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We know that dilute $\ce{ H2SO4}$ doesn't react with copper because copper has higher reduction potential ($0.33$) as compared to hydrogen ($0.00$).

However, it reacts with conc. $\ce{H2SO4}$ and gives $\ce{CuSO4}$ with $\ce{SO2}$ and $\ce{H2O}$.

Why does this reaction take place?

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  • $\begingroup$ Use $\ce{<insert-compound>}$ to format chemical compounds over <sub> (insert compound)</sub>, It's easier and looks better $\endgroup$
    – Safdar Faisal
    Commented Apr 14, 2021 at 9:48
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    $\begingroup$ You can imagine concentrated sulfuric acid as an oxidising agent which supplies [O]. The copper takes the [O] and become CuO while sulfuric acid become sulfur dioxide. Then CuO reacts with remaining sulfuric acid to form $\ce{Cu2SO4}$. This is a grossly simplified version of what probably happens, but it gives you a sense of why conc. $\ce{H2SO4}$ can react with Cu. $\endgroup$
    – S R Maiti
    Commented Apr 14, 2021 at 9:59
  • $\begingroup$ You can a frame the nersnst equation , and see that the term coreesponding to concentration of SO4 2- comes in denominator so increasing concentration might make E cell >0 . $\endgroup$
    – Rajat Golechha
    Commented Apr 14, 2021 at 12:08
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    $\begingroup$ @RajatGolechha, Nernst equation should never be applied on concentrated solutions. $\endgroup$
    – ACR
    Commented Apr 14, 2021 at 15:49
  • $\begingroup$ chemistry.stackexchange.com/questions/145740/… $\endgroup$
    – Mithoron
    Commented Apr 14, 2021 at 16:14

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