VCE Chemistry Question Thread

[miyukiaura]

Do we need to know how to write redox equations under alkaline conditions? Also, would vcaa ask us to write equations for the reactions in fuel cells with alkaline, liquid, molten carbonate, or solid oxide electrolytes?

[ArtyDreams]

Do we need to know how to write redox equations under alkaline conditions? Also, would vcaa ask us to write equations for the reactions in fuel cells with alkaline, liquid, molten carbonate, or solid oxide electrolytes?

Yes, you definitely need to know how to balance/write redox reactions in alkaline conditions.
I’m pretty sure that they can have any type of electrolyte and ask you to create equations based off it. I remember having to write ones in ‘ceramic’ (oxide) and liquid electrolytes. Usually they have a diagram that makes it clear what’s happening though.
Hope this helps!

[keltingmeith]

Do we need to know how to write redox equations under alkaline conditions? Also, would vcaa ask us to write equations for the reactions in fuel cells with alkaline, liquid, molten carbonate, or solid oxide electrolytes?

First bit - it's never been asked before, it's a safe no. However, it's actually pretty easy to do - first step to making an alkaline cell, simple balance it assuming it's in acidic conditions. Eg, let's say I have \(\ce{MnO4- + I- \to I2 + Mn^2+}\) Go through the steps:

Half-reaction 1: \(\ce{MnO4- \to Mn^2+}\)
1. Balance atoms that aren't oxygen or hydrogen: \(\ce{MnO4- \to Mn^2+}\)

2. Balance the oxygens by adding water: \(\ce{MnO4- \to Mn^2+ + 4H2O}\)

3. Balance the hydrogens by adding protons: \(\ce{MnO4- + 8H+ \to Mn^2+ + 4H2O}\)

4. Balance the charge: \(\ce{MnO4- + 8H+ + 5e- \to Mn^2+ + 4H2O}\)

Half-reaction 2: \(\ce{I- \to I2}\)
1. Balance atoms that aren't oxygen or hydrogen: \(\ce{2I- \to I2}\)

2. Balance the oxygens by adding water: \(\ce{2I- \to I2}\)

3. Balance the hydrogens by adding protons: \(\ce{2I- \to I2}\)

4. Balance the charge: \(\ce{2I- \to I2 + 2e-}\)

5. Combine the equations. I'm going to do 2*HR1 and 5*HR2:
\(\ce{2MnO4- + 16H+ + 10e- + 10I- \to 2Mn^2+ + 8H2O + 5I2 + 10e-}\)

Of course, the electrons cancel off:
\(\ce{2MnO4- + 16H+ + 10I- \to 2Mn^2+ + 8H2O + 5I2}\)

And you'd normally be done if this was acidic. For alkaline, now all you need to do is add the same amount of hydroxide to both sides of the equation as there are free protons:
\(\ce{2MnO4- + 16H+ + 10I- + 16OH- \to 2Mn^2+ + 8H2O + 5I2 + 16OH-}\)

NOTE: I've not really "added" anything new to the equation. Free hydroxide exists in all solutions of water, remember that water is in constant equilibrium with hydronium and hydroxide ions. I can add the hydroxide for the same reason I can add the acid.

And this is now in alkaline conditions! Do note that the protons and hydroxides on the same side should combine into water, and then cancel off any extra waters from the other side:
\(\ce{2MnO4- + 16H2O + 10I- \to 2Mn^2+ + 8H2O + 5I2 + 16OH-}\)
\(\ce{2MnO4- + 8H2O + 10I- \to 2Mn^2+ + 5I2 + 16OH-}\)

And you're done - alkaline condition redox reaction, with just one extra step of adding those hydroxides. Otherwise, see ArtyDreams for some good responses. Normally I wouldn't respond since they have, but I really wanted to showcase that the alkali version is not complicated, it's just 1 extra step

[miyukiaura]

Thanks so much!

