Question

Gold is produced electrochemically from a basic solution of Au(CN)6-1.Gold metal and O2(g) are produced at the electrodes: Au(CN)6-1 + 3e à Au(s) + 6CN-1(aq) E0cell = 0.960 V O2(g) + 2 H2O + 4e à 4OH-1(aq) E0cell = 0.402 V

1) write the overall reaction (balanced net equation)

2) what volume of pure O2 is produced at 25 oC and 740 torr for every kilogram of Gold produced? {Use PV = nRT R = 0.0821 l.atm/mol.K }

Answer 1

Au(CN)6-1 + 3e ----------->Au(s) + 6CN-1(aq)       E0cell = 0.960 V

O2(g) + 2 H2O + 4e ------> 4OH-1(aq) E0cell = 0.402 V

4Au(CN)6-1 + 12e ----------->4Au(s) + 24CN-1(aq)      

3O2(g) + 6 H2O + 12e ------> 12OH-1(aq)

-----------------------------------------------------------------

4[Au(CN)6]^-1 + 12OH- ---------> 4Au(s) + 24CN^- + 3O2

no of moles of Au     = 1000/197 = 5.076 moles

4 moles of Au --------------> 3 moles of O2

5.076 moles of Au ---------> 3*5.076/4   = 3.807moles of O2

P   = 740torr   = 740/760   =0.974 atm

T    = 25C⁰    = 25 + 273 = 298K

PV = nRT

V    = nRT/P

     = 3.807*0.0821*298/0.974   = 95.63 L