Problem 26 - Chemistry in 30 days

Question

In a cell that utilizes the reaction \(\mathrm{Zn}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{~g})\) addition of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) to cathode compartment will (a) Lower the \(E\) and shift equilibrium to the left (b) Increase the \(E\) and shift equilibrium to the left (c) Increase the \(E\) and shift equilibrium to the right (d) Lower the \(E\) and shift equilibrium to the right

Step-by-Step Solution

Answer

(c) Increase the \(E\) and shift equilibrium to the right.

Step 1: Understand the Cell Reaction

The given reaction is \( \mathrm{Zn} + 2 \mathrm{H}^{+} \rightarrow \mathrm{Zn}^{2+} + \mathrm{H}_{2} \). In this equation, the zinc metal goes from a zero oxidation state to \( \mathrm{Zn}^{2+} \), while \( \mathrm{H}^{+} \) gets reduced to \( \mathrm{H}_{2} \) gas. The oxidation occurs at the anode, and reduction occurs at the cathode.

Step 2: Explore the Effect of Adding \( \mathrm{H}_{2} \mathrm{SO}_{4} \)

Adding \( \mathrm{H}_{2} \mathrm{SO}_{4} \) to the cathode compartment increases the concentration of \( \mathrm{H}^{+} \) ions because sulfuric acid dissociates to give \( \mathrm{H}^{+} \) ions. This influences the reaction at the cathode.

Step 3: Predict the Effect on Cell Potential \( E \)

The Nernst equation states \[ E = E^0 - \frac{RT}{nF} \ln Q \], where \( Q \) is the reaction quotient. Increasing \( \mathrm{H}^{+} \) concentration decreases \( Q \), which increases \( E \), the cell potential.

Step 4: Apply Le Chatelier's Principle

By increasing \( \mathrm{H}^{+} \), Le Chatelier's principle states that the equilibrium will shift to the right to counter the change. More hydrogen ions support the forward reaction (more production of \( \mathrm{H}_{2} \) gas).

Step 5: Match the Effects to the Given Options

From Steps 3 and 4, adding \( \mathrm{H}_{2} \mathrm{SO}_{4} \) increases the cell potential \( E \) and shifts the equilibrium to the right. Therefore, the correct answer is (c).

Key Concepts

Nernst Equation

The Nernst Equation is a vital tool in electrochemistry for understanding how the cell potential of a reaction changes with concentration. It allows us to see how the potential varies from its standard state. The equation is given by:\[ E = E^0 - \frac{RT}{nF} \ln Q \]Where:

• E is the cell potential at non-standard conditions.
• E⁰ is the standard cell potential.
• R is the universal gas constant.
• T is the temperature in Kelvin.
• n is the number of moles of electrons transferred in the reaction.
• F is Faraday's constant.
• Q is the reaction quotient, representing the ratio of the products to reactants, raised to their stoichiometric coefficients.

In this context, adding sulfuric acid to the cathode compartment increases the hydrogen ion concentration, affecting the value of Q.This results in a higher E as the logarithm of a decreased Q becomes less negative. The cell thus experiences an increased potential as predicted by the Nernst Equation.

Le Chatelier's Principle

Le Chatelier's Principle helps chemists predict how changing conditions influence a chemical system at equilibrium. It is based on the idea that if an external change is applied to a system at equilibrium, the system will counteract the change to re-establish equilibrium. When applied to our electrochemical reaction:Adding \(\mathrm{H}_2\mathrm{SO}_4\) boosts the concentration of \(\mathrm{H}^+\), which acts in the same capacity as a product change. Le Chatelier's Principle suggests that to decrease the effect of additional \(\mathrm{H}^+\) ions, the equilibrium will shift to the right. This shift means more \(\mathrm{H}_2\) gas will be produced, aiding the cell reaction to proceed more efficiently. By understanding and applying Le Chatelier's Principle, we can predict how equilibrium adjusts and impacts the overall electrochemical cell behavior.

Cell Potential (E)

Cell potential, or electromotive force (EMF), is the measure of the potential difference between two half-cells in an electrochemical cell. It tells us how much voltage is pushed through the cell.The potential of a galvanic or voltaic cell shifts based on various factors:

• Concentration of reactants and products
• Temperature
• Pressure, if gases are involved

The experiment demonstrates that changing concentrations, like adding \(\mathrm{H}_2\mathrm{SO}_4\), directly affects the cell's potential. As more hydrogen ions are present, the reaction quotient Q in the Nernst equation decreases, raising the potential E. Being aware of these factors is crucial for understanding the effectiveness and efficiency of battery-based devices.

Oxidation-Reduction Reactions

Oxidation-reduction (redox) reactions are at the core of electrochemistry. They involve the transfer of electrons between substances, causing one element to be oxidized and another to be reduced.In our context, zinc is oxidized while hydrogen ions (\(\mathrm{H}^+\)) are reduced:

• Oxidation: \(\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+} + 2\mathrm{e}^-\)
• Reduction: \(2\mathrm{H}^+ + 2\mathrm{e}^- \rightarrow \mathrm{H}_2 \)

Understanding the movement of electrons allows us to predict reaction behaviors in cells like batteries.Redox reactions not only involve electron flow but also create potential differences that indicate how electrons will move. This understanding is crucial for developing efficient energy systems and optimizing chemical processes, making redox reactions fundamental to both academic studies and practical applications.