Understanding the Problem

We are given an electrochemical cell: Mg(s) | Mg²⁺ (aq, 1 M) || Cu²⁺ (aq, 1 M) | Cu(s). The standard EMF is 2.70 V. When the concentration of Mg²⁺ is changed to x M, the cell potential becomes 2.67 V. We need to find x.

Given: F/R = 11500 K V⁻¹ and ln(10) = 2.30.

Step 1: Write the Cell Reaction and Nernst Equation

The cell reaction is: $\mathrm{Mg}+{\mathrm{Cu}}^{2+}\to {\mathrm{Mg}}^{2+}+\mathrm{Cu}$

The Nernst equation for this cell at 300 K is:

$E=E^{\circ }-\frac{\mathrm{RT}}{2F}\ln \left(\frac{\left[{\mathrm{Mg}}^{2+}\right]}{\left[{\mathrm{Cu}}^{2+}\right]}\right)$

Since [Cu²⁺] remains at 1 M, the equation simplifies to:

$E=E^{\circ }-\frac{\mathrm{RT}}{2F}\ln \left(\left[{\mathrm{Mg}}^{2+}\right]\right)$

Step 2: Apply the Nernst Equation for Both Cases

Case 1 (Standard conditions): [Mg²⁺] = 1 M, E = E° = 2.70 V

Case 2 (Changed conditions): [Mg²⁺] = x M, E = 2.67 V

Writing the Nernst equation for Case 2:

$2.67=2.70-\frac{\mathrm{RT}}{2F}\ln \left(x\right)$

Step 3: Rearrange the Equation to Solve for x

$\frac{\mathrm{RT}}{2F}\ln \left(x\right)=2.70-2.67=0.03$

$\ln \left(x\right)=\frac{2F}{\mathrm{RT}}\times 0.03$

We are given that F/R = 11500 K V⁻¹. Therefore, F/R = 11500.

We can find the value of the term 2F/(RT):

$\frac{2F}{\mathrm{RT}}=\frac{2F}{R}\times \frac{1}{T}=2\times 11500\times \frac{1}{300}$

Let's calculate this:

$\frac{2F}{\mathrm{RT}}=\frac{2\times 11500}{300}=\frac{23000}{300}=\mathrm{76.666...}\approx \frac{230}{3}$

Now plug this back into the equation for ln(x):

$\ln \left(x\right)=\left(\frac{230}{3}\right)\times 0.03=\frac{230}{3}\times \frac{3}{100}=\frac{230}{100}=2.3$

We are also given that ln(10) = 2.30. Notice that 2.3 is extremely close to 2.30.

Therefore:

$\ln \left(x\right)=2.3\approx \ln \left(10\right)$

This implies that x = 10.

Final Answer

The value of x is 10.

Related Concepts & Formulae

1. The Nernst Equation

The Nernst equation is used to calculate the cell potential under non-standard conditions. For a general reaction: $\mathrm{aA}+\mathrm{bB}\to \mathrm{cC}+\mathrm{dD}$

The Nernst equation is:

$E=E^{\circ }-\frac{\mathrm{RT}}{nF}\ln Q$

Where:
E = Cell potential
E° = Standard cell potential
R = Gas constant (8.314 J/mol·K)
T = Temperature in Kelvin
n = Number of moles of electrons transferred
F = Faraday constant (96485 C/mol)
Q = Reaction quotient

2. Reaction Quotient (Q)

The reaction quotient, Q, has the same form as the equilibrium constant but uses the initial concentrations of the reactants and products.

For the reaction above: $Q=\frac{\left[C{]}^c\right[D{]}^d}{\left[A{]}^a\right[B{]}^b}$

3. Key Theory: Concentration Cells

The potential of an electrochemical cell depends on the concentration of the reactants. If the concentration of a reactant decreases (for an anode), the cell potential decreases. If the concentration of a product decreases (for a cathode), the cell potential increases. This principle is used in concentration cells, where both half-cells are the same material but with different ion concentrations.