The equation \(\Delta G^{\circ}=-n F \mathscr{E}^{\circ}\) also can be applied to halfreactions. Use standard reduction potentials to estimate \(\Delta G_{\mathrm{r}}^{\circ}\) for \(\mathrm{Fe}^{2+}(a q)\) and \(\mathrm{Fe}^{3+}(a q) .\left(\Delta G_{\mathrm{f}}^{\circ} \text { for } \mathrm{e}^{-}=0 .\right)\)

Short Answer

Expert verified
To estimate the standard Gibbs free energy change (\(\Delta G_{\mathrm{r}}^{\circ}\)) for \(\mathrm{Fe}^{2+}(aq)\) and \(\mathrm{Fe}^{3+}(aq)\), we first gather the standard reduction potential \(\mathscr{E}^{\circ}_{1} = 0.77V\) for the half-reaction \(\mathrm{Fe}^{3+}(aq) + e^{-} \rightarrow \mathrm{Fe}^{2+}(aq)\). In this half-reaction, only 1 electron is exchanged, so \(n = 1\). Using the Faraday's constant (\(F\)), we then apply the given equation \(\Delta G^{\circ} = -n F \mathscr{E}^{\circ}\) to obtain \(\Delta G_{\mathrm{r}}^{\circ} = - (7.4289 \times 10^4 J \cdot mol^{-1})\). Thus, the estimated standard Gibbs free energy change for the given half-reaction is approximately \(-7.43 \times 10^4 J \cdot mol^{-1}\).

Step by step solution

01

Gather standard reduction potentials

To estimate the standard Gibbs free energy change for \(\mathrm{Fe}^{2+}(aq)\) and \(\mathrm{Fe}^{3+}(aq)\), we need to first gather the standard reduction potentials. You can find these values in a textbook or online reference. For our problem, we have: \(\mathrm{Fe}^{3+}(aq) + e^{-} \rightarrow \mathrm{Fe}^{2+}(aq)\), with a standard reduction potential \(\mathscr{E}^{\circ}_{1} = 0.77V\).
02

Calculate the number of electrons exchanged

To apply the given equation, we need to find the number of electrons (n) exchanged in the half-reaction. In our case, the reaction is: \(\mathrm{Fe}^{3+}(aq) + e^{-} \rightarrow \mathrm{Fe}^{2+}(aq)\) From this equation, we can see that only 1 electron is exchanged (as indicated by the stoichiometry of the balanced equation). Therefore, \(n = 1\).
03

Calculate the Faraday's constant

In the given equation, \(F\) represents the Faraday's constant, which is the amount of electric charge in one mole of electrons. The value for \(F\) is approximately \(9.6485 \times 10^4 C \cdot mol^{-1}\).
04

Calculate the standard Gibbs free energy change

Now that we have gathered all required information, we can calculate the standard Gibbs free energy change for the half-reaction using the equation: \(\Delta G^{\circ} = -n F \mathscr{E}^{\circ}\). For our half-reaction, we have: \(\Delta G_{\mathrm{r}}^{\circ} = -(1) \times (9.6485 \times 10^4 C \cdot mol^{-1}) \times (0.77V)\)
05

Solve the equation

Solving the equation, we get: \(\Delta G_{\mathrm{r}}^{\circ} = - (7.4289 \times 10^4 J \cdot mol^{-1})\) Therefore, the estimated standard Gibbs free energy change for the given half-reaction \(\mathrm{Fe}^{2+}(aq)\) and \(\mathrm{Fe}^{3+}(aq)\) is approximately \(-7.43 \times 10^4 J \cdot mol^{-1}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Reduction Potential
Understanding the standard reduction potential is crucial for predicting the feasibility of electrochemical reactions. It measures the tendency of a chemical species to gain electrons and be reduced. Measured in volts, the standard reduction potential is determined under standard conditions: a 1 molar solution, a pressure of 1 bar, and a temperature of 25 degrees Celsius (298 K).

In electrochemistry, each half-reaction has an associated standard reduction potential. When comparing different half-reactions, the one with the higher positive standard reduction potential will tend to be reduced, acting as the cathode. Conversely, a lower, often negative, standard reduction potential indicates a propensity to lose electrons and be oxidized, thus serving as the anode in an electrochemical cell.

