Uses of Standard Electrode Potentials

Uses of the standard electrode potentials

So now that you understand this useful table of reduction potentials, it is important that you can use these values to calculate the potential energy differences. The following worked examples will help you do this. In all of these cases it is important that you understand what the question is asking.

Tip:

In this worked example you are given two half-reactions. Both are presented as shown in the table of standard reduction potentials, but in reality only one metal is being reduced, the other is being oxidised. The question being asked is: Which metal is being oxidised and which is being reduced?

Example:

Question

The following reactions take place in an electrochemical cell:

$\text{Cu}^{2+}(\text{aq}) + 2\text{e}^{-}$ $\rightleftharpoons$ $\text{Cu}(\text{s})$

$\text{Ag}^{+}(\text{aq}) + \text{e}^{-}$ $\rightleftharpoons$ $\text{Ag}(\text{s})$

Determine which reaction is the oxidation half-reaction and which is the reduction half-reaction in this cell.

Step 1: Determine the electrode potential for each metal

From the table of standard electrode potentials:

$\text{Cu}^{2+}(\text{aq}) + 2\text{e}^{-}$ $\rightleftharpoons$ $\text{Cu}(\text{s})$ (E° = $\text{+0.34}$ $\text{V}$)

$\text{Ag}^{+}(\text{aq}) + \text{e}^{-}$ $\rightleftharpoons$ $\text{Ag}(\text{s})$ (E° = $\text{+0.80}$ $\text{V}$)

Step 2: Use the electrode potential values to determine which metal is oxidised and which is reduced

Both values are positive, but silver has a larger, positive electrode potential than copper. Therefore silver is more easily reduced than copper, and copper is more easily oxidised than silver.

Step 3: Write the reduction and oxidation half-reactions

The E° values are taken from the table of standard reduction potentials. Therefore the reduction half-reaction is as seen in the table.

The reduction half-reaction: $\text{Ag}^{+}(\text{aq}) + \text{e}^{-}$ $\to$ $\text{Ag}(\text{s})$ (reduction is a gain of electrons)

Oxidation is a loss of electrons, and so would be written in reverse.

The oxidation half-reaction: $\text{Cu}(\text{s})$ $\to$ $\text{Cu}^{2+}(\text{aq}) + 2\text{e}^{-}$.

Example:

Question

Is magnesium able to displace silver from a solution of silver nitrate?

Step 1: Find appropriate reactions on the table of standard electrode potentials

The reaction involves magnesium and silver.

$\text{Mg}^{2+}(\text{aq}) + 2\text{e}^{-}$ $\rightleftharpoons$ $\text{Mg}(\text{s})$

$\text{Ag}^{+}(\text{aq}) + \text{e}^{-}$ $\rightleftharpoons$ $\text{Ag}(\text{s})$

Step 2: Determine the electrode potential for each metal

From the table of standard electrode potentials:

$\text{Mg}^{2+}(\text{aq}) + 2\text{e}^{-}$ $\rightleftharpoons$ $\text{Mg}(\text{s})$ (E° = $-\text{2.37}$ $\text{V}$)

$\text{Ag}^{+}(\text{aq}) + \text{e}^{-}$ $\rightleftharpoons$ $\text{Ag}(\text{s})$ (E° = $\text{+0.80}$ $\text{V}$)

Step 3: Which metal is more likely to be reduced, which is more likely to be oxidised?

Silver has a positive E°, while magnesium has a negative E°. Therefore silver is more easily reduced than magnesium, and magnesium is more easily oxidised than silver. The following reactions would occur:

Reduction half-reaction: $\text{Ag}^{+}(\text{aq}) + \text{e}^{-}$ $\to$ $\text{Ag}(\text{s})$

Oxidation half-reaction: $\text{Mg}(\text{s})$ $\to$ $\text{Mg}^{2+}(\text{aq}) + 2\text{e}^{-}$

It can be concluded that magnesium will displace silver from a silver nitrate solution so that there will be silver metal and magnesium ions in the solution.

Tip:

If magnesium is able to displace silver from a solution of silver nitrate, this means that magnesium metal will form magnesium ions and the silver ions will become silver metal. In other words, there will now be silver metal and a solution of magnesium nitrate. This will only happen if magnesium has a greater tendency than silver to form ions. In other words, what this worked example is asking is whether magnesium or silver will form ions more easily.

