Paper42/Nov/2009 /Question 8(d)(i) the marking scheme stated that dioxin and fur an as the answer. why is this so? Based on 1993 and 1994 the accumulated dioxin and furan for the upstream show a decrease whereas the accumulated dioxin and fu ran show a slight increase. This means that the dioxin and furan were responsibl e for the increase. Why is PCB not the answer. The reason being that the PCB con tent in the upstream and the PCB in the downstream are both showing an increase which indicate that it could come from other sources other than the paper mill. Anothe past year question( I cannot recalled the year) raised by Sin Li is a NMR question based on C7H8Oz. This question is tricky. My answer was a benzene ring , side group -OH and side group -CH3. However, the answer is not -CH3 but -OCH3. It was my mistake for not checking the value of the chemical shift for -OCH3. S o z=2 and not 1 as I assumed. Need to be vigilant when facing this type of quest ion. O/N/04 4)b)(ii) Predict, with a reason, the variation in the densities of the transitio n elements from Cr to Cu. The answer written is Increase, because even though the mass has been increase, the volume stays the same. My question: How could volume stays the same when each element is different with number of electron, proton, charge, and mass. Can I say that the volume has only small variation increase as the shielding eff ect increase, it cancel off the effect of charge increase? ANSWER:The learning outcome states that the atomic radii for the transition is r elatively invariant. That means the volume is relatively invariant as Vol is dir ectly proportional to r3. The examiner is looking for this minimum answer. Howev er, your answer exceeded the minimum requirement. However, your statement of "th e effect of charge increase" can be misunderstood. It should be stated as " the extra electron added to the 3d subshell which shield the 4s electron". O/N/05 2)d) Use the data from data booklet to predict the product of the reaction btw I 2(aq) and tin metal, writing a balance eq for the rxn. My answer is I2 + Sn -> SnI2 E: =0.54V+0.14= +0.68V But the answer in marking scheme is SnI4. Is it because I2 only has high reducing power so it oxidise Sn to +4 state ? But I thought it's less stable for Sn to stay in +4 state ? And how I know I should use Sn --> Sn 4+ + 4e- instead of Sn --> Sn2+ + 2e- ? ANSWER:I2 is the oxidising agent. It is able to oxidise Sn2+ to Sn4+ according t o the E0 values. Need to be vigilant to the extent the oxidation can proceed to completion. O/N/06 2)c)iii) A buffer solution containing equal concentrations of the two sodium pho sphate salts has a pH of 7.2. Calculate the pH of pharmaceutical preparation containing 0.002moldm-3 of NaHPO4 and 0.005moldm-3 of NaH2PO4. since H2PO4 would dissociate partially to HPO4 + H+, so it is acid ... but I can't get it about why we take pKa = 7,2 as the concentration of acid and base are different in this case..(0,002 and one is 0.005).
