A-level chem (old thread, expired)
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Standard electrode potnetial is the emf measured between a test electrode and a standard hydrogen electrode under standard conditions of temperature, presuure and concentartion.
Then what is standard cell potential?? and what is standard redox potential??
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Originally posted by hoay:
Standard electrode potnetial is the emf measured between a test electrode and a standard hydrogen electrode under standard conditions of temperature, presuure and concentartion.
Then what is standard cell potential?? and what is standard redox potential??
A level students only need to memorize the definition of standard electrode potential.For cell potential, just state and apply the formula Cell potential = Reduction potential @ Cathode + Oxidation potential @ anode.
Many schools (ie. Singapore JCs) wrongly teach the cell potential formula as "E-nought reduction - E-nought oxidation", or "E-nought cathode - E-nought anode". These versions are silly and technically wrong, even though you get the same correct final answer. Cambridge only marks the final answer though, so no worries here.
For redox potential, there are two types : reduction and oxidation. For instance, looking at silver, we say :
The standard reduction potential of Ag+ to Ag is +0.80V
The standard oxidation potential of Ag to Ag+ is -0.80V
The more positive the redox (ie. oxidation or reduction) potential, the more the position of equilibrium lies to the right hand side (of the oxidation or reduction half-equation).
Standard molarities refer to 1.0M for aqueous species on both sides.
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In measuring the E for Fe2+/Fe3+ system the electrode is Pt, while for measuring the E for Fe/Fe2+ system it is Fe metal....Can't we use Fe for the Fe2+/Fe3+ system?? I think the Fe electrodeis the conductor of electrons for which Pt works equally well but why Pt for Fe2+/Fe3+ system?
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Originally posted by hoay:
In measuring the E for Fe2+/Fe3+ system the electrode is Pt, while for measuring the E for Fe/Fe2+ system it is Fe metal....Can't we use Fe for the Fe2+/Fe3+ system?? I think the Fe electrodeis the conductor of electrons for which Pt works equally well but why Pt for Fe2+/Fe3+ system?
Because if you had used an Fe electrode against the SHE, then the Fe electrode itself would be oxidized to Fe2+, and hence the voltmeter reading would be for the oxidation potential of Fe to Fe2+, not Fe2+ to Fe3+.
Platinum is inert, which means it would not be oxidized, and instead Fe3+ would be reduced to Fe2+ (assuming you started out with equal molarities of Fe2+ and Fe3+ in the electrolyte solution of the half-cell connected to the SHE), and this reduction potential can be obtained from the voltmeter reading.
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okay okay.....why did'nt i consider that!!
Another query is regarding salt bridge.
When we connect two half-cells via salt bridge let say Cu/Cu2+ and Zn/Zn2+ cells then electrons flows from Zn half cell to external circuit to Cu half cell. Therefore Zn2+ and 2e both transfer to Cu half cell where Cu2+ is reduced so the [Cu2+] deceases which is balanced by the Zn2+ cations. On the other hand both electrons and [Zn2+] decareases in the zn half-cell, at this moment salt bridge which contains aqueous soltuion of KCl releases K+ [I think 2 K+/ loss of 1 Zn2+]....This restores the cations in Zn half-cell.
Finally at the end of the process when wires are disconnected we have two solutions left. In the Zn half-cell we have aqueous solution of K2SO4 since ZnSO4 was usedas electrolyte. In the Cu half-cell e have aqueous solution of ZnSO4 left.
Is there any detail i am missing....
is there any balancing of electrons needed since elctrons are being removed from Z half cell to cu half-cell??....please mention.
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Originally posted by hoay:
okay okay.....why did'nt i consider that!!
Another query is regarding salt bridge.
When we connect two half-cells via salt bridge let say Cu/Cu2+ and Zn/Zn2+ cells then electrons flows from Zn half cell to external circuit to Cu half cell. Therefore Zn2+ and 2e both transfer to Cu half cell where Cu2+ is reduced so the [Cu2+] deceases which is balanced by the Zn2+ cations. On the other hand both electrons and [Zn2+] decareases in the zn half-cell, at this moment salt bridge which contains aqueous soltuion of KCl releases K+ [I think 2 K+/ loss of 1 Zn2+]....This restores the cations in Zn half-cell.