Tip:

Remember that solid metal will not always be formed when an ion is reduced. For example $\text{Sn}^{4+} + 2\text{e}^{-}$ $\to$ $\text{Sn}^{2+}$.

Example: Determining Overall Reactions

Question

For a zinc ($\text{Zn}$) and gold(III) oxide ($\text{Au}_{2}\text{O}_{3}$) cell in solution of $\text{KOH}$ determine the:

oxidation and reduction half-reactions
overall balanced chemical equation
standard cell notation for the cell

Step : Find appropriate reactions on the table of standard electrode potentials

The reaction involves zinc and gold.

$\text{Zn}^{2+}(\text{aq}) + 2\text{e}^{-}$ $\rightleftharpoons$ $\text{Zn}(\text{s})$

$\text{Au}^{3+}(\text{aq}) + 3\text{e}^{-}$ $\rightleftharpoons$ $\text{Au}(\text{s})$

Step : Determine the electrode potential for each metal

From the table of standard electrode potentials:

$\text{Zn}^{2+}(\text{aq}) + 2\text{e}^{-}$ $\rightleftharpoons$ $\text{Zn}(\text{s})$ (E° = $-\text{0.76}$ $\text{V}$)

$\text{Au}^{3+}(\text{aq}) + 3\text{e}^{-}$ $\rightleftharpoons$ $\text{Au}(\text{s})$ (E° = $\text{+1.50}$ $\text{V}$)

Step 1: Which metal is more likely to be reduced, which is more likely to be oxidised?

Zinc has a negative E°, while gold has a positive E°. Therefore gold is more easily reduced than zinc, and zinc is more easily oxidised than gold. The following reactions would occur:

Reduction half-reaction: $\text{Au}^{3+}(\text{aq}) + 3\text{e}^{-}$ $\to$ $\text{Au}(\text{s})$

Oxidation half-reaction: $\text{Zn}(\text{s})$ $\to$ $\text{Zn}^{2+}(\text{aq}) + 2\text{e}^{-}$

Step 2: Compare the number of electrons in each equation

There are $\text{3}$ electrons in the reduction half-reaction, and $\text{2}$ electrons in the oxidation half-reaction.

Oxidation: $\color{red}{\times 3}$: $\phantom{\rule{6.pt}{0ex}}$ $\color{red}{3}$$\text{Zn}(\text{s})$ $\to \color{red}{3}$$\text{Zn}^{2+}(\text{aq}) +$ $\color{red}{\textbf{6}}\textbf{e}^{-}$

Reduction: $\color{red}{\times 2}$: $\color{red}{2}$$\text{Au}^{3+}(\text{aq}) +$ $\color{red}{\textbf{6}}\textbf{e}^{-} \to \color{red}{2}$$\text{Au}(\text{s})$

Step 3: Combine the equations into one equation

$2\text{Au}^{3+}(\text{aq}) + 6\text{e}^{-} + 3\text{Zn}(\text{s})$ $\to$ $2\text{Au}(\text{s}) + 3\text{Zn}^{2+}(\text{aq}) + 6\text{e}^{-}$

Step 4: Add the spectator ions and remove the electrons from the equation

$\text{Au}_{2}\text{O}_{3}(\text{aq}) + 3\text{Zn}(\text{s})$ $\to$ $2\text{Au}(\text{s}) + 3\text{ZnO}(\text{aq})$

Step 5: Which material is the anode and which is the cathode?

Oxidation is loss of electrons at the anode, therefore $\text{Zn}(\text{s})$ is the anode.

Reduction is gain of electrons at the cathode, therefore $\text{Au}_{2}\text{O}_{3}$ is the cathode.

Step 6: Give the standard cell notation for this reaction

The anode is always written first (on the left): $\text{Zn}(\text{s})|\text{ZnO}(\text{aq})$

The cathode is always written second (on the right): $\text{Au}_{2}\text{O}_{3}(\text{s}),\text{Au}(\text{s})$

Therefore the standard cell notation is:

$\text{Zn}(\text{s})|\text{ZnO}(\text{aq})||\text{Au}_{2}\text{O}_{3}(\text{s}),\text{Au}(\text{s})$

Optional Experiment: Displacement experiment

Aim

To demonstrate the effect of reacting a halogen with a halide.