ANSWER:Ka does not depend on the concentration of the species in equilibrium. He nce, pKa is also constant. M/J/08 1)b)ii) A solution of iron (III) chloride is used to dissolve unwanted copper fr om printed circuit boards. When a copper coated printed circuit board is immersed in FeCl3(aq), the solutio n turns pale blue. Suggest an eq for the rxn btw copper and iron ( III ) chloride and use the data booklet to calculate the E value. Errrr...I know this sounds stupid but I don'r know why can't I write Cu + Cl2 --> 2Cl- + Cu 2+ instead of 2Fe3+ + Cu --> 2Fe2+ + Cu2+....? ANSWER:Cl2 is not present. Moreover, it is a redox reaction as given by the info rmation " a copper coated printed circuit board is immersed in FeCl3(aq), the solution tur ns pale blue". Cu2+ must be pale blue species. 42/M/J/10 4)b)i) Describe what observations you would make when dilute KMnO4(aq) is added slowly and with shaking to an acidified solution of FeSO4(aq) until the KMnO4 is in a large excess. I'd construct a balanced ionic eq. to make you more convenient,, 5Fe2+ + MnO4- + 8H+ ---> Mn2+ + 4H2O + 5 Fe3+ so Fe 2+ and MnO4 - both are coloured when they are in aq form... so what should we write ? purple or pale green ? Then the products, Mn2+ is faint pink and Fe3+ is yellow in color, so what shoul d I write ? The solution isn't gonna change from faint pink to yellow right ? ANSWER:Purple colour due to MnO4- decolourised when added to Fe2+(aq). It is dif ficult to see pale pink colour of Fe2+(aq). Purple colour is very pronouced. Sin ce it is added to excess pale pink would be the likely observation. It is diffic ult to see a pale yellow due to Fe3+(aq). Random questions: - when we talk about the solubility of the group II sulphate going down the grou p, the lattice energy is decrease not much is it because the charge for SO4 2- ( and cation) and radius for SO4 2- remain unchange, only the radius is increasing , so the value of L.E actually have not much changed ? ANSWER:Solubility of the group II sulfate is defined by the enthalpy of solution which is related to the lattice enthalpy and the enthalpy of hydration of the c ation and the enthalpy of hydration of the sulfate ion. Please read page 266 in the chemistry coursebook. - if we want to predict the likely oxidation state of a transition element, why we ignore +1 and start from +2 ? if only one electron is left in 4s orbital, the electron is very unstable ? ANSWER:+1 oxidation is less likely although possible since 4s can lose two elect rons to form stable ions in a compound. We need to obtain a born-haber cycle to see why +2 is more likely.
- Last time when I asked u why we add equimolar of Sn2+ and Sn4+ to I2, sir you told me that I2 is oxidising agent, it can oxidise Sn2+ to +4 right ? But then again, we do we still add Sn 4+ initially since Sn2+ would be oxidise t o +4 ? ANSWER:This question was related to a mixture of Sn2+ and Sn4+ that was initiall y present. 43/O/N 2010 5) Chlorine is manufactured by the electrolysis of brine, NaCl(aq). At the catho de, H2(g) and OH-(aq) are produced, but the product at the anode depends on the [NaCl(aq)] in the solution. Either O2 or Cl2 is produced. (a) The eq for the cathode rxn is 2H2O + 2 e- ---> H2 (g) + 2OH- (aq) Starting from neutral NaCl (aq), write equations for the production at the anode of (i) O2 The answer written is 2H20 ---> 4H+ + O2 + 4e But I write 4OH- --> O2 + 2H2O + 4ehow should I know which one should I write ? ANSWER:Since it is mentioned neutral NaCl(aq) the solution is non-alkaline. I wo uld assume that the [H2O] is much larger than [OH-(aq)]. Water is ionised to for m H+ and and OH-. H+ is discharged as H2 . OH- remain in the solution. OH- will not be discharged as OH- is at the cathode(negative electrode) and thus remain u nchanged near the cathode. (ii) Hence explain why the Cl2 : 02 ratio increases as [NaCl(aq)] increases. I totally confused now. Either O2 or Cl 2 will produce at the anode isn't it ? And the product at the cathode is H2 only isn't it ? How the two gases release at the same time ( Cl2 & O2). ANSWER:At the anode(positive electrode), both gases are produced. O2 comes from the discharged of OH- and Cl2 from the discharged of Cl-.At low [Nacl(aq)] less Cl2 will be discharged since Cl2+2e--> 2Cl- is E0 is +1.23V. At higher [NaCl(aq) ] the equilibrium shifts towards the left and Eo value will drop causing more Cl - to discharge.AT the anode OH- remains in the solution. Only H2 is produced.Bot h gases are