To calculate the standard Gibbs free energy change \(\Delta G^\circ\) for a reaction, you can use the standard reduction potentials of the half-reactions involved. The relationship between Gibbs free energy and electrochemical cell potential is given by the equation \(\Delta G^\circ = -nF\mathscr{E}^\circ\), where \(n\) is the number of moles of electrons exchanged in the reaction.
Faraday's Constant
Faraday's constant, represented by the symbol \(F\), is a fundamental physical constant used in electrochemistry. It represents the total electric charge carried by one mole of electrons. The exact value of Faraday's constant is approximately \(9.6485 \times 10^4 C \cdot mol^{-1}\) (coulombs per mole).

This constant is pivotal in linking chemical reactions with electricity. When calculating the standard Gibbs free energy change \(\Delta G^\circ\) for an electrochemical reaction, Faraday's constant provides the bridge between the energy change and the electrochemical potential. It encapsulates the concept that energy change in a system is proportional to the amount of electric charge transferred and the potential difference the charge moves through.

For instance, if you know the number of electrons transferred in a half-reaction and the standard reduction potential, Faraday's constant allows you to calculate the amount of work that can be produced or the energy that can be stored or released in the reaction.
Half-Reaction Electrochemistry
Half-reactions are at the heart of electrochemistry, as they help dissect the overall redox reaction into two simpler parts: one that involves oxidation (loss of electrons), and one that involves reduction (gain of electrons). Each half-reaction occurs at a separate electrode in an electrochemical cell: the anode for oxidation and the cathode for reduction.

Understanding half-reactions is essential for calculating the standard Gibbs free energy change of a process because it allows us to isolate the contribution of each species. For the calculation, we focus on the reduction half-reactions and their associated standard reduction potentials. The free energy change of the overall reaction is the sum of the changes for the individual half-reactions.

When we balance half-reactions, it is important to ensure that the number of electrons lost in the oxidation half-reaction equals the number gained in the reduction half-reaction. The stoichiometry of electrons involved in the half-reactions directly affects the calculation of Gibbs free energy change using the number of moles of electrons (\(n\)) and Faraday's constant (\(F\)). This approach helps students to connect chemical changes with observable quantities like the electric potential of a cell.

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Most popular questions from this chapter

Assign oxidation numbers to all the atoms in each of the following: a. \(\mathrm{HNO}_{3}\) b. \(\mathrm{CuCl}_{2}\) c. \(\mathrm{O}_{2}\) d. \(\mathrm{H}_{2} \mathrm{O}_{2}\) e. \(C_{6} H_{12} O_{6}\) f .\(\mathrm {Ag} \) g. \(\mathrm{PbSO}_{4}\) h. \(\mathrm{PbO}_{2}\) i. \(\quad \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) j . \(\mathrm{CO}_{2}\) k. \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Ce}\left(\mathrm{SO}_{4}\right)_{3}\) 1\. \(\mathrm{Cr}_{2} \mathrm{O}_{3}\)

Consider the galvanic cell based on the following halfreactions: $$ \begin{array}{ll} \mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn} & \mathscr{E}^{\circ}=-0.76 \mathrm{V} \\ \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe} & \mathscr{E}^{\circ}=-0.44 \mathrm{V} \end{array} $$ a. Determine the overall cell reaction and calculate \(\mathscr{E}_{\text {cell. }}\) b. Calculate \(\Delta G^{\circ}\) and \(K\) for the cell reaction at \(25^{\circ} \mathrm{C}\). c. Calculate \(\mathscr{C}_{\text {cell }}\) at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{Zn}^{2+}\right]=0.10 \mathrm{M}\) and \(\left[\mathrm{Fe}^{2+}\right]=1.0 \times 10^{-5} \mathrm{M}\)

A factory wants to produce \(1.00 \times 10^{3} \mathrm{kg}\) barium from the electrolysis of molten barium chloride. What current must be applied for \(4.00 \mathrm{h}\) to accomplish this?

An aqueous solution of an unknown salt of ruthenium is electrolyzed by a current of 2.50 A passing for 50.0 min. If 2.618 g Ru is produced at the cathode, what is the charge on the ruthenium ions in solution?

Hydrazine is somewhat toxic. Use the half-reactions shown below to explain why household bleach (a highly alkaline solution of sodium hypochlorite) should not be mixed with household ammonia or glass cleansers that contain ammonia. $$\begin{array}{c} \mathrm{ClO}^{-}+\mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{OH}^{-}+\mathrm{Cl}^{-} \quad \quad \quad \mathscr{E}^{\circ}=0.90 \mathrm{V} \\ \mathrm{N}_{2} \mathrm{H}_{4}+2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{NH}_{3}+2 \mathrm{OH}^{-} \quad \mathscr{E}^{\circ}=-0.10 \mathrm{V} \end{array}$$

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