Problem 31 - Chemistry

Question

Calculate the standard potential of the cell consisting of the \(\mathrm{Zn} / \mathrm{Zn}^{2+}\) half-cell and the SHE. What will the emf of the cell be if \(\left[\mathrm{Zn}^{2+}\right]=0.45 \mathrm{M}, P_{\mathrm{H}_{2}}=2.0 \mathrm{~atm},\) and \(\left[\mathrm{H}^{+}\right]=1.8 M ?\)

Step-by-Step Solution

Answer

The standard potential is 0.76 V, and the emf is approximately 0.806 V.

Step 1: Write the Half-Reaction Equations

The half-reaction for zinc is \( \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \) with a standard reduction potential of \( E^{\circ}_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \text{ V} \). The standard hydrogen electrode (SHE) is represented by \( 2\text{H}^+ + 2e^- \rightarrow \text{H}_2 \) with \( E^{\circ}_{\text{SHE}} = 0 \text{ V} \).

Step 2: Calculate the Standard Cell Potential

The standard potential of the cell \( E^{\circ}_{\text{cell}} \) is calculated as the difference between the potentials of the cathode (SHE) and anode (zinc). Thus, \( E^{\circ}_{\text{cell}} = E^{\circ}_{\text{SHE}} - E^{\circ}_{\text{Zn}^{2+}/\text{Zn}} = 0 - (-0.76) = 0.76 \text{ V} \).

Step 3: Use the Nernst Equation

The Nernst equation is: \[ E = E^{\circ}_{\text{cell}} - \frac{RT}{nF} \ln Q \] where \( R \) is the gas constant (8.314 J/mol K), \( T \) is the temperature in Kelvin (assume 298 K), \( n \) is the number of moles of electrons (2 moles), \( F \) is Faraday's constant (96485 C/mol). \( Q \) is the reaction quotient given by \( \frac{[\text{Zn}^{2+}]}{[\text{H}^+]^2} \times \frac{1}{P_{\text{H}_2}} \).

Step 4: Calculate the Reaction Quotient Q

Substitute the given values into the expression for \( Q \): \[ Q = \frac{0.45}{1.8^2} \times \frac{1}{2} = \frac{0.45}{3.24} \times \frac{1}{2} = \frac{0.45}{6.48} \approx 0.0694. \]

Step 5: Substitute Values into Nernst Equation

Substitute the values into the Nernst equation: \[ E = 0.76 - \frac{8.314 \times 298}{2 \times 96485} \ln 0.0694. \] Calculating the logarithm: \( \ln 0.0694 \approx -2.6697 \). Thus, \[ E = 0.76 + \frac{8.314 \times 298}{2 \times 96485} \times 2.6697 \approx 0.806 \text{ V}. \]

Step 6: Final Answer

The emf of the cell is approximately \( 0.806 \text{ V} \) when \([\text{Zn}^{2+}] = 0.45 \text{ M}, P_{\text{H}_2} = 2.0 \text{ atm},\) and \([\text{H}^+] = 1.8 \text{ M} \).

Key Concepts

Standard Cell Potential

The standard cell potential, denoted as \( E^{\circ}_{\text{cell}} \), is an essential concept in electrochemistry. It represents the potential difference between two electrodes in a galvanic cell when all species are in their standard states. This means that solids and liquids are in their pure forms, gases have a partial pressure of 1 atm, and solutions have concentrations of 1 M.
Determining this potential involves comparing the standard reduction potentials of two half-cells making up the cell. The higher potential is typically the cathode (reduction occurs here), while the lower one is the anode (oxidation occurs here). The standard potential of the cell is the difference between these two values:
\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \]
In our example, we have a galvanic cell with \( \text{Zn}^{2+}/\text{Zn} \) as the anode with a standard reduction potential of \(-0.76 \text{ V}\) and the standard hydrogen electrode (SHE) as the cathode with a potential of \(0 \text{ V}\). The calculated \( E^{\circ}_{\text{cell}} \) is \(0.76 \text{ V}\). This value tells us how much work can be done with this cell when starting from standard conditions.

Nernst Equation

The Nernst equation allows us to calculate the cell potential under non-standard conditions. It is particularly useful when concentrations or pressures deviate from their standard state. The equation is a modification of the standard cell potential expression and takes into account the reaction quotient, temperature, and the number of electrons transferred.
The Nernst equation is represented as:
\[ E = E^{\circ}_{\text{cell}} - \frac{RT}{nF} \ln Q \]
Here,

• \( E \) is the cell potential under non-standard conditions,
• \( E^{\circ}_{\text{cell}} \) is the standard cell potential,
• \( R \) is the ideal gas constant (\(8.314 \text{ J/mol K}\)),
• \( T \) is the temperature in Kelvin (often assumed as \(298 \text{ K}\)),
• \( n \) is the number of moles of electrons transferred (2 moles for our cell),
• \( F \) is Faraday's constant (\(96485 \text{ C/mol}\)), and
• \( Q \) is the reaction quotient.

By substituting the given concentrations and pressures in the reaction and solving, we adapted the potential, arriving at \( E = 0.806 \text{ V} \), indicating that even under non-standard conditions, the cell still produces potential.

Reaction Quotient

The reaction quotient, \( Q \), is a measure of the relative amounts of products and reactants at any point in a reversible reaction. It's similar to the equilibrium constant but applies to conditions that are not necessarily equilibrium. This quotient is necessary for calculations of cell potential using the Nernst equation.
For the electrochemical cell under study, the reaction quotient \( Q \) can be determined using the formula:
\[ Q = \frac{[\text{Zn}^{2+}]}{[\text{H}^+]^2} \times \frac{1}{P_{\text{H}_2}} \]
In this calculation, we plugged in the specific values given in the exercise:

• \([\text{Zn}^{2+}] = 0.45 \text{ M}\),
• \([\text{H}^+] = 1.8 \text{ M}\), and
• \( P_{\text{H}_2} = 2.0 \text{ atm}\).

This gave us a \( Q \) value of approximately \(0.0694\). This value reveals the relative position of the reaction concerning its equilibrium and informs us how far the system is from equilibrium, directly impacting the cell's potential.

Half-cell Reactions

Half-cell reactions are the backbone of electrochemistry, representing the oxidation and reduction processes that occur in each separate half of an electrochemical cell. Each half-cell consists of a metal-electrode submerged in a solution containing a metal ion:
- The **anode** reaction involves oxidation and is shown as: \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \)
- The **cathode**, often the SHE in many reactions, involves reduction: \( 2\text{H}^+ + 2e^- \rightarrow \text{H}_2 \).
These reactions denote electron flow, with the anode being where electrons are lost (oxidized), and the cathode being where they are gained (reduced).
The potential of these individual half-reactions can be determined through standard reduction potentials. The difference between these two potentials directly impacts the cell's overall voltage and drives the operation of the electrochemical cell, as shown by their contribution to the standard cell potential a discussed previously.