produced since both OH- and Cl- could be selected for discharge base d on the concentration of Cl-. M/J 09 8)b)ii) PCB residues have been found in the breast milk of Inuit mothers in Northern Can ada. Food, such as oily fish, seal and whale meat, which are high in fat, form a n important part of the Inuit diet. Based on the information provided, what can you say about the partition coeffici ent between fat and water for PCB residues? Ans: PC more than 1. ANSWER:what does it means when Partition coefficient more than 1 ? Ratio of the Kd=[PCB]fat/[PCB]water. Since, PCB is more soluble in fat, [PCB]fat >[PCB]water, hence the ratio is greater than 1. M/J 08 10)b) * Diagram show a capsule containing nanospheres and inside the nanosphere, the d
rug is bounded by thick hydrogel coat * (iii) Suggest two ways in which the nanosphere shown in the diagram can be modif ied to change the rate of drug release. Ans : Different thickness of hydrogel ( to release drug over time) Different chemical composition ( for different breakdown time) ANSWER:The mark scheme suggested that you could use hydrogels of various thickne sses. Obviously the thicker the material of the "bubble", the more slowly it wou ld release the drug. Otherwise, you could change the composition of the polymer to make it more resis tant or less resistant to breaking down. Or you could make a bubble with holes o r pores in its walls. Changing the thickness of the bubble wall is fairly obvious. I didn't feel that it was quite as obvious that you could change the material or somehow design hol es into it. This question may well never come up again, but looking at it in det ail might give you hints about how to answer future similar questions. Hmmm... seriously ? We can control when do we want to release the drug inside ou r body ? ANSWER:The condition of the disease cells may provide the kind of control such a s the rate of drug release. For example, the high[H+] concentration induce the h ydrogels to release the drugs. O/N/11 (43) 5)b)(iv) Refer to the file attached? I don't know how to get the answer -_ANSWER:This is a mathematical problem involving probability. Since there is only 1 hydrogen attached directly to tertiary carbon in J the relative rate is 21. T here are 9 equivalent hydrogens attached to the primary carbon in K the relative rate at a single primary carbon is 9x1. Hence, the ratio of the relative rate o f J to K is 21:9. 8)a) (i) Why soluble form will be most effective ? I thought it'd be digested by gast ric juice in stomach ? ANSWER:The drug in liquid form would diffuse into the blood faster that the soli d form. The solid form would remain much longer in the digestive tract and diffu se into the blood slowly. (ii) ANSWER:The physical form of the drug is the focus of the comparison. (iii) Weird enough. It already stated that the coating is digestible. Why the ma rking scheme write 'stop being broken down' since it is digestible ? ANSWER:Drug may be broken down at extreme pH. The protective coating keep the co ntent from being destroyed at the early stage in the stomach. The protective cas ing would be digested slowly. Slow enough to keep the drug intact until it reach the intestine. Random questions, as always :D 1) I always write 'the lone pair of oxygen delocalised in pi bonds of the benzen
e ring' while I realised somewhere else write 'pi electron system'. Can my answer be accepted? The electrons are delocalised in pi BONDS right ? ANSWER: pi bonds in benzene ring are delocalised(the 6 lone pairs of carbon over lap with each pair)to form the pi electron system. The lone pair of oxygen overl ap with the pi electron system in benzene.This is delocalisation. 2) When we need to calculate the lattice energy, the arrow of first electron aff inity always pointing downward in energy diagram. But then for second electron a ffinity, the arrow is pointing downward again(even though it is endorthemic ! ), I wonder why the energy of 1st and 2nd electron affinity we need to point the a rrow downward ? Gain of electron making the anion lose energy ? ANSWER:Are you referring to the electron affinity of oxygen. Gaining the first e lectron is exothermic(arrow pointing downwards). Gaining the second electron is endothermic(arrow pointing upwards. 3) tripeptide means 3 amino acids right ?ANSWER:Yes but joined by amide bonds. M ore accurately, 3 amino acid residues joined together by amide bonds. 