# Hydrogen Oxygen Fuel Cell - Electrochemistry

In this video we want to discuss the hydrogen-oxygen fuel cell. We want to learn how to pull the relevant half equations from the Data Booklet without the need to memorise the reactions.

## Determine which species is oxidised and reduced

First we need to determine who is oxidised and reduced. This is simply done by comparing the change in oxidation states for hydrogen and oxygen.

The oxidation state for H increases from 0 in H_{2} to +1 in H_{2}O. Therefore H_{2} is oxidised.

The oxidation state for O decreases from 0 in O_{2} to -2 in H_{2}O. Hence O_{2} is reduced.

The half-equations in the Data Booklet are all written in reduction form.

This means that for the half-equations that we are looking for:

H_{2} is oxidised hence it is on the right-hand side of the half equation.

O_{2} is reduced hence it is on the left-hand side of the half equation.

## Acidic Electrolyte H_{2}SO_{4}

For oxidation of H_{2} we can find 2 half equations where H_{2} is on the right-hand side:

Since the medium is acidic, we need to choose the half equation which contains H^{+}, ie the first one.

Therefore the oxidation of H_{2} in acidic medium will be as follows:

For reduction of O_{2} we can find 4 half equations where O_{2} is on the left-hand side:

Again we should be choosing the half-equation which contains H^{+}, ie the first one.

Therefore the reduction of O_{2} in acidic medium will be:

The third equation is rejected since O_{2} will not spontaneously be reduced to H_{2}O_{2} where oxidation state of oxygen is -1 and unstable.

Finally we can combine both half-equations together and the processes at the anode and cathode in acidic electrolyte are as follows:

E_{cell} is calculated to be +1.23V which is a very easy number to remember. Hence we can use this as a way to verify the half equations we have chosen.

## Alkaline Electrolyte KOH

For oxidation of H_{2} we can find the same 2 half equations where H_{2} is on the right-hand side:

Since the medium is alkaline, we need to choose the second half equation which contains OH^{-}.

Therefore the oxidation of H_{2} in alkaline medium will be as follows:

For reduction of O_{2} we can find the same 4 half equations where O_{2} is on the left-hand side:

Again we should be choosing the half-equation which contains OH^{-}, ie the second one.

Therefore the reduction of O_{2} in alkaline medium will be:

The fourth equation is rejected since O_{2} will not spontaneously be reduced to HO_{2}^{-} where oxidation state of oxygen is -1 and unstable.

Finally we can combine both half-equations together and the processes at the anode and cathode in alkaline electrolyte are as follows:

E_{cell} is calculated to be +1.23V which has the same value in acidic medium.

This is not surprising since the overall redox reaction is the same.

It also reinforces the point that we can easily choose the relevant half equations in acidic and alkaline medium from the Data Booklet instead of blindly memorising them.

Topic: Electrochemistry, Physical Chemistry, A Level Chemistry, Singapore

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