Animated Solution for Chemistry - Electrochemistry: For the electrochemical cell, $\text{Mg(s)}|{\text{Mg}}^{2+}(\text{aq},1\text{M})||{\text{Cu}}^{2+}(\text{aq},1\text{M})|\text{Cu(s)}$ the standard emf of the cell is $2.70\text{ V}$ at $300\text{ K}$. When the concentration of ${\text{Mg}}^{2+}$ is changed to $x\text{ M}$, the cell potential changes to $2.67\text{ V}$ at $300\text{ K}$. The value of $x$ is______. (given, $\frac{F}{R}=11500{\text{ K V}}^{-1}$, where $F$ is the Faraday constant and $R$ is the gas constant, $\ln (10)=2.30$)

Imagine you are looking at a classic galvanic cell setup. On one side, we have a magnesium electrode dipping into a solution of magnesium ions. On the other side, a copper electrode sits in a solution of copper ions. The two are connected by a wire and a salt bridge, allowing electrons to flow and generate a voltage.

The problem tells us that the standard electromotive force (emf) of this cell is $2.70\text{ V}$. This is the voltage we would measure if both solutions were exactly $1\text{ M}$ in concentration. However, the actual measured cell potential is slightly lower, at $2.67\text{ V}$. This drop in voltage is our primary clue. It tells us that the concentrations are not standard. Specifically, the concentration of the magnesium ions has been changed to an unknown value, $x\text{ M}$, while the copper ion concentration remains at $1\text{ M}$. Our mission is to find this mysterious value of $x$.

Nernst

To bridge the gap between cell potential and ion concentration, we must call upon the Nernst equation. This powerful formula allows us to calculate the voltage of an electrochemical cell under non-standard conditions.

The Nernst equation is given by:

Here, $E_{\text{cell}}$ is the non-standard cell potential, $E_{\text{cell}}^{\circ }$ is the standard cell potential, $R$ is the universal gas constant, $T$ is the temperature in Kelvin, $n$ is the number of moles of electrons transferred in the balanced redox reaction, $F$ is the Faraday constant, and $Q$ is the reaction quotient.

First, let's determine $n$ and $Q$. The overall cell reaction is:

From this, we can clearly see that magnesium loses two electrons to become ${\text{Mg}}^{2+}$, and copper gains those two electrons. Therefore, $n=2$.

The reaction quotient $Q$ is the ratio of the concentrations of the aqueous products to the aqueous reactants. Solid metals do not appear in this expression.

Now, we substitute all our known values into the Nernst equation. We know $E_{\text{cell}}=2.67\text{ V}$, $E_{\text{cell}}^{\circ }=2.70\text{ V}$, and $T=300\text{ K}$.

Let's rearrange this equation to isolate the natural logarithm term. Subtracting $2.70$ from both sides gives:

The negative signs gracefully cancel out. Now, we solve for $\ln x$:

The problem provides a very specific and helpful ratio: $\frac{F}{R}=11500{\text{ K V}}^{-1}$. Let's plug this directly into our rearranged equation.

Let's do the math carefully. $0.03\times 2=0.06$.

We have arrived at $\ln x=2.30$. The problem statement also generously informs us that $\ln (10)=2.30$.

By direct comparison, the unknown concentration $x$ must be:

$$x=10$$

The concentration of the magnesium ions is $10\text{ M}$. This higher concentration of product ions pushes back against the forward reaction, which perfectly explains why the cell potential dropped from its standard value of $2.70\text{ V}$ to $2.67\text{ V}$.