You know how you consider the standard hydrogen half cell, in electrolysis if the electrolyte is acidic? If the electrolyte is basic would you consider the
4OH- (aq) ——-> O2(g) + 2H2O(l) + 4e-
Half cell? Or is that not a thing?
Hey
I don't fully understand what you are asking, but I will have a crack at answering it anyways - appologies if I don't quite address what you wanted.
The half cell reaction you have given is correct assuming you have replaced the acidic electrolyte (H+ solution) with a basic OH- solution. However, this reaction no-longer is a standard hydrogen half-cell/electrode. A standard hydrogen half cell exclusively refers to a platinum electrode exposed to H2(g) and a 1M H+ solution at slc. This standard hydrogen half cell is used as a reference point where the standard electrode potentials for all half cell reactions shown in your electrochemical series is the voltage/potential difference between the half cell in question and the standard hydrogen half cell.
The half cell reaction you have written above is not equivilant to the standard hydrogen half cell as each reaction involves different chemical species, and thus has a different standard electrode potential. The reaction you have written (the oxidation of hydroxide ions) is actually included in the electrochemical series as:
 + H_2 O(l) + 4e^- \leftrightarrow 4OH^- (aq) E= +0.40 )
Hopefully that might clear up any confusion, but feel free to point out anything I didn't explain clearly or anything else you need clarified
Hi,
I was struggling with these 2 questions and would really appreciate some help with them
For Q 3, I think I got how to do it for methane, but I got stuck on ethane and ethene.
For Q7, I feel like it's related to the homologous series but I'm still not sure whether propane or butane has the higher boiling point.
Thanks in advance!
q.3 lewis structures show the distribution of valence electrons in an atom/molecule where electrons are shown as dots around an atom. Some of these electrons might be bonding electrons, and some might be non-bonding (lone pairs). A good way to aproach drawing these molecules is by first drawing out all atoms as they would be as dictated by their strucural diagram, although leaving blank space between them where bonds would otherwise be. Then fill in the space between molecules with pairs of bonding electrons (a single bond will be one pair of electrons with one electron contributed by each atom, a double bond will consist of two pairs of electrons, two from each atom). Sometimes (however it is not essential) electrons from one type of atom will be denoted by an x, while electrons from another will be denoted by a dot (i.e. hydrogens electrons might be denoted by x's and carbon's by dots). Then after all atoms are connected by the apropriate bonding electrons, any pairs of valence electrons that are not participating in bonding can be drawn as a pair of dots (a lone pair). For both hydrogen and carbon here, all of their valence electrons should be pariticipating in bonding, so there should be no lone pairs.
Let me know if this helps - but if you still can't get the answer I am happy to work through the diagrams further for you.
q.7
The boiling point (as well as melting point) is determined by the intermolecular forces of attraction (the attractive forces existing between molecules). As such, for this question we need to first identify the intermolecular forces of attraction acting between the molecules of each substance.
There are 3 types of intermolecular attractions
1. dispersion forces
- due to temporary and instantaneous uneven distribution of charge. Even in non-polar molecules electrons might at a given instant be slightly unevenly distributed around a molecule, causing it to be slightly polar at points resulting in a weak attractive force that acts between molecules. As a molecule increases in size, the strength of the dispersion forces will increase (essentially there are more electrons and a greater potential for them to form these instantaneous poles)
2. dipole-dipole interactions
- polar molecules with a positive and negative dipole, will posess these dipole-dipole attractions between the a dipole of one polecule (i.e. the positive dipole) and a complimentary dipole of another molecule (i.e. a negative dipole). This interaction is stronger than dispersion forces, but only occurs in polar moleules
3. hydrogen bonding
- a special type of dipole-dipole interaction which only occurs between an oxygen, nitrogen, or flourine and a hydrogen which is covalently bonded to an oxygen, nitrogen, or flourine atrom. While not a true bond (despite its name) hydrogen bonds are stronger than ordinary dipole dipole attractions.
The molecules we are considering (butane and propane) are non-polar, thus we can eliminate dipole-dipole interactions and hydrogen bonding as forces of intermolecular atraction. Thus dispersion forces are the only forces acting between each type of molecule. Remembering the dispersion forces are stronger between longer molecules, since butane is a longer molecule (4 carbons to propanes 3) butane has stronger dispersion forces. This means that molecules of butane are held together with greater force than molecules of propane, thus there is a greater energy requirement to overcom,e the attraction between butanes molecules compared to propanes. This means that butane will have a greater boiling/melting point than propane.
Hopefully this helps
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