A half-reaction is an incomplete transfer of electrons. In an oxidation half-reaction, a substance loses some free electrons. In a reduction half-reaction, a substance gains some free electrons. However, in neither case do any electrons completely transfer from one chemical to another.
Each half-reaction has a standard reduction potential. The word "potential" comes from the fact that this value measures the potential a half-reaction has to create electricity. The half-reaction's standard reduction potential is measured for the reduction form of the half-reaction, and is denoted by \(E^{0}_{red}\). The "red" stands for reduction.
Example 1: If you look at a standard reduction potential table, you will see that
$$Mg^{2+} + 2e^{-} \rightarrow Mg$$
has a standard reduction potential of \(-2.36 \; V\). However, the reverse reaction
$$Mg \rightarrow Mg^{2+} + 2e^{-}$$
is not a reduction reaction, so its voltage cannot be found this way.
If we know the standard reduction potential of a reduction half-reaction, then the oxidation half-reaction formed by reversing the reduction half-reaction is \(-E^{0}_{red}\).
This is important to consider for half-reactions, especially when you design entire galvanic cells with two half-reactions.
The standard reduction potential of a particular half-reaction will have a sign--negative or positive. That has a rather large significance. If the value is positive, then that is how much electricity the half-reaction will contribute to a full galvanic cell. However, if the value is negative, then its magnitude is how much voltage you need to supply to trigger the half-reaction on its own. However, this is typically not done on its own since there is no transfer of electrons. Either way the value plays a role in full galvanic cells, as the two standard reduction potentials for the half-reactions need to be combined to get an overall potential voltage for the battery.
Given what we know what the signs mean, we can compare the voltages of galvanic cells conceptually, based on the standard reduction potentials of half-reactions.
Example 2: Consider two galvanic cells. The anode \(A\) for the two galvanic cells is the same. Now, for the first galvanic cell, the cathode half-reaction is
$$Cr^{3+} + 3e^{-} \rightarrow Cr$$
For the second galvanic cell, the cathode half-reaction is
$$Cl_2 + 2e^{-} \rightarrow 2Cl^{-}$$
We can compare the galvanic cells based on the standard reduction potentials of the two cathode half-reactions. The anodes are irrelevant because they are the same and thus have the same standard reduction potentials. Therefore their influence on the entire cells' voltage outputs will be the same regardless of the cathode.
We can look up the standard reduction potentials of the reduction half-reactions in a table online or in a textbook. Here are the standard reduction potentials for the reduction half-reactions:
\(Cr^{3+} + 3e^{-} \rightarrow Cr\): \(E^{0}_{red} = -0.74 \; V\)
\(Cl_2 + 2e^{-} \rightarrow 2Cl^{-}\): \(E^{0}_{red} = 1.36 \; V\)
Since the chlorine half-reaction has a stronger standard reduction potential, and both anodes are the same (this is very important), the battery with the chlorine cathode is expected to have a higher potential overall.
There are some "traps" that fall when the anodes are different or if the half-reactions above are for the anodes themselves, but those will not be covered here. The important thing to realize is how important it is to interpret standard reduction potentials for the half-reactions, before combining half-reactions to design an entire galvanic cell.
Reference for data, courtesy of
"Chemical Principles: The Quest for Insight 2nd Edition"
Atkins, Peter and Jones, Loretta