Represent the cell in which the following reaction takes place:
$Mg_{(s)} + 2Ag^{+}(0.0001 \, M) \rightarrow Mg^{2+}(0.130 \, M) + 2Ag_{(s)}$
Calculate its $E_{cell}$ if $E^{\Theta}_{cell} = 3.17 \, V$.
$Mg_{(s)} + 2Ag^{+}(0.0001 \, M) \rightarrow Mg^{2+}(0.130 \, M) + 2Ag_{(s)}$
Calculate its $E_{cell}$ if $E^{\Theta}_{cell} = 3.17 \, V$.
(N/A) The cell representation is $Mg | Mg^{2+}(0.130 \, M) || Ag^{+}(0.0001 \, M) | Ag$.
Using the Nernst equation: $E_{cell} = E^{\Theta}_{cell} - \frac{0.0591}{n} \log \frac{[Mg^{2+}]}{[Ag^{+}]^2}$.
Here,$n = 2$,$[Mg^{2+}] = 0.130 \, M$,and $[Ag^{+}] = 0.0001 \, M$.
$E_{cell} = 3.17 - \frac{0.0591}{2} \log \frac{0.130}{(10^{-4})^2}$.
$E_{cell} = 3.17 - 0.02955 \log \frac{0.130}{10^{-8}} = 3.17 - 0.02955 \log (1.3 \times 10^7)$.
$E_{cell} = 3.17 - 0.02955 \times 7.1139 = 3.17 - 0.210 = 2.96 \, V$.
Using the Nernst equation: $E_{cell} = E^{\Theta}_{cell} - \frac{0.0591}{n} \log \frac{[Mg^{2+}]}{[Ag^{+}]^2}$.
Here,$n = 2$,$[Mg^{2+}] = 0.130 \, M$,and $[Ag^{+}] = 0.0001 \, M$.
$E_{cell} = 3.17 - \frac{0.0591}{2} \log \frac{0.130}{(10^{-4})^2}$.
$E_{cell} = 3.17 - 0.02955 \log \frac{0.130}{10^{-8}} = 3.17 - 0.02955 \log (1.3 \times 10^7)$.
$E_{cell} = 3.17 - 0.02955 \times 7.1139 = 3.17 - 0.210 = 2.96 \, V$.
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