Question

Calculate \(\mathscr{E}^{\circ}\) for the following half-reaction: $$\mathrm{AgI}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{I}^{-}(a q)$$ (Hint: Reference the \(K_{\mathrm{sp}}\) value for AgI and the standard reduction potential for \(\mathrm{Ag}^{+}\).)

Short answer

The standard EMF (\(\mathscr{E}^{\circ}\)) for the given half-reaction is approximately \(0.536\) V.

Step-by-step solution

Write down the given half-reaction and constants

The given half-reaction is: $$\mathrm{AgI}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{I}^{-}(a q)$$ The constants we need to reference are: 1. \(K_{\mathrm{sp}}\) for AgI 2. Standard reduction potential for \(\mathrm{Ag}^{+}/\mathrm{Ag}\)

Write the half-reaction for the reduction potential of \(\mathrm{Ag}^{+}\)

We need the \(\mathrm{Ag}^{+}/\mathrm{Ag}\) half-reaction to determine the standard reduction potential for the given reaction. The half-reaction is: $$\mathrm{Ag}^{+}(aq) + \mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)$$

Use the Nernst equation to find the relationship between \(\mathscr{E}^{\circ}\) and \(K_{\mathrm{sp}}\)

The Nernst equation is given as: $$\mathscr{E}=\mathscr{E}^{\circ}-\frac{RT}{nF} \ln{Q}$$ For this problem, the reaction quotient (Q) will be the solubility product constant \(K_{\mathrm{sp}}\). The number of electrons transferred (n) in the half-reaction is 1. At standard conditions, the equation can be simplified to: $$\mathscr{E}^{\circ}=\mathscr{E}_{\mathrm{Ag}^{+}/\mathrm{Ag}}-\frac{RT}{F} \ln{K_{\mathrm{sp}}}$$

Calculate \(\mathscr{E}^{\circ}\) using the given constants

At this point, we need the values for \(K_{\mathrm{sp}}\) of AgI and \(\mathscr{E}_{\mathrm{Ag}^{+}/\mathrm{Ag}}\). $$K_{\mathrm{sp}}(\mathrm{AgI})=8.5\times10^{-17}$$ $$\mathscr{E}_{\mathrm{Ag}^{+}/\mathrm{Ag}}=0.7996\,\mathrm{V}$$ Now, we just need to plug in the values in the equation we derived in Step 3: $$\mathscr{E}^{\circ}=0.7996-\frac{RT}{F} \ln{8.5\times10^{-17}}$$ At standard conditions (298 K), \(R=8.314\,\mathrm{J}\,\mathrm{mol}^{-1}\,\mathrm{K}^{-1}\), and \(F=96,485\,\mathrm{C}\,\mathrm{mol}^{-1}\). $$\mathscr{E}^{\circ}=0.7996-\frac{(8.314\,\mathrm{J}\,\mathrm{mol}^{-1}\,\mathrm{K}^{-1})(298\,\mathrm{K})}{(96,485\,\mathrm{C}\,\mathrm{mol}^{-1})} \ln{8.5\times10^{-17}}$$

Solve for \(\mathscr{E}^{\circ}\)

Calculate the result: $$\mathscr{E}^{\circ} \approx 0.536\,\mathrm{V}$$ Hence, the standard EMF (\(\mathscr{E}^{\circ}\)) for the given half-reaction is approximately \(0.536\) V.

Key concepts

Nernst Equation

The Nernst Equation is a vital tool in electrochemistry that helps us calculate the electromotive force (EMF) of a cell under non-standard conditions. It relates the EMF of a cell to the concentrations of the reacting species. The equation is expressed as follows: \[ \mathscr{E} = \mathscr{E}^{\circ} - \frac{RT}{nF} \ln{Q} \] Here:

• \( \mathscr{E} \) is the cell potential at non-standard conditions.
• \( \mathscr{E}^{\circ} \) is the standard cell potential.
• \( R \) is the gas constant, equivalent to \( 8.314 \, \mathrm{J} \, \mathrm{mol}^{-1} \, \mathrm{K}^{-1} \).
• \( T \) is the temperature in Kelvin.
• \( n \) denotes the number of moles of electrons transferred in the reaction.
• \( F \) is Faraday's constant, approximately \( 96,485 \, \mathrm{C} \, \mathrm{mol}^{-1} \).
• \( Q \) is the reaction quotient, a ratio of product concentrations to reactant concentrations.

In our context, \( Q \) has been replaced with \( K_{\mathrm{sp}} \), the solubility product constant. The Nernst equation allows us to derive \( \mathscr{E}^{\circ} \) from known values of \( K_{\mathrm{sp}} \) and \( \mathscr{E}_{\mathrm{Ag}^{+}/\mathrm{Ag}} \).

Solubility Product Constant

The Solubility Product Constant, denoted as \( K_{\mathrm{sp}} \), is a crucial value in determining the extent to which a compound will dissolve in water. It is especially significant in sparingly soluble salts, such as AgI. The \( K_{\mathrm{sp}} \) for AgI, as given in the problem, is \( 8.5 \times 10^{-17} \). This extremely small number indicates that very little AgI dissolves in water. To understand \( K_{\mathrm{sp}} \), consider the dissociation of a salt like \( \mathrm{AgI}(s) \) in water: \[ \mathrm{AgI}(s) \rightleftharpoons \mathrm{Ag}^{+}(aq) + \mathrm{I}^{-}(aq) \] The \( K_{\mathrm{sp}} \) expression is: \[ K_{\mathrm{sp}} = [\mathrm{Ag}^{+}][\mathrm{I}^{-}] \] Since \( K_{\mathrm{sp}} \) is used within the Nernst Equation for electrochemical calculations, understanding its role provides insight into how changes in solubility affect the electrochemical potential of a system.

Reduction Potential

Reduction potential is a measure of the tendency of a chemical species to gain electrons and hence, be reduced. It is often compared to a standard hydrogen electrode under standardized conditions, which is set at zero volts for reference. For the half-reaction \( \mathrm{Ag}^{+}(aq) + \mathrm{e}^{-} \rightarrow \mathrm{Ag}(s) \), the standard reduction potential \( \mathscr{E}_{\mathrm{Ag}^{+}/\mathrm{Ag}} \) is \( 0.7996 \, \mathrm{V} \). This means that silver ions readily accept electrons to form silver metal. Reduction potentials are helpful in predicting whether the overall cell reaction will occur spontaneously. If \( \mathscr{E}^{\circ} \) for the cell reaction is positive, the reaction is spontaneous. By integrating reduction potentials in Nernst equation calculations, we can evaluate the directionality and feasibility of electrochemical reactions under various conditions.

Half-Reaction

Half-reactions are chemical equations that show either the reduction or the oxidation process separately in a redox reaction. They are broken down into oxidation half-reactions, where electrons are lost, and reduction half-reactions, where electrons are gained. For example, in the given exercise: \[ \mathrm{AgI}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{I}^{-}(aq) \] This reaction represents only the reduction process, where \( \mathrm{AgI} \) is reduced to \( \mathrm{Ag} \) and \( \mathrm{I}^{-} \). Separating into half-reactions is crucial for applying the Nernst equation since it clarifies the number of electrons transferred (\( n \)). Half-reactions also facilitate determining standard electrode potentials and balancing redox equations. By focusing on individual half-reactions, students can more easily grasp the complex interactions in redox chemistry.