4) how should I write a balance equation for a condensation polymer ? E.g . n COOHCH2NH2 ----> -OHCH2NHCOCH2NH- n + 2n -1 H20 ? There'll always have one end that the COOH group and NH2 remain right ? ANSWER:n COOHCH2NH2 ----> -(COCH2NH)- n + n H20 Two different functional groups in each monomer. 5)ANSWER:These are short lengths of DNA, radioactively labelled with phosphorus32 atoms. These are designed to bind with the repeating sequences in the DNA fra gments. I will be away from 8th Nov-15th Nov. 2012. I will try to answer during this pee riod but I cannot promise because of the poor internet connection. In the bond energy calculation, we usually use the bond break energy minus bond forming energy right? However, is there any case that we do not need to break ev ery bond in the reactant? I encounter a question in the past year: Given the enthalpy change of formation of the following compound: PbCl2(s) ----- (-359 kJ mol-1) PbCl4(l) ------ (-329 kJ mol-1) Find the enthalpy change of the following using data booklet and these data: PbCl2(s) + Cl2(g) ------> PbCl4(l) The answer given in the answer scheme is +30kJmol-1 which it didn't include the breaking of chlorine, why is that so?? And also, calculating the E value using half equation from data booklet, the answ er supposingly will be positive if the reaction work right?? But there are one q uestion in the past year which the reaction work but the value of calculating th e E is negative in the answer scheme.. Sir, I found a question on which half equation to use that I asked you previously. The question says use the E value to predict whether a reaction will occur. The reactant given are Cr2+ and I2.in the formula booklet,there are 2 possible h alf equation of Cr2+. I use the one with E value of -0.41 because it will form Cr 3+ thus can react with I- form from I2. So then the reaction can occur. Is my wa y correct? Cause if I use the Cr2+ half equation with -0.91, the reaction will b e unlikely to happen.
According to the syllabus you only need to remember the complex of copper. No ot
her element is required. MnO4- is not a complex ion. If you are asked about the complex of other transition element you will be given the information about thei r colours. ANSWER: Actually the Cl-Cl does break so that it can form Pb-Cl bond. However, t he bond enthalpy Pb-Cl is not given in the data booklet. Therefore, we cannot re ly on bond forming and bond breaking to find the enthalpy change. In the case of carbon tetrachloride the bond enthalpy of C-Cl is given in additi on to C-C and Cl-Cl in the data booklet. We can therefore use bond forming and b ond breaking for carbon tetrachloride to calculate the enthalpy change. Subject: Re: Enthalpy change Sir, Regarding the Hess law you sent me, does that means that we no need to break the bond of chlorine? There's branch of this question. This question is number 4 b) in Oct/Nov 2004 paper 4. The first equation which involves the carbon tetracholride need to break all the bonds to obtain the answer,why is these two equation differs??
9701/04/MJ/08 Question 1a(ii)
ANSWER: You need to observe the 2 equations carefully to arrive at the enthalpy change you are investigating.
Write down the equations Pb(s) +Cl2(g)----> PbCl2(s) delta H = -359 kJ mol-1 eqn(1) Pb(s) + 2Cl2-----> PbCl4(l) delta H = -329 kJ mol-1 eqn(2) then PbCl2(s) + Cl2 (g) -----> PbCl4(l) delta H of reaction eqn(3) Draw a Hess's cycle to using eqn(1), eqn(2) AND eqn (3) This is the way to approach this kind of problem
Attached is the Hess's Diagram
Could you state the questions and year for the question on electrochemistry? I n eed more information to answer your question.
In the bond energy calculation, we usually use the bond break energy minus bond forming energy right? However, is there any case that we do not need to break ev ery bond in the reactant? I encounter a question in the past year: Given the enthalpy change of formation of the following compound: PbCl2(s) ----- (-359 kJ mol-1) PbCl4(l) ------ (-329 kJ mol-1) Find the enthalpy change of the following using data booklet and these data: PbCl2(s) + Cl2(g) ------> PbCl4(l) The answer given in the answer scheme is +30kJmol-1 which it didn't include the breaking of chlorine, why is that so?? And also, calculating the E value using half equation from data booklet, the answ er supposingly will be positive if the reaction work right?? But there are one q uestion in the past year which the reaction work but the value of calculating th e E is negative in the answer scheme..
