Determination of an equivalent mass by electrolysis...?

Question

In an electrolysis experiment, a student observed that his unknown metal anode lost 0.208g while a total volume of 96.30 mL of H2 was being produced. The temperature in teh laboratory was 25 degrees Celcius and the barometric pressure was 748 mm Hg. At 25 degrees Celcius the vapor pressure of water is 23.8 mm Hg. To find the mass of his metal, he filled in the blanks below. Fill in the blanks as he did.

PH2=Pbar-VPh2o =_______mm Hg=______atm
VH2=____mL=_______L
T=______K
nH2=_____moles nH2=PV/RT (where P=PH2)

1 mole H2 requires passage of ____________faradays

No. of faradays passed = ___________

Loss of mass of metal anode =_______g

No. grams of metal lost per faraday passed = no. grams lost/no. faradays passed=__________g=EM

The student was told that his metal anode was made of iron.

MM Fe=__________g. The charge n on the Fe ion is therefore__________.

Can anyone help? I'm confused....

Answer 1

First we need to establish the number of moles of hydrogen produced. To do this we use the equation PV = nRT, which gives n = PV / RT.

P(H2) = P_bar - VP_H2O = 748 - 23.8 = 724 mm Hg = 0.953 atm.
V(H2) = 96.30 mL = 0.09630 L (given).
T = 25°C = 298K

Since we're explicitly converting to L and atm, I assume we're using R = 0.082057 L atm / mol / K. So we get
n(H2) = PV / RT = (0.953) (0.09630) / ((0.082057) (298))
= 3.75×10^-3 mol.

Now we have to convert moles of hydrogen to moles of metal. The intermediary is electrons, so we work out how many faradays (moles of electrons) we used in the electrolysis.

1 mole of H2 requires 2 moles of electrons, i.e. 2 faradays (the relevant half-equation is 2H(+) + 2e(-) --> H2).
So the number of faradays used is 2(3.75×10^-3) = 7.50×10^-3.

Now we have to establish the ratio between the number of moles of metal and the number of moles of electrons. This will tell us the charge on the metal ion. If the number of moles of electrons is 3 times the number of moles of metal, the metal must have had a +3 charge, and so on.

Loss of metal mass = 0.208g (given).
Grams of metal lost per faraday = 0.208 / 7.50×10^-3 = 27.7 g.

MM Fe = 55.85, so for each mole of electrons we have lost 27.7/55.85 = 0.497 moles of Fe. Flipping this, each mole of Fe requires 2.01 moles of electrons, so we can conclude that the charge on the Fe ion is +2.

Answer 2

thanks so much very helpful