Lurkers, register with SgForums and ask ur H2 Chem qns here!

• UltimaOnline

  20 Sep 15, 05:20

  Originally posted by BCME:

  Hi, the following qns are asked with reference to VJC_P3_2014

  1. 1(a)(iii) not very sure how to calculate the ratio 

  2. 2(d) how do we know how to construct the equation between P and NaOH? 

  3. 3(i) why KMnO4 cannot be used?

  4. 4(b) I know that J contains COOH. However when it is reduced won't it be converted to primary alcohol and alcohol will react with conc sulfuric acid so I am a bit confused. 

  Thanks

  
  Q1. At equilibrium, by definition, no electric current flows through the circuit, Gibbs free energy change = Cell Potential = 0 volts.
  
  Kc = [Products] / [Reactants]
  
  Hence Kc = [Mg2+] / [Cu2+]
  
  Since Mg2+ is the RHS product and Cu2+ is the LHS reactant.
  
  Q2. This is a disproportionation reaction in which highly reactive and unstable P4, undergoes (base-promoted) hydrolysis, and is simultaneously oxidized to oxo-acidic hypophosphorous acid (immediately deprotonated into its conjugate base salt), and reduced to hydridic-basic phosphine.
  
  Note the electronegativities of O > P > H, which will help you to better understand the oxidation-to-acidic and reduction-to-basic nature of the hydrolysis reaction.
  
  The chemistry and mechanism for this reaction is, of course, beyond the A level syllabus. But sufficient clues are given in the question, specifically data that enables you to calculate the empirical formula of species C, and of course, the textual description of the reaction of A + NaOH + H2O* --> B + C.
  
  * Whenever there is (aq), eg. NaOH(aq), the astute or distinction student will be open to the possibility of hydrolysis (ie. reaction with H2O), especially in acidic or alkaline conditions. Eg. In Organic Chemistry, whenever KMnO4 / H+ or KMnO4 / OH- is used, don't just think oxidation, but open yourself to the possibility of acidic or alkaline hydrolysis of any ester / amide / nitrile / etc groups simultaneously occurring as well, in challenging deductive elucidation questions.
  
  Q3. Heating under reflux with acidified KMnO4, the alkyl side chains of the benzene rings (there are 2) would be oxidized, yielding ortho-phthalic acid (from benzene ring on the left), benzoic acid (from benzene ring on the right), 2-methylpropandioic acid (centre fragment), and (from the methyl group beta to the benzene ring on the left) carbonic(IV) acid (which decomposes into CO2(g) and H2O(l) with thermodynamically favourable positive entropy change, and as predicted by Le Chatelier's principle).
  
  Q4. While J is a primary alcohol, its only beta C atom happens to be a tertiary C atom, ie. no beta H atom (required for elimination together with the alpha OH group during dehydration of alcohols to alkenes) is present. This clues you in on the branched structure of fragment J, and hence of parent G as well.
• Mrworry

  21 Sep 15, 04:59

  H202 + 2I- + 2H+ →2H202 + I2

  Initial [I-] = 0.001 mol/dm3

  [H202] = 0.05 mol/dm3

  Half life = 90s 

  Rate = k [H202] [I-]

  Predict half life iodide in expt 2 where  [H202]= 0.10 mol/dm3 and [I-] = 0.001. 

  Why the half life is deduced to 45 seconds? 

  Isnt halflife constant to concentration of reactant if it is 1st order to the said reactant? 

   

• UltimaOnline

  21 Sep 15, 05:25

  Originally posted by Mrworry:

  H202 + 2I- + 2H+ →2H20 + I2

  Initial [I-] = 0.001 mol/dm3

  [H202] = 0.05 mol/dm3

  Half life = 90s 

  Rate = k [H202] [I-]

  Predict half life iodide in expt 2 where  [H202]= 0.10 mol/dm3 and [I-] = 0.001. 

  Why the half life is deduced to 45 seconds? 

  Isnt halflife constant to concentration of reactant if it is 1st order to the said reactant? 

  
  Half-life of 1st order reactant iodide ions, is halved because molarity of other reactant, 1st order reactant H2O2, is doubled in the new experiment.
  
