Calculate the concentration of $Cu^{2+}$ in a $Cu$ plate kept in a $0.2 \ M$ $CuSO_4$ solution when the potential becomes $0.0 \ V$.
When the potential becomes $0.0 \ V$,then $E_{cell} = 0.0 \ V$ and $E^{\circ}_{cell} = +0.34 \ V$.
The reaction is: $Cu^{2+}(x \ M) + 2e^{-} \rightarrow Cu_{(s)}$.
Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.059}{n} \log \frac{1}{[Cu^{2+}]}$.
Substituting the values: $0.0 = 0.34 - \frac{0.059}{2} \log \frac{1}{x}$.
$0.34 = 0.0295 \log (\frac{1}{x})$.
$\log (\frac{1}{x}) = \frac{0.34}{0.0295} = 11.5254$.
$\log x = -11.5254$.
$x = \text{Antilog } (-11.5254) = 2.99 \times 10^{-12} \ M$.
Thus,the concentration of $Cu^{2+}$ is $2.99 \times 10^{-12} \ M$.
The reaction is: $Cu^{2+}(x \ M) + 2e^{-} \rightarrow Cu_{(s)}$.
Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.059}{n} \log \frac{1}{[Cu^{2+}]}$.
Substituting the values: $0.0 = 0.34 - \frac{0.059}{2} \log \frac{1}{x}$.
$0.34 = 0.0295 \log (\frac{1}{x})$.
$\log (\frac{1}{x}) = \frac{0.34}{0.0295} = 11.5254$.
$\log x = -11.5254$.
$x = \text{Antilog } (-11.5254) = 2.99 \times 10^{-12} \ M$.
Thus,the concentration of $Cu^{2+}$ is $2.99 \times 10^{-12} \ M$.
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$A$ $Daniel$ cell is made at $25\,^oC$ by connecting $Zn/Zn^{2+} (0.1\ M, 1\ L)$ and $Cu/Cu^{2+} (0.9\ M, 1\ L)$ electrodes. The cell is discharged until its $emf$ reaches $1.10\ V$,then it is charged (reversing the discharge process) by passing $0.6\ F$ charge.
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Select the incorrect option.
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Select the incorrect option.
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