stonecold's chem questions :)

[Greggler]

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.

Mao, I must ask. You say balance the equations. Which equations exactly are we meant to balance?
Will they say something like 'in a lead acid accumulator Pb is oxidised or whatever to .... - thus write a balanced equation to represent this'??

If this is the case what your saying is a lot clearer, and very helpful

[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...

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.

Yepp, but as you mentioned, it doesn't affect the answer, because n(Cu2+)=0.200

Water loss is also 0.200 mol, which is 3.6g and assuming density of 1g/mL, volume lost is 3.6 mL

Cu2+(Initial - reacted)=1.00-0.200=0.80 mol

(2 sig figs here is that right? textbook says it is 3.  you're supposed to keep the least number of d.p.s I believe)

conc=0.80/0.9964=0.803 M which is the same answer as if you just use 1 L, as 2 sig figs makes it just 0.80 M

Textbook gives it to 3 sig figs though which from what I can see looks wrong.

[Mao]

I am talking about balancing redox equations in general. In most cases involving non-metals, balancing these equations involve adding water and H+. This is the case in the lead acid accumulator, reduction of permanganate and dichromate, oxidation of alcohol to carboxylic acid, half equation for combustion reactions [etc]. You should be very familiar with this process. In most cases, you will know what the product is (either from prior knowledge or stated by the question, all reactions from the prior list are assumed knowledge).

For fuel cells, the electrolyte (ions present in the system) may not be H+, thus reactions containing H+ may not be suitable. In the case where OH- or O2- are the electrolytes, you can simply 'transform' the acidic redox equations into an alkaline/oxide equation.

[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.

Yepp, but as you mentioned, it doesn't affect the answer, because n(Cu2+)=0.200

Water loss is also 0.200 mol, which is 3.6g and assuming density of 1g/mL, volume lost is 3.6 mL

Cu2+(Initial - reacted)=1.00-0.200=0.80 mol

(2 sig figs here is that right? textbook says it is 3.  you're supposed to keep the least number of d.p.s I believe)

conc=0.80/0.9964=0.803 M which is the same answer as if you just use 1 L, as 2 sig figs makes it just 0.80 M

Textbook gives it to 3 sig figs though which from what I can see looks wrong.

You are exactly right. Well done.

[stonecold]

A nickel teapot, with a surface area of 0.0900 m2, is to be silver plated.

If the plating is to be 0.005 00 cm thick and the density of silver is 10.5 g cm–3, how long should the pot be put in a silver-plating cell with a current of 0.500 A?


Step 1   Calculate the volume of Ag required, using volume = area x thickness.
      V(Ag)   = (0.0900 x 10 000) m2 x 0.00500 cm
            = 4.5 cm3

Where did the random 10,000 value come from?

[fady_22]

A nickel teapot, with a surface area of 0.0900 m2, is to be silver plated.

If the plating is to be 0.005 00 cm thick and the density of silver is 10.5 g cm–3, how long should the pot be put in a silver-plating cell with a current of 0.500 A?


Step 1   Calculate the volume of Ag required, using volume = area x thickness.
      V(Ag)   = (0.0900 x 10 000) m2 x 0.00500 cm
            = 4.5 cm3

Where did the random 10,000 value come from?

Used to convert m2 to cm2, as the surface area is given in m2.

[stonecold]

Oh yeah thanks.  I totally forgot that you have to square it like an idiot.

[kenhung123]

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?

I'm wondering also, do you do stoichiometric calculations using the overall equations of half equations?

[Blakhitman]

if you use overall, there should be no electrons in the equation.

[kenhung123]

Well you can balance it without cancelling the e's

[stonecold]

yeah, i agree ken.  it is confusing.  sometimes they use the half equation, and other times they add the half equations and use the overall equation.  it lacks consistency, which is what i am all about.

[iffets12345]

But in this case you do need the overall equation cause your making chlorine from a reaction with potassium chloride. Hence, the balancing makes the faraday double since K only gives off one electron..

[stonecold]

yeah, i just need more practice.  is there any particular reason why sometimes the solutions will undergo redox reaction, and other times it is the anode and cathode themselves?  or is it all dependent on the ECS?

and also, how do you know if the metal in the solutions will dissolve somewhat and produce ions?

[Mao]

Well you can balance it without cancelling the e's

What? No way, free electrons do not exist unless you have ridiulously high temperatures (i.e. plasma phase, >10000 K)

Electrons must always be balanced.

[kenhung123]

I mean for calculations, put the electrons both sides so that you know the ratio