how many ampere of current should be passed when aq solution of Na2SO4

1.	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» 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

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»

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