[RCircles]

For fuels, what do we need to know in terms of the energy content, renewability and environmental impacts related to
sourcing and combustion of fuels

[Harrycc3000]

Hi guys, q about IR spectroscopy
For IR spectroscopy I'm confused by the logic (and most of IR spectroscopy tbh) for why lighter bonds absorb IR light with a higher wavenumber. Online articles have said 'lighter atoms vibrate more than heavier atoms' so they'd absorb with higher frequencies. I don't get how 'vibrates more' correlates with a higher wavenumber. With the explanation for why stronger bonds need higher wavenumbers, its because they would need more energy to bend and stretch them, so they'd need to absorb light that have more energy (light with higher wavenumbers). This makes sense but it doesn't have much to do with vibrations. I understand that IR measures the different vibrational energy levels in bonds of molecules but I don't get why the fact that some atoms vibrate more means they need higher wavenumbers. (Is it cos they vibrate more so it means they're vibrational energy level is higher -> need light with more energy?) Also, why do lighter atoms vibrate more in the first place anyway?
Any clarification would be welcome thanks!

[Chocolatepistachio]

If someone could help with this question
50 grams of dextrose was dissolved in water and the volume was made up to 100 ml. What is the molar concentration of the dextrose solution. Molar mass of dextrose= 180g

[p0kem0n21]

If someone could help with this question
50 grams of dextrose was dissolved in water and the volume was made up to 100 ml. What is the molar concentration of the dextrose solution. Molar mass of dextrose= 180g

Let's first establish the units of molar concentration: mol L^-1. Thus, we need to find both the amount of dextrose in moles and the volume of the solution in litres.

Finding the volume in litres is simple with the given info: we just convert 100 mL to 0.1 L.

Now we need to find the amount of dextrose in moles. We are given the mass (50g) and the molar mass (180 g mol^-1). Using trusty n=m/M, we get n=5/18 moles.

All we have to do now is chuck our info into c=n/V, or (5/18)/(0.1). Using a calculator, we can express the molar concentration as 2.78 M (or mol L^-1), to three sig figs (not quite sure if that's the right amount of sig figs though  ).

[Chocolatepistachio]

For d why is NH2 a higher priority than the carbon with the double bond oxygen and the OH isnt it based on molar mass so the one with 2 oxygens would be higher

[Owlbird83]

For d why is NH2 a higher priority than the carbon with the double bond oxygen and the OH isnt it based on molar mass so the one with 2 oxygens would be higher

You stop at the first atoms bonded unless there are two the same. So at first just look at the N, C, C, H, where you immediately can give N the highest priority. Then you can look at what's bonded to the carbons to determine the order of 2nd and 3rd.

Hope that makes sense!

[Chocolatepistachio]

Thanks

For a why would the 5th carbon be chiral because it has 2 CH which are the same but the 6th carbon isn't and for b why wouldn't the second carbon be chiral doesn't it have 4 different things attached

[Owlbird83]

Thanks

For a why would the 5th carbon be chiral because it has 2 CH which are the same but the 6th carbon isn't and for b why wouldn't the second carbon be chiral doesn't it have 4 different things attached

a) The 5th carbon does have 4 different things attached because for this (unlike the previous posted question) you don't stop at the first atoms bonded you look at the whole thing of each thing coming off the carbon, eg the 5th carbon has a -H, -CH3, -CH(CH3)2, -CH2CH(Cl)CH2COOH, which are all different, making it chiral. 6th carbon has two -CH3 bonded so it's not chiral.

b) The dashed line going back would be a -CH3, and so would the other skeletal line coming out to the right, so two -CH3

[Chocolatepistachio]

ok thank youu

For this question how do you know its SN1 or SN2 for b

[Ruchir]

Hello
Does any one know where I can download chemistry Heinmann units 1 and 2 worked solutions for free?
Thanks, help is really appreciated

[sweetiepi]

Hello
Does any one know where I can download chemistry Heinmann units 1 and 2 worked solutions for free?
Thanks, help is really appreciated

Hey! Sadly those solutions are copyrighted - and asking for copyrighted materials is against forum rules