Problem 41 - Chemistry Matter and Change

Question

Calculate the cell potential for the following voltaic cells. $$ \begin{array}{l}{\text { a. } 2 \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{Pb}(\mathrm{s}) \rightarrow \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})} \\ {\text { b. } \mathrm{Mn}(\mathrm{s})+\mathrm{Ni}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{Ni}} \\ {\text { c. } \mathrm{I}_{2}(\mathrm{aq})+\mathrm{Sn}(\mathrm{s}) \rightarrow 2 \mathrm{I}^{-}(\mathrm{aq})+\mathrm{Sn}^{2+}(\mathrm{aq})}\end{array} $$

Step-by-Step Solution

Answer

The cell potentials for the given voltaic cells are: \(a. E_{cell} = 0.93V\), \(b. E_{cell} = 0.93V\), and \(c. E_{cell} = 0.68V\).

Step 1: Identifying the half-reactions

First, need to identify the half-reactions (oxidation and reduction) for each given voltaic cell: a. Oxidation: \(\mathrm{Pb (s) \rightarrow Pb^{2+} (aq) + 2e^{-}}\) Reduction: \(\mathrm{2Ag^{+}(aq) + 2e^{-} \rightarrow 2Ag (s)}\) b. Oxidation: \(\mathrm{Mn(s) \rightarrow Mn^{2+}(aq) + 2e^{-}}\) Reduction: \(\mathrm{Ni^{2+}(aq) + 2e^{-} \rightarrow Ni(s)}\) c. Oxidation: \(\mathrm{Sn(s) \rightarrow Sn^{2+}(aq) + 2e^{-}}\) Reduction: \(\mathrm{I_{2}(aq) + 2e^{-} \rightarrow 2I^{-}(aq)}\)

Step 2: Determining standard reduction potentials

Now, let's determine the standard reduction potentials (\(E^0\)) for each half-reaction using a standard reduction potential table: a. \(E^0 (\mathrm{Pb^{2+}/Pb}) = -0.13V\) \(E^0 (\mathrm{Ag^{+}/Ag}) = 0.80V\) b. \(E^0 (\mathrm{Mn^{2+}/Mn}) = -1.18V\) \(E^0 (\mathrm{Ni^{2+}/Ni}) = -0.25V\) c. \(E^0 (\mathrm{Sn^{2+}/Sn}) = -0.14V\) \(E^0 (\mathrm{I_2/I^{-}}) = 0.54V\)

Step 3: Calculating cell potentials

Finally, we can calculate the cell potentials for each voltaic cell using the Nernst equation: a. \(E_{cell} = E_{cathode} - E_{anode} = 0.80V - (-0.13V) = 0.93V\) b. \(E_{cell} = E_{cathode} - E_{anode} = -0.25V - (-1.18V) = 0.93V\) c. \(E_{cell} = E_{cathode} - E_{anode} = 0.54V - (-0.14V) = 0.68V\) The cell potentials for the given reactions are: a. \(E_{cell} = 0.93V\) b. \(E_{cell} = 0.93V\) c. \(E_{cell} = 0.68V\)

Key Concepts

Cell Potential

In a voltaic cell, the cell potential is the driving force behind the movement of electrons from one half-cell to another. This potential, also called electromotive force (EMF), reflects the energy difference between the oxidizing agent being reduced and the reducing agent being oxidized in an electrochemical reaction. Imagine it as the voltage or electric pressure that propels electrons in the wire connecting the two metal plates (electrodes). The higher the cell potential, the greater the tendency for the redox reaction to proceed spontaneously.

To calculate the standard cell potential, combine the standard reduction potentials of the two half-reactions involved. Always keep in mind:

• Identify the cathode (where reduction happens) and anode (where oxidation occurs).
• Use the formula: \(E_{cell} = E_{cathode} - E_{anode}\).
• Sign and magnitude matter: a positive \(E_{cell}\) indicates a spontaneous reaction.

Thus, for any voltaic cell, determining the cell potential is crucial to understanding its ability to perform electrical work.

Standard Reduction Potentials

Standard reduction potentials are fundamental to predicting the direction of redox reactions. Each element or compound has its characteristic reduction potential, denoting how easily it gains electrons compared to the standard hydrogen electrode, which is set at 0 V. These potentials are presented in tables, making it straightforward to determine half-reaction tendencies.

In a tabulated format:

• A more positive value suggests a stronger oxidizing agent that readily undergoes reduction.
• A more negative value indicates a stronger reducing agent capable of easily releasing electrons.

Thus, when seeking to determine a cell's direction and spontaneity, reference these standard reduction potentials. You'd align half-reactions by alloting the more positive reduction potential to the cathode for a standard cell potential computation. This fundamental concept underpins not only chemistry but also fields such as battery technology and electroplating.

Nernst Equation

The Nernst equation refines our understanding of cell potentials, especially when conditions deviate from standard (1M concentration, 1 atm pressure). It allows us to calculate the actual cell potential by accounting for ion concentrations, temperature, and other real-time parameters. The equation is expressed as:\[E = E^0 - \left(\frac{RT}{nF}\right) \ln Q\]Where:

• \(E\) is the cell potential at non-standard conditions.
• \(E^0\) is the standard cell potential.
• \(R\) is the universal gas constant.
• \(T\) is the absolute temperature in Kelvin.
• \(n\) is the number of moles of electrons exchanged.
• \(F\) is Faraday's constant.
• \(Q\) is the reaction quotient, rarely equating to 1 in practical scenarios.

Through this equation, one can determine how shifts in concentration or temperature affect electrochemical efficiency. It epitomizes the intersection of chemistry with thermodynamics, solidifying our understanding of non-standard electrochemical processes.

Oxidation-Reduction Reactions

Oxidation-reduction reactions, or redox reactions, serve as the foundation for electrochemistry and are integral to our understanding of voltaic cells. These reactions involve the transfer of electrons from the reducing agent, which gets oxidized, to the oxidizing agent, which undergoes reduction. Here are some key points about these reactions:

• Oxidation involves a loss of electrons, increasing the oxidation state of an element.
• Reduction involves a gain of electrons, decreasing the oxidation state of an element.
• The complete redox process includes two half-reactions, one for oxidation and one for reduction.

When analyzing or designing voltaic cells, identifying these half-reactions allows us to predict which substances will serve as effective electrodes and the overall direction of electron flow. Ultimately, understanding oxidation-reduction reactions equips us to manipulate these processes for technological advancements, such as in power generation and metal extraction.