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In a concentration cell, one can measure a voltage because in one of the half cells (the one with lower concentration electrolyte), the atoms tend to dissipate more into the electrolyte as ions, leaving more electrons behind (compared to the other half cell), thus creating a current.

I am not sure if I understand, why a lower concentration causes the atoms to dissipate more. Is it because the atoms of the metal have more contact to water?

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  • $\begingroup$ Think about it as the net effect of 2 opposite processes. $\endgroup$
    – Poutnik
    Jan 18, 2021 at 23:17

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It is not primarily that the process $\ce{M(s) -> M^n+(aq) + n e-}$ is faster for more diluted solutions.

The main reason is that the opposite process $\ce{M^n+(aq) + n e- -> M(s)}$ is slower for more diluted solutions.

As the net effect, there is dissolution of $\ce{M^n+(aq)}$ ions and the electrode gaining more negative potential, until the rate of both processes gets equal and the electrode reaches its equilibrium potential.

Because the rate of ion dissolution grows with increased potential, while the rate of ion depositions grows with decreased potential.

The final effect is, the electrode in then more diluted ion solution has the more negative potential.

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