# Why the reduction potential of the reaction AgCl (aq) /Ag + Cl- is +0.22V

The point is $\ce{AgCl_{(aq)}}$ is the same as $\ce{Ag+}$ + $\ce{Cl-}$ (again, in water) So my reduction reaction would be $$\ce{Ag+ + Cl- + e- -> Ag + Cl-}$$ then the equation is simplified by removing $\ce{Cl-}$ from both sides (Spectator Ion).

Now there is $\ce{Ag+ + e- -> Ag}$ which has a potential of 0.80V, so why the original equation is 0.22V.

PS: I know it takes more energy to break the bonds and thus Ag is more stable with Cl and less likely to reduce. But the subscribed (aq) doesn't mean the bonds have already been broken by water?

## 1 Answer

I don't know what your source is, but it should definitely be an $\ce{AgCl(s)}$ in your first equation. Wikipedia gets it right and if you find any textbook that doesn't explicitly state the phase of $\ce{AgCl}$, you can be pretty darn sure they meant solid because talking about aqueous $\ce{AgCl}$ makes little to no sense precisely because it is so insoluble in water.

So, essentially you've already answered your own question; precipitation of $\ce{Ag(I)}$ as $\ce{AgCl}$ stabilises it towards reduction and lowers the standard reduction potential relative to $\ce{Ag+/Ag}$.