Answer: The concentration of [tex]Zn^{2+}[/tex] ion at cathode is 0.704 M
Explanation:
The half reactions for the cell is:
Oxidation half reaction (anode): [tex]Zn(s)\rightarrow Zn^{2+}+2e^-[/tex]
Reduction half reaction (cathode): [tex]Zn^{2+}+2e^-\rightarrow Zn(s)[/tex]
In this case, the cathode and anode both are same. So, [tex]E^o_{cell}[/tex] will be equal to zero.
To calculate cell potential of the cell, we use the equation given by Nernst, which is:
[tex]E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}_{anode}]}{[Zn^{2+}_{cathode}]}[/tex]
where,
n = number of electrons in oxidation-reduction reaction = 2
[tex]E_{cell}[/tex] = 25.0 mV = 0.025 V (Conversion factor: 1 V = 1000 mV)
[tex][Zn^{2+}_{cathode}][/tex] = ? M
[tex][Zn^{2+}_{anode}][/tex] = 0.100 M
Putting values in above equation, we get:
[tex]0.025=0-\frac{0.0592}{2}\log \frac{0.100}{[Zn^{2+}_{anode}]}[/tex]
[tex][Zn^{2+}_{anode}]=0.704M[/tex]
Hence, the concentration of [tex]Zn^{2+}[/tex] ion at cathode is 0.704 M
