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Topic: corrosion of tin by oxygen dissolved in water.  (Read 3326 times)

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Offline kli1

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corrosion of tin by oxygen dissolved in water.
« on: May 17, 2008, 08:22:08 AM »
Having a bit of trouble understanding how corrosion of tin by oxygen dissolved in
happens:

Sn | Sn2+ (1 mol dm−3) || O2 (10-6 atm), H+ (10−7 mol dm−3) | Pt

The question I have asks me to sketch what this cell looks like and the role of the PT electrode and also to write overall redox equaiton.

Would I just draw two beakers one containing Pd in 1M of solution of its ions and one containing water, both connected with a voltmeter and a saltbridge between them.
I am not quite sure what to do as I have only come across when there have been 2 metals present.
Is the pd electrode just to compare the equilibrium against that of the O2?

Any help appreciated!

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