  Half-life for 1st order reactant is indeed constant regardless of how its own molarity changes, but *ONLY IF* the molarities of all other non-zero order reactants remain unchanged (here molarity of other reactant H2O2 changes), *OR* if it is overall 1st order, *OR* if it is overall pseudo 1st order (ie. having any other non-zero order reactants present in large excess so any changes in the molarities of these other reactants as the reaction proceeds are negligible).
  
  If you drastically change the molarity of another non-zero order reactant going from experiment 1 to experiment 2, such as by doubling it, even if already present in large excess in experiment 1, the rate of reaction will still increase (eg. doubled if 1st order, or quadrupled if 2nd order, etc) and hence the half-life of the original 1st order reactant will still be halved (or quartered, etc) such as in this question you've typed out.
• BCME

  23 Sep 15, 11:28

  Hi, with reference to YJC_P3_2014

  2 (b)(iii) Why will NaBr and conc H2SO4 react with the alkene group? 

• UltimaOnline

  23 Sep 15, 11:53

  Originally posted by BCME:

  Hi, with reference to YJC_P3_2014

  2 (b)(iii) Why will NaBr and conc H2SO4 react with the alkene group? 

  
  Because alkenes, being electron-rich Lewis bases, are both nucleophilic and Bronsted-Lowry basic.
  
  Mechanism : alkene double bond attacks H+, then Br- attacks C+.
• BCME

  28 Sep 15, 16:19

  Hi

  With reference to TJC_2014_P3, 

  6(a)(iv) Will KMnO4 or 2,4-DNPH work as well?

  Thanks a lot 

• UltimaOnline

  28 Sep 15, 17:00

  Originally posted by BCME:

  Hi

  With reference to TJC_2014_P3, 

  6(a)(iv) Will KMnO4 or 2,4-DNPH work as well?

  Thanks a lot 

  
  It's P2 not P3.
  
  No cannot. 2,4-DNPH will be negative (no ppt) for both. KMnO4 (color change) will be positive for both. Only acceptable answers are the 2 methods given in the mark scheme.
• BCME

  2 Oct 15, 11:29

  Hi,

  With reference to PJC_P2_2009

  4(d)(ii) Why will the ester group react with KMnO4 and what happened to the middle section of compound T? 

  Thanks a lot 

• UltimaOnline

  2 Oct 15, 16:22

  Originally posted by BCME:

  Hi,

  With reference to PJC_P2_2009

  4(d)(ii) Why will the ester group react with KMnO4 and what happened to the middle section of compound T? 

  Thanks a lot 

  
  Ester group will undergo acidic hydrolysis. Of the C atoms in the middle, the C atom on the LHS (after 1-amino-benzene-2,4-dioic acid is generated) will be oxidized to methanoic acid, while the C atoms on the RHS (between the alkene double bond and the ester group hydrolysed) will be oxidized to first to ethandial, then further oxidized to ethandioic acid. Both the methanoic acid and the ethandioic acid will then be further oxidized to (total 3 moles of) carbonic(IV) acid, which decomposes into (3 moles of) CO2 and H2O (with thermodynamically favourable positive entropy change, and as predicted by Le Chatelier's principle, gaseous products leave the reaction mixture, pulling the position of equilibrium over to the RHS).
• Light5

  3 Oct 15, 08:10

  Prelim 2010 Raffles institution P1 Q25

  What is the best way to approach this question?

  obviously the chloroalkane will be secondary or tertiary and we can deduce that

  that the ketone will have a CH3CO group...but how can i use this information to deduce the correct structure. Do i simply do trial and error ie drawing the alkenes and ketones produced from each option or is there another way since on MCQ papers we usually do not have a lot of time. Also, is it better to draw condensed structural formula on the papers while solving organic chem mcqs or skeletal formulas to help us answer the question?

  Thank You.

   

• UltimaOnline

  3 Oct 15, 10:15

  Originally posted by Light5:

  Prelim 2010 Raffles institution P1 Q25

  What is the best way to approach this question?

  obviously the chloroalkane will be secondary or tertiary and we can deduce that

  that the ketone will have a CH3CO group...but how can i use this information to deduce the correct structure. Do i simply do trial and error ie drawing the alkenes and ketones produced from each option or is there another way since on MCQ papers we usually do not have a lot of time. Also, is it better to draw condensed structural formula on the papers while solving organic chem mcqs or skeletal formulas to help us answer the question?

  Thank You.

