Question

Gold is produced electrochemically from an aqueous solution of Au(CN)2- containing an excess of CN-. Gold metal and oxygen gas are produced at the electrodes. What amount (moles) of O2 will be produced during the production of 3.00 mol gold?

Amount of O2 = mol

Answer 1

The balanced chemical reaction anode and cathode are

Anode(oxidation): 4[Au(CN)₂]^-(aq) + 4e^-  --------- > 4Au(s) + 8CN^-(aq)

Cathode(reduction): 4OH^-(aq) --------- > 2H2O + O₂(g) + 4e^-  

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Overall reaction: 4[Au(CN)₂]^-(aq) + 4OH^-(aq) --------- > 4Au(s) + 8CN^-(aq) + 2H2O + O₂(g)

From the above redox reaction it is clear that 1.00 mole of O2 is produced when 4.00 mol of Au(s) is formed.

Hence the amount of O2 produced during the production of 3.00 mol Au is

= 3.00 mol Au x (1 mol O2 / 4 mol Au) = 0.75 mol O2 (answer)