Answer:
the concentration of the Zn²⁺ (aq) ion at the cathode is 0.255 M
Explanation:
The voltage generated by the zinc electric cell that is discribed by the following relation;
[tex]Zn(s)\mid 0.1MZn^{2+(aq)}\parallel 0.2 Zn^{2+}(aq)\mid Zn(s)[/tex]
The Nernst equation is given as follows;
[tex]E = E^0 - \frac{RT}{nF} ln(\frac{a^b_B}{a^a_A} )[/tex]
[tex]E_{anode} = E^0 - \frac{0.0591}{2} logZn^{2+}[/tex]
[tex]E_{anode}-E_{cathode} =-\frac{0.0591}{2} log\frac{[Z^{2+}]}{[x]}[/tex]
[tex]0.012 =-\frac{0.0591}{2} log\frac{0.1}{[x]}[/tex]
x = 0.255 M.
Therefore the concentration of the Zn²⁺ (aq) ion at the cathode = 0.255 M.