   

  
  For all papers, always draw skeletal structural formulae unless otherwise indicated by the qn (ie. "structural formula" means Cambridge will accept either skeletal structural formula, condensed structural formula, or displayed structural formula, or a hybrid of these).
  
  For MCQ qns however, you should *not* attempt deductive elucidation qns from scratch. Always use the 4 options to compare against the data given. For this MCQ, if you've mastered the H2 syllabus and are sufficiently prepared for the A levels, testing each option A B C D against the data given in the qn, should take you no more than 10 seconds per option. Since the answer for this qn is C, that's a total of 30s for this MCQ, which is well within the 1 min allocated per MCQ.
• Light5

  11 Oct 15, 02:39

  Paper:

  http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_s12_qp_51.pdf

  Mark Scheme http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_s12_ms_51.pdf

  ^Q1.) In the above paper in Q1 part(c) which is an experimental planning question, the examiner asks us to give the range of concentrations appropriate for this experiment. In the mark scheme, he has accepted from 0.8 to 1.39 mol/dm^3 .....but why cant we use a solution with conc less than 0.8 such as 0.5 or 0.2

  Furthermore, i am not sure how we will practically make this solution of aqueous copper II sulfate in the lab....we have to ensure each solution is 100 cm^3 in volume but the process is not mentioned in detail in the mark scheme. Please provide me with the necessary steps required to prepare these solutions.

  Lastly, in the c(iii) , we are asked to find the molar concentration of ONE of the solutions of copper sulfate but havent we already listed the range of concentrations in part (i)...isnt this part contradictory...if not, please explain how to solve it.

  Q2.) I wanted to know from where i can learn about mechanisms in INORGANIC chemistry. How can i predict which reactions will occur and which will not based on the structure, shape,bonding,etc of reactants and stability of products formed....and ultimately how can i deduce the most favorable product that can be formed in a reaction especially when we are asked to "Predict the reaction".

  Moreover, how can i understand why certain reactions actually occur,for example by explaing why SO3 + H2O makes H2SO4...which pretty much means learning the mechanisms for the reactions (which i dont know where to learn from)

  Also, from where can i read about why certain compounds or elements are present in the way they are for example S8 instead of O2, why SiO2 is giant covalent...i feel these are important because the examiners usually like testing anomalies (of a certain trend ). 

  Thank You !

• UltimaOnline

  11 Oct 15, 06:17

  Originally posted by Light5:

  Paper:

  http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_s12_qp_51.pdf

  Mark Scheme http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_s12_ms_51.pdf

  ^Q1.) In the above paper in Q1 part(c) which is an experimental planning question, the examiner asks us to give the range of concentrations appropriate for this experiment. In the mark scheme, he has accepted from 0.8 to 1.39 mol/dm^3 .....but why cant we use a solution with conc less than 0.8 such as 0.5 or 0.2

  Furthermore, i am not sure how we will practically make this solution of aqueous copper II sulfate in the lab....we have to ensure each solution is 100 cm^3 in volume but the process is not mentioned in detail in the mark scheme. Please provide me with the necessary steps required to prepare these solutions.

  Lastly, in the c(iii) , we are asked to find the molar concentration of ONE of the solutions of copper sulfate but havent we already listed the range of concentrations in part (i)...isnt this part contradictory...if not, please explain how to solve it.

  Q2.) I wanted to know from where i can learn about mechanisms in INORGANIC chemistry. How can i predict which reactions will occur and which will not based on the structure, shape,bonding,etc of reactants and stability of products formed....and ultimately how can i deduce the most favorable product that can be formed in a reaction especially when we are asked to "Predict the reaction".

  Moreover, how can i understand why certain reactions actually occur,for example by explaing why SO3 + H2O makes H2SO4...which pretty much means learning the mechanisms for the reactions (which i dont know where to learn from)

  Also, from where can i read about why certain compounds or elements are present in the way they are for example S8 instead of O2, why SiO2 is giant covalent...i feel these are important because the examiners usually like testing anomalies (of a certain trend ). 

  Thank You !

  
  I can't help you much with your questions in your post, for a couple of reasons.
  
