Generalkorn12 Question Thread.

[generalkorn12]

Pb (s) + PbO2 (s) + 4H+ (aq) + 2 SO4 (aq) -> 2PbSO4 (s) + 2H2O (l),

Sorry about that , copying and pasting seemed to have removed that arrow. -_-

No worries! I assumed that's where the arrow would be, just wanted to confirm.

Pb(s) + SO4/2-(aq) --> PbSO4(s) + 2e-

PbO2(s) + SO4/2-(aq) + 4H+(aq) + 2e- --> PbSO4(s) + 2H2O(l)

In the first Pb goes from 0 to +2 ergo oxidised and in the second Pb goes from+4 to +2 ergo reduced.

Ahh!! I get it now, thanks! I've just got one query about this, for the sulphate ions, when is it alright to place them in both half equations?

[charmanderp]

Pb (s) + PbO2 (s) + 4H+ (aq) + 2 SO4 (aq) -> 2PbSO4 (s) + 2H2O (l),

Sorry about that , copying and pasting seemed to have removed that arrow. -_-

No worries! I assumed that's where the arrow would be, just wanted to confirm.

Pb(s) + SO4/2-(aq) --> PbSO4(s) + 2e-

PbO2(s) + SO4/2-(aq) + 4H+(aq) + 2e- --> PbSO4(s) + 2H2O(l)

In the first Pb goes from 0 to +2 ergo oxidised and in the second Pb goes from+4 to +2 ergo reduced.

Ahh!! I get it now, thanks! I've just got one query about this, for the sulphate ions, when is it alright to place them in both half equations?

I don't understand your question, sorry!

[generalkorn12]

For the question, 'A solution calorimeter was calibrated with 105ml of water instead of 100ml. The calorimeter was emtpied, and then used to determine the Heat of reaction of Powdered Zinc in 100ml of HCL, it can be deduced from this information that....'

I'm having trouble understanding why the answer is, 'lower and heat of reaction should be lower'.

[charmanderp]

Ooook. So let's start from the beginning.

CF is equal to E/dT. Energy is calculated by SHC*m*dH. Since in this case we're using a mass of 105g rather than 100g we're going to have a greater CF than what we should.

Therefore when we try to calculate energy for heat of reaction (HoR=E/n) for the HCL reaction we're going to obtain a larger value for E than what we should, when we multiply an overestimated CF by dT. Hence when you divide your overestimated E by n you'll get a higher value for heat of reaction that you should.

[generalkorn12]

Thanks for that! Makes more sense now.

Another question, is there a reason we can electroplate objects  using Silver nitrate solution? I'm looking at the Electrochemical Series and assumed water would be preferentially reduced.

[stephanieteddy]

Hello!

Since galvanic cells require a negative gradient on the electrochemical series for a reaction to occur,

Does that mean electrolysis requires a positive gradient?

[Howzat]

^, Yeah, electrolysis does need a positive gradient.

(Just adding more stuff to make my post seem more complete ^_^)

Think about if the oxidant was below the reductant, then a spontaneous reaction would not occur.

Applying electricity to this will force a reaction to occur and hence, the oxidant will be forced to undergo reduction whilst the reductant will be forced to undergo oxidation!

, It definitely makes sense now that I think about it.

[generalkorn12]

For the following reaction:
CO(g) + 2H2(g) -> CH3OH(g), decreasing the pressure would have what effect on the mole/concentration of H2(g)?

I always thought that if we decreased pressure, system moves backwards to increase pressure, hence there will be more mol of H2 gas and a higher concentration. Yet the ans says that the mol would increase but the concentration decreases, I always based my poor logic ( on the formula c=n/v, and if mole is increased, concentration increases.

[charmanderp]

The number of moles increases but a change in pressure comes about due to a change in volume. Hence if you decrease the pressure you must have increased the volume, and even if you then increase the number of moles it's possible to still have a concentration lower than the original, and this will indeed be the case as concentration, when being re-established to equilibrium, never exceeds what it was originally. Always best to draw graphs for things like this if you don't understand.

[Bhootnike]

ideal gas equation - pV=nRT
if pressure decreases, then volume decreases.
if pressure or volume decrease, then n increases.

thats how i do it.
btw did the answer mention anything about which reaction would be favoured?
in response to you saying it moves backwards, i thought that if n(h2) increases, then system acts as to partially offset this initial increase in (h2) by subsequently decreasing (h2), which is achieved by favouring the forward reaction, since h2 is consumed.

[generalkorn12]

For the self-ionisation of water, why is it that, increasing temperature would increase concentration of Hydronium and therefore decrease pH, yet the solution remains neutral still?

[Hancock]

The solution remains neutral because the [H3O+] = [OH-]. Even though the ph =/= 7, it can still be neutral.

[generalkorn12]

For concentration-time graphs, does the addition of a catalyst, increase or decrease the amounts of the reactants?

[nisha]

A catalyst has no effect on the amounts of reactants, it merely increases the reation rate.

[generalkorn12]

Hi agian, I'm having trouble understanding the following questions:

a) Steady current is passed on three cells, containing solutions, Pb(NO3), AgNO3 and Al(NO3), the molar ratio of n(Pb):n(Ag):n(Al) deposited at the negative electrode is:

b) 0.5mol of Cu2+ and 1mol of Ag+ are added to an electrolytic cell, the quantity of electricity to deposit all of the copper and silver are?
(I've always thought, we'd determine the electron requirements for both, and add, as that would be the total required)

c). 400mL of 0.2M Sulfuric and 400ml of 0.2 KOH released 4.56kJ. The energy released for 400mL of 0.2M Sulfuric Acid is mixed with 800mL of 0.2M KOH is...?
(For this question, would it be 4.56kJ, as Sulfuric Acid is the limiting reagent?)