Apparatus

Bleach (approximately $\text{10}$ $\text{cm$^{3}$}$), bromine water, aqueous solutions of sodium chloride ($\text{NaCl}$), sodium bromide ($\text{NaBr}$) and sodium iodide ($\text{NaI}$), paraffin, concentrated $\text{HCl}$.
5 test tubes, 2 plastic droppers.

Method

Warning:

Concentrated $\text{HCl}$ can cause serious burns. We suggest using gloves and safety glasses whenever you work with an acid. Remember to add the acid to the water and to avoid sniffing the acid. Handle all chemicals with care.

Label the test tubes A, B, C, D and E.
Place $\text{1}$ $\text{cm$^{3}$}$ $\text{NaCl}$ solution into test tube A.
Place $\text{1}$ $\text{cm$^{3}$}$ $\text{NaBr}$ solution into both test tubes B and C.
Place $\text{1}$ $\text{cm$^{3}$}$ $\text{NaI}$ solution into both test tubes D and E.
Activate $\text{10}$ $\text{cm$^{3}$}$ of the bleach by adding $\text{2}$ $\text{cm$^{3}$}$ of the concentrated $\text{HCl}$.

Observe the liquid and note what happens on adding $\text{HCl}$, record your observations. You have formed a solution of chlorine in water.
Using a plastic dropper transfer approximately $\text{1}$ $\text{cm}$ height of the chlorine water into the test tubes labelled A, B and D.

Note any changes to the test tube. Record all observations.
Pour $\text{1}$ $\text{cm$^{3}$}$ of bromine water into the test tubes labelled C and E.

Note any changes to the test tube. Record all observations.
Using a plastic dropper transfer approximately $\text{2}$ $\text{cm}$ height of the paraffin into each test tube. Use a cork or rubber stopper to close the test tube, hold it firmly in place with your thumb, and shake the mixture.
Use the redox table to write overall net ionic equations for the reactions in test tubes B, D and E.
Using your understanding of solubility rules (like dissolves like) explain why the layer of paraffin became coloured in test tubes B, D and E. Explain what caused the paraffin to become coloured in test tube C.
Why was there no reaction and no colour change in test tube A?

Results

In test tube A $\text{Cl}_{2}$ is present, but is not coloured and so no change in the colour of the paraffin is observed.
In test tubes B and D the chlorine displaces the $\text{Br}^{-}$ and $\text{I}^{-}$ ions. $\text{Br}_{2}$ and $\text{I}_{2}$ are formed.

$\text{Br}_{2}$ is a brown colour in paraffin, while $\text{I}_{2}$ is a purple colour in paraffin.
In test tube C the $\text{Br}_{2}$ was already present and coloured the paraffin a brown colour.
In test tube E the $\text{Br}_{2}$ displaced the $\text{I}^{-}$ ions to form $\text{I}_{2}$ which would turn the paraffin a purple colour.

Conclusion

Halogen molecules are non-polar. They will therefore dissolve in a non-polar solvent such as paraffin. The paraffin layer will become the colour of the halogen present in the solution. Chlorine is the most likely of these three halogens to be reduced, followed by bromine, and then iodine. This can be seen on the standard electrode potential table as chlorine has the largest, positive electrode potential of the three halogens.

Uses of Standard Electrode Potentials

Uses of the standard electrode potentials

Tip:

Example:

Question

Step 1: Determine the electrode potential for each metal

Step 2: Use the electrode potential values to determine which metal is oxidised and which is reduced

Step 3: Write the reduction and oxidation half-reactions

Example:

Question

Step 1: Find appropriate reactions on the table of standard electrode potentials

Step 2: Determine the electrode potential for each metal

Step 3: Which metal is more likely to be reduced, which is more likely to be oxidised?

Tip:

Tip:

Example: Determining Overall Reactions

Question

Step : Find appropriate reactions on the table of standard electrode potentials

Step : Determine the electrode potential for each metal

Step 1: Which metal is more likely to be reduced, which is more likely to be oxidised?

Step 2: Compare the number of electrons in each equation

Step 3: Combine the equations into one equation

Step 4: Add the spectator ions and remove the electrons from the equation

Step 5: Which material is the anode and which is the cathode?

Step 6: Give the standard cell notation for this reaction

Optional Experiment: Displacement experiment

Aim

Apparatus

Method

Warning:

Results

Conclusion

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