Problem 114 - Chemistry: The Central Science in SI Units, Global Edition

Question

A voltaic cell is based on \(\mathrm{Cu}^{2+}(a q) / \mathrm{Cu}(s)\) and \(\mathrm{Br}_{2}(l) /\) \(\mathrm{Br}^{-}(a q)\) half-cells. (a) What is the standard emf of the cell? (b) Which reaction occurs at the cathode and which at the anode of the cell? (c) Use \(S^{\circ}\) values in Appendix \(\mathrm{C}\) and the relationship between cell potential and free-energy change to predict whether the standard cell potential increases or decreases when the temperature is raised above \(25^{\circ} \mathrm{C}\). (Thestandard entropy of \(\mathrm{Cu}^{2+}(a q)\) is \(\left.S^{\circ}=-99.6 \mathrm{~J} / \mathrm{K}\right)\)

Step-by-Step Solution

Answer

The standard emf of the cell is 1.41 V, with the Cu2+ + 2e- -> Cu reaction occurring at the cathode and the 2Br- -> Br2 + 2e- reaction occurring at the anode. The effect of temperature on the standard cell potential is positive, meaning that the standard cell potential increases when the temperature is raised above 25 °C.

Step 1: Find the Standard Reduction Potentials

Consult a table of standard reduction potentials to find the potentials for the cu 2 /cu and Br2/Br- half-cells. Cu2+ + 2e- -> Cu, E° = +0.34 V (Reduction) Br2 + 2e- -> 2Br-, E° = +1.07 V (Reduction)

Step 2: Determine the Anode and Cathode Reactions

In a voltaic cell, the reduction with the highest potential occurs at the cathode and the one with the lowest potential occurs at the anode. Cathode: Cu2+ + 2e- -> Cu Anode: 2Br- -> Br2 + 2e- (Reverse the reduction reaction)

Step 3: Calculate the Standard Emf

Next, we will calculate the standard emf (E°) of the cell by subtracting the standard potential of the anode reaction from the standard potential of the cathode reaction: E° = E°(cathode) - E°(anode) = 0.34 V - (-1.07 V) = 1.41 V Standard emf of the cell is 1.41 V.

Step 4: Calculate the Standard Free Energy Change

We will use the relationship between cell potential and free-energy change to find the standard free energy change (ΔG°) for the reaction. ΔG° = -nFE° n = 2 (number of electrons in each half-reaction) F = 96485 C/mol (Faraday's constant) ΔG° = -(2)(96485 C/mol)(1.41 V) = -271632 J/mol

Step 5: Calculate the Entropy Change

Calculate the standard entropy change (ΔS°) for the cell reaction using the standard entropy (S°) values. ΔS° = S°(products) - S°(reactants) Entropy of products: S°(Cu) + S°(Br2) = 33.2 J/mol K + 152.3 J/mol K Entropy of reactants: S°(Cu2+) + S°(2Br-) = -99.6 J/mol K + (2 * (10.4 J/mol K)) ΔS° = (33.2 J/mol K + 152.3 J/mol K) - (-99.6 J/mol K + 20.8 J/mol K) = 263.3 J/mol K

Step 6: Determine the Effect of Temperature

Now, we will use the Gibbs-Helmholtz equation to relate the standard free energy change, standard entropy change, and temperature: ΔG° = ΔH° - TΔS° ΔH° = ΔG° + TΔS° We can then find the effect of temperature (dT) on ΔG° and E°: d(ΔG°)/dT = d(ΔH°)/dT - ΔS° d(E°)/dT = -d(ΔG°)/dT/nF = d(ΔH°)/dT/nF - ΔS°/nF Since ΔS° > 0, the term (ΔS°/nF) will be positive and d(E°)/dT will be positive. This means that the standard cell potential increases when the temperature is raised above 25 °C.

Key Concepts

Standard EMF

The standard electromotive force (emf) of a voltaic cell is crucial because it indicates the energy available from the cell's reactions under standard conditions. The standard emf is determined by the difference between the reduction potentials of the cathode and the anode. For the cell in question, the half-reactions are:

• Reduction at the cathode: \[ \mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu}, \quad E^\circ = +0.34 \text{ V} \]
• Reduction at the anode:\[ \mathrm{Br}_2 + 2e^- \rightarrow 2\mathrm{Br}^-, \quad E^\circ = +1.07 \text{ V} \]

To find the emf, we subtract the anode potential from the cathode potential. This can be illustrated as follows:\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.34 \text{ V} - (-1.07 \text{ V}) = 1.41 \text{ V} \]This positive value of 1.41 V indicates a spontaneous reaction under standard conditions.

Cathode and Anode Reactions

Understanding cathode and anode reactions helps determine the flow of electrons in a voltaic cell. In a voltaic cell, reduction occurs at the cathode while oxidation occurs at the anode. The reaction at the cathode with the highest standard potential tends to occur naturally because it is energetically favorable. For this cell:

• At the cathode, the copper (\(\mathrm{Cu}^{2+}\)) ions gain electrons to form solid copper (\(\mathrm{Cu}\)):\[ \mathrm{Cu}^{2+}(aq) + 2e^- \rightarrow \mathrm{Cu}(s) \]
• At the anode, bromide ions (\(2\mathrm{Br}^-\)) are oxidized to produce bromine (\(\mathrm{Br}_2\)):\[ 2\mathrm{Br}^- \rightarrow \mathrm{Br}_2 + 2e^- \]

This arrangement ensures that electrons flow from the anode (where oxidation occurs) to the cathode (where reduction takes place), providing electrical energy.

Free Energy Change

The free energy change (\(\Delta G^\circ\)) is a vital thermodynamic quantity that determines whether a reaction can occur spontaneously. It is directly related to the standard emf through the equation:\[ \Delta G^\circ = -nFE^\circ \]where:

• \(n\) is the number of moles of electrons transferred,\(n = 2\) here.
• \(F\) is Faraday's constant, approximately 96485 C/mol.
• \(E^\circ\) is the standard emf of the cell (1.41 V for this cell).

Substituting these values in:\[ \Delta G^\circ = -(2)(96485 \text{ C/mol})(1.41 \text{ V}) = -271632 \text{ J/mol} \]The negative \(\Delta G^\circ\) indicates that the cell reaction is spontaneous under standard conditions.

Gibbs-Helmholtz Equation

The Gibbs-Helmholtz equation is an important relationship in thermodynamics, connecting free energy, enthalpy, and entropy changes. It helps us understand how the standard emf of a cell can vary with temperature:\[ \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \]where:

• \(\Delta H^\circ\) is the standard enthalpy change.
• \(T\) is the temperature in Kelvin.
• \(\Delta S^\circ\) is the standard entropy change.

For this cell, the entropy change was calculated as \(\Delta S^\circ = 263.3 \text{ J/mol K}\). Since \(\Delta S^\circ > 0\), it implies that the standard cell potential \(E^\circ\) increases as temperature rises. The term \(\Delta S^\circ/nF\) is positive, meaning higher temperature favors greater reaction spontaneity, enhancing the cell's capability to do work.