  The 1st reason is that, assuming you're taking the Nov 2015 A level paper, you're asking all these questions a little too late. Learning allotropic details for elements such as sulfur, or reaction mechanisms (particularly for Inorganic Chemistry), requires going beyond the A level syllabus, and thus requires a lot more time that you currently have. You only have a month left, and (assuming you're a student, not a teacher or private tutor), you should spend the remaining 1 month left revising your within-syllabus content, not trying to learn details and mechanisms beyond the A level syllabus.
  
  The 2nd reason has to do with experimental planning. The mark scheme naturally doesn't give a model answer, but it *does* specify the details required, which you are free to phrase in your own way. Whether you're a student in the UK, Singapore, or some other country taking the CIE exams, your school should have given you Planning notes with sample experiments. If you're think you really need all the marks you can get, including from Planning, then you should buy this book (if it's available locally in the UK or wherever you are, there might still be enough time to buy and study it before your Planning paper).
  
  http://www.worldscientific.com/worldscibooks/10.1142/9565
  
  But my exam-smart advice to all A level Chemistry students (particularly Singapore H2 Chemistry students), is not to worry too much about Planning, or spend too much time preparing for it, because not only is it futile to try to 'spot' or predict the Planning question, the nature of Planning questions and the marks allocated for it, means it's not worthwhile to memorize dozens and dozens of sample Planning experiments.
  
  Skip the Planning question first, complete the rest of the Paper first, then go back to the Planning question and write whatever you can, based on common sense and chemistry sense.
  
  Theoretically, if your chemistry is strong enough to get (almost) full marks for all the theory papers (excluding Planning qn), you can still get your A grade, even if you skipped the Planning qn. So focus on making sure your chemistry is strong, and don't worry about Planning.
  
  Edited : Almost forgot to tell you, for that particular A level exam question you pointed out in the link, you misinterpreted the Cambridge Mark Scheme; it's not that the molarity must be between 0.8M to 1.39M, it's that you must have a range of concentrations over at least 0.8 mol dm–3, which must cover 1.0 mol dm–3, that is to say, you need the maximum molarity to be at least 0.8M higher than the minimum molarity, and that 1.0M must be included. Eg. 0.5M, 0.6M, 0.7M, 0.8M, 0.9M, 1.0M, 1.1M, 1.2M, 1.3M. Notice that 1.3M - 0.5M = 0.8M, which thus satisfies the Mark Scheme requirements.
  
  The 3rd reason has to do with Inorganic Chemistry allotropes. The full reasons for why various elements exist in particular allotropes (ie. varying thermodynamic stabilities and feasibilities under varying conditions), go significantly beyond the A levels, and should not concern you at A levels. Sulfur for instance, has over 30 allotropes. Allotropes of carbon would be more relevant to the A level syllabus, eg. Cambridge could set question on fullerenes and their applications.
  
  If asked about allotropes, Cambridge will certainly give you hints, or at least phrase the question in a way which will be sufficient to enable you to link the questions to various parts of the A level syllabus. For instance, Cambridge *could* ask you to compare and contrast the molecular bond angles and bond strengths between N2 and P4 (since both are elements of Group V), or between CO2 and SIO2 (since both are oxides of Group IV), and *therefore* suggest reasons why nitrogen exists as N2 instead of N4, or why phosphorus exists as P4 instead of P2, or why carbon dioxide exists as simple covalent molecular, or why silicon oxide exists as a giant covalent lattice, etc. Here, the relevant concepts are molecular strain due to angle strain, effectiveness of different electron shell orbital overlaps, preference of forming sigma vs pi bonds, thermodynamic stabilities and feasibilities, etc. Don't worry too much about this, since this isn't in the basic A level syllabus, and thus Cambridge doesn't expect you to memorize such info, but apply your A level knowledge to make reasonable suggestions, with sufficient hints included in the question.
  
  The last reason, (linked to the 1st reason), is that while it is a laudable goal for A level students to learn and apply reaction mechanisms to Inorganic Chemistry (which I do teach to my own BedokFunland JC students), this is something that cannot and should not be learnt at the last minute before the A levels, and certainly not online over a forum. You need personalized guidance for this, and since most school teachers (I assume UK teachers are overworked and underpaid just like in Singapore) won't be willing to teach mechanisms beyond the basic A level syllabus, it is mostly up to private tutors to do so. But it's too late for you to get a private tutor now, 1 month left before the A levels. And if you already had a private tutor, you would be asking him/her, instead of posting these questions here.
  
