Question

In each of the following examples, sketch a voltaic cell that uses the given reaction. Label the anode and cathode; indicate the direction of electron flow; write a balanced equation for the cell reaction; and calculate \(E_{\mathrm{cell}}^{\circ}\). (a) \(\mathrm{Cu}(\mathrm{s})+\mathrm{Fe}^{3+}(\mathrm{aq}) \longrightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Fe}^{2+}(\mathrm{aq})\) (b) \(\mathrm{Pb}^{2+}(\mathrm{aq})\) is displaced from solution by \(\mathrm{Al}(\mathrm{s})\) (c) \(\mathrm{Cl}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{g})+\mathrm{H}^{+}(\mathrm{aq})\) (d) \(\mathrm{Zn}(\mathrm{s})+\mathrm{H}^{+}+\mathrm{NO}_{3}^{-} \longrightarrow \mathrm{Zn}^{2+}+\) \(\mathrm{H}_{2} \mathrm{O}(1)+\mathrm{NO}(\mathrm{g})\)

Short answer

In the reaction, the copper electrode acts as the anode and is being oxidized whereas the iron ion is the cathode and is being reduced. Electrons flow from the copper electrode to the iron ions. The balanced equation for the cell reaction is \(\mathrm{Cu(s) + Fe^{3+}(aq) \rightarrow Cu^{2+}(aq) + Fe^{2+}(aq)}\). The standard electromotive force for the cell reaction is +1.11V.

Step-by-step solution

Identify the Half-Reactions

For the provided reaction, the half-reactions that lay the groundwork for the voltaic cell are: (1) Oxidation - \(\mathrm{Cu(s)} \rightarrow \mathrm{Cu^{2+}}(\mathrm{aq}) + 2e^-\) and (2) Reduction - \(\mathrm{Fe^{3+}}(\mathrm{aq}) + e^- \rightarrow \mathrm{Fe^{2+}}(\mathrm{aq})\).

Determine the Anode and Cathode

The anode is where oxidation occurs. Therefore, in this case, the copper electrode is the anode. The cathode is where the reduction takes place, so the iron ion is being reduced at the cathode.

Indicate the Direction of Electron Flow

In a voltaic cell, electrons always flow from anode to cathode. Therefore, in this case, electrons would be flowing from the copper electrode to the iron ions.

Write a Balanced Equation for the Cell Reaction

The balanced equation for the cell reaction is given in the problem as: \(\mathrm{Cu(s) + Fe^{3+}(aq) \rightarrow Cu^{2+}(aq) + Fe^{2+}(aq)}\).

Calculate \(E_{\mathrm{cell}}^{\circ}\)

To calculate \(E_{\mathrm{cell}}^{\circ}\), the standard reduction potentials of the half-reactions are needed. The standard reduction potentials are as follows: for Fe3+ to Fe2+, \(E^{\circ}_{cathode}=+0.77 V\) and for Cu to Cu2+, \(E^{\circ}_{anode}=+0.34 V\). However, as Cu to Cu2+ is acting as an anode (oxidation), its \(E^{\circ}\) will change the sign, thus \(E^{\circ}_{anode}=-0.34 V\). The standard electromotive force \(E_{\mathrm{cell}}^{\circ}\) for the reaction is calculated by: \(E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = +0.77 V - (-0.34 V) = +1.11 V\).

Key concepts

Electrode Reactions

In a voltaic cell, chemical reactions occur at electrodes and are split into two distinct processes: oxidation and reduction. Each electrode hosts a different half-reaction. The anode is the site of oxidation, where the loss of electrons occurs. At the cathode, reduction takes place as electrons are gained.

For example, in the reaction \(\text{Cu(s)} + \text{Fe}^{3+}(\text{aq}) \rightarrow \text{Cu}^{2+}(\text{aq}) + \text{Fe}^{2+}(\text{aq})\), oxidation happens at the copper electrode. Copper metal (\(\text{Cu}\)) loses electrons to form copper ions (\(\text{Cu}^{2+}\)). On the other hand, reduction takes place at the iron electrode, as iron ions (\(\text{Fe}^{3+}\)) gain electrons to form iron ions of a lower oxidation state (\(\text{Fe}^{2+}\)).

By separating these half-reactions and connecting them via an external wire, we can harness the flow of electrons to produce electricity. This separation is the foundational principle of a voltaic cell.

Standard Cell Potential

The standard cell potential is an important aspect of a voltaic cell, describing the voltage difference between two half-cells when they are connected. To find it, we look at the standard reduction potentials (\(E^{\circ}\)) of each half-reaction, which are tabulated values showing how prone certain ions are to being reduced under standard conditions: 1 molar concentrations, 1 atmosphere pressure, and 25°C.

The cell potential, \(E_{\text{cell}}^{\circ}\), is calculated using:

• The reduction potential for the cathode (where reduction occurs).
• The oxidation potential derived from the anode, which is the negative of the reduction potential since the reversed reaction happens.

In our example, the reaction involves the reduction of \(\text{Fe}^{3+}\) to \(\text{Fe}^{2+}\) with a potential of \(+0.77 \text{ V}\), and the oxidation of \(\text{Cu}\) to \(\text{Cu}^{2+}\) with a potential of \(+0.34 \text{ V}\). Therefore, the standard cell potential is:
\[E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}\]\[= 0.77 \text{ V}-(-0.34 \text{ V}) = 1.11 \text{ V}\].

A positive \(E_{\text{cell}}^{\circ}\) indicates a spontaneous reaction—meaning it can do work and generate electricity.

Electron Flow

Electron flow is the movement of electrons from one electrode to another through an external circuit. It's this flow that allows a voltaic cell to generate electricity. Electrons are driven from the anode, where oxidation releases electrons, toward the cathode, where reduction accepts electrons.

In a voltaic cell designed from our sample reaction, electrons move through the external wire from the copper electrode (anode) to the iron electrode (cathode).

Understanding electron flow is crucial. It helps us grasp how energy transformations happen in electrochemical cells. In addition, electron flow direction is always from high to low potential, effectively translating chemical energy into electrical energy we can use.

• Remember: Electrons always travel from the anode to the cathode.
• They help balance charge differences created by the oxidation and reduction processes.

Oxidation and Reduction

Oxidation and reduction are two complementary processes in a chemical reaction. Their interplay is essential for the operation of a voltaic cell. In simple terms, oxidation involves the loss of electrons, while reduction involves gaining electrons.

Consider our example reaction once again: Copper is oxidized as it loses electrons to transform into copper ions:\[\text{Cu(s)} \rightarrow \text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \].
Simultaneously, the iron ions undergo reduction as they gain electrons to transform into a less oxidized state:\[\text{Fe}^{3+}(\text{aq}) + \text{e}^- \rightarrow \text{Fe}^{2+}(\text{aq})\].

Balancing these two reactions is crucial to ensure charge conservation and accurate representation of the electron movements.

• Oxidation is always paired with reduction. Together, they are known as redox reactions.
• The anode and cathode are connected by a wire and a salt bridge, ensuring both charge balance and electron flow.

Understanding redox reactions is fundamental to both electrochemistry and everyday chemical processes.