MCHEM30: L2 - Standard Electrode Potential & Corrosion (BIG PIC & SUMMARY): How is E⁰ used in Electrochemistry?

How is E⁰ used in Electrochemistry?

It is possible to experimentally measure the standard reduction potential in each half-cell by replacing the other side of the cell with a hydrogen reference half-cell since the hydrogen potential is exactly 0.00V. Once this is done, we will be able to measure the voltage and the direction that the current is flowing.

When the hydrogen half-cell has more reduction potential

The picture on the right shows a cell with aluminum and platinum electrodes, with hydrogen gas in the platinum half-cell that has a possibility of acting as an electrode. All three possible electrodes have to be considered.

First step is to write out the cell in cell notation,

Using the data booklet, the following half-reactions are considered,

In this case, the aluminum is by itself, so it must be one of the electrodes. In the other half-cell, both platinum and hydrogen needs to be considered. Since the aluminum already has a negative standard reduction potential (anode), it is going to act as the reducing agent. In the other half-cell, hydrogen gas will bond with the platinum in a half-reaction as the hydrogen ions act as the oxidizing agent (cathode) for the half-cell reaction. Click on the half-reaction picture on the right to take a closer look at what is going on.

Therefore, the cell reaction is the difference between the aluminum and the hydrogen,

E^o_r = 1.66 V

When the hydrogen half-cell has less reduction potential

The picture of the right shows a cell with platinum and copper electrodes, but since the hydrogen gas in the platinum half-cell also can act as an electrode, all three electrode potential have to be considered.

First step is to write out the cell in cell notation,

Using the data booklet, the following half-reactions are considered,

In this case, only copper is in one half-cell, so that has to act as one of the electrodes. In the other half-cell there is both platinum and hydrogen to consider. Since the copper already has a positive standard reduction potential (cathode), it is going to act as the oxidizing agent, the more negative reduction potential present in the reaction is the hydrogen (anode). Therefore the platinum acts as an inert electrode.

Therefore the potential of the cell is the difference between the copper and the hydrogen,

E^o_r = 0.34 V

Note: In both cases, platinum is acting as an inert electrode, which is why it is not found on the Standard Electrode Potential Table in your data booklet. If it is ever not inert, the question (or somewhere else on the diploma) will have to provide the platinum half-cell reduction potential.

Example - Hg & Mg

A galvanic cell is constructed with one compartment that contains a mercury electrode immersed in a 1 M aqueous solution of mercuric acetate $H g (C H_{3} C O_{2})_{2}$ and one compartment that contains a strip of magnesium immersed in a 1 M aqueous solution of $M g C l_{2}$ . When the compartments are connected, the following half-reactions occur:

cathode: Hg²⁺(aq) + 2e⁻ → Hg(l)
anode: Mg(s) → Mg²⁺(aq) + 2e-

What is the galvanic cell's potential now that the compartments are connected?

After using the data booklet, both reduction potentials are determined.

Hg²⁺(aq) + 2e⁻ → Hg(l) E^o_r = +0.85

Mg(s) → Mg²⁺(aq) + 2e- E^o_r = -2.37

Then the cell's potential is calculated.

E^o_cell = E^o_cathode - E^o_anode

= 0.85 - (-2.37)

= 3.22 V

Additional Question

There are practice questions on page 631 that provide additional sample questions to try

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