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Chemical cell

Metal P is the cathode, and it should attract anions. In this cell, anions present are Cl- and OH-, but they are reducing agents, reduction should occur at cathodes. So will H+ ions undergo reduction at the Cathode, even though it is a cation?

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2 Answers 2

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It it was a galvanic cell and if the copper electrode served as anode, the metal P would have to be a metal with higher standard reduction potential like silver and the electrolyte would have to be some soluble salt of that metal, like $\ce{AgNO3}$.

Then, there would be

dissolution of the copper anode:

$\ce{Cu^(s) -> Cu^2+(aq) + 2 e-}$

metallic silver deposition on the cathode:

$\ce{Ag+(aq) + e - -> Ag(s)}$


If is was an electrolytic cell and if the copper electrode served as anode, then

there would be dissolution of the copper anode again:

$\ce{Cu^(s) -> Cu^2+(aq) + 2 e-}$

but evolution of the gaseous hydrogen on the cathode:

$\ce{2 H+(aq) + 2 e- -> H2(g)}$


Note that any static potential difference of electrodes causes very tiny migration of ions in the opposite directions until the potential gradient in the solution is cancelled by the equal counter-gradient due charge dislocations.

Note that both anions and cations are present at both electrodes. the potential difference makes just very tiny difference in that.

If the is ongoing net oxidation/reduction on electrodes, it disrupts the above equilibrium, and there is ongoing ion migration, caused by electrostatic gradient and and concentration gradient.

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Poutnik's answer is perfectly correct. But the three following informations should be added.

$1)$ It must be mentioned that, in order to get a galvanic cell with an $\ce{Cu/Cu^{2+}}$ anode and a $\ce{Ag/Ag+}$ cathode, the two electrode compartments must be separated by a membrane or a bridge. If these compartments are not separated, the following reaction will immediately occur on the copper plate $$\ce{Cu + 2 Ag+ -> Cu^{2+} + 2 Ag}$$ and the cell will not work at all.

$2$) Also the anodic electrolyte must not contain $\ce{MgCl2}$, because the silver ions $\ce{Ag+}$ present around the anode will react with the $\ce{Cl-}$ ions to produce a precipitate of $\ce{AgCl}$ according to the following equation : $$\ce{Ag+ + Cl- -> AgCl}$$

$3$) Anyway the cell will not work a long time, because after a while, some $\ce{Cl-}$ from the cathodic compartment will cross the membrane and react with the silver ions. So the $\ce{Ag+}$ ions will disappear for two reasons : a) normal behavior of the anode, b) precipitation of $\ce{AgCl}$.

So the cell description must be entirely modified to be able to deliver some current.

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