Question

A common shorthand way to represent a voltaic cell is $$ \text {anode} | \text {anode solution} | | \text {cathode solution} | \text {cathode} $$ A double vertical line represents a salt bridge or a porous barrier. A single vertical line represents a change in phase, such as from solid to solution. (a) Write the half-reactions and overall cell reaction represented by Fel Fe \(^{2+} \| \operatorname{Ag}^{+} | A g;\) calculate the standard cell emf using data in Appendix E. (b) Write the half-reactions and overall cell reaction represented by Zn \(\left|Z \mathrm{n}^{2+}\right| \mathrm{H}^{+} | \mathrm{H}_{2} ;\) calculate the standard cell emf using data in Appendix E and use Pt for the hydrogen electrode. (c) Using the notation just described, represent a cell based on the following reaction: $$ \begin{aligned} \mathrm{ClO}_{3}^{-}(a q)+3 \mathrm{Cu}(s)+6 \mathrm{H}^{+}(a q) & \\ \longrightarrow & \mathrm{Cl}^{-}(a q)+3 \mathrm{Cu}^{2+}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$ Pt is used as an inert electrode in contact with the ClO \(_{3}^{-}\) and \(\mathrm{Cl}^{-} .\) Calculate the standard cell emf given: \(\mathrm{ClO}_{3}^{-}(a q)+\) \(6 \mathrm{H}^{+}(a q)+6 \mathrm{e}^{-} \longrightarrow \mathrm{Cl}^{-}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l); E^{\circ}=1.45 \mathrm{V}\).

Short answer

For the given voltaic cells: (a) The overall cell reaction is \(Fe(s) + 2Ag^{+}(aq) \rightarrow Fe^{2+}(aq) + 2Ag(s)\) and the standard cell emf is \(1.24V\). (b) The overall cell reaction is \(Zn(s) + 2H^{+}(aq) \rightarrow Zn^{2+}(aq) + H_{2}(g)\) and the standard cell emf is \(0.76V\). (c) The cell is represented as \(Cu|Cu^{2+}||ClO_{3}^{-}, Cl^{-}|Pt\), the overall cell reaction is \(ClO_{3}^{-}(aq) + 3Cu(s) + 6H^{+}(aq) \rightarrow Cl^{-}(aq) + 3Cu^{2+}(aq) + 3H_{2}O(l)\), and the standard cell emf is \(1.11V\).

Step-by-step solution

Write half-reactions

In this voltaic cell, the two half-reactions are the reduction of Ag+ and the oxidation of Fe: Oxidation half-reaction (anode): \(Fe(s) \rightarrow Fe^{2+}(aq) + 2e^-\) Reduction half-reaction (cathode): \(Ag^{+}(aq) + e^- \rightarrow Ag(s)\)

Write overall cell reaction

Now, we will balance the overall cell reaction by combining the two half-reactions, making sure the electrons are balanced. Overall cell reaction: \(Fe(s) + 2Ag^{+}(aq) \rightarrow Fe^{2+}(aq) + 2Ag(s)\)

Calculate the standard cell emf

Using the standard electrode potentials from Appendix E, we can calculate the emf of the cell: \(E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}\) The values for the given half-reactions are: \(E_{Fe^{2+}/Fe}^{\circ} = -0.44 V\) \(E_{Ag^{+}/Ag}^{\circ} = 0.80 V\) Thus, emf of the cell: \(E_{cell}^{\circ} = 0.80V - (-0.44V) = 1.24V\) (b)

Write half-reactions

In this voltaic cell, the two half-reactions are the reduction of H+ and the oxidation of Zn: Oxidation half-reaction (anode): \(Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-\) Reduction half-reaction (cathode): \(2H^{+}(aq) + 2e^- \rightarrow H_{2}(g)\)

Write overall cell reaction

Now, combine the two half-reactions, making sure the electrons are balanced: Overall cell reaction: \(Zn(s) + 2H^{+}(aq) \rightarrow Zn^{2+}(aq) + H_{2}(g)\)

Calculate the standard cell emf

Using the standard electrode potentials from Appendix E: \(E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}\) The values for the given half-reactions are: \(E_{Zn^{2+}/Zn}^{\circ} = -0.76 V\) \(E_{H^{+}/H_{2}}^{\circ} = 0.00 V\) Thus, emf of the cell: \(E_{cell}^{\circ} = 0.00V - (-0.76V) = 0.76V\) (c)

