Linda's Question Thread

[destain]

Q1) It is NOT possible to produce aluminium using an aq solution of aluminium compound, and this is because H2O is a stronger oxidant and thus will react first. (Just reading off the electrochemical series, I have no knowledge of these things in general), I was wondering if something like Mn was the reductant what would happen but that doesn't work because it doesn't exist or something but..

What if it was something else that was lowest than Al, like Mg2+?? Would it still be unable to react because water is a stronger oxidant? Even though there are other things in between water and Mg that could act as a reductant? Because apparently you don't need to know the reductant, and what would happen if there was a reductant in between, confusing me :S

2) Also with predicting the products from electrolysis of a mixture of aqueous solution containing Cu(NO3)2, Fe(NO3)2 and FeCl2 using inert electrodes,

I got six equations from the series, but i'm confused as the Fe3+ equation (0.77V) has both Fe3+ and Fe2+, so...HOW DOES THAT WORK?

3) Exothermic reactions release energy and so the energy content of the product molecules is lower than the energy content of the reactant molecules. Thus the bonds within the product molecules ( 2H2) molecules) are stronger than the bonds within the reactant molecules (2 H2 molecules and 1 O2 molecules)

Can someone explain this bond explanation?

4) LisaChem 2008, Question 4 in multiple choice. If anyone understands this.

[destain]

+2

1) Whenever you write out the equations from the electrochemical series, you use single arrows and thus the mol ratio is relevant right?

2) Why is the Ka value only used for weak acids and not strong acids?

[Conic]

+2

1) Whenever you write out the equations from the electrochemical series, you use single arrows and thus the mol ratio is relevant right?

2) Why is the Ka value only used for weak acids and not strong acids?

1.Yes
2. Strong acids fully ionise, so the Ka value is not relevant AFAIK.

[destain]

ah ok and are the calculations for K equlibrium value and K concentration fraction different?
As in...if the reaction is reversed, you just put a negative in front for K e but for Kcf, you do 1/Kcf when the reaction is reversed?

[Lasercookie]

ah ok and are the calculations for K equlibrium value and K concentration fraction different?
As in...if the reaction is reversed, you just put a negative in front for K e but for Kcf, you do 1/Kcf when the reaction is reversed?

Negative in front of the K? I'm not sure what you mean. Also I'm going to use the symbol Q for concentration fraction (personal preference, I find it less confusing).

The calculations for K and Q are the same.

The difference is that value of K is only for when the system is at equilibrium. Q is at any time. You can say that the concentration fraction and the equilibrium constant is the same at equilibrium.

I normally wouldn't have so many subscripts, but I'll stick them in to make what I'm trying to say clear. The f refers to forward reaction, the r refers to the reverse reaction. I like the write the eq on the equilibrium constants to make it clear that it's the concentration of those at equilibrium.

and

And then yeah, for the reverse reaction the "products of the forward reaction" are the "reactants of the reverse reaction". The "reactants of the forward reaction" are the "products of the reverse reaction".

So



edit: actually, I'm not sure if the concentration fraction actually has an inverse relationship. It's seems quite conceivable that, for example, the reverse reaction could go slower than the forward reaction. This would mean that they'd have different concentrations after proceeding for a particular time.

We can state it for the equilibrium constant because the concentration of the reactants and products always ends up going back to that same value.

[destain]

yes they do i came across it in a question/solutions and when the coefficient of the equation was halved, they did RootKf?

and can someone please answer the first few questions ;P really bugging me, so annoying to not understand things and having to skip them hahah

[destain]

the mole ratio can still be used in some way in equilibrium reactions?
if the coeffcient is..

2H --> 4J

You could say J is being produced at half the speed of H or something?
But you can't double the value of the mole or half to get one or the other?

Also.
In A+2B --> C and K = 0.318
[A] = 0.532 [C] = 0.0914
I can see from the info given that you have to work B out using the K formula and rearranging but I just wanted to ask how come you can't just use the ratio of A:2B

[destain]

yeah it would be