BedokFunland JC's A Level H2 Chemistry Qns (Part 2)

last SG · 2010-07-26T23:40:35+00:00

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http://themetapicture.com/only-very-smart-people-will-understand-these-jokes/

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The hybridisation of sulfur in anion SO42- is given to be sp3. Why is it so? Sulfur has two double bond and two single bond with four oxygen atoms, meaning there are a total of 12 electrons surrounding sulfur. sp3 only accomodates up to 8 electrons. besides, sp3 means that we are assuming there are only four single S-O bonds, which is not the case since resonance structure for it exists.

Also given that we know resonance structure exists for the sulfate ion, it must mean there is an unhypridised p orbital on sulfur. if sulfur were to be sp3 hybridised, there will not be any unhybridised p orbital to overlap with other p orbitals of oxygen. shouldn't the hybridisation of sulfur have beeen spd2? two unhybridised p orbitals, as well as four spd2 orbitals, so there will be 4 hybridised orbital to overlap with the ones in oxygen, and 2 overlap of p orbitals to form two double bonds?

Edit: in order for resonance structure for sulfate ion exists, does it mean that one sulfur atom would need to have 4 unhybridised p orbitals to overlap with each p orbitals from 4 oxygen atom? or would 1 unhybridised p orbital for sulfur will do? [likewise for CO32-, NO3-]

As the electron geometry about the S atom in SO4 2- is tetrahedral (in order to maximize stabilities by minimizing electron pair repulsions), hence sp3 hybridized orbitals are required for the sigma bonding about the S atom.

At A levels, the usual major resonance contributor taught to students, is that of an expanded octet with no separation of formal charges (at least in part because formal charges are not usually taught to most H2 students in the first place), ie. Fig 1 below.

But experimental evidence has indicated that such a resonance contributor, is actually only a minor resonance contributor. The major resonance contributor, is actually that of a dipositively formal charged S atom, singly bonded to 4 O atoms with uninegative formal charges, ie. Fig 2 below.

The (almost negligible) pi bonding that exists (for the minor resonance contributors), utilize the d orbitals of the S atom to overlap (poorly) with the p orbitals of the O atoms.

In other words, the bonding in the SO4 2- ion, is actually covalent with significant ionic character (due to the formal charges present in the major resonance contributor).

Experimental evidence has shown that the actual charge on the S atom is only slightly less than its formal dipositive charge, as such an S 2+ atom would accordingly be strongly electron-withdrawing by induction, withdrawing electron-density from the singly bonded O- atoms, towards the central S 2+ atom.

Concordantly, consequently and most interestingly, you could say that the (slight) partial double bond character (ie. in terms of bond length and bond strength) of the S=O bonds in the SO4 2- ion, is partly (only slightly) due to resonance, and partly (somewhat moreso) due to induction.

1 unhybridized p orbital would be sufficient, but because only one unhybridized p orbital is available, the sideways overlap with the p orbitals of the surrounding atoms to form pi bonds is only *partial*. This is concordant with the observation that in the resonance hybrid, each C=O bond in CO3 2-, and each N=O bond in NO3 -, only has *partial* double bond character.

One last point : for elements beyond period 2, while their atoms often appear to achieve an expanded octet (using vacant, energetically accessible 3d orbitals), which is what is taught at A levels, however experimental evidence has shown this to be inaccurate. In point of fact, all molecules/ions with central atoms that appear to have an expanded octet, actually have as their major resonance contributor, the central atom with only a stable octet, and (accordingly) with separation of formal charges. In other words, in the actual structure or resonance hybrid of the molecule/ion, the bonding is always covalent with significant ionic character.

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The electronegativity of Nitrogen is 3.0 and of chlorine is 3.2. I am not sure about the exact values. anybody knows the actual values.

There are multiple systems of electronegativity ratings using different experimental systems and calculation methods. As such, you will come across different systems (eg. Pauling, Mulliken, Allred–Rochow, Sanderson, Allen, etc) giving you different electronegativity values for each element.

For A level purposes, N is certainly regarded as more electronegative than Cl, and as such, only NH groups (in addition to HF and OH groups), and not HCl, are considered capable of donating H bonds.

