Unit 4 Questions MEGATHREAD!

[tony3272]

I'm going to say yes. But that's because with different concentrations, you would have different cell potentials, which would develop an electric current.

[Mao]

I'm going to say yes. But that's because with different concentrations, you would have different cell potentials, which would develop an electric current.

This will be most likely. The system will try to go towards equilibrium where the two cells have the same potentials (i.e. the 5M solution will decrease in concentration, the 0.5M solution will increase in concentration).

[HarveyD]

yeah thats right, i was a bit confused about the concentration thing, so thanks for that

[newayc]

Came across an interesting question in the STAV 2010 paper, MC Q16. Basically, a solution containing 0.010M KCl and 0.010M FeCl2 is undergoing electrolysis using inert platinum electrodes. Question is what is the product at the anode? A) Potassium metal B) Iron(III) ions C) Oxygen gas D) Iron metal.

The suggested solution gives B) because Fe2+ ions are the strongest reductants in the solution and hence get oxidised.

But I'm not so sure about this because if the anode is positive during electrolysis, and wouldn't that repel Fe2+ ions away from it? so that Fe2+ ions can't react at the anode? and so that water is still the one being oxidised and producing oxygen? (hence C is the correct answer?) 

[generalkorn12]

How do E° values come into determing whether the reaction occurs at the Anode or Cathode or is being oxidised or reduced?

[b^3]

The reaction with the higher E^o value will move forward and be reduced (at the cathode) leaving the lower E^o value reaction to move backwards and be oxidised (at the anode).

[generalkorn12]

Thanks for that

[b^3]

Thanks for that

I probably should just add that is for galvanic cells, for eletrolytic cells the higher value reaction will move backwards and be oxidised (at the anode +ve) and the lower reaction will move forward and be reduced (at the cathode -ve)

[pi]

Question regarding Le Chatelier's Principle.

For the reaction 2SO2 + O2 <--> 2SO2 deltaH = -ve, and I increased the temperature. When drawing a temperature-time graph following this change, two question:
1) do I assume all species are at same temp? (I'm thinking 'yes')
2) would the graph for all species show the same 'asymptotic' nature as per C-t graphs after a change?

If someone could draw one up ,that would be excellent

Thanks

[b^3]

It will be a bit different depending on the inital concentrations but should look something like this.

1) I would say yes aswell
2) It should as long as equlibrium is restablished, the rate at which the rates change will decrease so it should level out.

Keep in mind the amount of change in conc (i.e. the vertical change) in relation to the mole ratios in the equation.

[b^3]

Question regarding Le Chatelier's Principle.

For the reaction 2SO2 + O2 <--> 2SO2 deltaH = -ve, and I increased the temperature. When drawing a temperature-time graph following this change, two question:
1) do I assume all species are at same temp? (I'm thinking 'yes')
2) would the graph for all species show the same 'asymptotic' nature as per C-t graphs after a change?

If someone could draw one up ,that would be excellent

Thanks

Wait is that the right equation, there is SO2 on both sides?

Moderator action: removed real name, sorry for the inconvenience

[thushan]

SO3 in RHS.

[b^3]

SO3 in RHS.

That would make sense, other than that the working (graph + answers) should be fine right?

[pi]

SO3 in RHS.

That would make sense, other than that the working (graph + answers) should be fine right?

My bad, yeh should have been SO3 on RHS. C-T graph is right.

1) I would say yes aswell
2) It should as long as equlibrium is restablished, the rate at which the rates change will decrease so it should level out.

By 'should', do you mean that it will 'approach' the original temp prior to change, but never get there; or do you mean it will get to the original temp again? Still a bit confused/pedantic, sorry

[Mao]

Question regarding Le Chatelier's Principle.

For the reaction 2SO2 + O2 <--> 2SO2 deltaH = -ve, and I increased the temperature. When drawing a temperature-time graph following this change, two question:
1) do I assume all species are at same temp? (I'm thinking 'yes')
2) would the graph for all species show the same 'asymptotic' nature as per C-t graphs after a change?

If someone could draw one up ,that would be excellent

Thanks

1) Yes. Temperature is a measurement of the whole system. Therefore, if the system is at isothermal conditions (i.e. temperature is held constant), the temperature of all species in the system are equal.

2) The asymptotic behaviour will be the same for all species (though some are increasing, some are decreasing, some increase more than the others due to mole ratio, etc). The time taken to reach equilibrium will be the same for all species. This is because every reacting species is 'coupled' with other reacting species via the mole ratio, therefore they must increase/decrease at fixed ratio, the asymptotic behaviour must be the same.

Moderator action: removed real name, sorry for the inconvenience