Writing of cell reaction

The cell reaction corresponding to a cell notation is written on the assumption that the right hand side electrode is the cathode (+) and the left hand side electrode is the anode (-).

As mentioned in section 5.4.2, the oxidation half reaction occurs at the anode and the reduction half reaction at the cathode. It therefore follows that in a galvanic cell the oxidation half reaction takes place on the left hand side electrode and the reduction half reaction on the right hand side electrode.

The following steps are followed to write the cell reaction.

i. Write the oxidation half reaction for the left hand side electrode (anode) and the reduction half reaction for the right hand side electrode (cathode).

ii. Add the two electrode half reactions to get the overall cell reaction. While adding, the electrons must be balanced — for this purpose it may be necessary to multiply one or both the half reactions by a suitable numerical factor (or factors). No electrons should appear in the overall reaction.

iii. It is important to note that the individual half reactions may be written with one or more electrons. For example, half reactions for $\mathrm{H_2}$ gas, whether written as $\mathrm{2H^+\,(aq) + 2e^- \longrightarrow H_2\,(g)}$ or $\mathrm{H^+\,(aq) + e^- \longrightarrow \tfrac{1}{2}\,H_2\,(g)}$, make no difference. In writing the overall cell reaction, the electrons must be balanced.

Consider the cell,

(the book draws an $e^-$ arrow above the notation — electrons flow from the Al anode through the external circuit to the Ni cathode). The oxidation at the anode is

The reduction half reaction at the cathode is

$\mathrm{Ni^{2+}\,(1M) + 2e^- \longrightarrow Ni\,(s)}$ …