A level students are not expected to memorize electronegativity values from any particular system. Cambridge may give electronegativity values (from any of the different electronegativity systems) in the A level question itself, to examine if the A level candidate is able to interpret the relative values of the various elements. For instance :

Carbonyl sulfide has the formula O=C=S. If the electronegativity of O, C and S are 3.5, 2.5 and 2.4 respectively, which bond is more polar? The O=C bond or the C=S bond? Indicate the partial charges and dipoles on the O=C=S molecule.

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UltimaOnline posted :
Every year, many thousands of JC students with all straight As, and hundreds of Poly students with perfect GPAs, have to compete tooth and nail for just 200 places in NUS+NTU Medical School. Every year (for last few decades), hundreds of these straight As students are forced to study Medicine abroad (and these are the lucky ones whose family can afford the $1 million fees + expenses to study Medicine overseas), and the fact that recognized Medical schools overseas (including many even better than NUS) accept them, proves they're actually good enough to be medical doctors. And yet, every year PAP says "Jialat lah! we not enuff medical doctors in our hospitals leh! must import a lot more FT doctors!" and keep bringing in thousands of FT doctors every year. When our own Singaporean students who are good enough to study Medicine, are deprived of the opportunity to study Medicine locally.

mojito posted :
A poor sentiment, my friend. You see, if the PAP did not import more doctors, we would have an acute shortage of doctors as the number of patients spiked in the recent years. Medicine is an important vocation, too important in fact, to allow local students without the necessary years of housemanship to take on key roles as specialists. As you must now realize, the PAP had to make difficult trade offs. Without them we would have let our emotions get in the way of doing what is best for the people. For that, you have to thank the PAP.

halsey02 posted :
You weren't a medical student back, in the 70's & 80's...if not, you do not know what you are talking about....the PAP f$#$$^$ all those wanting to study for medicine here...many have to go abroad to study that...I know a couple classmates...they never returned... So, try not to bluff too much huh!.... they are people like me of that generation who went through the system....& still around & many still around scattered all over the world...

freedalas posted :
I have 2 nephews who were scored straight As for the A levels and yet were denied admission to our medical school here. Both had to study abroad and their respective university assessed them to so very good after their first and second year that they were both granted scholarships for the remaining years of the course. Now both of them are practicing overseas and really doing very well. Asked them if they ever want to come back to practice in Spore, they both said "tell the PAP government to go fuck itself". They very much prefer to contribute to the overseas country that treated them so much better than Singapore. My 2 nephews are just tip of the iceberg. I am sure many many more Singapore students who even better than them have been denied entry by PAP to medicine school here. The PAP would instead open their assholes to freely let Pinoys, Ah Nehs and Burmese with fake degrees in. And as a side point, if one ever get attended by a Pinoy doctor, be very very careful as almost 100% he or she is a quack. Of all foreign trash here, the Pinoys are the biggest bull shitter, even worse than Ah Nehs.

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BedokFunland JC H2 Chemistry Qn

State the formula and colours of the manganate(VI) and manganate(VII) ions, and suggest with explanation, whether the Mn-O bond length in manganate(VI) ion is expected (ie. don't Wikipedia the answer immediately, but first think over it critically based on what you've learnt in the H2 syllabus) to be longer or shorter, compared to the manganate(VII) ion.

Hints :
Draw the first couple of resonance contributors, then the resonance hybrid, for the 2 ions. Hence state the bond order (eg. 1.5 bonds, or 1.75 bonds, etc) for each individual Mn-O bond in the resonance hybrid for the 2 ions.

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I've always been supportive of the private candidacy route, as I've always been of the opinion that the majority of time and energy spent in JC life is wasted (eg. how many students really learn and benefit from JC lectures? Lectures should be replaced completely by tutorials), majority of JC students suffer from perpetual chronic sleep deprivation (it's a medically proven fact that everyone on average requires a minimum of 8 hours of uninterrupted sleep, and daily sleep deprivation of less than 8 hours of sleep results in accumulative damage to your brain and body resulting in damaged intellectual faculties and semi-permanently lower IQ (damaged can be partially reversed if you successfully manage 8 hours of sleep daily from now on) and consequently significantly poorer A level results, and it's also therefore no wonder that Singaporeans have the highest incidence of diabetes, heart attacks and strokes, the risk of these diseases greatly exacerbated by chronic lack of sleep) and the combined (mental, emotional, psychological) stress from all the different JC subjects, CCAs, projects and other committments, can actually do more harm than good (and in extreme cases can almost ruin one's entire life, even for students in or perhaps especially students from top JCs, such as Kevin Wee of RJC, and XinLin of HCJC, etc. Hope Kevin and XinLin are both doing well in their lives now).

