Problem 41 - Chemistry: An Atoms-Focused Approach

Question

Starting with standard potentials listed in Appendix 6 calculate the values of \(E_{\text {cell }}^{\circ}\) and \(\Delta G^{\circ}\) of the following reactions. a. \(\mathrm{Cu}(s)+\mathrm{Sn}^{2+}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{Sn}(s)\) b. \(\mathrm{Zn}(s)+\mathrm{Ni}^{2+}(a q) \rightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Ni}(s)\)

Step-by-Step Solution

Answer

Question: Calculate the standard cell potential and standard Gibbs free energy for the following redox reactions: a. Cu(s) + Sn^2+(aq) → Cu^2+(aq) + Sn(s) b. Zn(s) + Ni^2+(aq) → Zn^2+(aq) + Ni(s) Answer: a. The standard cell potential, \(E_{\text {cell }}^{\circ} = -0.477\,\text{V}\), and the standard Gibbs free energy, \(\Delta G^{\circ} = 92020\,\text{J/mol}\). b. The standard cell potential, \(E_{\text {cell }}^{\circ} = 0.503\,\text{V}\), and the standard Gibbs free energy, \(\Delta G^{\circ} = -97106\,\text{J/mol}\).

Step 1: Identify the half-reactions

Identify the half-reactions for each reaction: a. Oxidation: \(\mathrm{Cu}(s) \rightarrow \mathrm{Cu}^{2+}(a q)+2 e^{-}\) Reduction: \(\mathrm{Sn}^{2+}(a q) +2 e^{-} \rightarrow \mathrm{Sn}(s)\) b. Oxidation: \(\mathrm{Zn}(s) \rightarrow \mathrm{Zn}^{2+}(a q)+2 e^{-}\) Reduction: \(\mathrm{Ni}^{2+}(a q)+2 e^{-} \rightarrow \mathrm{Ni}(s)\)

Step 2: Locate the standard reduction potentials

Look up the standard reduction potentials (\(E^\circ\)) for the half-reactions in Appendix 6. For part a, we have: \(E_{\text {Cu}^{2+}/\text {Cu}}^{\circ} = 0.337\,\text{V}\) (reduction potential for Cu) \(E_{\text {Sn}^{2+}/\text {Sn}}^{\circ} = -0.14\,\text{V}\) (reduction potential for Sn) For part b, we have: \(E_{\text {Zn}^{2+}/\text {Zn}}^{\circ} = -0.763\,\text{V}\) (reduction potential for Zn) \(E_{\text {Ni}^{2+}/\text {Ni}}^{\circ} = -0.26\,\text{V}\) (reduction potential for Ni)

Step 3: Calculate the standard cell potentials

Now we can calculate \(E_{\text {cell }}^{\circ}\) for each reaction using the formula \(E_{\text {cell }}^{\circ} = E_{\text {cathode }}^{\circ} - E_{\text {anode }}^{\circ}\). For part a: \(E_{\text {cell }}^{\circ} = E_{\text {Sn}^{2+}/\text {Sn}}^{\circ} - E_{\text {Cu}^{2+}/\text {Cu}}^{\circ} = -0.14\,\text{V} - 0.337\,\text{V} = -0.477\,\text{V}\) For part b: \(E_{\text {cell }}^{\circ} = E_{\text {Ni}^{2+}/\text {Ni}}^{\circ} - E_{\text {Zn}^{2+}/\text {Zn}}^{\circ} = -0.26\,\text{V} - (-0.763\,\text{V}) = 0.503\,\text{V}\)

