In basic solution, \(\mathrm{Se}^{2-}\) and \(\mathrm{SO}_{3}^{2-}\) ions react spontaneously: \(2 \mathrm{Se}^{2-}(a q)+2 \mathrm{SO}_{3}^{2-}(a q)+3 \mathrm{H}_{2} \mathrm{O}(I) \longrightarrow\) $$2 \mathrm{Se}(s)+6 \mathrm{OH}^{-}(a q)+\mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q) \quad E_{\mathrm{cell}}^{\circ}=0.35 \mathrm{~V}$$ (a) Write balanced half-reactions for the process. (b) If \(E_{\text {sulfite }}^{\circ}\) is \(-0.57 \mathrm{~V},\) calculate \(E_{\text {selenium }}^{\circ}\)

In redox reactions, we split the overall reaction into two parts: oxidation and reduction half-reactions. Each half-reaction shows either the loss or gain of electrons. This is crucial for balancing complex redox reactions because it helps us track where electrons are going.
For example, in the given problem, we have:

• Oxidation half-reaction: \(\text{2 Se}^{2-} \rightarrow \text{2 Se} + 4 e^{-}\)
• Reduction half-reaction: \(\text{2 SO}_{3}^{2-} + 4 H_{2}O + 4 e^{-} \rightarrow \text{S}_{2}O_{3}^{2-} + 6 OH^{-}\)

Each half-reaction is balanced separately for mass and charge before combining them to form the overall balanced equation.

The standard electrode potential (\(E^\text{\textdegree}\)) is a measure of the tendency of a chemical species to be reduced. It's measured under standard conditions (25°C, 1 M concentration, 1 atm pressure).
Standard electrode potential is critical in calculating the voltage of electrochemical cells. The problem involves given potentials: \( E_{\text{sulfite}}^{\text{\textdegree}} = -0.57 \text{ V} \) and the overall cell potential \( E_{\text{cell}}^{\text{\textdegree}} = 0.35 \text{ V} \). Using the equation \( E_{\text{cell}}^{\text{\textdegree}} = E_{\text{cathode}}^{\text{\textdegree}} - E_{\text{anode}}^{\text{\textdegree}} \), we can solve for the unknown standard electrode potential for selenium, finding it to be \( E_{\text{selenium}}^{\text{\textdegree}} = -0.92 \text{ V} \).

Electrochemistry is the study of chemical processes that produce electricity or are driven by electricity. These processes are involved in batteries, electrolysis, and corrosion.
In our problem, we deal with spontaneous redox reactions occurring in a basic solution. Understanding how electrons move between reactants in half-reactions helps us determine the overall cell potential, which tells us if a reaction is spontaneous.
Calculating potentials for half-reactions and the overall cell provides insights into the energies involved and the feasibility of redox processes.

Oxidation involves the loss of electrons from an atom or molecule. This process increases the oxidation state of the species.
In our problem, selenium ions (\text{Se}^{2-}) are oxidized to elemental selenium (\text{Se}). The oxidation half-reaction is expressed as: \(\text{2 Se}^{2-} \rightarrow \text{2 Se} + 4 e^{-}\). Here, each ion loses 2 electrons, showing how selenium undergoes oxidation.

Reduction is the gain of electrons by an atom or molecule, resulting in a decrease in its oxidation state.
In the problem we study, sulfite ions (\text{SO}_{3}^{2-}) are reduced to thiosulfate (\text{S}_{2}O_{3}^{2-}). The reduction half-reaction is: \(\text{2 SO}_{3}^{2-} + 4 H_{2}O + 4 e^{-} \rightarrow \text{S}_{2}O_{3}^{2-} + 6 OH^{-}\).
Here, sulfite ions gain electrons, which is a hallmark of reduction.