Question

how many ampere of current should be passed when aq solution of Na2SO4 is electrolysis between graphite electrode at 300k temprature and 250ml/min of O2 gas obtain under 1 bar pressure​

Answer 1

Answer:

6.44 Amps

Explanation:

$2O^{2-} => O_{2} + 4e^{-}$

Convert the initial conditions to standard conditions (temperature and pressure)

P1V1/T1 = P2V2/T2

P1 = 1 bar = 750mmHg

V1 = 250ml = 250 cm^3

T1= 300K

P2 = 760mmHg

V2 = ?

T2 = 273 K

(750x250)/300 = (760xV2)/273

V2 = (750X250X273)/(300X760)

V2 = 224.51 cm^3

1 mole of gas occupies 22400cm^3 at STP

Moles of Oxygen = 224.51/22400 = 0.001 moles

One mole of oxygen is discharged by 4 moles of electrons

Moles of electrons = 4 x 0.001 = 0.004

Convert moles to Coulombs

1 moles of electrons = 96485 Coulombs

Number of Coulombs = 0.004 x 96484 = 386.8C

Coulombs (Q) = Current (I) x Time (t)

Time = 1 minute = 60 seconds

Current (Amps) = Q/t = 386.8/60

Current = 6.44 Amps