Page 124 - [B._MURPHY,_C._MURPHY,_B._HATHAWAY]_A_working_meth

108 Chapter 7
9. Apply Faraday’s Second Law of Electrolysis:

2~20,aq) + 02(g) + 4~+(aq) + 4e
4 F 4 1 mol of oxygen; 1 mol of oxygen -+ 4 x 96500 C;
24.8 dm3+4 x 96 500 C; 1 dm3 4 (4 x 96 500)/24.8 = 15564.52 C;
2.645 dm3 + 41 168.16 C.
Now t = Q/I = 41 168.16 /0.475 = 86669.81 s = 24.07 h.
Answer: 24.07 h

Example No. 4: An aqueous solution of gold(II1) nitrate,
Au(NO~)~, was electrolysed with a current of 219 mA, until 150
cm3 of oxygen gas was liberated at the anode, at 1 bar pressure and
298 K. Draw a fully labelled diagram of the cell, indicating clearly
the cathode, the anode, the direction of current, the direction of
electrons and ion flow. Write down the cathode, anode and cell
reactions. Find (a) the duration of the experiment and (b) the mass
of gold deposited at the cathode during electrolysis. (F = 96 500 C
mol-’; 1 mole of an ideal gas at 25 “C and 1 bar pressure occupies
24.8 dm3; molar mass (M) of Au = 196.97 g mol-I).
~~
Solution:

1. Type of cell: electrolytic.
2. Type of electrolysis: electrolysis of aqueous gold(II1) nitrate,
ANN03)3.
3. Identify all species present: Au3 NO3-(,,), H20.
4. Cathode: - ve electrode (‘CNAP’): Au3 + (as), H20.
Au is below Zn in the electrochemical series; therefore gold metal
is deposited at the cathode:
Au3+(aq) + 3e + Au(~)
5. Anode: +ve electrode (‘CNAP’): NO3-(aq), H20. But N03-(aq)
is not easily oxidised according to its position in the electro-
chemical series:

I- > OH- > ci- > NO^- > SO^^-
II

Ease of Oxidation

Therefore water is oxidised at the anode, evolving oxygen gas,
according to the half-reaction: 2H,O(,d -+ 02(*) + 4H i- + 4e
6. Write down the two half-reactions: