The reaction \(\mathrm{Cu}^{2+}+\mathrm{I}_{2} \rightarrow \mathrm{Cu}^{+}+2 \mathrm{I}^{-}\) is not possible as written. Assign an oxidation number to each atom and explain what is wrong with this reaction.

A fundamental concept in chemistry is the oxidation number, which helps us understand redox (reduction-oxidation) reactions. An oxidation number is a positive or negative number assigned to an atom in a compound that indicates its degree of oxidation or reduction. In simple terms, it represents the number of electrons an atom gains or loses when it forms a compound. For example, in the copper (II) ion, Cu²⁺, copper has an oxidation number of +2 because it has lost two electrons. Similarly, in molecular iodine, I₂, each iodine atom has an oxidation number of 0 because they are in their elemental state and have not gained or lost electrons.

To determine the correctness of a redox reaction, we compare the changes in oxidation numbers before and after the reaction takes place. If the oxidation numbers of the reactants and products don't reflect the correct transfer of electrons, we know the reaction is imbalanced.

Understanding oxidation and reduction is crucial when investigating redox reactions. Oxidation is the process where an atom or ion loses electrons, which increases its oxidation number. Conversely, reduction is the gain of electrons by an atom or ion, leading to a decrease in its oxidation number. When copper in the reaction Cu²⁺ becomes Cu⁺, its oxidation number decreases from +2 to +1, showcasing a reduction process as it gains an electron. On the other hand, elemental iodine (I₂) changes to iodide ions (I^-), each gaining an electron, reflecting oxidation as the oxidation number changes from 0 to -1.

In the heart of a redox reaction lies electron transfer, which is the movement of electrons from one species to another. This process is what defines oxidation and reduction activities within the reaction. For instance, the reduction half-reaction \[ \mathrm{Cu}^{2+} + 1e^- \rightarrow \mathrm{Cu}^{+} \] indicates that a copper ion (Cu²⁺) accepts one electron (e^-) and is reduced to Cu⁺. Meanwhile, in the oxidation half-reaction \[ \mathrm{I}_{2} + 2e^- \rightarrow 2 \mathrm{I}^- \], molecular iodine (I₂) is the electron acceptor, transforming into two iodide ions. Each molecule or ion involved either donates or accepts electrons, thus changing its oxidation state. In a balanced redox reaction, the number of electrons lost by the oxidizing agent must be equal to the number of electrons gained by the reducing agent.

The conservation of charge is a principle stating that in any chemical reaction, the total electric charge must remain constant. Therefore, when writing redox reactions, it is imperative to ensure that the total charge on both sides of the equation is balanced. In the reaction in question, \[ \mathrm{Cu}^{2+} + \mathrm{I}_{2} \rightarrow \mathrm{Cu}^{+} + 2\mathrm{I}^{-} \], there is an imbalance. The reduction half-reaction loses only one electron, while the oxidation half-reaction gains two electrons. This inconsistency means the reaction, as written, violates the law of conservation of charge and cannot occur under normal conditions. To rectify this, the reaction must be balanced so that the same number of electrons are exchanged, ensuring the total charge is conserved and thus allowing the reaction to proceed.