Question

A common shorthand way to represent a voltaic cell is anode|anode solution $\Vert$ cathode solution $\mid$ 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 $\mathrm{Fe}|{\mathrm{Fe}}^{2+}\Vert {\mathrm{Ag}}^{+}|\mathrm{Ag};$ sketch the cell. (b) Write the half-reactions and overall cell reaction represented by $\mathrm{Zn}|{\mathrm{Zn}}^{2+}\Vert {\mathrm{H}}^{+}|{\mathrm{H}}_2$; sketch the cell. (c) Using the notation just described, represent a cell based on the following reaction: $$\begin{array}{rl}{\mathrm{ClO}}_3^{-}(aq)+3\mathrm{Cu}(s)+6{\mathrm{H}}^{+}(aq)& {\mathrm{Cl}}^{-}(aq)+3{\mathrm{Cu}}^{2+}(aq)+3{\mathrm{H}}_2\mathrm{O}(l)\end{array}$$ $\mathrm{Pt}$ is used as an inert electrode in contact with the ${\mathrm{ClO}}_3^{-}$ and ${\mathrm{Cl}}^{-}$. Sketch the cell.

Short answer

(a) The half-reactions and overall cell reaction for the $Fe\left|F{e}^{2+}||A{g}^{+}\right|Ag$ cell are: Anode: $Fe(s)\to F{e}^{2+}(aq)+2e^{-}$ Cathode: $A{g}^{+}(aq)+e^{-}\to Ag(s)$ Overall reaction: $Fe(s)+2A{g}^{+}(aq)\to F{e}^{2+}(aq)+2Ag(s)$ The cell consists of an iron electrode in the Fe2+ solution on the left side and a silver electrode in the Ag+ solution on the right side, separated by a salt bridge or porous barrier. (b) The half-reactions and overall cell reaction for the $Zn\left|Z{n}^{2+}||H^{+}\right|H_2$ cell are: Anode: $Zn(s)\to Z{n}^{2+}(aq)+2e^{-}$ Cathode: $2H^{+}(aq)+2e^{-}\to H_2(g)$ Overall reaction: $Zn(s)+2H^{+}(aq)\to Z{n}^{2+}(aq)+H_2(g)$ The cell consists of a zinc electrode in the Zn2+ solution on the left side and an electrode in the H+ solution with hydrogen gas evolved on the right side, separated by a salt bridge or porous barrier. (c) The half-reactions and overall cell reaction for the cell with Pt as an inert electrode are: Anode: $2Cl{O}_3^{-}(aq)\to 2C{l}^{-}(aq)+3H_{2}O(l)+12e^{-}$ Cathode: $3Cu(s)+6H^{+}(aq)+12e^{-}\to 3C{u}^{2+}(aq)+6H_{2}O(l)$ Overall reaction: $2Cl{O}_3^{-}(aq)+3Cu(s)+6H^{+}(aq)\to 2C{l}^{-}(aq)+3C{u}^{2+}(aq)+3H_{2}O(l)$ The cell consists of a platinum electrode in a solution containing both $Cl{O}_3^{-}$ and $C{l}^{-}$ ions on the left side and a copper electrode in the Cu2+ solution on the right side, separated by a salt bridge or porous barrier.

Step-by-step solution

Write the half-reactions

For the anode (the left side), we have: $$Fe(s)\to F{e}^{2+}(aq)+2e^{-}$$ For the cathode (the right side), we have: $$A{g}^{+}(aq)+e^{-}\to Ag(s)$$

Write the overall cell reaction

Combine the half-reactions with the correct number of electrons and write the overall cell reaction: $$Fe(s)+2A{g}^{+}(aq)\to F{e}^{2+}(aq)+2Ag(s)$$

Sketch the cell

On the left side, there would be an iron electrode immersed in Fe2+ solution. On the right side, there would be a silver electrode immersed in Ag+ solution. The two solutions are separated by a salt bridge or porous barrier. #Part b: Zn|Zn2+ || H+|H2 cell#

Write the half-reactions

For the anode (the left side), we have: $$Zn(s)\to Z{n}^{2+}(aq)+2e^{-}$$ For the cathode (the right side), we have: $$2H^{+}(aq)+2e^{-}\to H_2(g)$$

Write the overall cell reaction

Combine the half-reactions with the correct number of electrons and write the overall cell reaction: $$Zn(s)+2H^{+}(aq)\to Z{n}^{2+}(aq)+H_2(g)$$

