Answer: The concentration of [tex]Zn^{2+}[/tex] ion at cathode is 0.295 M
Explanation:
The half reactions for the cell is:
Oxidation half reaction (anode): [tex]Zn(s)\rightarrow Zn^{2+}(0.100M,aq.)+2e^-[/tex]
Reduction half reaction (cathode): [tex]Zn^{2+}(?M,aq.)+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] = 14.0 mV = 0.014 V (Conversion factor: 1 V = 1000 mV)
[tex][Zn^{2+}]_{anode}[/tex] = 0.100 M
[tex][Zn^{2+}]_{cathode}[/tex] = ? M
Putting values in above equation, we get:
[tex]0.014=0-\frac{0.0592}{2}\log \frac{0.100M}{[Zn^{2+}]_{cathode}}[/tex]
[tex][Zn^{2+}]_{cathode}=0.295M[/tex]
Hence, the concentration of [tex]Zn^{2+}[/tex] ion at cathode is 0.295 M
The concentration of Zn²⁺ ion at cathode is 0.295 M
The redox half equations are those of oxidation and reduction
Oxidation half reaction: Zn(s) ---> Zn²⁺ + 2 e⁻ [Zn⁺] = 0.100 M
reduction half reaction: Zn²⁺ (aq) + 2 e⁻ ----> Zn (s) [Zn⁺] = y
The Ecell = 14.0mV = 0.014 V
Using the Nernst equation:
where n is number moles; n = 2
0.014 = 0 - 0.0592/2 * log (0.1/y)
y = 0.295 M
Therefore, the concentration of Zn²⁺ ion at cathode is 0.295 M
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