Question

Is 12.0 mv at 25 °c. calculate the concentration of the zn2 (aq) ion at the cathode?

Answer 1

the concentration of the Zn²⁺ (aq) ion at the cathode is 0.255 M

Step-by-step explanation:

The voltage generated by the zinc electric cell that is discribed by the following relation;

`Zn(s)\mid 0.1MZn^(2+(aq))\parallel 0.2 Zn^(2+)(aq)\mid Zn(s)`

The Nernst equation is given as follows;

`E = E^0 - (RT)/(nF) ln((a^b_B)/(a^a_A) )`

`E_(anode) = E^0 - (0.0591)/(2) logZn^(2+)`

`E_(anode)-E_(cathode) =-(0.0591)/(2) log([Z^(2+)])/([x])`

`0.012 =-(0.0591)/(2) log(0.1)/([x])`

x = 0.255 M.

Therefore the concentration of the Zn²⁺ (aq) ion at the cathode = 0.255 M.