Problem 25 - Chemistry and Chemical Reactivity

Question

Calculate the potential delivered by a voltaic cell using the following reaction if all dissolved species are \(2.5 \times 10^{-2} \mathrm{M}\) and the pressure of \(\mathrm{H}_{2}\) is 1.0 bar. $$\begin{aligned} \mathrm{Zn}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(\ell)+2 \mathrm{OH}^{-}(\mathrm{aq}) & \rightarrow \\ &\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g}) \end{aligned}$$

Step-by-Step Solution

Answer

The cell potential is approximately \(-0.04 \, \mathrm{V}\).

Step 1: Identify the Half-Reactions

Write down the oxidation and reduction half-reactions involved in the given overall reaction. For this cell, the oxidation half-reaction is \( \mathrm{Zn}(\mathrm{s}) + 4\mathrm{OH}^- \rightarrow \left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-} + 2e^- \) and the reduction half-reaction is \( 2\mathrm{H}_2O + 2e^- \rightarrow \mathrm{H}_2(g) + 2\mathrm{OH}^- \).

Step 2: Find Standard Electrode Potentials

Use a standard reduction potential table to find the standard reduction potentials for each half-reaction. The standard potential \( E^0 \) for \( \mathrm{Zn}(\mathrm{s}) \rightarrow \mathrm{Zn^{2+}} \) is typically \( -0.76 \, \mathrm{V} \), and for \( \mathrm{H}_2O + 2e^- \rightarrow \mathrm{H}_2(g) + 2\mathrm{OH}^- \) it is often \( -0.83 \, \mathrm{V} \) under alkaline conditions.

Step 3: Calculate Standard Cell Potential

Calculate the standard cell potential, \( E^0_{cell} \), using the formula \( E^0_{cell} = E^0_{cathode} - E^0_{anode} \). Here, \( E^0_{cell} = -0.83 - (-0.76) = -0.07 \, \mathrm{V} \).

Step 4: Apply Nernst Equation

Since we know the concentrations and partial pressure, we apply the Nernst equation to find the actual cell potential: \[ E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln{Q} \]. Given \( T = 298 \, \mathrm{K} \), \( n = 2 \), \( R = 8.314 \, \mathrm{J/mol}\cdot K \), and \( F = 96485 \, \mathrm{C/mol} \).

Step 5: Calculate Reaction Quotient \(Q\)

The reaction quotient \( Q \) is determined using concentrations and pressures: \[ Q = \frac{[\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}]}{[\mathrm{OH}^-]^2[\mathrm{H}_2]} \]. \([\mathrm{OH}^-] = 2.5 \times 10^{-2} \, \mathrm{M} \), \([\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}] = 2.5 \times 10^{-2} \, \mathrm{M} \), and for pressure of \( \mathrm{H}_2(g) = 1.0 \, \mathrm{bar} \). Thus, \( Q = \frac{2.5 \times 10^{-2}}{(2.5 \times 10^{-2})^2 \times 1.0} \).

Step 6: Solve for \(E_{cell}\)

Substitute the values in the Nernst equation: \[ E_{cell} = -0.07 - \frac{8.314 \times 298}{2 \times 96485} \ln{\left(\frac{2.5 \times 10^{-2}}{(2.5 \times 10^{-2})^2 \times 1.0}\right)}, \] which simplifies to \( E_{cell} = -0.07 + 0.03 \approx -0.04 \, \mathrm{V} \).

Key Concepts

Voltaic cell calculation

In electrochemistry, a voltaic or galvanic cell is a device that converts chemical energy into electrical energy through spontaneous redox reactions. To calculate the voltage of a voltaic cell, you must first identify the half-reactions occurring at the anode (oxidation) and cathode (reduction). These half-reactions provide the basis for understanding the electron transfer responsible for generating electric current.

Starting with our reaction, we separate it into two half-reactions:

• Oxidation at the anode: \( \mathrm{Zn}(\mathrm{s}) + 4\mathrm{OH}^- \rightarrow \left[\mathrm{Zn}(\mathrm{OH})_4\right]^{2-} + 2e^- \)
• Reduction at the cathode: \( 2\mathrm{H}_2O + 2e^- \rightarrow \mathrm{H}_2(g) + 2\mathrm{OH}^- \)

Once the half-reactions are established, we can use standard electrode potentials (often found in tables) to predict the cell voltage. These potentials help us understand how the flow of electrons corresponds to potential differences—the key function of a voltaic cell.

The potential difference (voltage) calculation is a multi-step process, starting with finding the standard cell potential \(E^0_{cell}\) using the equation:\[E^0_{cell} = E^0_{cathode} - E^0_{anode}\]where the standard potentials are taken from known data.

Nernst equation applications

The Nernst equation is a fundamental relation in electrochemistry that connects the cell potential to the concentration of reactants and products, thereby enabling one to determine the cell potential under non-standard conditions. This equation adjusts the standard electrode potential by accounting for the influence of concentration changes, temperature, and partial pressures.

For the given exercise, once you've determined the standard cell potential, \(E^0_{cell}\), it is time to use the Nernst equation:\[E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln{Q}\]where:

• \(E_{cell}\) is the actual cell potential
• \(R\) is the universal gas constant \(8.314 \, \mathrm{J/mol}\, K\)
• \(T\) is the temperature in Kelvin
• \(n\) is the number of moles of electrons transferred
• \(F\) is the Faraday constant \(96485 \, \mathrm{C/mol}\)
• \(Q\) is the reaction quotient

The Nernst equation considers the logarithm of the reaction quotient, \(Q\), which in turn is calculated using the concentrations and partial pressures of the reaction species. In this scenario, \(Q\) was calculated based on the concentrations and pressure provided, allowing for the computation of the actual potential \(E_{cell}\) derived as \(-0.04 \, \mathrm{V}\).

Standard electrode potentials

Standard electrode potentials, represented as \(E^0\), are the measures of the potential of a given electrode reaction under standard conditions, which include all solutes at a concentration of 1 M, gases at a pressure of 1 bar, and a specified temperature (usually \(25^\circ C\) or \(298\, K\)). These values are crucial for deducing the trends in redox reactions and determining the feasibility and direction of chemical changes.

In the provided exercise, the standard reduction potential values, \(E^0\), were used to identify the tendency of half-reactions to gain or lose electrons. The more positive the electrode potential, the greater the substance's affinity to gain electrons and act as a good oxidizing agent. Conversely, a more negative potential suggests a stronger reducing capability.

With zinc's standard reduction potential near \(-0.76 \, \mathrm{V}\) and water's reduction to \(\mathrm{H}_2\) being around \(-0.83 \, \mathrm{V}\), these two metals' standard electrode potentials helped us arrive at a standard cell potential of \(-0.07 \, \mathrm{V}\) for the complete cell in this electrochemical setup. This foundational understanding is crucial for applying the Nernst equation and making further predictions about cell behavior under varied conditions.