Finally at the end of the process when wires are disconnected we have two solutions left. In the Zn half-cell we have aqueous solution of K2SO4 since ZnSO4 was usedas electrolyte. In the Cu half-cell e have aqueous solution of ZnSO4 left.
Is there any detail i am missing....
is there any balancing of electrons needed since elctrons are being removed from Z half cell to cu half-cell??....please mention.
Anode : Zn is oxidized to Zn2+ (with ZnSO4 as electrolyte)Cathode : Cu2+ is reduced to Cu (with CuSO4 as electrolyte)
Salt bridge (eg. containing KCl) releases K+ ions to the cathode, and releases Cl- ions to the anode. This maintains electrical neutrality for both cathode and anode, allowing for electrons to continue flowing from anode to cathode.
Towards the end of the reaction, the anode would contain Zn2+, SO4 2-, Cl- and a very small amount of cathodic ions (diffused through the salt bridge), while the cathode would contain K+, SO4 2-, remaining Cu2+, and a very small amount of anodic ions (diffused through the salt bridge).
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K+ ions are released to cathode since Cu2+ ions are being reduced to Cu i.e to main the positive charges. But why Cl- ions are released from salt bridge to anode?? Is it to balance the unit positive charge carried out by K+ ions to be counterbalanced by Cl- ions??
Also, at the cathode you mentioned that the remaining ions are K+, SO4 2-, remaining Cu2+....Why not all the Cu 2+ ions from CuSO4 will be reduced to Cu thereby making blue solution to colorless??
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Originally posted by hoay:
K+ ions are released to cathode since Cu2+ ions are being reduced to Cu i.e to main the positive charges. But why Cl- ions are released from salt bridge to anode?? Is it to balance the unit positive charge carried out by K+ ions to be counterbalanced by Cl- ions??
Also, at the cathode you mentioned that the remaining ions are K+, SO4 2-, remaining Cu2+....Why not all the Cu 2+ ions from CuSO4 will be reduced to Cu thereby making blue solution to colorless??
Cl- ions are released to the anode to balance the newly generated Zn2+ ions.Depending on who the limiting reactant is, yes it's theoretically possible for all the Cu2+ to be reduced, but the voltage would be so low before that happens, that the reaction would be considered effectively terminated.
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Originally posted by UltimaOnline:
A level students only need to memorize the definition of standard electrode potential.For cell potential, just state and apply the formula Cell potential = Reduction potential @ Cathode + Oxidation potential @ anode.
Many schools (ie. Singapore JCs) wrongly teach the cell potential formula as "E-nought reduction - E-nought oxidation", or "E-nought cathode - E-nought anode". These versions are silly and technically wrong, even though you get the same correct final answer. Cambridge only marks the final answer though, so no worries here.
For redox potential, there are two types : reduction and oxidation. For instance, looking at silver, we say :
The standard reduction potential of Ag+ to Ag is +0.80V
The standard oxidation potential of Ag to Ag+ is -0.80V
The more positive the redox (ie. oxidation or reduction) potential, the more the position of equilibrium lies to the right hand side (of the oxidation or reduction half-equation).
Standard molarities refer to 1.0M for aqueous species on both sides.
What wrongly teached,lol, just prevent wrong writting of signs lah. :) (Thanks for my teachers whom thought this. As data booklet is all in standard reduction potential and in the syllabus only have this
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standard cell potential definition is needed by syllabus
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NaCl is not hydrolyzed by water instaed water just hydrates it. AlCl3 is hydrolyzed by water producing acidic solution. CH3Cl in hydrolzsed by NaOH (aq) and by water to form alcohol.....
Hydrolysis is the reaction with water or NaOH(aq)?? Please calrify and define hydrolysis.
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Originally posted by hoay:
NaCl is not hydrolyzed by water instaed water just hydrates it. AlCl3 is hydrolyzed by water producing acidic solution. CH3Cl in hydrolzsed by NaOH (aq) and by water to form alcohol... Hydrolysis is the reaction with water or NaOH(aq)?? Please calrify and define hydrolysis.
