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Electrochemistry 11: Electrolytic Cells 99

(b) 2H,O(aq) + 2e + H2(g) + 20H-(aq) E" = -0.08 V.
I -11 0 -11 I

1 to 0 (reduction)
In order to determine which half-reaction will occur, the E"
values need to be considered. Since the standard electrode
potential for sodium is -2.714 V, and that for the water is
-0.08 V, this means that water, having the relatively greater E"
value, will be more easily reduced at the cathode than Na+(,,),
and therefore the half-reaction (b) will dominate.
At the anode: + ve electrode ('CNAP')/oxidation takes place here
('CROA'/'OILRIG').
There are two possible half-reactions:
(a) Cl-(aq) -, $C12(g) + e E" = - 1.36 V.
-I 0
(b) 2H20(,,) + O-qg) + 4H+(aq) + 4e E" = - 1.23 V.
I -11 0 I
I t
- 2 to 0 (oxidation)
This time, since E" for the chloride is - 1.36 V and E" for the
water is - 1.23 V, the values are so close that it will be concentra-
tion dependent as to which half-reaction will actually occur, i.e.
if the solution is concentrated sodium chloride, the concentration
of chloride ion will be large, and chlorine gas will be evolved at
the anode, but if the solution is very dilute, then oxygen gas will
be evolved at the anode. In this example, let us assume that the
sodium chloride is very concentrated:
Cathode reaction: 2H20,aq) + 2e --+ H2(g) + 20H-(aq)
Anode reaction: 2Cl-(aq) --+ C12(g) + 2e

Cell reaction: 2H20(aq) + 2Cl-(aq) + Hqg) + Clq?) + 20H-(aq)
In the electrolysis of concentrated aqueous sodium chloride,
hydrogen gas, H2(g) is evolved at the cathode along with hydro-
xide anion, OH- (link to pH), and chlorine gas, C1,(g), is evolved
at the anode.
5. Draw the cell (Figure 7.5). In this cell, the sodium cation,
Na+(,,) is termed a spectator ion, as it does not participate in the
reduction half-reaction.