The voltage generated by the zinc concentration cell described by,

Zn(s)|Zn2 (aq, 0.100 M)||Zn2 (aq, ? M)|Zn(s)

is 23.0 mV at 25°C. Calculate the concentration of the Zn2 (aq) ion at the cathode.

Can someone please breakdown how to do this so I can figure out the rest of these types of Questions on my own?
Thank you.

Ecell = (-0.592/2)log(dil/conc)

0.023 = (-0.0592/2)log(0.1/x)(
Solve for x = approximately 0.6 M

To calculate the concentration of Zn2+ ion at the cathode in the given zinc concentration cell, you can use the Nernst equation. The Nernst equation relates the cell potential (Ecell) to the concentrations of the species involved in the redox reaction.

The Nernst equation is given as follows:

Ecell = E°cell - (0.0592/n) * log(Q),

where E°cell is the standard cell potential, n is the number of moles of electrons transferred in the balanced equation, and Q is the reaction quotient.

In this case, the cell diagram is as follows:

Zn(s)|Zn2+ (aq, 0.100 M)||Zn2+ (aq, ? M)|Zn(s)

At the anode (Zn(s)|Zn2+ (aq, 0.100 M)), zinc is being oxidized to form Zn2+ ions. Therefore, the half-reaction at the anode is:

Zn(s) → Zn2+ (aq) + 2e-

At the cathode (Zn2+ (aq, ? M)|Zn(s)), Zn2+ ions are being reduced to form zinc. Therefore, the half-reaction at the cathode is:

Zn2+ (aq) + 2e- → Zn(s)

Since both half-reactions involve the same species (Zn2+ and Zn), the standard cell potential (E°cell) is 0 V. Therefore, the Nernst equation simplifies to:

Ecell = - (0.0592/n) * log(Q)

Given that the cell potential (Ecell) is 23.0 mV (0.023 V), we can substitute the values into the simplified Nernst equation:

0.023 V = - (0.0592/2) * log(Q)

Simplifying further:

log(Q) = -(0.023 V) * 2 / 0.0592

log(Q) = -0.7777

Now, we can solve for Q by taking the antilog (inverse logarithm):

Q = 10^(-0.7777)

Q ≈ 0.178

Finally, since Q is the ratio of product concentrations to reactant concentrations, we can set up the following equation:

Q = [Zn2+]cathode / [Zn2+]anode

Given that [Zn2+]anode is 0.100 M, we can solve for [Zn2+]cathode:

0.178 = [Zn2+]cathode / 0.100

[Zn2+]cathode ≈ 0.0178 M

Therefore, the concentration of Zn2+ ion at the cathode is approximately 0.0178 M.