Experiment 31 Data and Calculations: Determination of an Equivalent Mass by Elec

Question

Experiment 31 Data and Calculations: Determination of an Equivalent Mass by Electrolysis Mass of metal anode Mass of anode after electrolysis Initial buret reading 22ml Buret reading after first electrolysis Buret reading after refilling Buret reading after second electrolysis Ambient (atmospheric) pressure, Pambient Room temperature, t Vapor pressure of H2O at t (consult Appendix ID, VPh,0 Total volume of H2 produced, V Absolute temperature, T Pressure exerted by dry H2: P Pambient VPi. mL mL 0 mL 360 mm Hg mm Hg mL (ignore any pressure effect due to liquid levels in buret) Moles of H, produced, n (use the Ideal Gas Law, PV = nRT) moles

Explanation / Answer

According to Ideal gas law : PV = nRT

Anode : Oxiation half reaction : loss of electrons:

M(s) ---------> M+n(aq) +ne-

Cathode : Reduction half reaction : gain of electrons:

2H+(aq) + 2e- ------> H2(g)

As buret readings are 0 after 2 electrolysis then the total Volume of hydrogen produced will be 100 ml

V = 100 ml = 0.1 lt

Vapor Pressure of water at 21 C, Pwater = 18.7 mmHg

Barometric pressure Pbar = 760 mm Hg

Pressure of hydrogen PH = Pbar - Pwater = 760 mm Hg -18.7 mmHg = 741.3 mmHg = 741.3 mmHg/760 = 0.9754 atm

Ideal gas constant, R = 0.082057 lt atm/mol K

Absolute temperature, T = 21+273 = 294 K

PV = nRT

0.9754 atm x 0.1 lt = n x 0.082057 lt atm/mol K X 294 K

n = 0.09754 mol /24.125 = 0.00404

number of moles hydrogen produced = 0.00404

In the cathode reaction shows that each mole of hydrogen require 2 electrons i.e 2 faradays

So 0.0040 moles require 2 x 0.00404 = 0.00808 faradays/electrons

Weight of metal lost = 11.787 g- 11.663 g = 0.124 gms

Per faraday weight of metal lost = 0.124/0.00808 = 15.35 g/mol e- = equivalent mass of metal

Equivalent mass of a metal = 15.3 g/mol e-

Probably the metal be Iron with +3 charge

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