Question

A common shorthand way to represent a voltaic cell is anode \(\mid\) anode solution || 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 \(\mathrm{Fe}\left|\mathrm{Fe}^{2+}\right|\left|\mathrm{Ag}^{+}\right| \mathrm{Ag} ;\) calculate the standard cell emf using data in Appendix E. (b) Write the half-reactions and overall cell reaction represented by \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}\right|\left|\mathrm{H}^{+}\right| \mathrm{H}_{2} ;\) 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: $$ \begin{aligned} \mathrm{ClO}_{3}^{-}(a q)+3 \mathrm{Cu}(s) &+6 \mathrm{H}^{+}(a q) \\ & \longrightarrow \mathrm{Cl}^{-}(a q)+3 \mathrm{Cu}^{2+}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$ \(\mathrm{Pt}\) is used as an inert electrode in contact with the \(\mathrm{ClO}_{3}^{-}\) 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

The half-reactions and overall cell reactions for the given cells are: 1a. \(Fe(s) \rightarrow Fe^{2+}(aq) + 2e^-\), \(Ag^{+}(aq) + e^- \rightarrow Ag(s)\), and \(Fe(s) + 2Ag^{+}(aq) \rightarrow Fe^{2+}(aq) + 2Ag(s)\). The standard cell emf is \(1.24 V\). 1b. \(Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-\), \(2H^{+}(aq) + 2e^- \rightarrow H_2(g)\), and \(Zn(s) + 2H^{+}(aq) \rightarrow Zn^{2+}(aq) + H_2(g)\). The standard cell emf is \(0.76 V\). 1c. The cell is represented as \(Pt | ClO_3^-(aq), Cl^-(aq) || Cu^{2+}(aq) | Cu(s)\). The standard cell emf is \(-1.11 V\).

Step-by-step solution

1a. Identify half-reactions of Fe | Fe²⁺ || Ag⁺ | Ag cell

The anode is Fe, and the cathode is Ag. The anode half-reaction is the oxidation of Fe, and the cathode half-reaction is the reduction of Ag⁺: Anode half-reaction: \(Fe(s) \rightarrow Fe^{2+}(aq) + 2e^-\) Cathode half-reaction: \(Ag^{+}(aq) + e^- \rightarrow Ag(s)\)

2a. Identify overall cell reaction

Combine the half-reactions to get the overall cell reaction: Overall reaction: \(Fe(s) + 2Ag^{+}(aq) \rightarrow Fe^{2+}(aq) + 2Ag(s)\)

3a. Calculate the standard cell emf

Look up the standard reduction potentials in Appendix E (not provided here, but values can be found in any standard chemistry textbook) and use them to calculate the standard cell emf: \(E^∘_{Fe^{2+}/Fe} = -0.44 V\) \(E^∘_{Ag^{+}/Ag} = 0.8 V\) Standard cell emf is the difference between the cathode and anode potentials: \(E^∘_{cell} = E^∘_{cathode} - E^∘_{anode} = 0.8 V - (-0.44 V) = 1.24 V\)

1b. Identify half-reactions of Zn | Zn²⁺ || H⁺ | H₂ cell

The anode is Zn, and the cathode is H₂. The anode half-reaction is the oxidation of Zn, and the cathode half-reaction is the reduction of H⁺ using Pt as the hydrogen electrode: Anode half-reaction: \(Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-\) Cathode half-reaction: \(2H^{+}(aq) + 2e^- \rightarrow H_2(g)\)

2b. Identify overall cell reaction

Combine the half-reactions to get the overall cell reaction: Overall reaction: \(Zn(s) + 2H^{+}(aq) \rightarrow Zn^{2+}(aq) + H_2(g)\)

3b. Calculate the standard cell emf

Look up the standard reduction potentials in Appendix E and use them to calculate the standard cell emf: \(E^∘_{Zn^{2+}/Zn} = -0.76 V\) \(E^∘_{H^{+}/H_2} = 0 V\) Standard cell emf is the difference between the cathode and anode potentials: \(E^∘_{cell} = E^∘_{cathode} - E^∘_{anode} = 0 V - (-0.76 V) = 0.76 V\)