  Be that as it may, since you asked about SO3 + H2O --> H2SO4, I'll explain the mechanism here :
  
  SO3 is electrophilic or Lewis acidic, and the central S atom has a strong partial positive charge, due to the 3 electronegative, inductively electron-withdrawing O atoms, as well as the 3 single bonds resonance contributor with separation of formal charges (3+ charge on S atom, 1- charge on each O atom), due to ineffective sideways overlap between the 2p orbital of O and 3p orbital of S. In addition, the S atom has vacant, energetically accessible 3d orbitals to accommodate an expanded octet, allowing for the potential for reaction mechanisms with relatively low Ea. Concordantly and consequently, SO3 has the propensity to accept dative bonds, ie. behaving as an electrophile or Lewis acid.
  
  H2O is nucleophilic or Lewis basic, due to the 2:1 ratio of H to O, ie. the magnitude of partial -ve charge on O atom is 2x that of the magnitude of partial +ve charge on H atom. In addition, the O atom has 2 lone pairs. Concordantly and consequently, 1 of the 2 lone pairs on the O atom in H2O is available for donation as a dative covalent bond.
  
  Hence, the hydrolysis of non-metal, covalent, Lewis acidic oxides, generates Bronsted-Lowry acids such as H2SO4.
  
  1st step of mechanism : H2O nucleophile or Lewis base attacks (ie. nucleophilic addition) the SO3 electrophile or Lewis acid. Assuming you used the 3 double bonds resonance contributor, there will be 2 curved arrows : 1 arrow from lone pair of O in H2O, to the S atom in SO3. The 2nd arrow will be from one of the pi bonds, heterolytically cleaving away from S and towards O.
  
  2nd step of mechanism : Intramolecular Bronsted-Lowy acid-base proton transfer reaction. A proton or H+ ion is transferred from the +ve formal charged O atom, to the -ve formal charged O atom. Hence, the curved arrow is from a lone pair on the O- atom to the delta+ H atom, and a 2nd curved arrow from the O-H bond pair cleaving heterolytically towards the O, to get rid of the +ve formal charge.
  
  Concordantly and consequently, you obtain your sulfuric(VI) acid, which is Bronsted-Lowry acidic due to the sulfate(VI) conjugate base stabilized by having its negative formal charges (for either the uninegative or dinegative conjugate bases) delocalized by resonance over 3 or 4 (depending on which conjugate base HSO4- or SO4 2-) electronegative, electron-withdrawing O atoms.
• BCME

  13 Oct 15, 12:06

  With reference to JJC_2013_P2 7(d)(iii),

  Is it okay to say that the N in N2H4 does not have a lone pair instead of the ans given? I Don't really understand the answer provided. Thanks a lot 

• BCME

  13 Oct 15, 12:12

  Sorry i know there will be 2 lone pairs but i don't really understand why there will be a ring strain 

• UltimaOnline

  13 Oct 15, 16:40

  Originally posted by BCME:

  With reference to JJC_2013_P2 7(d)(iii),
  
  Sorry i know there will be 2 lone pairs but i don't really understand why there will be a ring strain 

  
  After each donor N atom has donated the lone pair in the form of a coordinate dative covalent bond unto the vacant orbital of the central metal cation, the electron geometry about the N atom is tetrahedral, ideally with bond angle 109.5 degrees, ie. requiring sp3 orbital hybridization.
  
  This is still possible using ethane-1,2-diamine (American common name ethylenediamine). But with hydrazine, if both N atoms each donate a dative covalent bond unto the central metal cation, you end up with a 3-membered ring, ie. a triangle consisting of the metal atom and the 2 N atoms.
  
  Mathematically, the bond angle within the triangle would be 60 degrees, which deviates significantly from the ideal 109.5 degrees bond angle (ie. angle strain).
  
  Consequently, the destabilizing inter-electron (bond pair - bond pair) repulsions as a result of the angle strain results in the ring being strained, and therefore destabilizes the entire coordination complex, making it thermodynamically unfeasible for hydrazine to function as a bidentate ligand.
• Mrworry

  15 Oct 15, 08:35

  Hi I know it is stupid to suddenly ask this but MnO4- from KMnO4 is not reduced to MnO2 (black/gray ) , but instead to Mn2+ (colourless) when KMnO4(aq) is used as oxidizing agent? 

