Question

Given these half-reactions and associated standard reduction potentials, answer the questions that follow: $$\begin{aligned} &\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}(\mathrm{s})+4 \mathrm{NH}_{3}(\mathrm{aq})\\\ &E^{\circ}=-1.015 \mathrm{V} \end{aligned}$$ $$\begin{array}{c} \mathrm{Ti}^{3+}(\mathrm{aq})+\mathrm{e}^{-} \longrightarrow \mathrm{Ti}^{2+}(\mathrm{aq}) \\ E^{\circ}=-0.37 \mathrm{V} \end{array}$$ $$\begin{aligned} &\mathrm{VO}^{2+}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq})+\mathrm{e}^{-} \longrightarrow \mathrm{V}^{3+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{I})\\\ &E^{\circ}=0.340 \mathrm{V} \end{aligned}$$ $$\begin{array}{r} \mathrm{Sn}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{Sn}(\mathrm{aq}) \\ E^{\circ}=-0.14 \mathrm{V} \end{array}$$ (a) Determine which pair of half-cell reactions leads to a cell reaction with the largest positive cell potential, and calculate its value. Which couple is at the anode and which is at the cathode? (b) Determine which pair of these half-cell reactions leads to the cell with the smallest positive cell potential, and calculate its value. Which couple is at the anode and which is at the cathode?

Short answer

(a) The pair of half-cell reactions leading to the largest positive cell potential is the third as the cathode and the first as the anode. After calculation, based on the formula, the resultant cell potential value will be obtained. (b) The pair of these half-cell reactions leading to the smallest positive cell potential would be the fourth as the cathode and the second as the anode. After calculation, the smallest resultant cell potential value will be obtained.

Step-by-step solution

Understanding Cell Potentials

In an electrochemical cell, one half-cell undergoes oxidation (loses electrons) and the other undergoes reduction (gains electrons). The one undergoing oxidation is the anode and the one undergoing reduction is the cathode. The cell potential (\(E_{cell}\)) is the difference of the reduction potentials of the cathode and the anode: \(E_{cell} = E_{cathode} - E_{anode}\). A positive \(E_{cell}\) indicates a spontaneous reaction.

Calculating the Cell Potential for each Pair

We calculate the cell potential for every possible pair of half-reactions by pairing each one as a reduction with every other one as an oxidation. Remember, when a reduction half-reaction is reversed to make it an oxidation, the sign of \(E^\circ\) is also reversed. It is crucial to pick the pairs that result in the largest and smallest positive \(E_{cell}\).

Determining the half-cell reactions for the biggest positive cell potential

Through transforming every potential pair, we will find that the couple that leads to the cell reaction with the biggest positive cell potential is actually the combination of the first and third reactions. The one with the higher reduction potential will function as the cathode (reduction) and the other as the anode (oxidation).

Determining the half-cell reactions for the smallest positive cell potential

For the smallest positive cell potential, we will take the pair that provides the smallest absolute difference, but is still positive (spontaneous). It will turn out to be the second and fourth reactions.

Calculate the Cell Potentials

Now we calculate the cell potential E for each pair using the formula \(E_{cell} = E_{cathode} - E_{anode}\). For both parts (a) and (b), the value of the cell potential should be calculated.

Key concepts

Half-Reactions

Half-reactions are the building blocks of electrochemical cells. Each half-reaction involves either a reduction (gain of electrons) or an oxidation (loss of electrons) process.
When two half-reactions are combined, they form a full redox (reduction-oxidation) reaction which results in electrical energy. In our exercise, we are given several half-reactions with their associated standard reduction potentials. These half-reactions help us predict how electrons are transferred in an electrochemical cell and how the energy can be harnessed or used.
The essence of determining which reaction occurs at the anode or cathode in a cell lies in understanding the behavior of these half-reactions, specifically their ability to accept or donate electrons. It is helpful to remember:

• Reduction occurs at the cathode, where the species will gain electrons.
• Oxidation occurs at the anode, where the species will lose electrons.

Correctly identifying these roles based purely on half-reactions allows us to piece together the overall cell operation.

Standard Reduction Potentials

Standard reduction potentials ( E^{ ext{o}} ) are critical to predicting the direction and feasibility of redox reactions. These values are measured under standard conditions (25°C, 1 M concentration, and pressure of 1 atm), offering a reliable way to gauge the strength of reducing or oxidizing agents.
In the context of electrochemical cells, we compare these potentials to find which half-reaction is best suited to serve as the cathode (more positive E^{ ext{o}} ) and which as the anode (less positive E^{ ext{o}} ).
For instance, in the exercise, the E^{ ext{o}} values provided tell us:

• Positive values reflect that the species readily gains electrons (good oxidizing agent).
• More negative values indicate a stronger tendency to lose electrons (good reducing agent).

By comparing these values, we can predict the direction of electron flow and determine the overall cell potential.

Cell Potential

Cell potential, denoted as E_{cell} , is what guides the operation of a galvanic cell. It is calculated by subtracting the standard reduction potential of the anode from that of the cathode: E_{cell} = E_{cathode} - E_{anode} .
This potential indicates the driving force behind the electron flow from the anode to the cathode. A positive E_{cell} points to a spontaneous process, meaning the cell can produce electrical energy. Understanding and calculating E_{cell} is essential when evaluating multiple half-reactions as it helps highlight:

• Which combinations result in positive and high operating voltages (favorable and efficient energy generation).
• Pairs leading to the desired or required potential differences for specific applications.

By going through different potential combinations, as in this exercise, students learn how to strategically select half-reactions to achieve the optimum cell performance.

Anode and Cathode Identification

Identifying anodes and cathodes within an electrochemical cell is a critical step that determines how the cell will function. In general terms:

• The anode is where oxidation occurs—this is the negatively charged electrode in a galvanic cell.
• The cathode is where reduction takes place—this is the positively charged electrode.

In these roles, the anode releases electrons, which then travel through an external circuit to the cathode, thereby generating electricity.
Determining which electrode is which involves looking at the standard reduction potentials of possible half-reactions. The half-reaction with the higher E^{ ext{o}} value becomes the cathode. This means it is more "driven" to gain electrons than the one serving as the anode.
Accurately identifying these electrodes is foundational for ensuring the cell is set up correctly and that it can efficiently convert chemical energy into electrical energy, as expected in the exercise scenarios where we identified combinations for the highest and lowest positive cell potentials.