Using a table, the standard reduction potential of a half-reaction can be found. The standard reduction potentials are calculated relativistic to each other such that this reduction half-reaction has a standard reduction potential of zero volts:
$$2H^{+} + 2e^{-} \rightarrow H_2$$
The standard reduction potential is the amount of a voltage a reduction half-reaction is expected to contribute to a galvanic cell. The notation used for it is \(E^0_{red}\). However, a galvanic cell will have an oxidation half-reduction as well, so how do you calculate the voltage of the entire galvanic cell? There is a simple formula to find the total voltage, where \(E^{0}\) is the total expected voltage, and \(E^{0}_{ox}\) is the standard oxidation potential of the anode half-reaction (where oxidation occurs):
$$E^{0} = E^0_{red} + E^0_{ox}$$
This formula will give you the voltage of the cell given both standard potentials, but since it is an equation, it can be used to solve either of the other variables.
Determining the standard oxidation potential of a half-reaction requires the standard reduction potential table. Here is the general process for finding the standard oxidation potential of an oxidation half-reaction:
Reverse the half-reaction.
Look up the standard reduction potential for this reduction half-reaction.
Change the sign of the standard reduction potential to get the standard oxidation potential of the original half-reaction.
This works because the reduction and oxidation half-reactions are exact opposites and have the exact opposite requirements to occur. If one half-reaction produces electricity, then the other one requires the same amount of electricity to occur.
Let’s see this process in use with an example.
Example 1: What is the standard oxidation potential for the half-reaction
$$V \rightarrow V^{2+} + 2e^{-}$$
Solution: Reverse the half-reaction to make a reduction half-reaction:
$$V^{2+} + 2e^{-} \rightarrow V$$
Now look up the standard reduction potential in the table. It is \(-1.19 \; V\). Finally, changing the sign to get the standard oxidation potential of the original half-reaction to get \(1.19 \; V\).
Here is one more similar example.
Example 2: Find the standard oxidation potential of the reaction
$$Li \rightarrow Li^{+} + e^{-}$$
Solution: We need the standard reduction potential of the opposite reaction:
$$Li^{+} + e^{-} \rightarrow Li$$
Looking it up in a table gives a voltage of \(-3.05 \; V\). Therefore the voltage of the original reaction is the opposite value, \(3.05 \; V\).
Now that we have become comfortable with finding the standard reduction potentials for oxidation half-reactions, we can find the total voltage of a galvanic cell using the aforementioned formula
$$E^{0} = E^0_{red} – E^0_{ox}$$
Example 3: A galvanic cell has a cathode half-reaction of
$$Br_2 + 2e^{-} \rightarrow 2Br^{-}$$
and an anode half-reaction of
$$Mg \rightarrow Mg^{2+} + 2e^{-}$$
Find the overall potential voltage of the cell.
Solution: The standard reduction potential of the bromine cathode is, according to a table, \(1.09 \; V\), and the standard reduction potential of the magnesium anode is \(-2.36 \; V\). Therefore, the standard oxidation potential of the anode is \(2.36 \; V\), and we can get the total voltage:
$$1.09 + 2.36 = 3.45 \; V$$
Let's do one more example.
Example 4: The overall reaction for a galvanic cell is
$$Fe^{3+} + 3K \rightarrow 3K^{+} + Fe$$
Find the total potential voltage for this galvanic cell.
Solution: We can tell by the full reaction that the half-reactions are
$$Fe^{3+} + 3e^{-} \rightarrow Fe$$ $$K \rightarrow K^{+} + e^{-}$$
In particular, the cathode reaction is the one involving iron, and the anode reaction is the one involving potassium. The standard reduction potential of the cathode reaction is \(-0.04 \; V\), and the standard reduction potential of the anode is \(-2.93 \; V\), so we can find the total potential:
$$-0.04 - (-2.93) = -0.04 + 2.93 = 2.89 \; V$$
Being able to use a standard reduction table is very important for calculating the potential voltage of an entire cell. The concept of oxidation and reduction half-reactions being opposites of each other is also important for understanding the formulas presented in this article, as well as basic terminology of electrochemistry such as cathode and anode.