How many grams of Cu can be produced from 1.0 L of 0.50 M aqueous Cu(NO3)2 when a current of 0.50 A is run through the solution for 2 hours?
1.186g
Explanation
Cu2+(aq) + 2e -------> Cu(s)
1mole of Cu is deposited by 2 moles of electrons
1A = 1coulombs per second
0.5A = 0.5 coulombs per second
cuurent duration = 2hours = 7200s
Total coulombs passed = 0.5C/s × 7200s = 3600C
Faraday constant = 96485Coulombs per mole of electrons
No of moles of electrons = 3600C/96485C/mol = 0.03731mol
No of moles of Cu deposited = 0.03731mol/2 = 0.01866mol
Molar mass of Cu = 63.55g/mol
mass of Cu deposited = 0.01866mol × 63.55g/mol = 1.186g
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