You are given the following three half-reactions: (1) Fe^3+(a q)+e^-⟹Fe^2+(a q) (2) Fe

step 1 Question 138 is a buffer solution calculation. With buffer solution calculations, we can use the

You are given the following three half-reactions: (1)...

You are given the following three half-reactions: $$ \begin{array}{l}{\text { (1) } \mathrm{Fe}^{3+}(a q)+\mathrm{e}^{-} \Longrightarrow \mathrm{Fe}^{2+}(a q)} \\ {\text { (2) } \mathrm{Fe}^{2+}(a q)+2 \mathrm{e}^{-} \Longrightarrow \mathrm{Fe}(s)} \\ {\text { (3) } \mathrm{Fe}^{3+}(a q)+3 \mathrm{e}^{-} \rightleftharpoons \mathrm{Fe}(s)}\end{array} $$ (a) Use $E_{\text { half-cell values for }}(1)$ and $(2)$ to find $E_{\text { half-cell }}^{\circ}$ for $(3)$ (b) Calculate $\Delta G^{\circ}$ for $(1)$ and $(2)$ from their $E_{\text { half-cell values. }}^{\circ}$ (c) Calculate $\Delta G^{\circ}$ for $(3)$ from $(1)$ and $(2)$ (d) Calculate $E_{\text { half cell }}^{\circ}$ for $(3)$ from its $\Delta G^{\circ} .$ (e) What is the relationship between the $E_{\text { half-cell values for }}(1)$ and (2) and the $E_{\text { half-cell }}^{\circ}$ value for $(3) ?$

You are given the following three half-reactions:
$$
\begin{array}{l}{\text { (1) } \mathrm{Fe}^{3+}(a q)+\mathrm{e}^{-} \Longrightarrow \mathrm{Fe}^{2+}(a q)} \\ {\text { (2) } \mathrm{Fe}^{2+}(a q)+2 \mathrm{e}^{-} \Longrightarrow \mathrm{Fe}(s)} \\ {\text { (3) } \mathrm{Fe}^{3+}(a q)+3 \mathrm{e}^{-} \rightleftharpoons \mathrm{Fe}(s)}\end{array}
$$
(a) Use $E_{\text { half-cell values for }}(1)$ and $(2)$ to find $E_{\text { half-cell }}^{\circ}$ for $(3)$
(b) Calculate $\Delta G^{\circ}$ for $(1)$ and $(2)$ from their $E_{\text { half-cell values. }}^{\circ}$
(c) Calculate $\Delta G^{\circ}$ for $(3)$ from $(1)$ and $(2)$
(d) Calculate $E_{\text { half cell }}^{\circ}$ for $(3)$ from its $\Delta G^{\circ} .$
(e) What is the relationship between the $E_{\text { half-cell values for }}(1)$ and (2) and the $E_{\text { half-cell }}^{\circ}$ value for $(3) ?$

Key Concepts

Standard Reduction Potential

This concept refers to the voltage associated with a half-cell reaction measured under standard conditions. It quantifies the tendency of a chemical species to gain electrons and be reduced. Measured against the standard hydrogen electrode, it provides a benchmark for comparing the oxidizing and reducing power of different half-reactions.

Gibbs Free Energy in Electrochemistry

Gibbs free energy represents the maximum reversible work obtainable from a chemical process, and in the context of electrochemistry, it is directly related to the electrode potential through the equation ?G° = -nFE°. This relationship connects thermodynamics with redox processes, allowing the prediction of reaction spontaneity based on electron transfer and voltage.

Combining Electrochemical Half-Reactions

When constructing an overall redox reaction from half-reactions, one must combine them in a way that conserves both mass and charge. Importantly, while standard potentials are intensive properties and cannot be directly added, free energy changes are extensive and can be summed. This process involves accounting for the number of electrons transferred in each half-reaction to correctly determine the overall energetics.

Electron Stoichiometry

Electron stoichiometry, or the n-factor, is critical in redox reactions as it denotes the number of electrons exchanged in a half-reaction. This factor is essential when applying the ?G° = -nFE° relationship, ensuring that the free energy change is accurately calculated and that the overall reaction is balanced in terms of electron transfer.

Intensive versus Extensive Properties in Electrochemistry

Understanding the difference between these two types of properties is vital. Electrode potentials are intensive properties, meaning they do not depend on the scale of the reaction, while free energy is an extensive property, scaling with the amount of substance involved. This distinction explains why potentiometric values cannot be simply added and must be converted to free energy values when combining reactions.

Transcript

00:01 Question 138 is a buffer solution calculation.

00:06 With buffer solution calculations, we can use the henderson -hasselbaltz equation, where ph is equal to pca plus the log of the moles of base over the moles of acid or the concentration of base over the concentration of acid.

00:20 If we rearrange this equation, then the change in ph that will occur, if we start with a buffer solution that has equal concentration, of base and acid, then ph equals pca.

00:36 If we want to determine what the shift in ph is, starting at a ph equal to pca, then simply moving this over to essentially subtracting this from both sides, so minus pka, minus pqa, we'll get a difference here, which will, and this will go away, which will just be equal to the log moles base over moles acid.

01:06 So again, if we're starting with this, then the change in ph is going to be the difference of the new ph minus the old ph, or the initial ph, which was equal to pca.

01:17 And that's what we're starting with.

01:19 We're starting with a solution that contains the same concentration of base as it does acid.

01:28 So the change in ph is simply going to be equal to the log of the new moles base over the new moles acid.

01:36 So if we don't want the ph to change more than negative 0 .05 when we add a strong acid, then we don't want this ratio to be equal to anything more negative than negative 0 .05.

01:53 So what we need to do is take the volume that will contain the buffer.

02:06 This is the total 1 liter volume of the medium to which we're adding the buffer, multiplied by its concentration, which we don't know, because we don't know how much of the .1 molar, h .a.

02:21 .1 molar a.

02:22 Minus buffer we're going to add...

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