) A concentration cell is based on the aqueous reaction
Cu2+(1.00 M)  Cu2+(0.0100 M)
The cell consists of copper electrodes dipping into solutions of Cu2+ ions. The anions present are sulfate ions. Draw a neat diagram to represent this cell, showing and labeling all necessary components including: anode, cathode, electron flow, cation flow and anion flow.

Kindly check the attachment for the diagram representing the cell and showing all necessary components including: anode, cathode, electron flow, cation flow and anion flow.

Explanation:

So, the reaction in the concentration cell is given below as;

Cu2+(1.00 M) → Cu2+(0.0100 M).--(1).

The anode = Cu2+(0.0100 M) because it has lesser Concentration, thus, lesser potential value.

Cathode = Cu2+(1.00 M) because it has higher Concentration, hence higher potential value.

It must be noted that in the digaram depicting the Reaction, the electrons moves from the anode part of the cell to the cathode part of the cell and this is done through an external circuit. The following are the things that happens at each electrode;

At the Anode: in here is where oxidation occurs and Cu^2+ is released into the solution.

At the cathode: in here is where the reduction occur and the Cu^2+ moves in the direction to where the Cu electrode is, thus, causing the deposition of Cu.

the answer is either b or c