[Mikeshandon]

Hi, I have some homewoek due tommorow for class but I am struggling to answer the questions. The sheet says:

An alumium anode oxidises slowly during use, it sets up a cell between itself and the iron structure of the ship it is attached to.In this cell oxygen in water is reduced to hydroxide ions on the iron's surface. Electrode potential for this is 0.4 V

Half equation is:

O2 + 2H2O + 4e- ---> 4OH-

Overall Cell Equation including the aluimium anode?

Then it says calculate the emf of the cell?

Later on it says that the ph of seawater varies from 7 to 8, give a few reasons why the emf of the cell you calculated will not be archieved in seawater


Thank you for reading, I would really appreciate any help

[thetada]

I would start by deducing a half equation for aluminium, which isn't as hard as it might seem. Think about how the aluminium starts off, think about the ion it is likely to form and then consider the electron transfer necessary to achieve that.

For example, in its elemental state, chlorine has an oxidation state of 0, but chlorine atoms usually accept one electron to become Cl^- ions. So we can write the half equation as follows:

Cl₂ + 2e^-  2Cl^-

What do you think the half equation for aluminium would be?