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292 ELECTROCHEMISTRY

There is a phase bound- To be a redox couple, the zinc ions will be in contact with the
zinc electrode, which we write as Zn 2+ (aq) |Zn (s) , the vertical line
ary between the Zn
and Zn 2+ because the emphasizing that there is a phase boundary between them. We can
write the other couple as Cu 2+ (aq) |Cu (s) , with similar reasoning. Note
Zn is solid but the Zn 2+
is dissolved within a that if the two electrodes are written at the extreme ends of the
liquid electrolyte. A cell schematic, then the redox ionic states must be located some-
similar boundary exists where between them. The schematic now looks like Zn (s) |Zn 2+ (aq) ...
in the copper half-cell. Cu 2+ (aq) |Cu (s) ⊕.
Finally, we note that the two half-cells must ‘communicate’ some-
how – they must be connected. It is common practice to assume that
a salt bridge has been incorporated, unless stated otherwise, so we
join the notations for the two half-cell with a double vertical line, as
Zn (s) |Zn 2+ (aq) ||Cu 2+ (aq) |Cu (s) ⊕.
An alternative way of
looking at the schematic
is to consider it as
‘the path taken by a SAQ 7.4 Write the cell schematic for a cell comprising
3+
2+
charged particle dur- the Fe ,Fe (positive) and Co ,Co (negative) couples.
ing a walk from one
electrode to the other’.
Worked Example 7.5 Write a cell schematic for a cell comprising
+
−
the couples Br 2 , Br and H ,H 2 . The bromine | bromide couple is
the more positive. Assume that all solutions are aqueous.
This is a more complicated cell, because we have to consider the involvement of more
phases than in the previous example.

Right-hand side: the bromine couple is the more positive couple, so we write it on the
−
right of the schematic. Neither Br 2 nor Br is metallic, so we need an inert electrode. By
convention, we employ platinum if no other choice is stipulated. The electrode at the far
right of the schematic is therefore Pt, as ... Pt (s) ⊕.
Br 2 and Br are both soluble in water – indeed, they are mutually
−
We write the oxidized soluble, forming a single-phase solution. Being in the same phase, we
−
−
form first if both redox cannot write a phase boundary (as either ‘Br 2 |Br ’oras‘Br |Br 2 ’),
−
states reside in the so we write it as ‘Br 2 , Br (aq) ’. Note how we write the oxidized form
same phase, sepa- first and separate the two redox states with a comma. The right-hand
−
rating them with a side of the schematic is therefore ‘Br 2 , Br (aq) |Pt (s) ⊕’.
comma.
Left-hand side: neither gaseous hydrogen nor aqueous protonic solu-
tions will conduct electrons, so again an inert electrode is required on
the extremity of the schematic. We again choose platinum. The left-hand side of the cell
is: Pt (s) .
Hydrogen gas is in immediate contact with the platinum inert electrode. (We bub-
ble it through an acidic solution.) Gas and solution are different phases, so we write the
+
hydrogen couple as H 2(g) |H (aq) , and the left-hand side of the schematic becomes ‘ Pt (s) |
+
H 2(g) |H (aq) ’.