In the electrolysis of molten \(\mathrm{NaBr}\) (a) What product forms at the anode? (b) What product forms at the cathode?

Oxidation is a key concept in chemistry, especially in electrolysis. When we talk about oxidation, we mean the process where an atom or ion loses electrons. For instance, in the electrolysis of molten \text{NaBr}\, the \text{Br}^{-}\ ions at the anode lose electrons to become bromine gas \text{Br}_2\. This process can be described by the half-reaction:
\[\text{2Br}^{-} \rightarrow \text{Br}_2 + 2e^{-}\]
Here, the bromine ions are undergoing oxidation because they are losing electrons.

In electrochemical terms, oxidation always happens at the anode, which is positively charged in an electrolytic cell. The anode attracts negative ions (anions) and it is at this point where the extraction of electrons occurs. Remember, oxidation is often remembered by the acronym 'OIL' in 'OIL RIG', where OIL stands for Oxidation Is Loss (of electrons).

Reduction is the opposite of oxidation. It involves the gain of electrons by an atom or ion. During the electrolysis of molten \text{NaBr}\, the \text{Na}^{+}\ ions at the cathode gain electrons to form sodium metal. The half-reaction for this process is:
\[\text{Na}^{+} + e^{-} \rightarrow \text{Na}\]
Here, the sodium ions are undergoing reduction because they are gaining electrons.

In an electrolytic cell, reduction takes place at the cathode, which is negatively charged. The cathode attracts positive ions (cations) and provides them with the electrons they need to become neutral atoms. Similar to oxidation, reduction can be remembered with the acronym 'RIG' in 'OIL RIG', where RIG stands for Reduction Is Gain (of electrons).

Understanding half-reactions is crucial for grasping how electrolysis works. A half-reaction shows either the oxidation or reduction process separately, highlighting the transfer of electrons. Let's break down the half-reactions in the electrolysis of molten \text{NaBr}\.

At the anode, we have the oxidation half-reaction:
\[\text{2Br}^{-} \rightarrow \text{Br}_2 + 2e^{-}\]
This half-reaction shows how each bromide ion loses one electron, which collectively forms bromine gas.

At the cathode, we have the reduction half-reaction:
\[\text{Na}^{+} + e^{-} \rightarrow \text{Na}\]
This half-reaction displays how each sodium ion gains one electron to become sodium metal.

By separately examining each half-reaction, we can better understand the overall process occurring during electrolysis. It also helps to identify the specific reactions happening at the anode and cathode. These half-reactions are essential for balancing the overall redox equations correctly.