (for which most JC students and even JC teachers often wrongly call "eeee-nought value"; that's not a "nought" symbol, it's a "standard" symbol! Different meaning!) don't truly realize or appreciate the meaning (ie. they just blindly memorize it!) the formula for calculating cell potentials. I encourage students to appreciate deeper meaning in their chemistry, their work, their life, their relationships. Most people don't, at all. Pity. ~~~ so, it means that the answer given is wrong?? since the cell potential becomes less positive if strictly following the question? and about the relation between the potential values and concentration, am i supposed to memorise that by hard?? and isn't the first equation supposed to be oxidation since carbon is oxidised?? ~~~ Don't bother with the question; it's far more important for you now, to expose yourself to, understand and appreciate the correct chemistry behind the carbon monoxide detector. Use Wikipedia to support your self-learning, Wikipedia - One of the Best Things that Happened to Mankind. As for redox potentials and concentration, just apply Le Chatelier's principle for a qualitative understanding (which is all that's required for 'A' levels). As for your last question, don't bother with it. Since the question is erroneous, it'll just confuse you further if you bother with it. But the idea is, by comparing redox potentials (if you know the reduction potentials, you also know oxidation potentials, and vice-versa), you can decide which is the reduction half-equation at the cathode, and which is the oxidation half-equation at the anode.

Charmy

22 Oct 09, 22:45

okay, i'm much clearer now. thanks a lot!

UltimaOnline

23 Oct 09, 00:03

With regards to Le Chatelier's principle, the more positive the redox potential (could be oxidation or reduction potential, depending on whether an oxidation or reduction half-equation is written next to the redox potential value), the more the position of equilibrium lies to the right. The more negative the redox potential, the more the position of equilibrium lies to the left. Hence, adding LHS species will result in the reduction/oxidation potential becoming more positive / less negative (as position of equilibrium shifts to the right); and adding RHS species will result in the reduction/oxidation potential becoming less positive / more negative (as position of equilibrium shifts to the left).