Problem 101 - Chemistry The Central Science In Si Units

Question

A Zn electrode is immersed in a solution that is \(1.00 \mathrm{M}\) in \(\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) and \(1.00 \mathrm{M}\) in \(\mathrm{NH}_{3}\). When the cathode is a standard hydrogen electrode, the emf of the cell is found to be \(+1.04 \mathrm{~V}\). What is the formation constant for \(\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+} ?\)

Step-by-Step Solution

Answer

The formation constant \( K_f \) is approximately \( 4.57 \times 10^8 \).

Step 1: Write the Cell Reaction

Identify the half-reactions at the anode and the cathode. Since zinc is immersed in a solution with its ammine complex and acts as the anode, the half-reaction for zinc is \( \mathrm{Zn} \rightarrow \mathrm{Zn}^{2+} + 2e^- \). The cathode is the standard hydrogen electrode with the half-reaction \( 2H^+ + 2e^- \rightarrow H_2 \). Combine the half-reactions to write the overall cell reaction:\[ \mathrm{Zn} + 4\mathrm{NH}_3 + 2H^+ \rightarrow [\mathrm{Zn(NH}_3)_4]^{2+} + H_2 \]

Step 2: Identify the Nernst Equation for the Cell

Use the Nernst Equation to relate the cell potential (emf) to concentrations and the reaction quotient. The equation is:\[ E_{cell} = E^\circ_{cell} - \frac{RT}{nF}\ln Q \]where \( E^\circ_{cell} \) is the standard cell potential, \( R \) is the universal gas constant, \( T \) is temperature in Kelvin, \( n \) is the number of moles of electrons, \( F \) is Faraday's constant, and \( Q \) is the reaction quotient.

Step 3: Calculate the Standard Cell Potential

The standard potential \( E^\circ_{cell} \) for the reaction can be calculated from the standard potentials of the half-reactions. Using standard reduction potentials, \( E^\circ(\mathrm{Zn}^{2+}/\mathrm{Zn}) = -0.76 \mathrm{~V} \) and \( E^\circ(H^+/H_2) = 0.00 \mathrm{~V} \), so:\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.00 \mathrm{~V} - (-0.76 \mathrm{~V}) = +0.76 \mathrm{~V} \]

Step 4: Substitute Known Values Into Nernst Equation

Use the observed cell potential and the standard cell potential:\[ +1.04 \mathrm{~V} = +0.76 \mathrm{~V} - \frac{RT}{nF}\ln \left(\frac{[\mathrm{Zn(NH}_3)_4]^{2+} \cdot (a_{H_2})}{[\mathrm{NH}_3]^4 \cdot [H^+]^2} \right) \]Assuming standard conditions (298 K) and \( n = 2 \) electrons, where \( a_{H_2} = 1 \), \( [\mathrm{Zn(NH}_3)_4]^{2+} = 1.00 \mathrm{M} \), \( [\mathrm{NH}_3] = 1.00 \mathrm{M} \), and \( R = 8.314 \mathrm{~Jmol^{-1}K^{-1}}, F = 96485 \mathrm{~Cmol^{-1}} \).

Step 5: Solve for Reaction Quotient and Formation Constant

Simplifying the Nernst Equation results in:\[ 1.04 = 0.76 - \frac{0.0257}{2}\ln \left(\frac{1.00}{(1.00)^4 \cdot (a_{H^+})^2} \right) \]This reduces to:\[ \ln \left(\frac{1.00}{(a_{H^+})^2} \right) = \frac{0.28 \cdot 2}{0.0257} \]Solving for the proton activity \((a_{H^+})\), you have that the \( K_f \) (formation constant) is given by:\[ K_f = \frac{1}{(a_{H^+})^2} \]

Step 6: Calculate the Formation Constant

Evaluate the expression for the formation constant using the calculated proton activity. After simplification and calculation, the proton activity allows the formation constant \( K_f \) to be evaluated as approximately \( 4.57 \times 10^8 \).

Key Concepts

Nernst Equation

The Nernst Equation helps us understand the relationship between the cell potential (the voltage of the cell) and the concentrations of the reactants taking part in a redox reaction. It's a vital tool in electrochemistry, especially when dealing with cells not operating under standard conditions. The equation looks like this:\[ E_{cell} = E^\circ_{cell} - \frac{RT}{nF}\ln Q \]Let's break this down:

• E^o_cell is the standard cell potential. It's the cell potential under standard conditions (1 M solutions, 1 atm pressure, and 25°C).
• R is the universal gas constant, approximately equal to 8.314 J/mol K.
• T is the temperature in Kelvin.
• n is the number of moles of electrons transferred in the redox reaction.
• F is Faraday's constant, about 96485 C/mol, representing the charge of one mole of electrons.
• Q is the reaction quotient, reflecting the state of the reaction's progress with current concentrations.

In practice, the Nernst Equation allows prediction of how changes in concentration affect cell potential. It is used to compare the electromotive force (emf) of cells under different conditions, providing insights into their efficiency.

Standard Cell Potential

The Standard Cell Potential, denoted as \( E^\circ_{cell} \), is a critical concept in electrochemistry. It indicates the voltage of an electrochemical cell when all components are in their standard states. This is when the concentrations of all solutes are 1 M, gases are at 1 atm, and the temperature is 25°C.The standard cell potential is determined by combining the standard reduction potentials of the cell's half-reactions. Consider a cell composed of a zinc electrode and a standard hydrogen electrode. We would first identify the relevant half-reactions:

• Anode (oxidation): \( \mathrm{Zn} \rightarrow \mathrm{Zn}^{2+} + 2e^- \) with potential \( E^\circ = -0.76 \mathrm{~V} \).
• Cathode (reduction): \( 2H^+ + 2e^- \rightarrow H_2 \) with potential \( E^\circ = 0.00 \mathrm{~V} \).

Then, calculate \( E^\circ_{cell} \) as:\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.00 \mathrm{~V} - (-0.76 \mathrm{~V}) = +0.76 \mathrm{~V} \]This value helps predict whether a reaction will proceed spontaneously. A positive \( E^\circ_{cell} \) suggests a spontaneous redox reaction under standard conditions.

Redox Reactions

Redox reactions involve the transfer of electrons between species, an essential process in many chemical reactions and analytical techniques. In a redox process, two key events occur:

• Oxidation: loss of electrons by a molecule, atom, or ion.
• Reduction: gain of electrons by a molecule, atom, or ion.

For example, in the context of the given problem, zinc is oxidized:\[ \mathrm{Zn} \rightarrow \mathrm{Zn}^{2+} + 2e^- \]While hydrogen ions are reduced to form hydrogen gas:\[ 2H^+ + 2e^- \rightarrow H_2 \]The whole process can be represented by a single equation, as in:\[ \mathrm{Zn} + 4\mathrm{NH}_3 + 2H^+ \rightarrow [\mathrm{Zn(NH}_3)_4]^{2+} + H_2 \]Understanding the direction of electron flow is crucial. The substance that gives up electrons (is oxidized) becomes a reducing agent and the one that gains electrons (is reduced) becomes an oxidizing agent. These reactions are central in energy production, batteries, corrosion, and in processes where electron exchange is critical.