In the following reaction: \(\mathrm{Cr}(\mathrm{OH})_{3}+\mathrm{OH}^{-}+\mathrm{IO}_{3} \longrightarrow \mathrm{CrO}_{4}^{2-}+\mathrm{H}_{2} \mathrm{O}+\mathrm{I}^{-}\) (a) \(\mathrm{IO}_{3}^{-}\) is oxidizing agent (b) \(\mathrm{Cr}(\mathrm{OH})_{3}\) is oxidized (c) \(6 \mathrm{e}^{-}\) are being taken per \(\mathrm{l}\) atom (d) All are true

Understanding oxidation and reduction is crucial when studying redox reactions in chemistry. Oxidation is the process where an atom, ion, or molecule loses electrons, resulting in an increase in its oxidation state. Conversely, reduction involves the gain of electrons, leading to a decrease in the oxidation state.

These processes always occur together; when one species is oxidized, another is reduced. This pairing is also referred to as a redox reaction, which stands for reduction-oxidation. To easily remember this concept, think of the mnemonic 'OIL RIG': Oxidation Is Loss, Reduction Is Gain of electrons.

In a redox reaction, oxidizing and reducing agents are key players in the transfer of electrons. An oxidizing agent, or oxidant, accepts electrons and is reduced in the process. On the flip side, a reducing agent, or reductant, donates electrons and is oxidized. How do you spot them? Look for the substance that undergoes reduction (gain of electrons) to identify the oxidizing agent, and find the one that is oxidized (loss of electrons) to pinpoint the reducing agent.

In our exercise, \(\mathrm{IO}_{3}^{-}\) is acting as the oxidizing agent because it is being reduced, while \(\mathrm{Cr}(\mathrm{OH})_{3}\) is the reducing agent as it is being oxidized.

To accurately determine what is being oxidized or reduced in a redox reaction, you must first find the oxidation states of the atoms involved. Oxidation state, sometimes called oxidation number, is a value assigned to an atom in a compound representing the atom's degree of electron possession or deficiency, compared to the pure element.

For example, in the given reaction, chromium goes from an oxidation state in \(\mathrm{Cr}(\mathrm{OH})_{3}\) to a different state in \(\mathrm{CrO}_{4}^{2-}\). Similarly, the oxidation state of iodine changes from \(\mathrm{IO}_{3}^{-}\) to \(\mathrm{I}^{-}\), indicating the transfer of electrons and helping us assess which atoms are oxidized and reduced.

Electron transfer is the backbone of redox chemistry. It's the literal movement of electrons from one atom or molecule (reducing agent) to another (oxidizing agent). The number of electrons transferred must be the same for both the oxidation and reduction to maintain electrical neutrality.

In the case of our exercise, each iodine atom in \(\mathrm{IO}_{3}^{-}\) receives a certain number of electrons to become \(\mathrm{I}^{-}\). The reaction shows that one chromium atom donates six electrons, confirming the answer in (c) of the exercise. These kinds of insights are fundamental for understanding the specifics of chemical reactions and for solving problems related to redox events.