Redox Equilibria

Oxidation and reduction are electron transfer reactions

Oxidation is the loss of electrons and reduction is the gain of electrons

In redox reactions the oxidation state of the elements involved changes.

When looking at redox reactions it is sometimes useful to look at half equations involved.

Na > Na+ + e- the sodium has lost an electron and has therefore been oxidised

2H2O + 2e- > H2 + 2OH- the hydrogen has lost an electron and has therefore been reduced.

These can be combined as follows: 2Na + 2H2O > NaOH + H2

Half equations are also useful when examining the reactions that occur during electrolysis

An easy way to deal with electrode potential calculations

Electrode potentials are always quoted as REDUCTIONS.

 

eg Zn2+ + 2e- → Zn Eo = - 0.76 V

 

To show that this requires no knowledge of the actual chemical involved, we can use two imaginary chemical substances, J and Q.

 

We can ask the question, “Does J2+ oxidise Q2+ to Q5+ by changing from J2+ to J+ ?”.

 

Let’s imagine J2+ + e- → J+ has Eo value of 0.7 V and

Q5+ + 3e- → Q2+ has Eo value of -0.2 V

 

We are thinking about the possible chemical equation:

 

3J2+ + Q2+ → Q5+ + 3J+

 

1. Construct the cell diagram.

We must have a reduction on the right…...

/ // J2+ / J+

And an oxidation on the left…..

Q2+ / Q5+ // J2+ / J+

Notice that we don’t include the “3” in the cell diagram.

2. Cell e.m.f. = EoRHS - EoLHS Use the Eo values as they are quoted.

Cell e.m.f. = -0.2 V – 0.7 V

Cell e.m.f. = -0.9 V

Notice that we don’t include the “3” in the cell e.m.f. calculation.

Since this has a positive value, this reaction cannot go spontaneously.

If the cell e.m.f. value had come out to be positive, we would have been able to say that the reaction was feasible but that it might not go for some other reason (such as it being a very slow reaction)

Notice that Cell e.m.f. calculations are different to the ΔH and ΔG calculations.

For a reaction to be possible we are looking for a negative value for ΔG.

In ΔH and ΔG calculations we need to include the multipliers for how many moles of each substance are being used. This does not happen in cell e.m.f. calculations.

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