Question

A dilute aqueous solution of Na2SO4 is electrolyzed between Pt electrodes for 3.70 h with a current of 2.77 A .

Part A
What volume of gas, saturated with water vapor at 25∘C and at a total pressure of 742 mmHg , would be collected at the anode? The vapor pressure of water at 25∘C is 23.8 mmHg.

My answer is Incorrect. Anyone know what went wrong?

1 Coulomb = 1 amp / second, so 2.77 A = 37395 Coulombs

total = O2 pressure + Water vapor pressure
O2 Pressure = 742 - 23.8 = 718.2 mm Hg

Mole fraction of water vapor = 23.8/742 = 0.0321

V=nRT/P = (0.1)(0.082057)(298.15K)/(742/760) = 2.39 L

Thanks a lot;)

Answer 1

To start the coulombs is not right.
2.77 A x 1 hr x (60 min/hr) x (60 sec/min) = ? and I quit reading there.

Answer 2

Get number or moles of electrons yung Q=It=nF then get number of moles of SO2. Use PV=nRT to get volume.