Answer:
The cell potential = +0.03 V
Step-by-step 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