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?

Step by step solution

01

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.

02

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 \(E_{cell}\).

03

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).

04

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.

05

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.

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