Question

A common shorthand way to represent a voltaic cell is to list its components as follows: anode|anode solution || cathode solution|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 \(\mathrm{Fe}\left|\mathrm{Fe}^{2+} \| \mathrm{Ag}^{+}\right| \mathrm{Ag}\); sketch the cell. (b) Write the half-reactions and overall cell reaction represented by \(\mathrm{Zn}\left|\mathrm{Zn}^{2+} \| \mathrm{H}^{+}\right| \mathrm{H}_{2} ;\) sketch the cell. (c) Using the notation just described, represent a cell based on the following reaction: \(\mathrm{ClO}_{3}^{-}(a q)+3 \mathrm{Cu}(s)+6 \mathrm{H}^{+}(a q)-\rightarrow\) $$ \mathrm{Cl}^{-}(a q)+3 \mathrm{Cu}^{2+}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) $$ Pt is used as an inert electrode in contactwith the \(\mathrm{ClO}_{3}^{-}\) and \(\mathrm{Cl}^{-}\). Sketch the cell.

Short answer

(a) Half-reactions and overall cell reaction for the Fe|Fe²+||Ag⁺|Ag cell: Oxidation half-reaction (anode): \(\mathrm{Fe}(s) \rightarrow \mathrm{Fe}^{2+}(aq) + 2 e^-\) Reduction half-reaction (cathode): \(\mathrm{Ag}^{+}(aq) + e^- \rightarrow \mathrm{Ag}(s)\) Overall cell reaction: \(\mathrm{Fe}(s) + 2 \mathrm{Ag}^{+}(aq) \rightarrow \mathrm{Fe}^{2+}(aq) + 2 \mathrm{Ag}(s)\) (b) Half-reactions and overall cell reaction for the Zn|Zn²+||H⁺|H₂ cell: Oxidation half-reaction (anode): \(\mathrm{Zn}(s) \rightarrow \mathrm{Zn}^{2+}(aq) + 2 e^-\) Reduction half-reaction (cathode): \(2\mathrm{H}^{+}(aq) + 2 e^- \rightarrow \mathrm{H}_{2}(g)\) Overall cell reaction: \(\mathrm{Zn}(s) + 2 \mathrm{H}^{+}(aq) \rightarrow \mathrm{Zn}^{2+}(aq) + \mathrm{H}_{2}(g)\) (c) A cell based on the ClO₃⁻(aq)||H⁺(aq)|H₂O(l)|Cu²+(aq)|Cu reaction is represented by the notation: \(\mathrm{Pt} | \mathrm{ClO}_{3}^{-} , \mathrm{Cl}^{-} || \mathrm{H}^{+} , \mathrm{H}_{2} \mathrm{O}(l) | \mathrm{Cu}^{2+} , \mathrm{Cu}\)

Step-by-step solution

1. Write half-reactions for the cell.

Identify the anode (oxidation) and cathode (reduction) half-reactions. From the cell notation, we know that the anode is Fe (s), and the cathode is Ag⁺ (aq). Write the following half-reactions: Oxidation half-reaction (anode): \(\mathrm{Fe}(s) \rightarrow \mathrm{Fe}^{2+}(aq) + 2 e^-\) Reduction half-reaction (cathode): \(\mathrm{Ag}^{+}(aq) + e^- \rightarrow \mathrm{Ag}(s)\)

2. Write the overall cell reaction.

Combine the half-reactions by balancing the number of electrons. Since the anode generates 2 electrons and the cathode requires 1 electron, multiply the cathode half-reaction by 2 to balance the electrons: Overall cell reaction: \(\mathrm{Fe}(s) + 2 \mathrm{Ag}^{+}(aq) \rightarrow \mathrm{Fe}^{2+}(aq) + 2 \mathrm{Ag}(s)\)

3. Sketch the cell.

Draw the anode (Fe) with its Fe²+ (aq) solution connected by a salt bridge to the cathode solution (Ag⁺ aq) and a cathode (Ag). Label the salt bridge and the electrodes on the sketch. #(b) Zn|Zn²+||H⁺|H₂ Cell#

1. Write half-reactions for the cell.

Identify the anode (oxidation) and cathode (reduction) half-reactions. From the cell notation, we know that the anode is Zn(s) and the cathode is H⁺(aq). Write the following half-reactions: Oxidation half-reaction (anode): \(\mathrm{Zn}(s) \rightarrow \mathrm{Zn}^{2+}(aq) + 2 e^-\) Reduction half-reaction (cathode): \(2\mathrm{H}^{+}(aq) + 2 e^- \rightarrow \mathrm{H}_{2}(g)\)