Represent the cell

Using the provided reaction, the cell may be represented as: \(Cu|Cu^{2+}||ClO_{3}^{-}, Cl^{-}|Pt\)

Write half-reactions

In this voltaic cell, the two half-reactions are: Oxidation half-reaction (anode): \(3Cu(s) \rightarrow 3Cu^{2+}(aq) + 6e^-\) Reduction half-reaction (cathode): \(ClO_{3}^{-}(aq) + 6H^{+}(aq) + 6e^- \rightarrow Cl^{-}(aq) + 3H_{2}O(l)\)

Write overall cell reaction

Combining these half-reactions, we obtain the overall cell reaction: \(ClO_{3}^{-}(aq) + 3Cu(s) + 6H^{+}(aq) \rightarrow Cl^{-}(aq) + 3Cu^{2+}(aq) + 3H_{2}O(l)\)

Calculate the standard cell emf

Now, we need to calculate the standard emf for the oxidation half-reaction. Given: \(E_{ClO_{3}^{-}/Cl^{-}}^{\circ} = 1.45V\) We need to find the standard emf for the oxidation half-reaction as follows: \(E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}\) Rearranging, we can find the standard emf for the anode: \(E_{anode}^{\circ} = E_{cathode}^{\circ} - E_{cell}^{\circ}\) Using the given values: \(E_{anode}^{\circ} = 1.45V - E_{Cu^{2+}/Cu}^{\circ}\) From Appendix E, we know: \(E_{Cu^{2+}/Cu}^{\circ} = 0.34 V\) Thus, the standard cell emf: \(E_{cell}^{\circ} = 1.45V - 0.34V = 1.11V\)

Key concepts

Understanding Half-Reactions in Voltaic Cells

In a voltaic cell, half-reactions show what happens at each electrode. They break down the complex processes into two simpler reactions: oxidation and reduction.
Oxidation occurs at the anode, where a substance loses electrons. For instance, in the reaction \(Fe(s) \rightarrow Fe^{2+}(aq) + 2e^-\), iron loses electrons to form iron ions.
Reduction takes place at the cathode, where a substance gains electrons. For example, in \(Ag^{+}(aq) + e^- \rightarrow Ag(s)\), silver ions gain electrons and become solid silver.
By analyzing each half-reaction, we can easily understand how electrons move in the whole cell.

Calculating Standard Cell EMF

The standard cell emf (electromotive force) measures the voltage or electrical potential difference between the two electrodes. It is calculated using the equation:

• \(E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}\)

You look up the standard electrode potentials, \(E^{\circ}\), for each half-reaction in a reference table. These values tell you how easily a substance can be reduced or oxidized.
For example, if \(E_{cathode}^{\circ} = 0.80 \text{ V}\) and \(E_{anode}^{\circ} = -0.44 \text{ V}\), the standard cell emf is \(1.24 \text{ V}\). This tells you that the cell can drive a current of 1.24 volts under standard conditions.

Understanding Electrode Potentials

Electrode potentials provide insight into how likely a substance will gain or lose electrons. They are expressed as \(E^{\circ}\) values and serve as a guide for predicting reactions.
A positive \(E^{\circ}\) value indicates a good oxidizing agent that readily gains electrons, while a negative \(E^{\circ}\) value suggests it easily loses electrons and acts as a reducing agent.
For instance, silver ion \(Ag^{+}/Ag\) has a potential of 0.80 V, whereas zinc \(Zn^{2+}/Zn\) is \(-0.76 \text{ V}\). Silver is more favorable for reduction compared to zinc, making it a strong oxidizing agent.

Analyzing Electrode Reactions

Electrode reactions are specific to where oxidation or reduction occurs in a cell, providing the pathway for electron flow.
The oxidation reaction at the anode releases electrons, such as \(Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-\).
Conversely, the reduction reaction at the cathode consumes these electrons, like \(2H^{+}(aq) + 2e^- \rightarrow H_{2}(g)\).
These reactions ensure that electrons flow from the anode to the cathode through an external circuit, allowing the cell to do electrical work.