Sketch the cell

On the left side, there would be a zinc electrode immersed in Zn2+ solution. On the right side, there would be an electrode immersed in H+ solution with hydrogen gas evolved. The two solutions are separated by a salt bridge or porous barrier. #Part c: Cell based on given reaction and Pt as an inert electrode#

Write the half-reactions

For the anode (the left side), we have: $$2Cl{O}_3^{-}(aq)\to 2C{l}^{-}(aq)+3H_{2}O(l)+12e^{-}$$ For the cathode (the right side), we have: $$3Cu(s)+6H^{+}(aq)+12e^{-}\to 3C{u}^{2+}(aq)+6H_{2}O(l)$$

Write the overall cell reaction

Since both half-reactions have the same number of electrons, the overall cell reaction is: $$2Cl{O}_3^{-}(aq)+3Cu(s)+6H^{+}(aq)\to 2C{l}^{-}(aq)+3C{u}^{2+}(aq)+3H_{2}O(l)$$

Sketch the cell

On the left side, there would be a platinum electrode immersed in a solution containing both $Cl{O}_3^{-}$ and $C{l}^{-}$ ions. On the right side, there would be a copper electrode immersed in the Cu2+ solution. The two solutions are separated by a salt bridge or porous barrier.

Key concepts

Electrochemical Cells

Electrochemical cells are the foundation of batteries and various forms of corrosion protection. They enable the conversion of chemical energy into electrical energy through redox reactions. An electrochemical cell consists of two electrodes, the anode and the cathode, immersed in electrolyte solutions. These electrodes are connected externally by a wire and internally by a salt bridge or a porous barrier. It's crucial to understand that in a spontaneous reaction in voltaic or galvanic cells, the anode is the site of oxidation (loss of electrons), while the cathode is where reduction (gain of electrons) occurs.

To enhance understanding, it’s beneficial to visualize an electrochemical cell in action. Imagine a setup with two beakers, each containing an electrode and an electrolyte solution. The electrodes are connected with a wire that allows electrons to flow, and the salt bridge completes the circuit by allowing ions to move and balance charge changes during the redox process. The cell notation as seen in the textbook exercise, using vertical lines to represent phase changes and double lines for salt bridges, provides a compact way of showing how such a cell is organized without needing to draw complex diagrams.

Half-Reactions

Half-reactions are valuable tools for deciphering the specifics of the redox process in an electrochemical cell. They facilitate the understanding of exactly what occurs at each electrode. A half-reaction equation represents either the oxidation or reduction process separately, and it helps to balance the number of electrons lost in the oxidation half with those gained in the reduction half.

In illustrating half-reactions, it's helpful to explicitly show the electron exchange, as in the textbook solution. For example, the anode half-reaction for the iron electrode could be visualized as iron metal atoms at the surface of the electrode losing electrons and becoming iron ions, which then dissolve into the solution. In the cathode half-reaction, silver ions in the solution gain electrons from the cathode and plate out as metallic silver. Simplifying and separating these reactions allows students to concentrate on one at a time, thereby improving their comprehension of the overall redox process.

Galvanic Cells

Galvanic cells, also known as voltaic cells, are a type of electrochemical cell which generate electricity by means of a spontaneous redox reaction. Understanding a galvanic cell's purpose in powering devices can clarify its practical applications. Students may not always grasp the significance of standard cell notation, such as the one provided in the textbook exercise. Thus, likening these cells to everyday batteries might make the concept more relatable.

For instance, devices like remote controls or flashlights are powered by galvanic cells where chemical reactions occur within the battery to produce electrical energy. The textbook example delineated an iron-silver galvanic cell, which can be imagined as a miniature battery with Fe and Ag electrodes. Each electrode undergoes its respective half-reaction, and together they drive the flow of electrons through an external circuit to provide power, just like common AA or AAA batteries.

Redox Reactions

The interplay of reduction and oxidation reactions, collectively termed 'redox' reactions, is pivotal in the operation of electrochemical cells. A redox reaction involves the transfer of electrons between two substances. Reduction refers to the gain of electrons, while oxidation is the loss of electrons. Comprehending the art of balancing redox reactions, where the number of electrons lost is equivalent to the number gained, is crucial for solving textbook exercises correctly.

For clearer understanding, imagine that oxidation is like giving away apples, and reduction is like receiving apples. In a redox process, the total number of apples given away must match the number of apples received; otherwise, the reaction wouldn't be balanced. In the context of our previous examples, iron giving up electrons (losing apples) and silver gaining them (gaining apples) need to be even. By teaching the student to visualize these reactions as exchanging items, the balancing act becomes more tangible and less abstract.