Better to leave now and go the private route and take back control of your life, than risk a severe psychological breakdown that will not only ruin your A level results anyway, but may even cost you your phsyical, emotional, mental and psychological health, and in tragically worst case scenarios (there have been several cases in the last few years), it may even cost you your life.

If you can afford it, go for tuition (group tuition is always a lot cheaper than solo tuition) for your H2 subjects. But even if you can't afford any tuition at all, even solely self-study will still (in all likelihood) gain you a better performance at the A levels, compared to if you were to suffer a psychological breakdown in the unhealthy JC environment.

Final disclaimer : obviously this is just my opinion. I'm *not* telling you to quit JC and take the private route (coz this is your own life and therefore you must make your own decisions for yourself). I'm just saying that if that's what you decide for yourself that it's you want to do, then as far as I'm concerned, there's certainly a lot of sense in doing so. Obviously there will be some JC and ex-JC students will say, "What a load of rubbish! I'm totally enjoying JC life! No stress at all for me! No sleep deprivation at all for me! I love JC life, which has been socially, mentally, emotionally and personally enriching and fulfilling for me in every way! I just wish it was at least 4 to 6 years instead of just 2 short years!", but just as there are such students on one end of the spectrum, there are bound to be students on the other end of the spectrum, for which private candidacy would make a lot of sense.

Decide for yourself, and no matter the outcome, you can respect yourself for taking back control and responsibility for your own life.

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Sure no prob.

Q1. Yes, that's correct.

Q2. No, that's not quite correct (you stated two facts correctly, but there's no direct link between them, or the link is indirect and tenuous at best). Different isomers (eg. structural, functional group, geometric, etc) have different thermodynamic stabilities and hence possess different potential enthalpies, and therefore their combustion enthalpies (ie. magnitude of enthalpy difference between the isomers, and the combustion products) are different. A good example at A levels would be terminal alkenes vs internal alkenes (internal alkenes have greater number of stronger C-C sigma bonds with the higher % s orbital character), or cis alkenes vs trans alkenes (cis alkenes have greater steric van der Waals strain or repulsion between bulky alkyl groups on same side of the double bond).

Basic idea is to draw an energy profile diagram with y-axis as energy/enthalpy level. Near the top would be the elements, near the bottom would be the combustion products, and in the middle (with varying energy/enthalpy levels and thermodynamic stabilities) would be the different isomers. By Hess Law, overall (formation + combustion) enthalpy would be the same. But if the isomer is less stable = more energy, its formation enthalpy would be less exothermic but its combustion enthalpy would be more exothermic; if the isomer is more stable = less energy, its formation enthalpy would be more exothermic but its combustion enthalpy would be less exothermic. Examples would be (as I mentioned earlier) internal vs terminal alkenes, and cis vs trans alkenes.

At O levels this question shouldn't, and wouldn't, be asked by Cambridge at all. Whoever asks this question at O levels (eg. school prelim paper, Cambridge, etc) should be shot, as the correct answer requires a minimum of A level Chemistry.

Q3. Correct, per unit volume of the fuel, there is a greater number of C and H atoms, and hence greater number of C and H atoms oxidized (exothermically via combustion) to CO2 and H2O, and concordantly, larger amount of heat released during combustion. An additional secondary effect, results from having stronger van der Waals attractions between molecules with larger number of atoms, allowing for closer packing of the molecules, ie. increasing density of fuel, hence resulting in greater number of fuel molecules per unit volume.

Hi ultimaonline, thank you for taking your time to reply, really appreciate it. I have a few o level questions to ask. 1) what are the difference between addition and condensation polymerisation? Can i state addition produces polymers that have the same empirical formula as the monomer itself whereas condensation does not? I am very certain about the other 2 differences.,

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Rachel Lee JiaLi asked BedokFunland JC this question 2 days before her 2014 A levels Paper 2 exam : "If you're given 25.0 cm3 of 0.040 M of CH3COOH with pKa = 4.76, and 50.0 cm3 of 0.050 M Ba(OH)2, and you carry out titration (ie. limiting reactant in conical flask, excess reactant drip down from burette), what's the pH at equivalence point?"

