A sucrose solution is prepared to a final concentration of 0.250 M . Convert this value into terms of g/L, molality, and mass %.
1 answer:
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Answer: a) Cathode(-):
Anode(+):
b) 0.640 grams of Ba will be deposited.
Explanation: a) The problem is based on Faraday law of electrolysis. Molten barium chloride has ion and
ion. Barium ion is reduced and chloride is oxidized. Reduction takes place at cathode and oxidation at anode. So, the half reactions will be:
Cathode(-):
Anode(+):
b) The question asks, how many grams of barium metal can be produced by supplying 0.50 ampere for 30 minutes.
From the Cathode half-reaction, 1 mole of Ba is deposited by 2 moles of electrons and we know that 1 mole of electron carries one Faraday that is 96485 Coulomb.
Coulombs for 2 moles of electrons will be = 2*96485 C = 192970 C
So, we can say that, 192970 C will deposit 1 mole of Ba metal.
Total available coulombs can be calculated using the formula:
q=i*t
where, q is electric charge in coulomb, i is current in ampere and t is time in seconds.
q = 900 C (note: 1 C = 1 A*sec)
Let's calculate how many moles of Ba will get deposited by 900 C.
= 0.00466 mole Ba
Convert the moles of Ba to grams and for this we multiply by molar mass of Ba which is 137.33 gram per mol.
= 0.640 g Ba
So, 0.50 A for 30 minutes will deposit 0.640 grams of Ba metal.