The voltage generated by the zinc concentration cell described by the line notation Zn ( s ) ∣ ∣ Zn 2 + ( aq , 0.100 M ) ∥ ∥ Zn 2 + ( aq , ? M ) ∣ ∣ Zn ( s ) is 25.0 mV at 25 °C . Calculate the concentration of the Zn 2 + ( aq ) ion at the cathode.

The concentration of ion at cathode is 0.704 M

Explanation:

The half reactions for the cell is:

Oxidation half reaction (anode):  

Reduction half reaction (cathode):  

In this case, the cathode and anode both are same. So, will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

where,

n = number of electrons in oxidation-reduction reaction = 2

= 25.0 mV = 0.025 V   (Conversion factor: 1 V = 1000 mV)

= ? M

= 0.100 M

Putting values in above equation, we get:

Hence, the concentration of ion at cathode is 0.704 M

wher is the rest of the question