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 electrode potential for the given half-reaction can be calculated using the Nernst Equation and the given values for \(E^{\circ}_{\text{Ag}^+}\) and \(K_{\text{sp}}\). First, calculate the equilibrium constant for the half-reaction using \(K = \frac{1}{K_{\mathrm{Ag}^{+}}} \times K_{sp}\). Then, use the Nernst Equation with n = 1: \(\mathscr{E}^{\circ} = -0.0592 \log K\). Plug in the calculated K value to find the standard electrode potential \(\mathscr{E}^{\circ}\).

Step-by-step solution

Write the related half-reactions for Ag+

We are given the standard reduction potential for \(\mathrm{Ag}^{+}\). Here are the standard half-reaction and its reverse reaction: $$ \mathrm{Ag}^{+}(aq) + \mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) \quad \quad \text{Reduction: E}^{\circ} = E_{\mathrm{Ag}^{+}} $$ $$ \mathrm{Ag}(s) \longrightarrow \mathrm{Ag}^{+}(aq) + \mathrm{e}^{-} \quad \quad \text{Oxidation} $$

Derive the given half-reaction from Ag-related reactions and calculate K

To get to our target half-reaction, take the reverse of the oxidation reaction and add it to the target half-reaction: $$ \mathrm{Ag}(s) \longrightarrow \mathrm{Ag}^{+}(aq) + \mathrm{e}^{-} $$ $$ \mathrm{Ag}^{+}(aq) + \mathrm{I}^{-}(aq) \longleftrightarrow \mathrm{AgI}(s) \quad \quad K = K_{sp} $$ Adding them, we get: $$ \mathrm{AgI}(s) + \mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) + \mathrm{I}^{-}(aq) $$ Now we have the given half-reaction, and we can calculate the equilibrium constant (K) for it using the equilibrium constants of the reactions we used: $$ K = \frac{1}{K_{\mathrm{Ag}^{+}}} \times K_{sp} $$

Apply Nernst Equation to calculate the standard electrode potential

Now we can use the Nernst Equation to find the standard electrode potential (E°) for the given half-reaction: $$ \mathscr{E}^{\circ} = \frac{-0.0592}{n} \log K $$ Here n = 1, as there is only one electron involved in the half-reaction. So, we can plug in the values we calculated in step 2: $$ \mathscr{E}^{\circ} = -0.0592 \log K $$ Finally, you can use the given values for \(E^{\circ}_{\text{Ag}^+}\) and \(K_{\text{sp}}\) to find the standard electrode potential for the given half-reaction.

Key concepts

Nernst Equation

The Nernst Equation is an essential formula in electrochemistry that allows you to calculate the potential of an electrochemical cell under non-standard conditions. It connects the concentrations of the chemical species involved with the cell potential and is especially helpful in real-world scenarios where standard conditions are not maintained.
The general form of the Nernst Equation is given by:

• \(E = E^{\circ} - \frac{0.0592}{n} \log Q \)
• Where \(E\) is the cell potential, \(E^{\circ}\) is the standard cell potential, \(n\) is the number of moles of electrons exchanged, and \(Q\) is the reaction quotient.

In the context of a half-reaction, the Nernst Equation becomes a valuable tool to adjust the standard potential by considering the concentration of reactants and products.
For example, in our exercise, you can find the standard potential for a reaction by using the relationship with the equilibrium constant \(K\) as shown:

• \(\mathscr{E}^{\circ} = \frac{-0.0592}{n} \log K \)

This helps students to relate chemical equilibrium with electrochemical potentials and to apply these concepts effectively to various chemical reactions.

Standard reduction potential

The standard reduction potential is a key concept in understanding how substances behave in electrochemical cells. It is the inherent potential of a half-reaction, measured in volts, under standard conditions, where all substances are at 1 M concentration, gases are at 1 atm pressure, and the temperature is typically set at 25°C.
Reduction potential represents the tendency of a chemical species to gain electrons and be reduced.
Substances with high positive values are more likely to gain electrons and thus act as oxidizing agents. Conversely, a negative standard reduction potential indicates a substance is prone to lose electrons, acting as a reducing agent.In the context of the given exercise, we use the standard reduction potential of silver ions \(\text{Ag}^+\), which informs us about its ability to be reduced to silver metal \(\text{Ag}(s)\). The given half-reaction being analyzed involves the combination of this standard reduction potential with the solubility product constant to calculate the potential under specific conditions. Understanding and applying the standard reduction potential is crucial for calculating the overall electrode potential in electrochemical reactions.

Solubility product constant

The solubility product constant, \(K_{sp}\), is a fundamental concept in chemistry used to describe the equilibrium between a solid and its ions in a solution. \(K_{sp}\) signifies the extent to which a solid can dissolve in water, thereby indicating its solubility.
A small \(K_{sp}\) value suggests that the compound is poorly soluble, whereas a larger \(K_{sp}\) value demonstrates higher solubility. For instance, silver iodide \(\text{AgI}\) is known for its low solubility, reflected in its \(K_{sp}\) value.In the given half-reaction, the solubility product constant is used in conjunction with the standard reduction potential of \(\text{Ag}^+\) to determine the standard potential for the specific reaction:

• The equation for \(\text{AgI}(s) \rightarrow \text{Ag}^+(aq) + \text{I}^-(aq)\) involves \(K_{sp}\).
• This connects back to using the Nernst equation to find the electrode potential, where \(K_{sp}\) helps calculate the equilibrium constant \(K\).

Ultimately, comprehending \(K_{sp}\) allows students to predict and calculate solubility-driven reactions and their effect on electrochemical processes effectively.