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I don't see why negative ions don't affect the voltage. I mean the electric field of a negative ion like $\ce{SO4^{-2}}$ is not like what $\ce{NO3-}$ produces that means that it should affect the overall voltage.

This is the same reason why for example for $\ce{Cu+^2}$ the reduction electrode potential is higher than for $\ce{Zn+^2}$

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  • $\begingroup$ What reaction? How is the cell constructed? Need some more information here in order to answer. $\endgroup$ – Burak Ulgut Jan 10 '17 at 7:55
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Well, assuming that you're talking about an aqueous cell, in order to get the anion into solution, you need a salt: eg $\ce{CuSO4}$ or $\ce{Cu(NO3)2}$. Yes, the local electric field around a single $\ce{SO4^2-}$ anion would be stronger than around a single $\ce{NO3^-}$. However, at the same concentration of $\ce{Cu^2+}$ you would have twice as many $\ce{NO3^-}$ ions so the overall solution would still be neutral.

In a microscopic level, each ion is surrounded by a solvation shell of water molecules that all have their partial positive charge on their hydrogens pointing at the anion (or oxygen towards the cation). For this reason they are also shielded from the electrode. Speculation: double charged anions would probably have a thicker solvation shell.

I don't see how this explains the difference in potential of $\ce{Cu^2+/Cu}$ vs $\ce{Zn^2+/Zn}$ since both have the same charge.

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  • $\begingroup$ Maybe on nano and micro lengths the charge distribution of electrons on both of them play to produce different electric field $\endgroup$ – user3733086 Jan 16 '17 at 15:54

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