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Why is the following balanced reaction not a proper redox reaction? $$ \mathrm{Fe}^{2+}(\mathrm{aq})+2 \mathrm{Br}^{-}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Br}_{2}(\ell) $$

Short Answer

Expert verified
The electron transfer is imbalanced; oxidation provides 1 electron, but reduction requires 2.

Step by step solution

01

Identify Oxidation Numbers

Determine the oxidation numbers of each element in the reactants and products. In the reactants, \( ext{Fe}^{2+}\) has an oxidation number of +2, and each \( ext{Br}^-\) has an oxidation number of -1. In the products, \( ext{Fe}^{3+}\) has an oxidation number of +3, and \( ext{Br}_2\) has an oxidation number of 0.
02

Check for Changes in Oxidation Numbers

Compare the oxidation numbers of each element in the reactants and products to see if there is any change. \( ext{Fe}^{2+}\) is oxidized to \( ext{Fe}^{3+}\), showing a change in oxidation number from +2 to +3. \( ext{Br}^-\) is reduced to \( ext{Br}_2\), with its oxidation number changing from -1 to 0, indicating reduction.
03

Balance Electron Transfer

Confirm electron transfer between oxidation and reduction processes. The oxidation of \( ext{Fe}^{2+}\) to \( ext{Fe}^{3+}\) involves the loss of one electron. The reduction of \( ext{Br}^-\) to \( ext{Br}_2\) involves the gain of two electrons per molecule of \( ext{Br}_2\) since two \( ext{Br}^-\) ions each lose one electron.
04

Identify Electron Discrepancy

Note the discrepancy in the electron transfer numbers: the oxidation only provides one electron per \( ext{Fe}^{2+}\), but the reduction requires that two electrons be gained by two \( ext{Br}^-\) ions to form \( ext{Br}_2\). This mismatch proves that the reaction cannot simultaneously satisfy both oxidation and reduction requirements.
05

Conclusion on Reaction Type

Conclude that because there is an imbalance in the electron transfer (one electron lost vs. two electrons needed), the reaction cannot be considered a proper redox reaction as written.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxidation Numbers
Oxidation numbers are essentially a bookkeeping tool used in redox reactions to determine how many electrons are gained or lost by atoms during the reaction. They provide a quick way to identify the changes in electron configuration of atoms as they transition from reactants to products.
Oxidation numbers follow a specific set of rules. For instance, in a simple ion, the oxidation number equals the charge of the ion. For molecules, the sum of oxidation numbers for all atoms in the molecule must equal the overall charge.
In the given reaction, Fe is in a +2 oxidation state as a reactant, which becomes +3 in the product, indicating that it has lost an electron. Br in the Br鈦 ion begins with an oxidation number of -1, and in the product Br鈧, each Br atom attains an oxidation number of 0, suggesting that Br has gained electrons.
Recognizing that Fe is oxidized and Br is reduced allows us to identify whether a reaction is a redox reaction. However, this alone is not enough; electron transfer must also be properly balanced.
Electron Transfer
In a redox reaction, electron transfer is the movement of electrons from one species to another. It is the core component of redox processes.
To identify an electron transfer in the reaction, note how electrons move between reactants and products. When Fe虏鈦 becomes Fe鲁鈦, it loses one electron. Conversely, when Br鈦 forms Br鈧, it gains electrons, requiring two Br鈦 ions, each gaining one electron, to combine into Br鈧.
For a balanced electron transfer, the number of electrons lost in oxidation must equal the number gained in reduction, ensuring that no electrons are left unaccounted for. This balance is crucial to proving a proper redox reaction.
In our reaction, Fe provides one electron upon oxidation, while forming Br鈧 requires two electrons for each pair of Br鈦. This imbalance in electron transfer demonstrates why the reaction as written fails to be a proper redox reaction.
Chemical Reaction Balancing
Balancing chemical reactions ensures that the same number of each type of atom appears on both sides of the equation, but for redox reactions, it also involves the careful balancing of electrons involved in the redox process.
The typical method involves separating the equation into two half-reactions, one for oxidation and one for reduction, and balancing each separately. Then, they are combined to ensure electrons are conserved on both sides of the equation.
The redox process for Fe and Br in our example is unbalanced because the electrons lost in oxidation do not match those needed for reduction. This electron imbalance prevents the reaction from being a proper redox reaction, despite having the correct stoichiometric balance of atoms.
Adjustments need to be made to achieve balance in both atoms and electrons, often involving coefficient changes to ensure that the number of electrons lost in oxidation equals those gained in reduction. This is how balanced redox reactions are correctly structured.

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Most popular questions from this chapter

Assign the oxidation numbers of all atoms in the following species. (a) \(\mathrm{ClO}_{3}^{-}\) (b) \(\mathrm{PF}_{3}\) (c) \(\mathrm{CO}\)

List four species that can oxidize \(\mathrm{Fe}^{2+}\) to \(\mathrm{Fe}^{3+}\).

At \(298 \mathrm{~K}\), the solubility product constant \(\mathrm{for} \mathrm{PbC}_{2} \mathrm{O}_{4}\) is \(8.5 \times 10^{-10}\), and the standard reduction potential of the \(\mathrm{Pb}^{2+}(\) aq \()\) to \(\mathrm{Pb}(\mathrm{s})\) is \(-0.126 \mathrm{~V}\). (a) Find the standard potential of the half-reaction $$ \mathrm{PbC}_{2} \mathrm{O}_{4}(\mathrm{~s})+2 \mathrm{e}^{-} \rightarrow \mathrm{Pb}(\mathrm{s})+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(\mathrm{aq}) $$ (Hint: The desired half-reaction is the sum of the equations for the solubility product and the reduction of \(\mathrm{Pb}^{2+}\). Find \(\Delta G^{\circ}\) for these two reactions and add them to find \(\Delta G^{\circ}\) for their sum. Convert the \(\Delta G^{\infty}\) to the potential of the desired half-reaction.) (b) Calculate the potential of the \(\mathrm{Pb} / \mathrm{PbC}_{2} \mathrm{O}_{4}\) electrode in a \(0.025 M\) solution of \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\)

A half-cell that consists of a silver wire in a \(1.00 M\) \(\mathrm{AgNO}_{3}\) solution is connected by a salt bridge to a \(1.00 \mathrm{M}\) thallium(I) acetate solution that contains a metallic T1 electrode. The voltage of the cell is \(1.136 \mathrm{~V}\), with the silver as the positive electrode. (a) Write the half-reactions and the overall chemical equation for the spontaneous reaction. (b) Use the standard potential of the silver half-reaction, with the voltage of the cell, to calculate the standard reduction potential for the thallium half-reaction.

Sphalerite is the naturally occurring mineral zinc sulfide, from which zinc metal is extracted. The ore is heated in oxygen to form zinc oxide and sulfur dioxide, followed by the reaction of metal oxide with elemental carbon to form CO. (a) Write a balanced equation for each of these reactions. (b) For each reaction, identify the element that is oxidized and the one that is reduced.

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