Suppose I bought a newly manufactured Daniell cell then in this situation, can someone explain what all will be going on at the zinc and copper electrode? (Since it is newly manufactured and not used in external circuit the question of charges being transferred from one electrode to another does not arise) So would there be negative charge accumulation at zinc and positive charge at copper electrode? (Actually I am confused over situation at cathode a. I have read it to be positively charged but I could not understand how as the electrons $\ce{cu^{2+}}$ gain to get reduced actually comes from zinc and not from copper electrode)

Can someone please clarify my both doubts?


If it may help you to understand how the Daniell cell works, you may use a language full of imagery and say that the zinc metal absolutely wants to loose its electrons, so as to go back to its initial state on Earth, that is $\ce{Zn^2+}$ like in the mineral $\ce{ZnS}$ called blende. To be allowed to carry out this transformation, $\ce{Zn}$ needs somebody or something that would accept its electrons. Fortunately, copper is not like $\ce{Zn}$. It has not the same tendency. Copper does not want so eagerly to loose its electrons. In case of necessity like here, $\ce{Cu}$ even accepts foreign electrons to discharge its own ions $\ce{Cu^{2+}}$ and make some new neutral atoms $\ce{Cu}$.

That is how the Daniell cell works. On the left-hand side, a plate of metal $\ce{Zn}$ is dipped into water. It looses its electrons, so that the ion $\ce{Zn^{2+}}$ could pass into solution. The electrons are sent through the outer circuit to the next plate, which can be in copper or in platinum or in silver. This is not important, The only important thing is that the copper or platinum plate dips in a $\ce{Cu^{2+}}$ solution, for example a solution made by dissolving $\ce{CuSO4}$ in water. When the electrons have reached the copper or silver plate, they attract $\ce{Cu^{2+}}$ ions, discharge them, and some Copper metal is deposited on the plate.

One should notice that there are two different solutions, one in contact with the $\ce{Zn}$ plate, one in contact with the copper or silver plate. These solutions should not be mixed. If the two solutions are mixed, the Daniell cell will not produce any current, because the zinc atoms would find next to them ions that accept their electrons. No need to send them through a long outer circuit.

The only thing happening then is the fate of the negative ion $\ce{SO4^{2-}}$ remaining in the solution near the copper plate. It cannot stay here. It is attracted by the $\ce{Zn^{2+}}$ ion newly appeared near the $\ce{Zn}$ place. There must be a wet junction or a bridge to let the negative ion move to the other compartment.

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    $\begingroup$ That all I know. What I am trying to understand is how positive charge buildup occurs on cathode? $\endgroup$ – Sharad Dec 20 '19 at 17:06

It does not matter if the cell was previously already used in circuit or not.( unless the cell is exhausted )

For both electrodes, there is ongoing bidirectional exchange of the positive metallic cations, what changes the electrode potential.

At equilibrium, both processes go by the equal rate with the equilibrium potential, given by the Nernst equation.

If the electrode potential is more positive then the equilibrium one, the net effect is metal dissolution. If it is more negative, the net effect is metal deposition.

That may happen by applying the external voltage ( electrolysis or charging ) or by galvanic electrode connection ( power provisioning ).

The conventional electrode positive/negative potential is not absolute, but relative wrt the standard hydrogen electrode(SME). The absolute SME potential is $\pu{+4.44 \pm 0.02 V}$. So both electrodes have the same charge sign.

Note the terms cathode/anode:

A cathode is the electrode from which a conventional current leaves a polarized electrical device. An anode is an electrode through which the conventional current enters into a polarized electrical device. 

Be aware that conventional current has for historical reasons the opposite direction than the electron flow.

It means electrons leaving an anode by wire during the primarily intended process(electrolysis or power provision), respectively coming by a wire to the cathode.

A help to remember is the Greek word anabasis=way upwards(for electrons), catabasis=way downwards.

That also means, cathode is the more negative electrode at electrolysis, but the more positive in cell providing power, as for both electrons "go downwards from a wire". And vice versa for an anode.

Naming for secondary cells stays the same as for primary ones.

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