A concentration cell is one in which both the anode and cathode are the same but with different concentrations. Calculate the cell potential with [Zn2+] = 0.10 M for the cathode and the [Zn2+] = 0.010 M for the anode?

The cell potential = +0.03 V

Explanation:

The half reactions for the cells are given below:

Oxidation half reaction (anode):   Zn(s) > Zn²⁺(aq) + 2e⁻

Reduction half reaction (cathode):  Zn²⁺(aq) + 2e⁻ > Zn(s)

To calculate cell potential of the cell, we use the equation given by Nernst, which is given as: Ecell = E°cell - (0.0591/n) log Q

where  Q = Concentration of dilute solution/Concentration of concentrated solution = 0.01/0.1

n = number of moles of electrons transferred = 2

Since, the cell is a concentration cell, E°cell = 0

therefore, the Nernst equation becomes, Ecell = - (0.0591/n) log Q

substituting the values;

Ecell = - (0.0591/2) log (0.01/0.10)

Ecell = +0.03 V

Therefore, the cell potential = +0.03 V

i think so it’s will be b yes i believe its