Technically, hydrolysis refers to the chemical reaction with water (don't worry about the "lysis" part, as either water or the other reactant will inevitably be broken up or "lysed" when the reaction occurs).However, because water is a weaker nucleophile, therefore to save time and money, we use a stronger nucleophile OH- in nucleophilic substitution reactions involving water, as the richer the electron-rich species, the stronger its nucleophilic strength, the lower the Ea required, the faster the rate of reaction.
Furthermore, the final intended product is the same. For instance :
R-X reacts with water to generate R-OH2+, which loses a proton to X-, hence generating R-OH and H-X as the final products. Higher Ea, slower reaction.
R-X reacts with Na+OH- to generate R-OH, with Na+ and X- as counter ions, generating R-OH and Na+X- as the final products. Lower Ea, faster reaction.
On a related note, notice that physical & inorganic chemists are lazier compared to organic chemists in some ways (eg. physical/inorganic chemists' quick dative bond arrowhead versus organic chemists' full curved-arrow mechanism), but are stricter in other ways (eg. physical/inorganic chemists say "hydration refers to physical interaction with water, while hydrolysis refers to chemical reaction with water"; but organic chemists say, "since dehydration means removing water from alcohol to alkene, then let's just call adding water to alkene as hydration; no need to specify hydrolysis rather than hydration").
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Dear all
I have a problem with crystal solid. I don't know what type of solid it is if they give me a fomula. For e.g. in Step by Step there are a question about Se02.I don't know why but the answer is Se02 is giant molecular
Can you help me to know exactly what type of solid if i know a fomula
Thx
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Originally posted by Theplayfulgirl brainy:
Dear all
I have a problem with crystal solid. I don't know what type of solid it is if they give me a fomula. For e.g. in Step by Step there are a question about Se02.I don't know why but the answer is Se02 is giant molecular
Can you help me to know exactly what type of solid if i know a fomula
Thx
Based on the position of Se in the periodic table, you'd expect SeO2 to be covalent rather than ionic. For covalent, there is simple molecular, versus giant covalent. Since SeO2 is a solid at rtp with a high melting point (such info would be inlcluded in a fair exam question), you can deduce it's giant covalent, to be precise, giant covalent polymeric (ie. from the auto-polymerization of individual SeO2 simple molecular monomers).
Structure :
http://en.wikipedia.org/wiki/Selenium_dioxide -
It means that i have to know what kind of bonding it is and then deduce type of solid. How about BeCl2 and BeF2. Be-F is ionic and Be-Cl is covalent right. My friends told me that it is because of the diffence of electronegativity but it is quite confusing. How can we know without any number
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Originally posted by Theplayfulgirl brainy:
It means that i have to know what kind of bonding it is and then deduce type of solid. How about BeCl2 and BeF2. Be-F is ionic and Be-Cl is covalent right. My friends told me that it is because of the diffence of electronegativity but it is quite confusing. How can we know without any number
For electronegativity values, you can look it up on the internet (though there are a few different systems giving different values).For A level purposes, you don't need to know the electronegativity values, you only need to memorize that as long as the cationic charge density is very high (eg. Be2+, Al3+, etc), the metal fluoride is ionic (since F is extremely electronegative, all metal fluorides are considered fully ionic), but all other metal halides (eg. Cl, Br, I) are considered covalent (on condition that the metal cation has high charge density).
If the cationic charge density is fairly high, but not too high (eg. Ag+), then you've to memorize to write that it has "significant covalent character", but is still primarily ionic in character.
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I had problem with the shapes of molecules. How can i determine it without learning by heart all types of shapes???
Thank you
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Originally posted by Theplayfulgirl brainy:
I had problem with the shapes of molecules. How can i determine it without learning by heart all types of shapes???
Thank you
Guys only like Girls, Guys don't like Guys.Electrons (negatively charged species) are guys. Due to inter-Guy (ie. inter-electron) repulsions, electron pairs want to spread out as far away from each other as possible.
NH3 - For N to have a stable octet but no formal charge, N must have 1 lone pair and 3 bond pairs. Hence electron geometry = tetrahedral (since that's the best shape that has 4 electron charge clouds away from each other as far as possible, but still bonded to the central atom). Since one of the four electron charge clouds is a lone pair, ie. treat it as invisible, hence molecular geometry = trigonal planar.
