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:
Should actually be Pb(NO3)2 and Al(NO3)3 but that's not strikingly important. Anyway, if you look at the electrochemical series you'll see that for each mole of electrons you get 0.5 mole of lead, 1 mole of silver and 1/3 mole of aluminium, hence the molar ratio of n(Pb):n(Ag):n(Al) deposited is 2:3:1.
BUT WAIT!!!!!! A wild standard reduction potential appeared! Al3+ is a weaker oxidant than water and hence H2O will always be reduced preferentially and you will not be able to obtain any aluminium. Hence the molar ratio is 1:2:0 (n(Pb):n(Ag):n(Al)).
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)
Since Ag is a stronger oxidant that will be reduced first. Calculate Q for Ag (96500C) and then calculate Q for Cu (also 96500C) and then add them together.
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?)
The first step is to write a balanced equation, like so:
H2SO4(aq) + 2KOH(aq) --> K2SO4(aq) + 2H2O(l)
n(H2SO4)=CV=0.08mol
n(KOH)=CV=0.08
KOH is actually the limiting reagent here. So divide 4.56 by 0.08/2 which is 0.04 which will give you the heat of reaction, 114kJ.
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