From the information provided, use cell notation to describe the following systems: (a) In one half-cell, a solution of \(\operatorname{Pt}\left(\mathrm{NO}_{3}\right)_{2}\) forms \(\mathrm{Pt}\) metal, while in the other half-cell, Cu metal goes into a \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) solution with all solute concentrations \(1 \mathrm{M}\) (b) The cathode consists of a gold electrode in a \(0.55 M \mathrm{Au}\left(\mathrm{NO}_{3}\right)_{3}\) solution and the anode is a magnesium electrode in \(0.75 M \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) solution. (c) One half-cell consists of a silver electrode in a \(1 \mathrm{M} \mathrm{AgNO}_{3}\) solution, and in the other half-cell, a copper electrode in \(1 M \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) is oxidized.

Electrode reactions are at the heart of how electrochemical cells function. An electrochemical cell consists of two electrodes: an anode and a cathode. At the anode, oxidation occurs; this involves the loss of electrons by a species. For example, in the system (a) from the exercise, the platinum ion in \( \rm Pt(NO_3)_2 \) loses electrons to form platinum metal (Pt). On the other hand, the cathode is the site of reduction, a process involving the gain of electrons. In the same system, copper (Cu) gains electrons to form \( \rm Cu(NO_3)_2 \).

Understanding these reactions requires familiarity with the concept of redox (reduction-oxidation) processes. Each half-cell in an electrochemical cell contains the electrode and a solution with ions that will participate in the redox reactions. The reactions at the electrodes involve a transfer of electrons, which is the basis for the flow of electric current through an external circuit connecting the anode and cathode.

Oxidation and reduction are two halves of a redox reaction. These processes are essential for generating power in electrochemical cells. Oxidation involves the loss of electrons from an atom, ion, or molecule. When discussing system (c), copper is oxidized at the anode when it transitions from Cu metal to \( \rm Cu(NO_3)_2 \) by losing electrons.

In contrast, reduction is the gain of electrons. In system (c), also, silver undergoes reduction at the cathode, going from \( \rm Ag^+ \) in \( \rm AgNO_3 \) solution to silver metal (Ag). To remember this concept, you can use the mnemonic 'OIL RIG' - Oxidation Is Loss, Reduction Is Gain, referring to the electron movement. The exercise improvement advice would recommend understanding the role electrons play in these processes as key to mastering the concepts of oxidation and reduction.

In cell notation, concentration plays a significant role as it can affect the cell's potential. Concentrations are included in the notation to provide a complete description of the half-cells. For instance, in system (b), the notation includes the concentration of the magnesium nitrate and gold nitrate solutions, \(0.75 M \rm Mg(NO_3)_2 \) and \(0.55 M \rm Au(NO_3)_3\), respectively.

The standard form to represent concentrations in cell notation is by listing them in parentheses after the formula of the anode or cathode solution, like \( \rm Mg(NO_3)_2(0.75 M) \) for the magnesium nitrate solution. A consistent concentration measurement, such as molarity (M), is crucial for accurate descriptions and for calculations involving the Nernst equation, which relates cell potential to concentration.