Subject:
ChemistryAuthor:
hezekiahgoodCreated:
1 year agoAnswer: The ratio of \frac{[Sn^{2+}_{(diluted)}}{[Sn^{2+}_{(concentrated)}]} in the cell is 4.07\times 10^{-4}
Explanation:
The half reactions for the cell is:
Oxidation half reaction (anode): Sn(s)\rightarrow Sn^{2+}+2e^-
Reduction half reaction (cathode): Sn^{2+}+2e^-\rightarrow Sn(s)
In this case, the cathode and anode both are same. So, E^o_{cell} will be equal to zero.
To calculate cell potential of the cell, we use the equation given by Nernst, which is:
E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Sn^{2+}_{(diluted)}}{[Sn^{2+}_{(concentrated)}]}
where,
n = number of electrons in oxidation-reduction reaction = 2
E_{cell} = 0.10 V
\frac{[Sn^{2+}_{(diluted)}}{[Sn^{2+}_{(concentrated)}]} = ?
Putting values in above equation, we get:
0.10=0-\frac{0.0592}{2}\log \frac{[Sn^{2+}_{(diluted)}}{[Sn^{2+}_{(concentrated)}]}
\frac{[Sn^{2+}_{(diluted)}}{[Sn^{2+}_{(concentrated)}]}=4.016\times 10^{-4}
Hence, the ratio of \frac{[Sn^{2+}_{(diluted)}}{[Sn^{2+}_{(concentrated)}]} in the cell is 4.016\times 10^{-4}
Author:
mitch68tw
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9The 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
Author:
díazhxpg
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