Observe that CH3COOH acid is monoprotic and Ba(OH)2 base is diprotic.

Moles of H+ available = 25/1000 x 0.04 = 1 x 10^-3 mol

Moles of OH- available = 2 x (50/1000 x 0.05) = 5 x 10^-3 mol

Hence limiting reactant is CH3COOH = 1 x 10^-3 mol

Hence moles of OH- required to reach equivalence = 1 x 10^-3 mol

Hence moles of diprotic Ba(OH)2 required to reach equivalence = 5 x 10^-4 mol

Hence to find volume of diprotic Ba(OH)2 required to reach equivalence :

Molarity = Moles / Volume

0.05M = 5 x 10^-4 mol / Volume dm3

Hence volume of diprotic Ba(OH)2 required to reach equivalence = (5 x 10^-4) / (0.05) = 1 x 10^-2 dm3

Hence total volume in conical flask at equivalence point

= volume of CH3COOH (originally in conical flask) + volume of Ba(OH)2 (added in from burette)

= (25/1000) dm3 of CH3COOH + (1 x 10^-2) dm3 of Ba(OH)2

= 3.5 x 10^-2 dm3

Moles of CH3COO- ions from the soluble salt Ba(CH3COO)2 at equivalence point = 1 x 10^-3 mol

(explanation : since that's how much CH3COOH acid we used ; don't let the stoichiometry of the formula of the barium ethanoate salt Ba(CH3COO)2 confuse you, focus on the moles and molarity of the active species which determines the pH, ie. the CH3COO- ion)

Hence molarity of CH3COO- ions at equivalence point

= moles of CH3COO- / total volume at equivalence point

= (1 x 10^-3) / (3.5 x 10^-2)

= 2.8571 x 10^-2 mol/dm3

From pKa of 4.76 for CH3COOH, we find pKb for CH3COO-

= 14 - 4.76

= 9.25

Hence Kb for CH3COO- = 10^-(pKb) = 5.7544 x 10^-10

Approximating equilibrium molarity of CH3COO- back to it's initial molarity (in the ICE table) since Kb value for the weak base is much smaller than initial molarity of the base CH3COO-, that is to say, X or [OH-] is negligible compared to 2.8571 x 10^-2 mol/dm3,

Kb = [CH3COOH] x [OH-] / [CH3COO-]

5.7544 x 10^-10 = X^2 / 2.8571 x 10^-2

X or [OH-] = square root of (1.6441 x 10^-11)

[OH-] = 4.0547 x 10^-6

Hence pOH = 5.392

Hence pH = 14 - 5.392

= 8.608 (Note : 8.608 is actually to 3 significant figures as following the data given in the question ; for pH, significant figures are counted only after the decimal point of the pH value).

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A few hours before 2014 A Level Paper 3, LeeJunEr asked BedokFunland JC, "If we need to convert para-methylnitrobenzene into para-aminobenzoic acid, does the order in which we carry out oxidation of the methyl group, versus the reduction of the nitro group, matter?"

BedokFunland JC to the rescue :

Yes, it does matter.

Although not taught in the H2 syllabus, KMnO4 can indeed oxidize phenylamine into nitrobenzene (at neutral pH), or azobenzene (at alkaline pH), or polyphenylamine (at acidic pH).

Therefore, the methyl group needs to be oxidized using acidified KMnO4 first (which doesn't affect the NO2 group), and thereafter the nitro group needs to be reduced using Sn and excess concentrated HCl (which doesn't affect the COOH group, since carboxylic acids can only be reduced by LiAlH4, and not by H2 gas generated by acid and metal reaction).

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AlphaEternala asked, "I don't like my school notes. How should I best answer H2 Chemistry questions in the A levels?"

No, there is no secret shortcut way to determine the perfect way to answer. As it is, Singapore JC teachers all give their own different answers, not just across JCs, but even within a JC.

It's common knowledge that after sharing Prelim papers across JCs, school teachers will open criticize other JCs' pelim paper questions and given answers in the mark schemes, saying to their students, "Don't follow their mark scheme answers, you should instead follow the notes I gave you".

