Problem 150 - Chemistry An Atoms First Approach

Question

Given the following two standard reduction potentials, $$\begin{array}{ll} \mathrm{M}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{M} & \mathscr{E}^{\circ}=-0.10 \mathrm{V} \\ \mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M} & \mathscr{E}^{\circ}=-0.50 \mathrm{V} \end{array}$$ solve for the standard reduction potential of the half-reaction $$ \mathbf{M}^{3+}+\mathbf{e}^{-} \longrightarrow \mathbf{M}^{2+} $$ (Hint: You must use the extensive property \(\Delta G^{\circ}\) to determine the standard reduction potential.)

Step-by-Step Solution

Answer

The standard reduction potential for the half-reaction \(M^{3+} + e^{-} \longrightarrow M^{2+}\) is \(0.7\,\mathrm{V}\).

Step 1: Write down the given half-reactions and their corresponding standard reduction potentials

Half-reaction 1: \(M^{3+} + 3e^{-} \longrightarrow M\), \(E^{\circ}_{1} = -0.10\mathrm{V}\) Half-reaction 2: \(M^{2+} + 2e^{-} \longrightarrow M\), \(E^{\circ}_{2} = -0.50\mathrm{V}\)

Step 2: Calculate the Gibbs Free Energy Changes for the given half-reactions

For Half-reaction 1: \( \Delta G^{\circ}_{1} = -n_{1}FE^{\circ}_{1} = -3F(-0.10\,\mathrm{V}) \) For Half-reaction 2: \( \Delta G^{\circ}_{2} = -n_{2}FE^{\circ}_{2} = -2F(-0.50\,\mathrm{V}) \)

Step 3: Obtain the equation for the desired half-reaction

We want to find the standard reduction potential for the following half-reaction: Half-reaction 3: \(M^{3+} + e^{-} \longrightarrow M^{2+}\) Notice that if we subtract the Half-reaction 2 from the Half-reaction 1, we will get the desired Half-reaction 3. Hence, we can also subtract the Gibbs Free Energy Changes of these half-reactions: \( \Delta G^{\circ}_{3} = \Delta G^{\circ}_{1} - \Delta G^{\circ}_{2} \)

Step 4: Calculate the Gibbs Free Energy Change for the desired half-reaction

Using the equations obtained in Step 2, we calculate \(\Delta G^{\circ}_{3}\) as follows: \( \Delta G^{\circ}_{3} = -3F(-0.10\,\mathrm{V}) - (-2F(-0.50\,\mathrm{V})) \)

Step 5: Calculate the standard reduction potential for the desired half-reaction

Now we can use the equation \(\Delta G^{\circ}_{3} = -n_{3}FE^{\circ}_{3}\) to determine the standard reduction potential for Half-reaction 3: \(-n_{3}FE^{\circ}_{3} = -3F(-0.10\,\mathrm{V}) - (-2F(-0.50\,\mathrm{V}))\) For Half-reaction 3, \(n_{3} = 1\), so we can rewrite the equation as: \(-FE^{\circ}_{3} = -3F(-0.10\,\mathrm{V}) - (-2F(-0.50\,\mathrm{V}))\) Now, divide both sides of the equation by \(-F\): \(E^{\circ}_{3} = 3(-0.10\,\mathrm{V}) - 2(-0.50\,\mathrm{V})\) After performing the arithmetic operation, we get the standard reduction potential for Half-reaction 3: \( E^{\circ}_{3} = 0.7 \,\mathrm{V} \) Thus, the standard reduction potential for the half-reaction \(M^{3+} + e^{-} \longrightarrow M^{2+}\) is \(0.7\,\mathrm{V}\).

Key Concepts

Gibbs Free Energy

Gibbs Free Energy (GFE) is a thermodynamic quantity that is pivotal in predicting the direction in which a chemical reaction will proceed. It's represented as \( \Delta G \) and provides critical information about the spontaneity of reactions. When \( \Delta G \) is negative, the process or reaction occurs spontaneously, indicating that it's thermodynamically favorable.

In the context of electrochemistry, the relationship between GFE and cell potential is expressed by the equation: \( \Delta G = -nFE \) where \( n \) is the number of moles of electrons transferred, \( F \) is the Faraday constant (approximately 96,485 coulombs per mole), and \( E \) is the cell potential in volts. It's useful to remember that any change in Gibbs Free Energy can be calculated using the standard reduction potentials of the half-reactions involved in the electrochemical cell.

Electrochemistry

Electrochemistry is the study of chemical processes that cause electrons to move. This movement of electrons is typically achieved through redox reactions, which involve the transfer of electrons from one substance to another. These reactions are fundamental for the operation of batteries, fuel cells, and electroplating.

An electrochemical cell consists of two electrodes immersed in an electrolyte. One electrode acts as the anode and undergoes oxidation (loss of electrons), while the other acts as the cathode and undergoes reduction (gain of electrons). The flow of electrons from the anode to the cathode through an external circuit generates an electric current, which can be used to do work.

Half-Reactions

In electrochemistry, a half-reaction is either the oxidation or reduction that occurs, treated as a separate event in an electrochemical cell. Writing a complex redox reaction as two simpler half-reactions helps in balancing the overall equation and understanding the electron transfer process.

Each half-reaction has an associated standard reduction potential, \( \mathscr{E}^\circ \), which is a measure of the tendency of the chemicals involved to be reduced. By separating a redox reaction into half-reactions, we can calculate the overall cell potential and gain insights into the Gibbs Free Energy changes for the reactions.

Cell Potential

The cell potential, also known as electromotive force (emf), is the driving force behind the movement of electrons through an electrochemical cell. It is determined by the difference in the standard reduction potentials of the cathode and anode in their half-reactions.

The cell potential can be measured in volts (\( V \) ) and provides a quantitative indication of a cell's ability to produce an electric current. A positive cell potential implies a spontaneous reaction, while a negative cell potential suggests a non-spontaneous reaction. Understanding cell potential is essential when dealing with the calculation of Gibbs Free Energy changes and the predictability of electrochemical processes.