Do a tripod stand with 3 fingers of your left hand, and use one finger of your right hand do the 4th and top electron charge cloud, since lone pairs are invisible, make your right hand invisible by removing it (although bear in mind the lone pair still remains there, otherwise the electron geometry would change into trigonal planar instead of tetrahedral), and what you have left, is called the molecular geometry, which is trigonal pyramidal.
In future, I recommend you start a new thread on the forum, with a suitable title eg. "[H2 Chem] - Qn on Chem Bonding", instead of posting in an old thread such as this one.
Also, if you can afford it, go for tuition sooner rather than later. Most JC1 students make the mistake of thinking, "I'll try to handle it myself first, only if I fail the mid year exams, then I'll look for tuition", which by then, it's a bit too late for the Promos, bearing in mind you have other subjects to catch up on as well.
Also, there are many intricacies and aspects to H2 Chem that I have to teach in person, which cannot be effectively done via an online forum. Regardless, feel free to continue posting your H2 Chem qns here. I'll reply whatever I can.
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Originally posted by Theplayfulgirl brainy:
I had problem with the shapes of molecules. How can i determine it without learning by heart all types of shapes???
Thank you
We can't totally do away with using a bit of memory.
But it helps if you can group them into five basic shapes ( Linear, Trigonal planar, Tetrahedral, Trigonal Bipyramidal , Octahedral).
For example, if a molecule has 3 bond pairs and 1 lone pair
1) its "basic shape" (sometimes refered to as electron group geometry) is 3+1 = 4, tetrahedral.
2) You visualise it as a tetrahedral shape with 4 branches, then "erase" 1 of the branches to account for it being a lone pair.
3) You have 3 branches left with that looks like a pryramid with a triangle base - trigonal pryrimidal.
Be careful of the branches to "erase" if you start off with trigonal bi-pyramidal.
And yes, it might be better if you create your own thread of questions next time.
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I totally get the shape of molecules. Anw, now, i 'm struggling with Electrochemistry. I confused about this question. I 'm wrong or the solution wrong?
In acidic solutionn, Mn04- ions oxidise Cl- ions to Cl2. The value of Eo for the reaction is 0.16V. They asks several things but i think we don't need to care about it.
There are 2 equation of this excercise:
MnO4- +8H+ +5e- => Mn2+ +4H20 Eo= 1.52 V
Cl2 +2e- => 2Cl- Eo = 1.36 V
1. If Mn04 oxidise Cl- to Cl2. I think it must be reduce or are oxidised.
2. The right equation must be opposite. It must be:
2Cl- => Cl2 +2e Eo= -1.36
3. In the end, Eo of the whole must be : =1.52 - - 1.36.
Some one help me plz
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Originally posted by Theplayfulgirl brainy:
I totally get the shape of molecules. Anw, now, i 'm struggling with Electrochemistry. I confused about this question. I 'm wrong or the solution wrong?
In acidic solutionn, Mn04- ions oxidise Cl- ions to Cl2. The value of Eo for the reaction is 0.16V. They asks several things but i think we don't need to care about it.
There are 2 equation of this excercise:
MnO4- +8H+ +5e- => Mn2+ +4H20 Eo= 1.52 V
Cl2 +2e- => 2Cl- Eo = 1.36 V
1. If Mn04 oxidise Cl- to Cl2. I think it must be reduce or are oxidised.
2. The right equation must be opposite. It must be:
2Cl- => Cl2 +2e Eo= -1.36
3. In the end, Eo of the whole must be : =1.52 - - 1.36.
Some one help me plz
Under standard conditions,
Reduction potential of MnO4- to Mn2+ is +1.52V
Oxidation potential of Cl- to Cl2 is -1.36V
Cell potential = Red@Cathode + Ox@anode = (+1.52) + (-1.36) = +0.16V
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Originally posted by Theplayfulgirl brainy:
Your pic link is broken. Anyways, as I've told you before, start a new thread. Don't post your new question in someone else's old thread. I'm thereby locking this thread.