Step 4: Calculate the standard Gibbs free energies

With the standard cell potentials known, we can use the formula \(\Delta G^{\circ} = -nFE_{\text {cell }}^{\circ}\) to determine the standard Gibbs free energy for each reaction. Faraday's constant, \(F\), equals \(96485\, \text {C/mol}\), and the number of moles of electrons transferred (\(n\)) is 2 for both reactions. For part a: \(\Delta G^{\circ} = -(2)(96485\,\text {C/mol})(-0.477\, \text{V}) = 92020\,\text{J/mol}\) For part b: \(\Delta G^{\circ} = -(2)(96485\,\text {C/mol})(0.503\,\text {V}) = -97106\,\text{J/mol}\) To summarize the results: a. \(E_{\text {cell }}^{\circ} = -0.477\,\text{V}\) and \(\Delta G^{\circ} = 92020\,\text{J/mol}\) b. \(E_{\text {cell }}^{\circ} = 0.503\,\text{V}\) and \(\Delta G^{\circ} = -97106\,\text{J/mol}\)

Key Concepts

Standard Reduction Potential

Standard reduction potentials, denoted as \( E^\circ \), are fundamental in electrochemistry and provide a measure of the tendency of a chemical species to gain electrons and be reduced. These values are typically listed in volts (V) and are measured under standard conditions of 25°C, 1 atm pressure, and 1 M concentration for all solutions.

To better understand these potentials, consider a half-reaction involving the gain of electrons which is displayed as: - Reduction of copper: \( \mathrm{Cu}^{2+} (aq) + 2e^- \rightarrow \mathrm{Cu}(s) \)- Corresponding \( E^\circ \) value: 0.337 V
The more positive the \( E^\circ \) value, the greater the species’ tendency to be reduced. This is why copper ions are good at gaining electrons, resulting in the formation of copper metal.

When evaluating reactions like the ones in electrochemical cells, we refer to the standard reduction potentials of both the oxidation and reduction reactions involved. These potentials help us understand which species will undergo oxidation and which will undergo reduction by comparing their respective \( E^\circ \) values.

Cell Potential Calculation

The cell potential, \( E_{\text{cell}}^\circ \), provides a measure of the voltage produced by an electrochemical cell and indicates the work done by the cell per unit charge. Calculating \( E_{\text{cell}}^\circ \) involves determining the difference between the potential of the cathode (reduction site) and the anode (oxidation site).

For any redox reaction in a cell, the formula is:- \( E_{\text{cell}}^\circ = E_{\text{cathode}}^\circ - E_{\text{anode}}^\circ \)- **Cathode**: Site of reduction reaction- **Anode**: Site of oxidation reaction

Here's an example for clarity:- For \( \mathrm{Sn}^{2+}/\mathrm{Sn} \) with \( E^\circ = -0.14 \text{ V} \)- For \( \mathrm{Cu}^{2+}/\mathrm{Cu} \) with \( E^\circ = 0.337 \text{ V} \)- Thus, \( E_{\text{cell}}^\circ = -0.14 \text{ V} - 0.337 \text{ V} = -0.477 \text{ V} \)

Positive values of \( E_{\text{cell}}^\circ \) indicate a spontaneous reaction under standard conditions, revealing a net gain in energy.

Gibbs Free Energy

Gibbs Free Energy is a thermodynamic function that indicates the amount of available energy capable of doing work during a reaction at constant pressure and temperature. It is represented by \( \Delta G^{\circ} \). In electrochemistry, it is directly linked to the cell potential via the equation:- \( \Delta G^{\circ} = -nFE_{\text{cell}}^\circ \)

**Where:**

• \( n \) = Number of moles of electrons exchanged
• \( F \) = Faraday’s constant (96485 C/mol)

Using this formula, you can assess whether a reaction is spontaneous or non-spontaneous:- **Negative \( \Delta G^{\circ} \)**: Spontaneous reaction- **Positive \( \Delta G^{\circ} \)**: Non-spontaneous reaction

For instance, in our previously discussed reactions:
- \( \Delta G^{\circ} = -(2)(96485 \text{ C/mol})(-0.477 \text{ V}) = 92020 \text{ J/mol} \) for the copper-tin reaction, indicating non-spontaneity.
- \( \Delta G^{\circ} = -(2)(96485 \text{ C/mol})(0.503 \text{ V}) = -97106 \text{ J/mol} \) for the zinc-nickel reaction, suggesting spontaneous behavior.
This computation reveals the interplay of chemical potential and electrons to drive electrochemical transformations.