Introduction to cell potentials

In this lesson, we will learn:

• The meaning of the term standard electrode potential.
• The conditions which apply to standard electrode potential values.
• To predict redox reactions based on the standard electrode (reduction) potentials of the species involved.

• We now know that redox involves a reduction and an oxidation working together. Redox reactions ALWAYS have one of each process happening.
  Because many substances are capable of either (you just need another species able to do the opposite process), chemists have established a data table showing the ability of a substance to get reduced.
  This is known as a standard electrode potential table, also known as a standard reduction potential table. You will find one in the back of any good chemistry textbook.
  Some key facts about it:
• A standard reduction potential table is a list containing species in order of their tendency to get reduced in a redox reaction. They are always written as a substance getting reduced, never oxidised.
  This means the higher up a substance is in this table, the greater its tendency to be reduced. This ordering also means the further down the table a species is, the greater its tendency to get oxidised.


• Substances are ordered by their standard electrode potential, symbol E⁰, measured in volts (V) at 298 K, 100 kPa pressure with the substances at a concentration of 1.00 M.


• The E⁰ values are a comparison to the reduction of hydrogen, which is defined as zero volts (V) for reference.
  Really, you can think of this as a position of equilibrium compared to a H⁺/H₂ equilibrium.
  A substance could be set up in a redox with H⁺ _(aq) ions. There are two equilibria: one between H⁺/H₂ and one between the substance’s reduced/oxidised form (for example Mg²⁺ / Mg). Whichever substance gets reduced will obtain the necessary electron(s) from what is being oxidised.
• The more positive the E⁰ value, the more the substance’s redox equilibrium is positioned to the right (i.e. you have Mg, the reduced form, and hydrogen is in the H⁺ oxidised form).
• The more negative the E⁰ value, the more the substance’s redox equilibrium is positioned to the left (i.e. you have Mg²⁺, the oxidised form, and hydrogen is in the H₂ reduced form).


• When predicting or running a redox reaction, you will see that the absolute values are not as important as the values relative to each other – the absolute value just compares to hydrogen, so you just need to compare the two values of what you are reacting.
  An example would be Na metal and dilute hydrochloric acid - in solution, this is H⁺ _(aq). Will there be a reaction between these two species?
  The two equilibria we need to see can be found in the table¹:
• 2H⁺ + 2e^-$\, \rightleftharpoons \,$ H₂ where E⁰ = 0.00 V
• Na⁺ + e^- $\, \rightleftharpoons \,$ Na where E⁰ = -2.71 V

The simplest way to look at these half equations is to check which value is more positive. That species is going to be reduced. That’s why the values are sometimes called reduction potentials!
You will also notice in the table that the hydrogen equation is further up – it has a greater tendency to be reduced than sodium metal.
Because hydrogen has a higher (more positive) electrode potential than sodium, the reaction between sodium metal and hydrogen ions is feasible.
You would expect so; alkali metals react vigorously with acid to produce salt and hydrogen gas. This is chemistry you learn quite early on.


• There is a simple calculation you can do to show the feasibility of a reaction using the reduction potential values. This is to calculate the cell potential:


E⁰_cell = E⁰_reduction + E⁰_oxidation
Where:
E⁰_reduction = the electrode potential of the substance being reduced.
E⁰_oxidation = the electrode potential of the substance being oxidised. This is found by taking the negative of the E⁰ value found in the table.
E⁰_cell = the cell potential measured in volts (V).

In feasible redox reactions, E⁰_cell must be greater than 0.

In the example with sodium metal and acid (H⁺) written above:
• E⁰_reduction would be 0.00V as it was H⁺ being reduced.
• E⁰_oxidation would be -1 * -2.71 V which is the negative of the E⁰ value for sodium, which is getting oxidised.

We would calculate E⁰_cell:

E⁰_cell = 0.00 V + -(-2.71 V) = +2.71 V
Because E⁰_cell is greater than zero, or is a positive value, we can conclude that the reaction between sodium metal and H⁺ ions is a feasible redox reaction.

We will see more examples of this in the next lesson.


¹ Source for data: ATKINS, P. W., & DE PAULA, J. (2006).?Atkins' Physical chemistry. Oxford, Oxford University Press.