• UltimaOnline

  15 Oct 15, 14:24

  Originally posted by Mrworry:

  Hi I know it is stupid to suddenly ask this but MnO4- from KMnO4 is not reduced to MnO2 (black/gray ) , but instead to Mn2+ (colourless) when KMnO4(aq) is used as oxidizing agent? 

  
  If acidic, you get Mn2+. If neutral or alkaline, you get MnO2.
• BCME

  15 Oct 15, 18:01

  With reference to YJC_2013_P2

  2(c)(i) For the steel electrode equation is it okay to write 2H20 + 2e --> H2 + 2OH-?

  3(a)(iii) The answer is trying to say that hydrolysis causes malic anhydride to be broken in a certain conformation so only the cis isomer may be formed right? 

  3(d)(ii) Gibbs free=0 because of the equal sign in the qns right? if it is a single arrow then Gibbs free will decrease is it?

  4(d)(ii) in step 2 why NH2CH2CH2NH2 must be limited 

• BCME

  15 Oct 15, 18:07

  The last 2 qns are with reference to JJC_2013_P3 

  Thanks a lot 

• UltimaOnline

  15 Oct 15, 19:20

  Originally posted by BCME:

  With reference to YJC_2013_P2

  2(c)(i) For the steel electrode equation is it okay to write 2H20 + 2e --> H2 + 2OH-?

  3(a)(iii) The answer is trying to say that hydrolysis causes malic anhydride to be broken in a certain conformation so only the cis isomer may be formed right? 

  
  The last 2 qns are with reference to JJC_2013_P3

  3(d)(ii) Gibbs free=0 because of the equal sign in the qns right? if it is a single arrow then Gibbs free will decrease is it?

  4(d)(ii) in step 2 why NH2CH2CH2NH2 must be limited 

  
  YJC 2013 P2 Q2c(i) Yes that's fine.
  
  YJC 2013 P2 Q3(a)(iii) It's not the hydrolysis mechanism that matters, but the fact that the *reactant* (being only a 5 membered ring) only exists in the cis isomer, it doesn't exist in the trans isomer, due to ring strain caused by angle strain (in the non-existent, thermodynamically unfeasible trans isomer).
  
  JJC 2013 P3 3(d)(ii) It's because the question states "Calculate the temperature at which the protein molecule *starts* to unfold", which means equilibrium, which means delta G = 0.
  
  JJC 2013 P3 4(d)(ii) Because the molar ratio of ethane-1,2-diamine to Q, to generate R, is 1:4. If ethane-1,2-diamine was in excess instead of limiting, do you think you can get R? What would you get as the major product instead?
• Mrworry

  16 Oct 15, 10:05

  Hi I know it is stupid to suddenly ask this but MnO4- from KMnO4 is not reduced to MnO2 (black/gray ) , but instead to Mn2+ (colourless) when KMnO4(aq) is used as oxidizing agent? 

  
  If acidic, you get Mn2+. If neutral or alkaline, you get MnO2.

   

  >>How to check Mn2+ will be oxidised to MnO2 and Mn2+ in electrochem qns?

• Mrworry

  16 Oct 15, 10:11

  Similiarly, how to choose whether Cu2+ will be reduced to Cu+ and Cu in electro chem qns? 

  i.e reaction between KI (aq) and Cu(SO4)2 (aq)?

• UltimaOnline

  16 Oct 15, 16:42

  Originally posted by Mrworry:

  How to choose whether Cu2+ will be reduced to Cu+ or Cu in electrochemistry questions? 

  i.e reaction between KI (aq) and Cu(SO4)2 (aq)?

  
  The standard reduction potential of Cu2+ to Cu is +0.34V, which is more positive compared to the standard reduction potential of Cu2+ to Cu+, which is +0.15V. Hence, always reduce Cu2+ to Cu (especially for electrochemistry questions, since regardless of whether the standard cell potential using Cu2+ to Cu+ reduction half-equation is feasible or not, the standard cell potential using Cu2+ to Cu reduction half-equation will always be more positive and hence more thermodynamically feasible, ie. Gibbs free energy more negative), *unless* otherwise indicated / hinted by the question (eg. in inorganic deductive-elucidation questions), or when it's for Fehling's and Tollen's, in which H2 Chem students are required to memorize and be able to write out the balanced redox equation in which the aldehyde* is oxidized to carboxylate ion, and Cu2+ is reduced to Cu+ in brick-red / brownish-red ppt of Cu2O(s)*.
  