So you should still follow yourself, write what you're comfortable with, even if your school lecture notes say otherwise. A general rule of thumb to guide you (ie. to make sure you will score most, if not all, the marks for the question in the A levels), is to ensure your answer includes all the main points and keywords found in reliable A level publications (ie. *not* school lecture notes, but quality A level books from CS Toh, George Chong, Chan Kim Seng & Jeanne Tan, etc). Combine the answers or keywords from 3 of your personal favourite reliable sources, and write it in your own style that you can be satisfied with (instead of reluctantly and blindly memorizing your school lecture notes answer which you personally disagree with).

To further allay your fears, rest assured that Cambridge allows a reasonable degree of flexibility in the phrasing of the answers. For instance, to a certain question, eg. "Why do transition metals form coloured compounds?", all 3 books CS Toh, George Chong, Chan Kim Seng & Jeanne Tan, would phrase their answers very differently. But Cambridge would award full marks to all 3 of them.

Moral of the story : so go ahead and combine their answers and keywords into your own unique answer that you're personally happy with. *Make it your own*. That's the secret to mastering Chemistry.

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The boiling point of CCl4 is higher than SiCl4. CCl4 has Vander waal forces as well as SiCl4 has but the b.p of SiCl4 is expected to be higher than CCl4 as the former has more spacious electron cloud. Can someone explain.

I actually posted the answer to this question on my BedokFunland JC website several years ago :

Due to the greater magnitude of electronegativity difference between Si and Cl, compared to C and Cl, there is a larger magnitude of partial negative charge on the Cl atoms in SiCl4 molecules (compared to CCl4 molecules), and consequently a greater extent of electrostatic van der Waals repulsions between SiCl4 molecules, and therefore SiCl4 has a lower boiling point, compared to CCl4.

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The following is aquestion from CIE A-level (9701) Nov 2010

G is NH2(CH2)3NH2 is titrated with HCl(aq).

A 0.10 mol dm–3 solution of G has a pH of 11.3. When 30 cm3 of 0.10 mol dm–3 HCl is added to 10 cm3 of a 0.10 mol dm–3 solution of G, the final pH is 1.6.

Using the following axes, sketch the pH changes that occur during this addition of

HCl(aq). [I cannot show the axes here]

No other data such as pka values were given. Its was a 2 marks question.

Solution : I only knew the starting pH that is 11.3 and the end pH 1.6. There will be two humps as the base is diacidic. The end-point i have no idea. Please help.

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Here is my BedokFunland JC answer to this 2010 Cambridge A level Exam Qn.

Since molarities of both acid and base are the same, the volumes of HCl required for 1st and 2nd equivalent points are 10cm3 and 20cm3.

Show a steeper and longer vertical portion for the 1st equivalence point, with slightly acidic pH, eg pH 6.

Show a less steep and shorter vertical portion for the 2nd equivalence point, with a strongly acidic pH. eg pH 2.

Rationale :

Although technically diprotic, but you can deduce that it in practice it is mostly monoprotic (ie. extremely small Kb2), because of 3 (inter-related) reasons :

1) Considering the unipositive conjugate acid, the unipositive formal charged N atom / unipositive BH+ ion, electrostatically repels the H+ / H3O+ cation. Hence kb2 would be extremely small.

2) Considering the unipositive conjugate acid, the unipositive formal charged N atom is electon-withdrawing by induction, making the lone pair on the other N atom (3 C atoms away) to be less available to accept protons. Hence kb2 would be extremely small.

3) Considering the dipositive conjugate acid, 2 positively formal charged N atoms with only 3 C atoms between them is highly destabilizing, due to inter-nuclei repulsions within the highly unstable high cationic charge density dipositive conjugate acid. The more unstable the conjugate acid, the weaker the conjugate base. Hence kb2 would be extremely small.

As such, regard 1,3-diaminopropane as mostly monoprotic, in which case you can (even without Kb values available) expect the 1st equivalence point to have a acidic pH (due to proton dissociation hydrolysis of the acidic ammonium salt).

Which is why we draw a steeper, longer vertical portion for the 1st equivalence point, and a less steep, shorter vertical portion for the 2nd equivalence point (since Kb2 will be extremely small).