1c. Represent the cell and calculate the standard cell emf for the given reaction

For the given reaction, chlorine is the anode and the copper is the cathode: Cell representation: \(Pt | ClO_3^-(aq), Cl^-(aq) || Cu^{2+}(aq) | Cu(s)\) Anode half-reaction: \(ClO_3^-(aq) + 6H^{+}(aq) + 6e^- \rightarrow Cl^-(aq) + 3H_2O(l)\) with \(E^∘_{anode}=1.45 V\) Cathode half-reaction: \(3Cu^{2+}(aq) + 6e^- \rightarrow 3Cu(s)\) Look up the standard reduction potential of the Cu²⁺ | Cu half-reaction in Appendix E to calculate the standard cell emf: \(E^∘_{Cu^{2+}/Cu} = 0.34 V\) Standard cell emf is the difference between the cathode and anode potentials, considering the standard reduction potential provided is for the anode half-reaction: \(E^∘_{cell} = E^∘_{cathode} - E^∘_{anode} = 0.34 V - 1.45 V = -1.11 V\)

Key concepts

Standard Cell Potential

The standard cell potential, often denoted as \(E^∘_{cell}\), is a measure of how much voltage (electrical potential) a voltaic, or galvanic, cell can produce under standard conditions. Specifically, these conditions involve all reactants and products at a concentration of 1 M, a pressure of 1 atm for gases, and a temperature of 25°C. The standard cell potential is determined using the standard reduction potentials of the half-reactions involved.

To calculate the standard cell potential, you'll typically need the standard reduction potentials from a reference, which are usually tabulated in appendices of chemistry textbooks. The cell potential is calculated by taking the difference between the reduction potential of the cathode (positive electrode) and the anode (negative electrode):

• \(E^∘_{cell} = E^∘_{cathode} - E^∘_{anode}\)

This calculation involves identifying both half-reactions occurring in the cell, converting them from their list of reduction potentials if necessary, and then calculating their difference to determine the overall cell potential.

A positive \(E^∘_{cell}\) indicates a spontaneous reaction, meaning it can do work on its own, as seen in cases like oxidative metals such as zinc or iron when coupled with a more noble metal like silver.

Reduction Potentials

Reduction potentials are specific to each half-reaction and measure the tendency of a substance to gain electrons in a chemical reaction. The more positive the standard reduction potential, the greater the substance's tendency to be reduced.

Reduction potentials are tabulated under standard conditions, much like the standard cell potentials. These tables are crucial for understanding which way the electrons will flow between two reactants in an electrochemical cell. The substance with the higher (more positive) reduction potential will act as the oxidizing agent because it tends to gain electrons more readily, serving as the cathode in a voltaic cell.

For example:

• The reduction potential for \(Ag^+\) to \(Ag\) is +0.80 V, whereas for \(Fe^{2+}\) to \(Fe\) it is -0.44 V. Hence, silver is more likely to be reduced compared to iron.

While looking at reduction potentials, switch the sign for calculations if the half-reaction is written as an oxidation (for anode reactions). The essence of using reduction potentials effectively lies in comprehending this switching between reduction and oxidation to get the proper direction and magnitude of electron flow.

Half-Reactions

Half-reactions are a way to split a full redox (reduction-oxidation) reaction into two separate processes: the oxidation and the reduction. Each involves either a gain or loss of electrons.

In an electrochemical cell, the oxidation half-reaction occurs at the anode, while the reduction half-reaction occurs at the cathode. By writing out these half-reactions, it's easier to balance redox equations, calculate cell potentials, and understand the movement of electrons in the cell.

• Anode (Oxidation): Electrons are released when a substance is oxidized. Example: \(Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-\).
• Cathode (Reduction): Electrons are gained by a substance being reduced. Example: \(2H^+(aq) + 2e^- \rightarrow H_2(g)\).

Combining these half-reactions gives the overall cell reaction, which can be observed in the voltaic cell shorthand notation. This method clarifies what each element in the equation is doing, making complex redox processes easier to understand and manipulate, especially when determining the flow of electrons and energy production in electrochemical cells.