Question

39. + 0/0.2 points Previous Answers 2/4 Submissions Used (a) In the electrolysis of aqueous NaClO3, how many liters of O2(g) (at STP) are generated by a current of 77.8 A for a period of 61.7 min? The unbalanced chemical reaction representing this electrolysis is shown below. NaClO3(aq) + H2O(1) <+ Cl2(g) + O2(g) + NaOH(aq) 33.421 x liters of O2(g) is generated by this electrolysis. (b) How many moles of NaOH(aq) are formed in the solution in this process? 2.984 x moles of NaOH(aq) are formed.

Answer 1

According to Faraday's law , when 96,500 coulomb passed through a electrolytic cell 1 equivalent of the substance/ electrolyte ( here NaClO3) undergo electrolysis...

Nacio, ag + H₂O - g) + Ogg & Naon la Balaneed et 4 NaoH equation - Nacio, + 2H₂0 Nacioz + 1 H20 Imol produces 202 1 l t 500 & Joe » & Nach [deviding by a] molo. & I mol Noor during electrolysis Now current passed , Q (coulomb ) = Amp (1 2 x See (t) - 7718 Amp X (61.7x60) See given) 2880515.6 Coulomb

Therefore 96500 coulomb = 1 mol of NaClO3

And. 288015.6 coulomb= (288015.6/96500) mol

=2.984 mol

Now 1 mol NaClO3 produces 5/4 mol O2 gas

2.984mol NaClO3 produces= (2.984×5)/5 mol O2 gas

= 3.73 mol O2 gas

Again 1 mol O2 = 22.4 liter at stp

Hence, 3.73 mol O2 = (22.4×3.73) lit = 83.552 liters of O2 gas (ANSWER)

Now 1 mol NaClO3 produces = 1mol NaOH

3.73 mol NaClO3 produces= 3.73 mol NaOH ( ANSWER)