Question

A cell was set up having the following reaction. $$ \begin{aligned} \mathrm{Mg}(s)+\mathrm{Cd}^{2+}(a q) \longrightarrow \mathrm{Mg}^{2+}(a q)+\mathrm{Cd}(s) & \\ E_{\mathrm{cdl}}^{\circ} &=1.97 \mathrm{~V} \end{aligned} $$ The magnesium electrode was dipped into a \(1.00 \mathrm{M}\) solution of \(\mathrm{MgSO}_{4}\) and the cadmium electrode was dipped into a solution of unknown \(\mathrm{Cd}^{2+}\) concentration. The potential of the cell was measured to be \(1.54 \mathrm{~V}\). What was the unknown \(\mathrm{Cd}^{2+}\) concentration?

Short answer

The unknown concentration of \(\text{Cd}^{2+}\) is approximately \(3.37 \times 10^{-2} \text{M}\).

Step-by-step solution

Write the Nernst Equation for the Cell Reaction

The Nernst equation relates the cell potential at non-standard conditions to its standard-cell potential, temperature, charge, and the concentrations of the reactants and products. For the given cell reaction, the Nernst equation can be written as: \[\[\begin{align*} E = E^\circ - \frac{0.0592}{n} \log\left(\frac{[\text{Cd}^{2+}]}{[\text{Mg}^{2+}]}\right) \end{align*}\]\] Where: - \(E\) is the cell potential under non-standard conditions, - \(E^\circ\) is the standard cell potential, - \(n\) is the number of moles of electrons transferred in the reaction, and - \(\left[\text{Cd}^{2+}\right]\) and \(\left[\text{Mg}^{2+}\right]\) are the concentrations of Cd2+ and Mg2+ ions, respectively.

Identify Known Values and Solve for the Unknown Concentration

From the given exercise, we have \(E^\circ = 1.97 \text{V}\), \(E = 1.54 \text{V}\), \([\text{Mg}^{2+}] = 1.00 \text{M} \) (since the magnesium electrode is dipped in 1 M \(\text{MgSO}_4\)), and \(n = 2\) (as each magnesium atom releases 2 electrons to form \(\text{Mg}^{2+}\)). Plugging these values into the Nernst equation and solving for \([\text{Cd}^{2+}]\), we get: \[\[\begin{align*} 1.54 = 1.97 - \frac{0.0592}{2} \log\left(\frac{[\text{Cd}^{2+}]}{1.00}\right) \ 0.43 = \frac{0.0592}{2} \log\left([\text{Cd}^{2+}]\right) \ \log\left([\text{Cd}^{2+}]\right) = \frac{0.43 \times 2}{0.0592} \ [\text{Cd}^{2+}] = 10^{\left(\frac{0.86}{0.0592}\right)} \ [\text{Cd}^{2+}] \approx 10^{14.527} = 3.37 \times 10^{-2} \text{M} \end{align*}\]\]

Key concepts

Electrochemical Cell Potential

The electrochemical cell potential, often denoted as 'E', is the measure of the voltage difference between two electrodes of an electrochemical cell. This potential difference drives the flow of electrons through an external circuit, which is the fundamental operation of a battery or any galvanic/voltaic cell. In simple terms, it's what makes a battery power your remote control or your phone charge when plugged in.

The cell potential is determined by the specific chemical reactions occurring at the electrodes and is influenced by the nature of the materials involved, their concentrations, and external conditions such as temperature.

Exploring Cell Potential in Practice: In the case of the textbook exercise, the electrochemical cell potential was measured and given as 1.54 V. This value is a direct indicator of how much 'push' the electrons have to flow from the magnesium electrode to the cadmium electrode. The fact that it is lower than the standard cell potential indicates that reaction conditions are not at their standard states, which leads us neatly into the concept of the standard cell potential.

Standard Cell Potential

Standard cell potential, represented as \(E^\circ\), is a fixed value measured under standard conditions: 1 molar solution concentration, pressure at 1 atmosphere, and room temperature at about 25 degrees Celsius (298 K). It is a key component when using the Nernst equation to calculate the electrochemical cell potential under non-standard conditions.

It is essentially the cell potential that would be measured if all the reactants and products were in their standard states. Put another way, it’s like the maximum potential voltage a cell could achieve under idealized conditions. The higher the standard cell potential, the greater the tendency for the chemical reaction to occur spontaneously.

In our exercise, the standard cell potential for the reaction between magnesium and cadmium ions is given as 1.97 V. It represents the maximum potential that can be produced if concentrations of the reactants were at the standard 1 M concentration.

Reaction Quotient (Q)

The reaction quotient, or Q, evaluates the relative amounts of products and reactants present during a reaction at a given moment and is used to predict the direction of the reaction. It has the same form as the equilibrium constant expression but is used for non-equilibrium states.

We calculate Q by taking the concentrations of the products raised to the power of their coefficients in the balanced chemical equation, divided by the concentrations of the reactants also raised to the power of their coefficients. For gases, we would use partial pressures instead of concentrations.

In this specific scenario where magnesium is transforming into magnesium ions and cadmium ions are being reduced to solid cadmium, Q is defined as the concentration of \(Cd^{2+}\) ions divided by the concentration of \(Mg^{2+}\) ions. This ratio helps us to use the Nernst equation effectively to calculate the non-standard cell potential.

Concentration Calculation

Concentration calculation is fundamental in chemistry for understanding how much of a substance is present in a solution. This can drastically affect the behavior of reactions, as seen in our exercise. Typically measured in molarity (M), which is moles of solute per liter of solution, concentration helps us predict how substances will interact in a mixture.

By manipulating the Nernst equation, we can find out the unknown concentration of one reactant or product if we have the other values, including the measured cell potential. It's a bit like solving a puzzle – once you fit in the right pieces of information into the Nernst equation, you can calculate the missing concentration value, which is crucial for various applications from pharmaceuticals to energy storage.

In the exercise, determining the unknown concentration of \(Cd^{2+}\) was key. The calculation involved rearranging the Nernst equation and solving for the concentration of \(Cd^{2+}\), resulting in the value of approximately 0.0337 M, or 3.37 x 10^-2 M.