ANSWER: The reactants are I2 and Cr2+. For redox reaction, One of the reactant(s pecies) must undergo reduction and the other one must undergo oxidation. The E0 value I2/I- being more positive must undergo reduction(the forward direction). R emember that there are plenty of I2 initially. Similarly, there are plenty of Cr 2+ initially, and since the E0 value of Cr3+/Cr2+ is less positive, it would pre fer to undergo oxidation(the reverse direction). You are therefore right in your explanation. Cr2+ cannot undergo reduction just as you suspected. Paper 43 Oct/Nov 2010 Question 2(c). Is there an assumption needed for the volum e of leach solution and the mass of leach solution in this question? The density of the leach solution was not given in this part of the question. The answer wa s expressed as a % of the leach solution. Is the student require to make assumpt ion in this type of question? My student could only express it in mol dm-3 in th e final answer. Is it correct to express two different quantities with different units ( mass of copper and the volume of leach solution) as a %? CIE Examiner reply: It looks to me as though one needs to assume either that the calculation is of % weight/volume(w/v) or that the solution has a density equal to that of water at 1 g/cm3 (which we tend to do, for example in calorimetry ca lculations using q = mc deltaT) 9701/42 May/June 2011 Question 5(c) Step 3 The marking scheme given for step 3 CH3COCl or (CH3CO)2O ( aq. negates). 4-amino ph enol has two functional groups, namely -OH and -NH2 , attached to the ring and t hey are capable of reacting with CH3COCl. Why is CH3COCl chosen and not CH3COOH in this case to form paracetamol? CIE forum moderator. using CH3COOH would presumably result in an acid-base reaction between the basic phenylamine and the acid to give a salt instead of the ester also the involveme nt of the lone pair on the N in the delocalised ring makes it a poor nucleophile so the more powerful acylating agents are needed i.e. acyl chloride or acid anh ydride. Why the -OH isn't esterified as well I can't explain so will need to give that s
ome more thought Another reply from chemistry forum participant. The mechanism involves nucleophiles OH and NH. Perhaps it is partly due to the n ucleophilicity of the N atom compared to the O atom. O is more electronegative t han N and less likely to give up it's lone pair as easily as nitrogen (this idea could also be used to explain why water is a weaker Lewis base than ammonia etc ....) In reality, a mixture of O- and N-acylated products would be formed...we can be easily misled to believe that the O- isomer does not form when it probably does as a minor product. In the majority of schemes in synthetic chemistry, we ignore all but the major product. Inert pair effect. There are various approaches of explanation on the inert-pair effect to account for the behavior of tin and lead elements. I would like to fi nd out whether this explanation that I read is acceptable to CIE. There is less shielding effect by the diffused inner orbital in tin and lead, namely, d and f orbital, hence making the nuclear attraction for the 2 outermost s-electrons to be held more strongly. Thus, there is less available electrons for forming bonds with chlorine or oxygen atom. The latest textbook explanation on inert-pair effect is based on chemical energe tics approach. CIE forum Moderator reply. A complicated topic and I will be interested to see o ther replies. Personally I would recommend the Chemguide site as a useful refere nce http://www.chemguide.co.uk/inorganic/group4/oxstates.html 9709/42 June/May2011 Question 2(b)(iii). The marking scheme give the following r emarks Other possible oxidants (Eo must be > +0.2V) include: S2O82, H2O2, Cl2, Br2, I2 and Ag+. No observations with the first three of these, but this should be stated explici tly, e.g. no colour change. Is chlorine water colourless so that there is no colour? Could chlorine water ap pear as green/yellow colour? CIE forum moderator reply. I think it's very unlikely any students will have seen chlorine water sufficient ly concentrated to give an obvious colour - but if an appropriate description wa s offered in an answer then I'm sure it would be allowed as the intention is alw ays to credit correct chemistry. Paper 43 Oct/Nov 2010 Question 5(b). For electrolysis to occur, the voltage applied to the cell must be at least as l arge as the E o cell, as calculated from standard electrode potentials. Use the Data Booklet to calculate E o cell for the production at the anode of (i) O2(g), ..................................................................... .................................................. (ii) Cl2(g). ................................................................... ................................................... Comment. There are two types of cell, namely electrochemical(battery) and electr
olytic cell. They are not the same. It is important that you know the difference otherwise you are going to get confused. E0 values from Data Booklet are referr ed to electrochemical cell. Do not use E0 to calculate the electrolytic cell. Fo r electrolysis reaction, the electrolytic cell need an electrochemical cell or b attery to be placed across the electrolytic cell. The positive terminal of the electrochemical cell is connected to the titanium e lectode in the electrochemical cell. This is electrode where both Chlorine and o xygen may be produced or you may say that the chloride and hydroxide ions are di scharged. Remember that anode is where oxidation or loss of electrons are happen ing. In this question you are asked to find the least E0 of the electrochemical cell not the electrolytic cell. This means you have to ask this question. What then i s happening at the positive terminal of the electrochemical cell. Oxidation is a lso happening at the positive terminal of the electrochemical cell. Earlier in the question The equation for the cathode reaction is 2H2O(l) + 2e -> H2(g) + 2OH (aq) was given. Note that this is not the equation of th e standard hydrogen electrode. The solution is not neutral as OH- (aq) is presen t. The reaction taking place at the cathode of the electrolytic cell is the posi tive terminal of the electrochemical cell. Hence, the reaction is the same as th e cathode of the electrolytic cell. Similar explantion goes for the reaction happening at the negative terminal of t he electrochemical cell for both oxygen and chlorine that is written in (a). If you get a negative value then it is wrong. This is because the E0 of the batt ery cannot be negative. A negative value means that the battery is not working. Hence, the E0 of the electrochemical cell must be positive. Sir, for O/N 2008, Question no. 4 (a) (ii), Why is the answer 8? Can u explain? What is mean by average molecule of the polymer? I don't really understand the q uestion. and for the same question, (b) (iii), how do we get the answer? why is it 4(c-c) - 2(c=c) ? For the same paper, question no. 6 (a), can we test using K MnO4 and say that the solution decolourised? Sir, for O/N 2008, Question no. 4 (a) (ii), Why is the answer 8? Can u explain? What is mean by average molecule of the polymer? I don't really understand the q uestion. and for the same question, (b) (iii), how do we get the answer? Answer:the monomer is CH2=CH(CH2CH2CH3). There are 5 Cs in this monomer. Since a verage no. of C in the polymer is 40. Hence there are 40/5 or 8 monomers per pol ymer. So does it mean that if they ask for average molecule of polymer, we take the to tal number of carbon contained in the polymer divide by the number of carbon in a monomer? (b) (iii), how do we get the answer? why is it 4(c-c) - 2(c=c) ? bonds broken: 4(C C) = 4 × 350 = 1400 kJ mol 1 bond formed: 2 (C=C) = 2 × 610 = 1220 kJ mol 1 ??H = +180 kJ mol 1 C40H82 ??? C16H34 + 2 C12H24 The answer in MS is based on this equation. Your equation may not be this one an d your answer may be different from this one. C1----C16-C17-C18---C28-C29-C30 ---C40. Need to break C-C bond at these points C16-C17, C17-C18, C28-C29 and C29-C30 ass uming that the C=C bond is formed at C17-C18 and C29 and C30. Remember also a CH bond is formed at C16, and C28 and C-H bond is broken at C18 and C30. These ca ncelled off each other.
