Problem 95 - Chemistry: Principles and Reactions

Question

Consider a voltaic cell in which the following reaction occurs. $$ \mathrm{Zn}(s)+\mathrm{Sn}^{2+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Sn}(s) $$ (a) Calculate \(E^{\circ}\) for the cell. (b) When the cell operates, what happens to the concentration of \(\mathrm{Zn}^{2+}\) ? The concentration of \(\mathrm{Sn}^{2+}\) ? (c) When the cell voltage drops to zero, what is the ratio of the concentration of \(\mathrm{Zn}^{2+}\) to that of \(\mathrm{Sn}^{2+}\) ? (d) If the concentration of both cations is \(1.0 \mathrm{M}\) originally, what are the concentrations when the voltage drops to zero?

Step-by-Step Solution

Answer

In summary, the process for solving this exercise involved the following steps: (a) The E° for the cell was calculated as 0.62V, based on the standard reduction potentials of the half-reactions. (b) As the cell operates, the concentration of Zn²⁺ increases while the concentration of Sn²⁺ decreases. (c) The ratio of the concentrations of Zn²⁺ to Sn²⁺ when the voltage drops to zero was found using the Nernst equation: [Zn²⁺]/[Sn²⁺] = e^((0.62V * 2F) / RT). (d) The equilibrium concentrations of Sn²⁺ and Zn²⁺ can be determined by solving the equation numerically. The resulting concentrations will depend on the given temperature and initial concentrations, which were not provided in the question.

Step 1: Find the standard reduction potentials

To calculate the E° for the given cell, you'll first need the standard reduction potentials for the half-reactions: Zn (s) → Zn²⁺(aq) + 2e⁻ (oxidation) Sn²⁺(aq) + 2e⁻ → Sn (s) (reduction) You can find these values in a table of standard reduction potentials. The standard reduction potentials are: E°(Zn²⁺/Zn) = -0.76 V (a negative value denotes oxidation) E°(Sn²⁺/Sn) = -0.14 V (a negative value denotes reduction)

Step 2: Calculate E° for the cell

Now that you have the standard reduction potentials for both half-reactions, you can calculate the E° for the cell by adding the two potentials together. E°cell = E°(reduction) - E°(oxidation) E°cell = (-0.14V) - (-0.76V) E°cell = 0.62V (a) The E° for the cell is 0.62V.

Step 3: Determine changes in concentration

As the cell operates, Zn (s) is oxidized and Zn²⁺ (aq) is produced, so the concentration of Zn²⁺ increases. Concurrently, Sn²⁺(aq) gains electrons and is reduced to Sn(s), so the concentration of Sn²⁺ decreases. (b) Concentration of Zn²⁺ increases; concentration of Sn²⁺ decreases.

Step 4: Calculate the ratio of concentrations when voltage drops to zero

When the cell voltage drops to zero, the reaction reaches equilibrium. You will use the Nernst equation to determine the ratio of concentrations: ΔE = E° - (RT/nF) lnQ Here, ΔE is the cell voltage, E° is the standard cell potential, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred (2 in this case), F is the Faraday constant, and Q is the reaction quotient. When the cell voltage drops to zero: 0 = 0.62V - (RT/2F) ln([Zn²⁺]/[Sn²⁺]) You can now rearrange the equation to find the ratio of concentrations: ln([Zn²⁺]/[Sn²⁺]) = (0.62V * 2F) / RT (c) The ratio of the concentration of Zn²⁺ to that of Sn²⁺: [Zn²⁺]/[Sn²⁺] = e^((0.62V * 2F) / RT)

Step 5: Calculate the concentrations when the voltage drops to zero

Given that the initial concentrations of both cations (Zn²⁺ and Sn²⁺) were 1.0M: [Zn²⁺]/[Sn²⁺] = x / (1 - x) Where x is the concentration of Sn²⁺ at equilibrium and (1 - x) is the concentration of Zn²⁺ at equilibrium. You can now substitute the ratio of concentrations previously calculated: x / (1 - x) = e^((0.62V * 2F) / RT) Now, solve for x and (1 - x) numerically to find the equilibrium concentrations of Sn²⁺ and Zn²⁺: (d) The concentrations of Zn²⁺ and Sn²⁺ when the voltage drops to zero are obtained by solving the above equation numerically.

Key Concepts

Standard Reduction Potential

In a voltaic cell, reactions occur in two separate half-cells, each featuring its own half-reaction. The standard reduction potential is a key concept in understanding these processes. It refers to the tendency of a substance to gain electrons, measured in volts under standard conditions (25°C, 1M concentration, and 1 atm pressure). Each half-cell has a standard reduction potential, which helps in determining the direction of electron flow. For instance, in the exercise provided:

• The zinc half-reaction ( Zn (s) → Zn²⁺(aq) + 2e⁻ ) has a standard reduction potential of -0.76 V, denoting oxidation.
• The tin half-reaction ( Sn²⁺(aq) + 2e⁻ → Sn (s) ) has a standard reduction potential of -0.14 V, indicating reduction.

The more positive the standard reduction potential, the greater the substance's tendency to be reduced. This driving force determines which half-reaction will proceed as a reduction and which as oxidation.

Cell Voltage

In a voltaic cell, the cell voltage is the difference in potential between the two half-cells. It represents the cell’s power to perform electrical work when discharging. Calculating the cell voltage involves using the standard reduction potentials of the half-reactions involved:

• The overall cell potential ( E°_{cell} ) is found by subtracting the oxidation potential from the reduction potential.
• It can be written as E°_{cell} = E°_{reduction} - E°_{oxidation} . In our example, this translates to E°_{cell} = -0.14V - (-0.76V) = 0.62V .

A positive cell voltage indicates a spontaneous reaction, suggesting that the cell can energetically drive the reaction forward and produce an electric current. The calculated voltage (0.62 V) shows a functioning voltaic cell at standard conditions.

Nernst Equation

The Nernst equation is used to calculate the cell potential under non-standard conditions by accounting for concentrations of the reactants and products. It is an important tool for assessing how cell voltage changes as the reaction proceeds toward equilibrium:

• The Nernst equation is given by ΔE = E° - (RT/nF) lnQ , where ΔE is the cell voltage, E° is the standard cell potential, R is the gas constant, T is temperature in Kelvin, n is the number of moles of electrons transferred, and F is the Faraday constant.
• Q is the reaction quotient, represented by the ratio of the concentrations of products to reactants, each raised to the power of their stoichiometric coefficients.

When the cell voltage reaches zero, the equation demonstrates the relationship between concentration changes and voltage change, balancing the driving forces of the reaction.

Reaction Equilibrium

In a voltaic cell, the reaction equilibrium occurs when the forward and reverse reactions proceed at the same rate, resulting in no net change in reactant or product concentration. This state is achieved when the cell voltage drops to zero:

• At equilibrium, the reaction quotient Q equals the equilibrium constant K , signifying that the concentrations of reactants and products have stabilized.
• In our exercise, with a cell voltage of zero, the Nernst equation comes into play to establish the equilibrium ratios. We use it to determine the ratio of concentrations ( [Zn²⁺]/[Sn²⁺] ) by solving it for ln([Zn²⁺]/[Sn²⁺]) = (0.62V * 2F) / RT .
• In equilibrium, both the concentration of oxidized zinc increases and reduced tin decreases until these forces are equally opposed by their respective reaction tendencies.

Understanding equilibrium in voltaic cells is crucial in designing efficient batteries and predicting how they'll behave over time.