The voltage generated by the zinc concentration cell described by, zn(s)|zn2 (aq, 0.100 m)||zn2 (aq, ? m)|zn(s),is 16.0 mv at 25 °c. calculate the concentration of the zn2 (aq) ion at the cathode.

The concentration cell is:
Zn(s) \ Zn²⁺(aq,0.100 M) // Zn²⁺(aq, x M) \ Zn(s)
voltage = 16 mV x (1V / 10³ mV) = 16 x 10⁻³ V
- In the cell notation, the concentration on the left is that of the anode and that on the right is that of the cathode.
- Oxidation takes place at the anode and reduction takes place at the cathode.
so [Zn²⁺]oxidation = 0.100 M
[Zn²⁺] reduction = x M
From Nernst equation:
Ecell = -0.0592 / n log [Zn²⁺] oxidation / [Zn²⁺]reduction
Number of electrons, n = 2. Substitute and solve for x:
16 x 10⁻³ V = - 0.0592 / 2 log (0.100 /x)
log 0.100 / x = - 0.54
0.100 / x = 0.288
x = 0.347
So the concentration of Zn²⁺ at the cathode = 0.406

 

Is there a diagram/picture you can provide? it seems this question goes along with a