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Electrochemistry I: Galvanic Cells 79
cathode and which electrode is acting as the anode in the cell,
write down the one-line representation of the cell, and follow
convention by keeping the cathode as the right-hand electrode:

Anode Cathode

Single vertical lines, (I) should be used to represent phase
boundaries or junctions and a double line (11) should be used in
the centre to separate the two half-cells, which can be a salt
bridge (e.g. KN03).
Write down the two electrode half-reactions, remembering:
(a) ‘CROA’: cathode-reduction, anode-oxidation
(b) ‘OILRIG’: oxidation is loss of electrons, reduction is gain of
electrons
(c) Reduction is a decrease in the oxidation number, oxidation
is an increase in the oxidation number (recall the two
vowels!)
It may now be necessary to balance these two half-reactions,
such that the number of electrons lost, or the number of
electrons gained, is identical for both. This will then yield v, the
number of electrons in the Nernst equation:
E = Eocell - (RT/vF) In K.
The two balanced electrode half-reactions should now be
added, to give the net cell reaction. The electrons should cancel
each other on both sides of the net cell reaction.
Draw the cell, and indicate clearly:
(a) The cathode (RHE) and the anode (LHE).
(b) The salt bridge, e.g. KN03 (if present).
(c) The direction of electron flow in the wire or solution
(remember electrons move from the anode to the cathode,
just remember the vowels again!).
(d) The spontaneous direction of current, I(simp1y the opposite
direction to the movement of electrons in the wire).
(e) The ion flow: -ve ions from the salt bridge migrate to the
anode and +ve ions from the salt bridge migrate to the
cathode.
Figure 6.8 shows a typical galvanic cell diagram.
Determine Eocell from the equation Eoce.l = EoRHE - EoLHE,