  *The following are exceptions (not taught in Singapore JCs to H2 Chem students) :
  
  Methanal is (the only aldehyde to be) able to reduce the Cu2+ in Fehling's solution to either Cu+ or Cu, and hence (depending on factors such as molarity) you may get either a red ppt of CuO(s), or a copper mirror of Cu(s), or both.
  
  Methanoic acid is (the only carboxylic acid to be) able to react and give a positive result with both Fehling's and Tollen's. This is because the oxidation potential of methanoic acid to carbonic(IV) acid is large and positive, concordant to the observation that H2CO3(aq) exists in equilibrium with, and hence can decompose into, CO2(g) and H2O(l) with thermodynamically favorable positive entropy change, and also as predicted by Le Chatelier's principle since CO2(g) leaves the reaction mixture, pulling the position of equilibrium over to the RHS.
  
  Ketones do not react with Fehling's and Tollen's. Alpha-(pri)hydroxy-ketones are (the only ketones to be) able to react and give a positive result with both Fehling's and Tollen's.
  
  (Bonus challenge : Draw the curved-arrow electron-flow mechanism in which the alpha-(pri)hydroxy-ketone prototropically tautomerizes first to an ene-diol, then further prototropically tautomerizes to a alpha-(sec)hydroxy-aldehyde ; the redox reaction with Fehling's and Tollen's pulls the position of equilibrium to the RHS, as predicted by Le Chatelier's principle).
  
  Benzaldehyde and its derivatives only give a positive result for Tollen's, but not Fehling's. There are 2 reasons for this :
  
  Firstly, the more alkaline Fehling's solution results in benzaldehyde undergoing the Cannizzaro disproportionation reaction instead (which occurs for aldehydes unable to undergo the aldol reaction due to lack of an enolizable acidic alpha proton, such as in benzaldehye ; however in other aldehydes (ie. those with an enolizable acidic alpha proton), the redox reaction with the oxidizing (and tartarate and citrate ligand coordination complexed to prevent Cu(OH)2 precipitation in Fehling's and Benedict's respectively) Cu2+ ion takes thermodynamic (but not kinetic, as heating is required) priority over Aldol reactions, and as predicted by Le Chatelier's principle since the redox reaction is largely irreversible, while the Aldol reactions involve reversible steps. The Aldol reaction may still occur as a minor side-reaction, but does not affect the observable generation of the red Cu2O ppt).
  
  Secondly, using my BedokFunland JC formula "Oxidation State = Formal Charge + Electronegativity consideration", notice that while the oxidation state (OS) of the aldehyde C atom in aliphatic aldehydes is +1, the OS of the benzaldehyde C atom in the original resonance contributor is +1, while in all other resonance contributors (with pi electrons within the benzene ring withdrawn out towards the electron-withdrawing carbonyl oxygen) is 0, and thus the OS of the benzaldehyde C atom in the resonance hybrid is much closer to 0 than +1. As such, it requires a greater magnitude of thermodynamic Gibbs free energy, and thus a stronger oxidizing agent (Tollen's is a stronger oxidizing agent compared to Fehling's, as the standard reduction potential of Ag+ to Ag is +0.80V, which is more positive compared to the standard reduction potential of Cu2+ to Cu, which is +0.34V, even as the redox potentials are tweaked in the presence of the ammonia and tartarate ligands in Tollen's and Fehling's respectively) to oxidize benzaldehyde (OS of carbonyl C atom = 0) to benzoic acid (OS of acyl C atom = +3).
• Mrworry

  17 Oct 15, 09:30

  Hi. Thanks for the detailed reply!

  How about these:

  #How to check Mn2+ will be oxidised to MnO2 and Mn2+ in electrochem qns?

   

  #Based on your description, Cu2+ will generally be reduced to Cu except with the exception you said above. However why in the following question, it is reduced to Cu+ instead:

  1. Show that reaction between CuSO4 (aq) and KI (aq) is unlikely to occur by looking at standard electrode potential from data booklet

  2  Suggest one reason why the reaction does in fact occur

  Xiexie