For the same paper, question no. 6 (a), can we test using KMnO4 and say that the solution decolourised? No. You have to differentiate J and K. Acidified KMnO4 will give the same negati ve result with J and K. Question:So does it mean that if they ask for average molecule of polymer, we ta ke the total number of carbon contained in the polymer divide by the number of c arbon in a monomer? Ans: The number of monomer units is 8. But the number of C in each monomer is 5 and the average number of Carbons in the polymer is 40. The reason for mentionin g average is because of the different length of molecules in the polymer. Some a re longer, others may be shorter. Some may be 45, 40, or 35. Paper 4 Oct/Nov 2008 Q 9(a)(ii) My question: The examiner report mentioned (ii) Quite a few suggested the PCR re action, or something along the way , before electrophoresis. Marking scheme suggest ed (ii) these undergo electrophoresis OR are placed on agarose gel. Why is PCR n ot acceptable as a stage after treatment with restriction enzymes? In the recent Paper 42 June 2011 Q 7 (a) , PCR was mentioned as a stage before carrying out e lectrophoresis. The examiner report mentioned (a)Most candidates appreciated tha t the production of genetic fingerprinting starts with the extraction of DNA, fo llowed by the use of restriction enzymes before using the polymerase chain react ion. Which is the correct stage to mention after treatment with restriction enzy mes? CIE reply: In 2008 it was an open-ended question and we discussed candidates answers and de cided that only answers which made sense in the context of the question and the syllabus would be permitted. This meant that we allowed PCR increasing the amoun t of DNA to investigate, in other words BEFORE treatment with restriction enzyme s. In 2011 the candidates had to place a list of stages in sequence and so we were rather tighter in the sequence since we had provided the information. I have some dought about the Oct/Nov 2007 Question 3. c.(ii) Use the Data Bookle t to identify relevant half equations and E o values for the production of chlor ine from the reaction between K2Cr2O7 and HCl.Use these equations to write the o verall full ionic equation for this reaction. According to the reduction potenti al value between chlorine and acidified potassium dichromate chlorine has to red uce and dicromate has to be oxidized but it is just reverse in markscheme . how is it possible? answer:As given in the markscheme for this question on Paper 4 we see: Eo data and half equations: Cr2O72 + 14H+ + 6e ? 2Cr3+ + 7H2O Eo = 1.33 V [1] Cl2 + 2e ? 2 Cl Eo = 1.36 V [1] overall ionic equation: Cr2O72 + 6Cl + 14H+ ? 2Cr3+ + 3Cl2 + 7H2O [1] Comparing the Eo values does suggest that the more favourable direction of chang e is the reverse of this reaction - but when the Eo values are so close together we can expect an equilibrium mixture to be produced Another point to remember is that the Eo values refer to standard conditions and , in this scenario, concentrated hydrochloric acid is being used so the concentr ation effects will have a significant impact on the position of equilibrium as w
ill the temparature. In the June 2011 paper 42 (A Level), there was a question which required student s to complete a table showing the expected discharge products of three aqueous s olutions. The marking scheme gave bromine as the anode product of aqueous magnes ium bromide. Students were asked to use information from the Data Booklet to get their answer s. Surely, using the numbers quoted in the Data Booklet, the hydroxide ion will be preferentially discharged to give oxygen. I understand that at higher bromide concentrations, it will be more likely to be discharged but the question says n othing about concentration. I don't see how, logically, students could be expected to pick bromine as the an ode product. This response was provided by the Principal Examiner for this paper. Candidates should know that chlorine is the anode product of electrolysing conce ntrated NaCl solutions (that's in the syllabus), despite the fact that the Cl2 E o is +1.36, way above the O2 one. This, it may have been explained to them, is partly due to the fact that [OH-] i s only 1 x 10(-7) in neutral water, and partly due to the overpotential of oxyge n. Since the Br2 Eo is 1.09, lower than that of Cl2, it would be expected that Br2 would be released even more readily than Cl2. The question was why the boiling point of ethanolamine is much higher than that of propylamine. Draw a diagram to illustrate your answer. The marking scheme just showed hydrogen bonding (1 mark) and for diagram via two possible orientations for hydrogen bonding (1 mark). H-bond from OH group to ei ther OH or NH2 This is a comparison question. The propylamine has hydrogen bonding too (C3H7NH2 ) although it has a longer non-polar propyl chain due to which it is fairly solu ble in water or less soluble than ethanolamine. Ethanolamine, on the other hand, has two groups OH and NH2 which can both take part in H-bonding as compared to only one NH2 group in propylamine resulting in ethanolamine having extensive H-b onding than propylamine. The diagram should be: NH2CH2CH2OH..............NH2CH2CH2OH........NH2CH2CH2OH Consequently both OH and NH2 should be accepted as the correct answer and not an y one of them. Please explain. Answer:The mark scheme for publication shows the anticipated answer and will oft en show the minimum required to gain the mark. Other answers are often also allo wed. In the example you give above, the diagram which shows H-bonding from both ends of the molecule rather than from one or the other as given in the mark scheme wo uld have been credited and is quite correct and a more complete explanation that the minimum required for the mark. Again, most candidates would have written more text than just 'hydrogen bonding' but the mark scheme highlights this as the minimum required for the mark.
