stonecold's chem questions :)

[stonecold]

Someone please enlighten me and say that we don't have to know the battery equations in chapter 27...

And also, what is the chemical reasoning behind the fact that to recharge a car battery, you have to connect the anode to the anode and the cathode to the cathode?

Is it an equilibrium thing?  Is it right to say that you basically pump the anode full of electrons, to drive the reaction in the half cell backwards whilst simultaneously removing electrons from the cathode again to push the reaction for that half cell backwards?

[kenhung123]

Hmm isn't it the other way around? Cathode to anode and anode to cathode? This is so the reactions are reversed?

Edit: actually you are correct and I don't understand. So the reaction is reversed but the polarities are unchanged, how does that work?

[Greggler]

mm yes i hope to god we dont have to memorise those specific equations regarding each cell type..

[iffets12345]

Someone please enlighten me and say that we don't have to know the battery equations in chapter 27...

And also, what is the chemical reasoning behind the fact that to recharge a car battery, you have to connect the anode to the anode and the cathode to the cathode?

Is it an equilibrium thing?  Is it right to say that you basically pump the anode full of electrons, to drive the reaction in the half cell backwards whilst simultaneously removing electrons from the cathode again to push the reaction for that half cell backwards?

Yea you're right I think. By pumping the anode full of electrons, whatever was oxidised will accept those electrons and reform the solid or whatever it was originally, thus free to oxidise again. The positive terminal extracts electrons to create the oxidant reactant again so that it can be free to receive electrons.

my crappy explanation.

[physics]

do we need to know the batteries at all? the equations and stuff!? :O?

[Whatlol]

Someone please enlighten me and say that we don't have to know the battery equations in chapter 27...

And also, what is the chemical reasoning behind the fact that to recharge a car battery, you have to connect the anode to the anode and the cathode to the cathode?

Is it an equilibrium thing?  Is it right to say that you basically pump the anode full of electrons, to drive the reaction in the half cell backwards whilst simultaneously removing electrons from the cathode again to push the reaction for that half cell backwards?

Yes, in the forward reactions the cathode is positively charged as it is accepting electrons, and the anode is negatively charged, as it is constantly "producing" electrons/ electrons are going to the cathode.

So in order to recharge the cell, you need to reverse these reactions to get more of the initial reactants. So the anode will now have electrons flowing into it and the cathode will have electrons withdrawn.

So what this means in terms of which terminal of the power source is connected where is that at the anode, the negative terminal of the battery is connected and the positive terminal is connected to the cathode.

So during recharging, electrons flow out from the cathode to the positive terminal of the battery and electrons flow from the negative terminal of the battery to the anode.

[Greggler]

I asked my teacher today and she said that we should know the equations for the various commerical cells.....

omfg

[iffets12345]

I heard we just had to be familiar, but I guess knowing wouldnt hurt.
Except, I have a sheet with 20 different kinds of fuel cells. FML

[Mao]

I heard we just had to be familiar, but I guess knowing wouldnt hurt.
Except, I have a sheet with 20 different kinds of fuel cells. FML

Balance the equation as you would for anything in an acidic environment (key element, water, H+, electrons)

Next, if it is an alkaline cell, add enough OH- to both sides to neutralize any H+ in the equation.
If it is a solid oxide fuell cell, add enough O2- to both sides to neutralize any H+ in the equation.

Voila, all the fuel cells learnt.

[iffets12345]

Legendary Mao comes to the rescue! Thanks, so this implies that they usually give you the starting and end products of the reaction?

[Mao]

Legendary Mao comes to the rescue! Thanks, so this implies that they usually give you the starting and end products of the reaction?

Yes

[stonecold]

Legendary Mao comes to the rescue! Thanks, so this implies that they usually give you the starting and end products of the reaction?

You can say that again! Thanks.


State the number of faradays of charge needed to produce 1 mole of chlorine molecules from molten potassium chloride.

I'm confused.  Why is it not 1 F but 2 F?

The reduction of K+ only needs one electron, and doesn't the oxidation of Cl- give off two electrons?
Or is it because you are supposed to balance everything first, which would then make it 2 F?

[stonecold]

A charge of 0.400 faraday was passed through 1.00 L of 1.00 M copper(II) sulfate solution using carbon electrodes.
a   Write equations for the reactions at each electrode.
b   Calculate the concentration of the copper ions in solution after electrolysis.

With this question, technically, shouldn't you take into account the water loss?  So you will end up with slightly less than 1L, because you will lose some to the oxygen gas.

I used 996.4 mL in the calculation.  Solutions didn't bother and kept the 1L initial value...

[iffets12345]

Legendary Mao comes to the rescue! Thanks, so this implies that they usually give you the starting and end products of the reaction?

You can say that again! Thanks.


State the number of faradays of charge needed to produce 1 mole of chlorine molecules from molten potassium chloride.

I'm confused.  Why is it not 1 F but 2 F?

The reduction of K+ only needs one electron, and doesn't the oxidation of Cl- give off two electrons?
Or is it because you are supposed to balance everything first, which would then make it 2 F?

Yea, its because you need to balance. You need 2 electrons for every mole of chlorine gas hence you double the K(s) and thus its 2 faraday instead of one.

[Mao]

A charge of 0.400 faraday was passed through 1.00 L of 1.00 M copper(II) sulfate solution using carbon electrodes.
a   Write equations for the reactions at each electrode.
b   Calculate the concentration of the copper ions in solution after electrolysis.

With this question, technically, shouldn't you take into account the water loss?  So you will end up with slightly less than 1L, because you will lose some to the oxygen gas.

I used 996.4 mL in the calculation.  Solutions didn't bother and kept the 1L initial value...

Took me a couple of reads to understand what you were saying. You are exactly right, that's very smart thinking. Some water would be lost, and I'd imagine you have taken into account the stoichiometric ratio and determined 0.2 mol of water was lost.

What you also need to consider is significant digits. 1.00mol/0.9964L=1.003M would still give 1.00M as the answer.