Answer :
Answer: The ratio of [tex]\frac{[Sn^{2+}_{(diluted)}}{[Sn^{2+}_{(concentrated)}]}[/tex] in the cell is [tex]4.07\times 10^{-4}[/tex]
Explanation:
The half reactions for the cell is:
Oxidation half reaction (anode): [tex]Sn(s)\rightarrow Sn^{2+}+2e^-[/tex]
Reduction half reaction (cathode): [tex]Sn^{2+}+2e^-\rightarrow Sn(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{[Sn^{2+}_{(diluted)}}{[Sn^{2+}_{(concentrated)}]}[/tex]
where,
n = number of electrons in oxidation-reduction reaction = 2
[tex]E_{cell}[/tex] = 0.10 V
[tex] \frac{[Sn^{2+}_{(diluted)}}{[Sn^{2+}_{(concentrated)}]}[/tex] = ?
Putting values in above equation, we get:
[tex]0.10=0-\frac{0.0592}{2}\log \frac{[Sn^{2+}_{(diluted)}}{[Sn^{2+}_{(concentrated)}]}[/tex]
[tex]\frac{[Sn^{2+}_{(diluted)}}{[Sn^{2+}_{(concentrated)}]}=4.016\times 10^{-4}[/tex]
Hence, the ratio of [tex]\frac{[Sn^{2+}_{(diluted)}}{[Sn^{2+}_{(concentrated)}]}[/tex] in the cell is [tex]4.016\times 10^{-4}[/tex]
The ratio of the Sn2+ concentrations in the two half-cells is 0.00042.
The cell reaction is; Sn(s) ------> Sn^2+(aq) + 2e
Using the Nernst equation;
Ecell = E°cell - 0.0592/n log Q
Ecell = 0.10 V
n = 2
Q = ?
E°cell = 0.00 V
0.10 V = 0.00 V - 0.0592/2 log Q
-(0.10 V × 2/0.0592) = log Q
-3.378 = log Q
Q = 0.00042
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