2. Write the overall cell reaction.

Combine the half-reactions by balancing the number of electrons. In this case, since both half-reactions involve 2 electrons, no further balancing is required: Overall cell reaction: \(\mathrm{Zn}(s) + 2 \mathrm{H}^{+}(aq) \rightarrow \mathrm{Zn}^{2+}(aq) + \mathrm{H}_{2}(g)\)

3. Sketch the cell.

Draw the anode (Zn) with its Zn²+ (aq) solution connected by a salt bridge to the cathode solution (H⁺ aq) and hydrogen gas electrode (H₂). Label the salt bridge and the electrodes on the sketch. #(c) ClO₃⁻(aq)||H⁺(aq)|H₂O(l)|Cu²+(aq)|Cu Cell#

1. Write half-reactions for the cell.

First, we must identify the anode and cathode half-reactions from the overall reaction provided: Oxidation half-reaction (anode): \(2 \mathrm{ClO}_{3}^{-}(aq) \rightarrow 2 \mathrm{Cl}^{-}(aq) + 6 e^-\) Reduction half-reaction (cathode): \( 6 \mathrm{H}^{+}(aq) + 3 \mathrm{Cu}(s) + 6 e^- \rightarrow 3 \mathrm{Cu}^{2+}(aq) + 3 \mathrm{H}_{2} \mathrm{O}(l)\)

2. Write the cell notation.

Now that we have the half-reactions, we can write the cell notation. Since Pt is an inert electrode in contact with ClO₃⁻ and Cl⁻, it will be in contact with both the anode and cathode side: \(\mathrm{Pt} | \mathrm{ClO}_{3}^{-} , \mathrm{Cl}^{-} || \mathrm{H}^{+} , \mathrm{H}_{2} \mathrm{O}(l) | \mathrm{Cu}^{2+} , \mathrm{Cu}\)

3. Sketch the cell.

Draw the anode side with the Pt electrode in contact with the ClO₃⁻(aq) and Cl⁻(aq) solution, connected by a salt bridge to the cathode side containing the H⁺(aq) and H₂O(l) solution and the Cu²+(aq) and Cu(s) electrode. Label the salt bridge and the electrodes on the sketch.

Key concepts

Electrochemical Cell Representation

Understanding the representation of electrochemical cells is crucial for interpreting and describing how voltaic cells operate. In a shorthand notation, components are listed sequentially from the anode to the cathode, separated by vertical lines. A single vertical line indicates a phase change, while a double vertical line represents a salt bridge or porous barrier.

For instance, the cell notation \(\mathrm{Fe}|\mathrm{Fe}^{2+} \|\| \mathrm{Ag}^{+}|\mathrm{Ag}\) signifies that iron (Fe) is the anode where oxidation occurs, and silver ions (Ag+) at the cathode are reduced. An important aspect of the notation is the clear representation of the reaction sites and substances involved. The anode, indicated first, is where the oxidation half-reaction takes place, and the cathode, listed last, is where reduction occurs.

Interpreting this shorthand allows students to visualize the cell setup, the flow of electrons, and the movement of ions within the electrolyte solutions. Furthermore, the notation succinctly shows the materials used for electrodes and the electrolytes in both compartments.

Half-Reactions in Voltaic Cells

Half-reactions are the essence of electrochemical processes. These reactions represent the oxidation and reduction events that occur at the electrodes of a voltaic cell. Oxidation, the loss of electrons, takes place at the anode. Meanwhile, reduction, the gain of electrons, occurs at the cathode.

For example, when iron oxidizes to iron(II) ions at the anode, the half-reaction is:\(\mathrm{Fe}(s) \rightarrow \mathrm{Fe}^{2+}(aq) + 2 e^-\). Conversely, the reduction of silver ions to solid silver at the cathode can be demonstrated by the half-reaction:\(\mathrm{Ag}^{+}(aq) + e^- \rightarrow \mathrm{Ag}(s)\).

Students must learn to identify the reactants and products in these half-reactions, including the electrons involved. This skill allows them to predict which direction the electrons will flow and helps in visualizing the complete electrochemical process.

Balancing Redox Reactions

Balancing redox reactions is fundamental to understanding and working with electrochemical cells. It ensures that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction. This balance is essential to maintain the conservation of charge.

Consider the redox reaction involving zinc and hydrogen ions:\(\mathrm{Zn}(s) + 2\mathrm{H}^{+}(aq) \rightarrow \mathrm{Zn}^{2+}(aq) + \mathrm{H}_{2}(g)\). In this example, the zinc anode provides two electrons through oxidation, which perfectly match the two electrons required for the reduction of hydrogen ions at the cathode. No additional balancing is required.

However, if the number of electrons differs, as in the case with iron and silver ions, the reduction half-reaction needs to be multiplied so that it balances with the oxidation half-reaction. Accurate balancing of redox reactions is not only necessary for theoretical understanding but is also crucial for practical applications, such as predicting the voltage output of a voltaic cell.