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Electrolysis question

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HEY guys.. can someone please help me with this question, have spend almost 2 hours on it but cant figure it out.

it's about aqueous electrolysis of CuSO4

At anode: h2so3 + h20 ---> so4^2-    +   4H+   + 2e-          (-0.17)

               Or  H2O  ----> 0.5O2 +2H2 +2e-                (-1.23)

 

Why does the second reaction occur at anode even thou the eletctrode potential for the first one is more positive??

(question says using graphite electrodes, does that has to do something with it?)

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This reaction is specified under 19.1 in the new syllabus. The short explanation is that the large amount (high concentration) of water is so much higher than that of sulfurous acid that the reaction with the lower oxidation potential proceeds faster. This is similar to the phenomenon that some reactions of higher equilibrium constants Kc may proceed slower than those of lower Kc The potential acts as the driving force, but if there is a large quantity then a smaller force can drive more moles of reaction. 

Indeed with a copper electrode instead, the oxidation reaction is Cu --> Cu2+ + 2e(ie, the reverse reaction of the reduction), with amount of water no longer able to compensate for more favored copper oxidation. Even in this case, sulfurous acid is not involved because it has a higher rate order than either oxidation of water of copper, but you don't need to explain sulfurous acid on the exam. 

 

 

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