Calculation of Magnesium Ion Concentration from Nernst Equation

To determine the value of $x$, we analyze the cell reaction and apply the Nernst equation.

The given electrochemical cell is: $\text{Mg}(s) \mid \text{Mg}^{2+}(\text{aq}) \parallel \text{Cu}^{2+}(\text{aq}) \mid \text{Cu}(s)$

The half-cell reactions are: $\text{Anode (Oxidation): } \text{Mg}(s) \rightarrow \text{Mg}^{2+}(\text{aq}) + 2e^-$ $\text{Cathode (Reduction): } \text{Cu}^{2+}(\text{aq}) + 2e^- \rightarrow \text{Cu}(s)$

The overall cell reaction is: $\text{Mg}(s) + \text{Cu}^{2+}(\text{aq}) \rightarrow \text{Mg}^{2+}(\text{aq}) + \text{Cu}(s)$

The number of moles of electrons transferred in the balanced equation is: $n = 2$

The reaction quotient $Q$ for this process is given by: $Q = \frac{[\text{Mg}^{2+}]}{[\text{Cu}^{2+}]}$

Initially, when $[\text{Mg}^{2+}] = 1\text{ M}$ and $[\text{Cu}^{2+}] = 1\text{ M}$, the cell potential is standard: $E_{\text{cell}}^\circ = 2.70\text{ V}$

When the concentration of $\text{Mg}^{2+}$ is changed to $x\text{ M}$ (keeping $[\text{Cu}^{2+}] = 1\text{ M}$), the reaction quotient becomes: $Q = \frac{x}{1} = x$

Applying the Nernst equation at temperature $T = 300\text{ K}$: $E_{\text{cell}} = E_{\text{cell}}^\circ - \frac{RT}{nF} \ln Q$

Substitute the known values into the equation: $2.67 = 2.70 - \frac{R(300)}{2F} \ln x$

Rearranging the terms: $2.70 - 2.67 = \frac{300}{2} \left(\frac{R}{F}\right) \ln x$ $0.03 = 150 \left(\frac{R}{F}\right) \ln x$

Given that $\frac{F}{R} = 11500\text{ K}\cdot\text{V}^{-1}$, we have: $\frac{R}{F} = \frac{1}{11500}\text{ V}\cdot\text{K}^{-1}$

Substituting this in: $0.03 = \frac{150}{11500} \ln x$ $\frac{3}{100} = \frac{3}{230} \ln x$

Solving for $\ln x$: $\ln x = \frac{3}{100} \times \frac{230}{3} = \frac{230}{100} = 2.30$

Given that $\ln(10) = 2.30$, we have: $\ln x = \ln(10) \implies x = 10$