Sam Davyson >> A2 Level >> Chemistry >> Topic 17

Page Contents: Redox Equilibria - Cell Diagrams - Standard Hydrogen Electrode - Anode, Cathode, e^- - Predicting Outcomes - Entropy Considerations - Odd Bits
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Redox Equlibria
The reactivity series classifies metals in the order from most to least reactive. Through considering a metal in the reaction:

It is clear the most reactive metals will be those that can lose electrons most readily. That is: those which oxidise most easily or the best reducing agents are the most reactive metals.

Displacement reactions now become competitions between the reducing power of two metals. Whichever is the best reducing agent will lose electrons (be oxidised) and form the relevant compound. The poorer reducing agents will suffer reduction from the electrons lost by the first metal.

Depending on the competition sometimes metals are reduced, sometimes they are oxidised. But importantly there is the capability for both of these processes. Forward and back. An equilibrium.

The position of equilibrium is obviously tied to the reactivity (or reducing power) of the metal, with the more reactive having the position further to the right. Again this is just the more reactive metals being more likely to lose their electrons and form ions. Importantly metals have varying tendencies for forming ions in solution.

Cell Diagrams
A half cell is simply a piece of metal dipped in a solution. Unfortunately you can't determine the EMF of a half cell as EMF is only ever a comparison of two potentials (potential difference). So to find the EMF we set things up with two half cells as if shown on the left. This is called a cell. The half cells are joined via a wire (through the voltmeter) and a salt bridge. The salt bridge is chromatography paper soaked in saturated KNO₃. This is chosen as potassium is a very strong reducing agent.

Cells can be represented as follows:

Note that the anode (that is the most reducing metal) goes first. And if there is no solid state of an element in the redox process, and a different electrode is used it is vital to ensure that the ion with the lowest oxidation state goes closest to the electrode (typically Pt). Adding arrows from left to right indicate how the reaction will proceed. If the value that results from subtracting the EMF of the right hand side from the left is negative this indicates that the reaction proceeds in the other direction (i.e. the anode has been wrongly chosen). The EMF is an indicator of the difference in reducing power between the electrodes.

Changing the concentration of an ion in solution changes the position of equilibrium and therefore the reducing power. Changing the temperature also causes changes to the equilibrium, and these depend on the sign of ΔH.

Standard Hydrogen Electrode
This is the standard which provides a level of reference for measuring the EMF of a half cell. It uses H_2(g) at 1 Atm (298K) and H⁺_(aq) 1 M at 298K. This standard is defined as being 0 V. All other electrode potentials are relative to this. So elements with a greater tendency to release electrons (that is a greater reducing power) than hydrogen have more negative electrode potential values. The hydrogen electrode is hardly the most practical thing ever made so often a secondary standard is used.

Anode, Cathode and e^-
In half cells the positive electrode is where the electrons are least likely to accumulate and at the negative electorde there is the greatest tendency for electrons to build up. In the full cell electrons flow to where there are few electrons which is obviously the positive electrode. This positive electrode therefore becomes the cathode and all the electrons that arrive are used in the reduction on the metal ions. At the other electrode electrons are being nicked off the most reducing ions (that is the ions are being oxidised) and fed round to the cathode. This (initially) negative electrode becomes the anode. (Another description of this is available here).

Predicting Outcomes
You can use electrode potentials to determine in which direction reactions will progress. For this there are four vital steps.

1. Write the two half equations as reduction. So with the electrons on the left.
2. Now list the equations by electrode potential with the least positive highest up the page.Add an anticlockwise circle between the equations.
3. The arrows on the circle indicate the direction of the reaction.

Ni2+(aq) + 2e- ⇔ Ni(aq)

I2(aq) + 2e- ⇔ 2I-(aq)

So will the following reaction proceed:

Ni²⁺_(aq) 2I^-_(aq) → Ni_(s) + I_2(aq)

Following steps one to three gives the cycle shown on the left which shows that Ni_(s) will be converted to nickel ions and I_2(aq) will be converted to iodide ions. This is not however what the original equation stated, so the reaction suggested will not proceed.

Now you can make a statement by using the acronym LORA which stands for Left Oxidises Right Above. In this case this means that the iodine oxidises the nickel metal. Looking at the half equations that the anticlockwise circle gives us this is clearly the case. The iodine gains electrons to become iodide ions (so is itself reduced) and this oxidises (takes electrons from) the nickel, forming nickel ions.

Entropy Considerations
When a metal atom is converted to a metal ion and an electron there is an increase in entropy and the metal ion is free* in the solution. The more concentrated the metal ion solution is the smaller the entropy increase is when a metal ion is formed as the equilibrium is already nearing the right, and there are fewer positions the metal ion can take. And as everytime a metal ion is formed the concentration of the metal ion solution rises, the increase in entropy for each successive metal ion forming gets smaller each time. The entropy changes will be greatest when there are many initial possible arrangements. That is the volume of solvent is large, and the concentration of ions is small.

	Wont Go	1:1	Completion
KC	10-10	1	1010
ΔStot	-200	0	+200
E	-0.6	0	+0.6

Only Δ S_tot can help us to judge the spontaneity of a reaction. However:

ΔS_tot α E_cell / T

This gives the values in the table which show the rough positions of equilibrium that you can expect with different values of K_C, ΔS_tot and E. Note that these numbers say absolutely nothing about the rate at which the equilibrium will be reached, only the position it will be in when it is reached.

Odd Bits
Potassium Haxacyanoferrate(III) tests for Fe²⁺ ions. It changes colour to prussian blue in a positive result. (Alternative text here).
Ions with the lowest oxidation state go closest to the platinum electrode.
Balancing equations for electrons is sometimes required first, remember to include water, and hydrogen ions to balance both sides after making the number of electrons in each half equation the same.

Notes
* Well 'free' is not totally accurate, because some metal ions form complexes with water. But they are certainly more free than when they were in the solid metal.

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