(a) In the electrolysis of aqueous NaNO3, how many
liters of O2(g) (at STP) are generated by a current of
50.2 A for a period of 60.4 min? The unbalanced chemical reaction
representing this electrolysis is shown below.
NaNO3(aq) + H2O(l) NO(g) + O2(g) + NaOH(aq)
liters of O2(g) is generated by this electrolysis.
(b) How many moles of NaOH(aq) are formed in the solution in this
process?
moles of NaOH(aq) are formed.
(a) volume of O2 = 31.7 Liters of O2(g) is generated by this electrolysis.
(b.) moles NaOH = 1.89 moles of NaOH(aq) are formed.
Explanation
The balanced reaction equation is : 4 NaNO3 + 2 H2O 4 NaOH + 3 O2 + 4 NO
Anode half reaction : NaNO3 Na+ + O2 + NO + e-
Cathode half reaction : 2 H2O + O2 + 4 e 4 OH-
(a) Current = 50.2 A
time = 60.4 min
time = 60.4 min * (60 s / 1 min)
time = 3624 s
charge = (current) * (time)
charge = (50.2 A) * (3624 s)
charge = 181924.8 C
moles of electron = (charge) / (Faraday's constant)
moles of electron = (181924.8 C) / (96485 C/mol)
moles of electron = 1.8855 mol
moles of O2 produced = (3/4) * (moles of electron)
moles of O2 produced = (0.75) * (1.8855 mol)
moles of O2 produced = 1.414 mol
volume of O2 produced at STP = (moles of O2 produced) * (molar volume at STP)
volume of O2 produced at STP = (1.414 mol) * (22.4 L/mol)
volume of O2 produced at STP = 31.68 L
(a) In the electrolysis of aqueous NaNO3, how many liters of O2(g) (at STP) are generated...
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