Radium-226 decays by alpha emission to radon-222. Suppose that \(25.0 \%\) of the energy given off by one gram of radium is converted to electrical energy. What is the minimum mass of lithium that would be needed for the voltaic cell \(\mathrm{Li}\left|\mathrm{Li}^{+} \| \mathrm{Cu}^{2+}\right| \mathrm{Cu}\), at standard conditions, to produce the same amount of electrical work \(\left(\Delta G^{\circ}\right) ?\)

Alpha decay is a type of radioactive decay in which an atomic nucleus emits an alpha particle, consisting of two protons and two neutrons. This process decreases the mass of the original nucleus and transforms it into a different element. In the case of the exercise above, Radium-226 undergoes alpha decay and transmutes into Radon-222.

Understanding alpha decay is crucial because it's central to many nuclear reactions and phenomena, including how certain types of nuclear batteries convert radioactive energy into electrical energy. In the context of our exercise, a gram of Radium-226 emits energy during this decay process, which can theoretically be harnessed to do electrical work.

A voltaic cell is an electrochemical cell that harnesses chemical energy and converts it into electrical energy through spontaneous redox reactions. It consists of two half-cells, each containing an electrode and an electrolyte. The two electrodes are connected by an external circuit and a salt bridge.

In our exercise, the voltaic cell is made of lithium and copper half-cells. When the lithium electrode loses electrons (oxidation), and copper ions in solution gain electrons (reduction), the cell generates an electromotive force, resulting in the flow of electrons through the circuit—the essence of electrical work in electrochemistry.

Electrical work refers to the energy transferred when electrical charge moves through a potential difference. It can be calculated using the formula:
\(W = qV\)
where \(W\) is the electrical work, \(q\) is the charge, and \(V\) is the voltage or potential difference. Electrical work is a significant concept because it represents the usable work that can be performed by electrical energy generated in various ways, including voltaic cells.

In solving our exercise, we first calculate the energy available from the radioactive decay of Radium-226 and then determine how much of that energy can be converted into electrical work. Understanding these calculations is vital for comparing energy conversions in different processes.

Gibbs free energy (\(G\)), is the energy in a system that can be used to do work at constant temperature and pressure. It is a critical concept in thermodynamics and electrochemistry. The change in Gibbs free energy (\(\Delta G\)) during a chemical reaction can indicate whether a reaction is spontaneous. A negative \(\Delta G\) signifies that the reaction occurs without the need for additional energy.

The electrical work obtained when a charge moves through a potential difference in a voltaic cell is a form of Gibbs free energy. This connection allows us to use the electrical work calculated in our previous steps (in J or joules) to determine the feasibility and spontaneous nature of electrochemical reactions.

The standard cell potential (\(E^\circ\)) is the voltage (potential difference) between two half-cells in a voltaic cell when all components are at standard conditions (1 M concentration and 298.15 K temperature). It is calculated by subtracting the standard reduction potential of the anode from the standard reduction potential of the cathode:
\(\Delta E^\circ = E^\circ_{cathode} - E^\circ_{anode}\)

This value helps quantify the driving force behind the electrochemical reaction and is a direct measure of a voltaic cell's ability to generate an electromotive force (EMF). In our exercise, we determine the standard cell potential to find out how many joules of electrical work can be expected from a cell, which aids in comparing it with the work potential of another energy source such as decaying Radium-226.