Question

Given: $$\begin{array}{ll} \mathrm{Br}_{2}(\ell)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Br}^{-}(\mathrm{aq}) & E^{\circ}=+1.07\mathrm{~V} \\ \mathrm{I}_{2}(\mathrm{~s})+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{I}^{-}(\mathrm{aq}) & E^{\circ}=+0.54 \mathrm{~V}\end{array}$$ determine whether this reaction is product-favored: $$\mathrm{Br}_{2}(\mathrm{~g})+2 \mathrm{I}^{-}(\mathrm{aq}) \longrightarrow 2 \mathrm{Br}^{-}(\mathrm{aq})+\mathrm{I}_{2}(\mathrm{~s})$$

Short answer

Yes, the reaction is product-favored as it has a positive cell potential of 0.53 V.

Step-by-step solution

Identify the half-reactions

The given reaction involves the reduction of bromine, \( \mathrm{Br_2} \), and the oxidation of iodide ions, \( \mathrm{I^-} \). The reactions provided are reduction potentials, so for iodide to be oxidized, we need to reverse its related reaction.

Write the oxidation half-reaction

Reversing the iodide reduction reaction gives us the oxidation half-reaction: \(2 \mathrm{I}^- \rightarrow \mathrm{I}_2 + 2 \mathrm{e}^-\). When we reverse this reaction, the new cell potential (\(E^{\circ}\)) becomes \(-0.54 \mathrm{~V}\).

Find net cell potential

Sum the standard potential of the two half-reactions. The reduction potential for \( \mathrm{Br_2} \) is \(+1.07 \mathrm{~V}\) and the oxidation potential for \( \mathrm{I^-} \) is \(-0.54 \mathrm{~V}\). So, the net cell potential is: \[E^{\circ}_{\text{cell}} = 1.07 \mathrm{~V} + (-0.54 \mathrm{~V}) = 0.53 \mathrm{~V}.\]

Determine if the reaction is product-favored

A positive cell potential indicates a spontaneous reaction in the forward direction. Since \( E^{\circ}_{\text{cell}} = 0.53 \mathrm{~V} \) is positive, the given reaction is product-favored.

Key concepts

Standard Reduction Potentials

Standard reduction potentials are crucial in understanding how likely a species is to gain electrons and be reduced. These potentials are measured under standard conditions, which include a concentration of 1M, a pressure of 1 atm, and a temperature of 25°C. In essence, the standard reduction potential, denoted as \(E^{ ext{°}}\), provides insight into the oxidizing or reducing strength of a chemical species.

The more positive the standard reduction potential, the stronger the oxidizing agent. This means the species has a higher tendency to gain electrons. Conversely, a more negative potential indicates a strong reducing agent that is more likely to lose electrons. For instance, in the given exercise, bromine has a standard reduction potential of \(+1.07 \mathrm{V}\), positioning it as a stronger oxidizing agent compared to iodine, which has a \(+0.54 \mathrm{V}\) potential. Understanding these values helps predict how species will behave in redox reactions and how electrons will transfer between them.

Knowing the standard reduction potentials of different chemical species allows us to predict the direction of electron flow and determine if a reaction will be spontaneous.

Redox Reactions

Redox reactions, short for oxidation-reduction reactions, involve the transfer of electrons between two chemical species. These reactions can be split into two processes: reduction and oxidation, occurring simultaneously. An easy way to remember this is through the acronym "OIL RIG" - Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons).

In these reactions, one species donates electrons (gets oxidized), while another accepts electrons (gets reduced). For the exercise we are examining, the iodide ions \(\mathrm{I^-}\) get oxidized to iodine \(\mathrm{I_2}\), losing electrons in the process. Conversely, bromine \(\mathrm{Br_2}\) gets reduced to bromide ions \(\mathrm{Br^-}\), gaining electrons.

Understanding which species is oxidized and which is reduced is key in writing balanced redox equations and necessary for calculating the net cell potential. Each half-reaction either shows a reduction or an oxidation, and they need to be balanced with respect to electrons to achieve overall neutrality in the reaction.

Cell Potential Calculation

Calculating the cell potential for a reaction involves combining the standard reduction potentials of the half-reactions for the species involved. The equation is given by: \[ E^{ ext{°}}_{ ext{cell}} = E^{ ext{°}}_{ ext{reduction}} + E^{ ext{°}}_{ ext{oxidation}} \]
It's important to note that when a half-reaction is reversed to represent oxidation, its potential will have the opposite sign.

For the provided reaction, bromine \(\mathrm{Br_2}\) maintains a reduction potential of \(+1.07 \mathrm{V}\). Meanwhile, by reversing the reduction reaction of iodide \(\mathrm{I^-}\), the oxidation potential becomes \(-0.54 \mathrm{V}\). Adding these potentials together, we find that:

\[E^{ ext{°}}_{ ext{cell}} = 1.07 \mathrm{V} + (-0.54 \mathrm{V}) = 0.53 \mathrm{V}.\]

This positive cell potential indicates that the reaction is spontaneous and, thus, is product-favored. It implies that, under standard conditions, the reaction will proceed in the forward direction without external energy input, generating products efficiently. Recognizing the importance of cell potential calculations lets chemists predict reaction spontaneity and electrochemical cell performance.