Question based on Faradays laws orf electrolysis

[hana]

Hi everyone,
I managed to do part 1 of the following question but I'm having trouble with the second part.I really hope someone can help me.

QUESTION: A current is passed for 45 minutes through 3 solutions in series using Pt electrodes.In one ,0.203g of Ag is deposited from AgNO3 solution.In the second one,H2 gas is evolved from a solution of HNO3 acid and in the third one,Pb is deposited from a solution of Pb(NO3)2.
Calculate, 1.the current passed(ans:0.067A)
2.the volume of H2 gas collected at 0.983*10^5 Pa and 18 degree celsius.(ans:2.31*10^-5 m3)


for the 1st part, 96500C discharge 1mol of Ag which is 108g
therefore the quantity of electricity required to discharge 0.203g=(96500*0.203)/108=181.38C

since Q= i*t
i= 181.38/(45*60)= 0.0672 A


2. Applying Faradays' 2nd law(I think?),
method 1
96500C gives 1mol of H2,
therefore 181.38 C gives 1.88*10^-3 mols H2 gas
so using PV=nRT , i get V= 5.78 *10^-7 m^3 ,which is incorrect.

Then I tried equating the mass ratios of Ag and H2 to their chemical equivalents(E),
method2
mH2/mAg = E H2/ E Ag
mH2/0.203 = (2/1) /108
I get mass of H2=3.76*10^-3, Here again I Get the no. of mols of H2 as
1.88*10^-3 .

Please help. thanks

[hana]

Could someone please check if my chemical equivalent for H2 is right.
I'm not very sure if we're supposed to substitute the Relative Molecular Mass of H2(i.e 2)/valency of H atom(i.e. 1) or the relative atomic mass of H(i.e 1)/valency(1)
Much thanks

[Russ]

Are you converting temperature / pressure / etc. to the right units for PV = nRT?

I just solved it with your values and got that answer.

[hana]

This is how I worked it,
nH2=1.88810^-3 mols
T=273+18=291K
R=8.314Jpermol per kelvin
P=0.983*10^5
Are the above values correct?
After substituting,
PV=nRT
V=nRT/P
=(1.88*10^-3 *8.314 *291)/(0.983*10^5)
=4.627*10^-5 m3
Could you please check this for me?
Thank you

[Russ]

I think that n(H2) should be half the value it is, since it's got a Molar mass of 2

Otherwise, you converted properly :)

[hana]

Hmmm,
are you referring to the Electrochemical equivalent of H2?
I understand that we get the answer if we halve the no. of moles,but I checked the definition of E,it was given as a.m.u/valency.I know its a vey basic question but what is amu exactly?
I've learnt that it is 1/12 th the mass of an isotope of C-12.I've also learnt in school that it's the same as the relative molecular mass of the respective element?I'm a bit confused.infact very confused:confused:.

the 1st method I've used also gives the wrond answer?But why?Is my interpretation of Faradays law inaccurate?
Please help.I really need to understand this.
THANK YOU

[hana]

2. Applying Faradays' 2nd law(I think?),
method 1
96500C gives 1mol of H2,
therefore 181.38 C gives 1.88*10^-3 mols H2 gas
so using PV=nRT , i get V= 5.78 *10^-7 m^3 ,which is incorrect.

I just realised my mistake here,
96500C actually gives 1mol of electrons
and so 131.38c gives 1.88*10^-3 mols of electrons
and... since 1mol of H2 gives out 2 mols of electrons , there's only 9.4*10^-4 mols of H2.Substituting this to PV=nRt ,I get V=2.31*10^-5 m^3

method2
mH2/mAg = E H2/ E Ag
mH2/0.203 = (2/1) /108
I get mass of H2=3.76*10^-3, Here again I Get the no. of mols of H2 as
1.88*10^-3 .

As for this method I know that the chemical equivalent(E) of H2 has to be 1 inorder to get the right answer,but I'm not sure if E H2 should be 2/2 or 1/1.
I know the valency of H atom is 1 but what I don't know, is, if the valency of a molecule like H2 is 2 or 1?
Can someone help clarify? please