The voltage generated by the zinc concentration cell (described by,

Zn(s)|Zn2 (aq, 0.100 M)||Zn2 (aq, ? M)|Zn(s)) is 27.0 mV at 25 °C. Calculate the concentration of the Zn2 (aq) ion at the cathode.

0.027 volts = -(0.05916/2)log(x/0.1)

Solve for x.

10^.91277= x/.1

answer= .818

To calculate the concentration of the Zn2+ ion at the cathode, you need to use the Nernst equation. The Nernst equation relates the cell potential (Ecell) to the concentrations ([Zn2+]) of the species involved in the concentration cell.

The Nernst equation is given by:

Ecell = E°cell - (0.0592 V/n) log [Zn2+]cathode / [Zn2+]anode

Where:
Ecell is the cell potential (27.0 mV or 0.0270 V)
E°cell is the standard cell potential
n is the number of electrons involved in the reaction (in this case, since Zn is being oxidized to Zn2+, n is 2)
[Zn2+]cathode is the concentration of Zn2+ at the cathode (the side where reduction occurs)
[Zn2+]anode is the concentration of Zn2+ at the anode (the side where oxidation occurs)

In this case, both half-reactions involve the Zn2+ ion, so the concentration at the anode is given as 0.100 M.

Since this is a concentration cell, the standard cell potential E°cell is zero since both half-cells contain the same species. Therefore, we can simplify the Nernst equation to:

Ecell = - (0.0592 V/2) log [Zn2+]cathode / 0.100

Now, rearrange the equation to solve for [Zn2+]cathode:

log [Zn2+]cathode = - (2 x 0.0270 V) / (0.0592 V) = -0.921

Take the antilog (10^x) of both sides to get rid of the logarithm:

[Zn2+]cathode = 10^-0.921 = 0.122 M

Therefore, the concentration of Zn2+ at the cathode is 0.122 M.