Hi The marking scheme in June 2003 Question 3 (a)(iii) states states that Sn(IV)O2 is a giant ionic structure or having ionic bonds but the Chemistry Coursebook(pa ge 354) states that it is a giant covalent structure having both covalent and a little ionic characters in the bonds. In another part of the marking scheme Ques tion 3 (b), it was mentioned that aqueous NaOH reacts with tin(IV) oxide. Howeve r, the Chemistry Coursebook(page 355) states that it reacts with hot concentrate d NaOH. Which of these statements given above are acceptable? Answer:SnO2 is amphoteric so shows some reactivity with both acids and alkalis. I suspect that the MS reference to 'aqueous' NaOH was simply slightly generous after all 'concentrated' is still aqueous. I haven't seen the coursebook but the Unit 8 Scheme of Work on the website state s: "SnO2 and PbO2 are most easily described as ionic, with the rutile structure, although some covalency occurs within the latt ice. Acidity increases from the +II to the +IV oxidation state, and decreases from C to Pb. Thus CO is neutr al whereas CO2 acidic; PbO amphoteric whereas PbO2 reacts more as an acidic than a basic oxide (e.g. it does not dissolve in aqueous acids, unless it oxidised them, but it does dissolve in concentrated NaO H(aq))." There are discrepancy in some references I referred to. Just wonder if I would be penalized for writing the product of electrophilic add ition to an alkene by aqueous bromine solution as a bromo-alcohol. Answer:Correct chemistry is never penalised The examiner report mentioned (ii) Quite a few suggested the PCR reaction, or so mething along the way, before electrophoresis. Marking scheme suggested (ii) these undergo electrophoresis OR are placed on agarose gel. Why is PCR not acceptable as a stage after treatment with restriction enzymes? In the recent Paper 42 Jun e 2011 Q 7 (a) , PCR was mentioned as a stage before carrying out electrophoresi s. The examiner report mentioned (a)Most candidates appreciated that the product ion of genetic fingerprinting starts with the extraction of DNA, followed by the use of restriction enzymes before using the polymerase chain reaction. Which is the correct stage to mention after treatment with restriction enzymes? Answer:I have discussed your query with the Principal Examiner for this paper an d he has provided the following comments: In 2008 it was an open-ended question and we discussed candidates answers and de cided that only answers which made sense in the context of the question and the syllabus would be permitted. This meant that we allowed PCR increasing the amoun t of DNA to investigate, in other words BEFORE treatment with restriction enzyme s. June 2012/P41/Question 2b(iii) The following mechanism has been put forward for this reaction. step 1 NO + NO ? N2O + O step 2 H2 + O ? H2O step 3 H2 + N2O ? N2 + H2O Step 1 is bimolecular, so the order is 2 Step 2 depends on reactant H2 and the intermediate O. Since O is not the reactan
t it is indirectly related to NO. Step 3 depends on H2 as the reactant and N2O which is an intermdiate form in ste p 1. There are 2 acceptable situations (i) H2 reacts with N2O (ii) H2 reacts with O t o form the products.