As for the strongly acidic pH at the 2nd equivalence point, as explained earlier, Kb2 is extremely small, meaning Ka1 (ie. proton dissociation for the dipositive conjugate acid) is extremely large, meaning that the dipositive conjugate acid is a strong acid. Hence at 2nd equivalence point, the dipositive diammonium salt is strongly acidic and will have a pH almost as acidic as the final pH of 1.6 (ie. with 10cm3 excess HCl). A reasonable estimate would be approx pH 2.

Thank you for the detailed answer.

Is there any book you would recommend for these type of pian-staking problems?

If you're referring specifically to acid-base equilibria problems involving polyprotic / multiprotic acids & bases, there's plenty of free materials available on the internet :
https://www.google.com.sg/#q=polyprotic+acid+base+equilibrium+problems

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My Original BedokFunland JC H2 Chemistry Challenge Qn (January 2015)

Draw the structure of the major product for each electrophilic aromatic substitution reaction.
[ Note : in all 3 cases, assume only 1 electrophile is substituted onto the benzene ring (ie. mono-substitution), and there is only 1 single most major product (ie. ignore the minor products and give only 1 answer). Hint : You have to simultaneously consider *both* steric and electronic (including *both* inductive and resonance) effects to arrive at the *single* most major product.]

Case 1 : 4-nitro-benzaldehyde : meta-directing moderate deactivator vs meta-directing strong deactivator, who wins? Identify all relevant factors (sterics / induction / resonance).

Case 2 : trichloromethylbenzene : is this substituent activating or deactivating? Is it ortho/para or meta-directing? Identify all relevant factors (sterics / induction / resonance).

Case 3 : 3-methyl-nitrobenzene : ortho/para-directing weak activator vs meta-directing strong deactivator, who wins? Identify all relevant factors (sterics / induction / resonance).

Case 4 : 3-bromo-nitrobenzene : ortho/para-directing weak deactivator vs meta-directing strong deactivator, who wins? Identify all relevant factors (sterics / induction / resonance).

Case 5 : 2-chloro-5-bromo-nitrobenzene : a 3 cornered fight, who wins? What's the single most major product? Identify all relevant factors (sterics / induction / resonance).

Special Note : I will not reveal the answers here ; ask your own school teacher or private tutor for their answers, but regard their proposed answers critically with a pinch of salt and ask yourself if you're truly satisfied with their explanations, as I wrote these tricky BedokFunland JC questions myself, and many school teachers and private tutors may not get their anwers correct.

On a related note, check out the following Yahoo Q&As :
https://answers.yahoo.com/question/index?qid=20100430074722AAB69Ac
https://answers.yahoo.com/question/index?qid=20111228172010AAonefP
but be warned : both the chemical engineer and the organic chemist got their answers wrong. Can you / your school teacher / your private tutor figure out what's wrong with their answers, and what the correct answers are?

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The difference between Chemical Biology versus BioChemistry :
http://www.brandeis.edu/departments/chemistry/docs/Chemical%20Biology.pdf

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The by-product of electrolysis of brine is sodium hydroxide. The reaction that is quoted is 2H2O + 2e - H2 + 2OH- .....which occur at cathode. The problem is that, surely at the cathode cations are discharged such as here H+ ions, should be discharged. No idea how water being a neutral species migrates to cathode ?? Please explain.

Within the H2O molecule, because O atoms are more electronegative than H atoms, hence the H atoms, having a partial positive charge, are electrostatically attracted to, and accept electrons from, the cathode. For every H2O molecule, 1 H atom accepts 1 electron, generating an OH- ion and a H atom, which readily combines with another H atom from the reduction of the next H2O molecule, thus generating H2 gas (with positive, thermodynamically favorable entropy change).

If you prefer, an alternative acceptable way to look at it, is that H+ ions (or H3O+ ions) from the auto-dissociation of H2O molecules, are electrostatically attracted to, and accept electrons from, the cathode, to be reduced to H atoms, which readily combines with the next H atom to generate H2 gas (again with positive, thermodynamically favorable entropy change), thereby causing the position of equilibrium of the auto-dissociation of water to shift to the RHS as predicted by Le Chatelier's principle, thus generating more OH- ions as a byproduct.

Both ways of looking at this are correct, use whichever you prefer, whichever helps you understand better, and Cambridge certainly won't ask for such explanatory details at A levels in any case.

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