Question

Using the standard reduction potentials listed in Table 19.1 and the Handbook of Chemistry and Physics, show that the following reaction is favorable under standardstate conditions: $$ 2 \mathrm{Ag}(s)+\mathrm{Pt}^{2+}(a q) \longrightarrow 2 \mathrm{Ag}^{+}(a q)+\operatorname{Pt}(s) $$ What is the equilibrium constant of this reaction at \(25^{\circ} \mathrm{C} ?\)

Short answer

The reaction is favorable with an extremely large equilibrium constant.

Step-by-step solution

Identify Half-Reactions and Potentials

First, identify the half-reactions involved. The oxidation half-reaction is \( 2\ \mathrm{Ag}(s) \to 2\ \mathrm{Ag}^{+}(aq) + 2e^- \) with a potential \( E^\circ = -0.80\ \text{V} \). The reduction half-reaction is \( \mathrm{Pt}^{2+}(aq) + 2e^- \to \mathrm{Pt}(s) \) with a potential \( E^\circ = +1.20\ \text{V} \).

Calculate Standard Cell Potential

The standard cell potential \( E^\circ_{\text{cell}} \) is calculated using the equation:\[E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 1.20\ \text{V} - (-0.80\ \text{V}) = 2.00\ \text{V}\]

Determine Reaction Favorability

Since the standard cell potential \( E^\circ_{\text{cell}} = 2.00\ \text{V} \) is positive, the reaction is favorable under standard conditions.

Use Nernst Equation for Equilibrium Constant

The equilibrium constant \( K \) can be determined using the relation:\[\Delta G^\circ = -nFE^\circ_{\text{cell}} = -RT\ln K\]Rearrange to find:\[\ln K = \frac{nFE^\circ_{\text{cell}}}{RT}\]Where \( n = 2 \), \( F = 96485 \ \text{C/mol} \), \( R = 8.314 \ \text{J/mol}\,\text{K} \), and \( T = 298 \ \text{K} \).

Calculate Equilibrium Constant

Substitute values:\[\ln K = \frac{2 \times 96485 \times 2.00}{8.314 \times 298}\]\[\ln K = 155.8\]Calculate \( K \):\[K = e^{155.8}\]

Final Evaluation of Equilibrium Constant

The value for \( K \) is extremely large (much greater than 1), indicating a very favorable reaction.

Key concepts

Standard reduction potentials

Standard reduction potentials are crucial in electrochemistry for predicting how electrons will flow in a redox reaction. Each half-reaction in an electrochemical cell has a standard reduction potential, denoted as \(E^\circ\). These potentials are measured in volts and referenced to the standard hydrogen electrode, which is assigned a potential of 0.00 V.

Here are a few key points about standard reduction potentials:

• High positive \(E^\circ\) values indicate a strong tendency to gain electrons and undergo reduction.
• Conversely, low or negative \(E^\circ\) values suggest a greater likelihood to lose electrons and undergo oxidation.
• By comparing reduction potentials, you can determine which substances will be reduced or oxidized in a redox reaction.

In this particular reaction, the \(E^\circ\) values for silver and platinum are used to calculate the standard cell potential, providing insight into which pathway the electrons prefer.

Equilibrium constant

The equilibrium constant, denoted as \(K\), quantitatively describes the balance of products and reactants in a chemical reaction at equilibrium. For redox reactions, \(K\) provides valuable insights into the reaction's favorability under standard conditions.

Certain important facts about the equilibrium constant include:

• If \(K > 1\), the reaction is product-favored, indicating a predominance of products over reactants.
• If \(K < 1\), the reaction is reactant-favored, meaning reactants are more prevalent than products.
• The larger \(K\) is, the more the reaction proceeds toward completion.

In this problem, calculating \(K\) for the given reaction shows its tendency to dramatically favor the formation of products, since the value is very large. This confirms the reaction is favorable under standard-state conditions at \(25^{\circ} \mathrm{C}\).

Nernst equation

The Nernst equation is a fundamental tool in electrochemistry used to relate the standard cell potential to the cell potential under non-standard conditions. It also helps in determining the equilibrium constant \(K\).

The Nernst equation is as follows:\[E = E^\circ - \frac{RT}{nF}\ln Q\]Where:

• \(E\) is the cell potential at non-standard conditions
• \(E^\circ\) is the standard cell potential
• \(R\) is the universal gas constant \(8.314\ \text{J/mol}\,\text{K}\)
• \(T\) is the temperature in Kelvin
• \(n\) is the number of electrons exchanged in the reaction
• \(F\) is Faraday's constant \(96485\ \text{C/mol}\)
• \(Q\) is the reaction quotient

In practice, the Nernst equation can calculate how \(E\) changes as the reaction progresses and even determine \(K\) by setting \(E = 0\) when the reaction is at equilibrium.

Cell potential

Cell potential, represented as \(E_{\text{cell}}\), is the driving force behind the flow of electrons in an electrochemical cell. It quantifies the voltage (or electric potential difference) between two electrodes.

Key aspects of cell potential include:

• \(E^\circ_{\text{cell}}\) is the standard cell potential, calculated by subtracting the standard potential of the anode from the cathode: \[E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\]
• A positive \(E^\circ_{\text{cell}}\) indicates a spontaneous reaction under standard conditions.
• The cell potential also determines the maximum electric work attainable from an electrochemical reaction.

For the exercise in question, the calculated \(E^\circ_{\text{cell}} = 2.00\ \text{V}\) implies this reaction is highly favorable, allowing it to serve as a powerful energy source.

Reaction favorability

Reaction favorability in electrochemistry is often inferred from the standard cell potential. This concept is crucial for predicting whether a chemical reaction will spontaneously occur in a given direction.

To determine reaction favorability:

• Calculate the standard cell potential \(E^\circ_{\text{cell}}\). A positive \(E^\circ_{\text{cell}}\) suggests a spontaneous reaction.
• Examine the value of the equilibrium constant \(K\). If \(K\) is significantly greater than 1, the reaction strongly favors product formation.

In this exercise, both a positive \(E^\circ_{\text{cell}}\) and a large \(K\) confirm the reaction proceeds favorably from reactants to products under standard-state conditions. This means the reaction does not require external energy input to occur.