# Electrolysis

Electrolysis is a process that uses electric current to do work on a chemical against its electric potential—in essence, the opposite of a galvanic cell.

A galvanic cell, a system where a chemical reaction produces energy as flow of electrons, produces electricity by allowing a redox reaction to happen spontaneously. In contrast, an electrolytic cell is an electrochemical cell that requires electrical energy to drive a nonspontaneous redox reaction. Electrolysis, then, is the process of using electric current to do work on a chemical cell against its electric potential. Electrolytic cells are used to split the components of an electrolyte into constituent ions.

In an electrolytic cell, as in a galvanic cell, reduction—gaining of electrons—takes place at the cathode. Oxidation—loss of electrons—takes place at the anode. However, because the flow of electrons is reversed compared to a galvanic cell, the signs of the electrodes are reversed as well. Therefore, in an electrolytic cell, the anode is positive while the cathode is negative. The sign of the cell potential is also reversed. An additional difference between galvanic and electrolytic cells is that the two half-cells are discrete, with a salt bridge connecting them, while in an electrolytic cell the electrodes are both immersed in a single molten electrolyte. The electrolyte must be molten in order for electrolysis to occur, because the ions of the electrolyte must be free to move and thus separate from one another.

Electrolysis has many applications. For example, electrolysis underlies electroplating, in which a substrate is covered with a thin film of metal. Electrolysis is also used to isolate nonmetals such as hydrogen, calcium, and fluorine. In order to apply electrolysis in this manner, engineers must know precisely how much electricity to pass through the electrolytic cell. They do so by using the Faraday constant, the charge of one mole of electrons (96,485 C/mol). The charge Q of n moles of electrons can be given as $Q=nF$, where n is the number of moles of electrons and F is the Faraday constant. The charge Q can also be given as the strength of the current for the time it takes the current to pass through the electrolyte. This is written as $Q=I(t)$, where I is the current in amperes and t is the time in seconds. Combining the equations for charge (Q) produces the equation $I(t)=nF$. Engineers can use this equation to determine precisely how to apply current over what time to get the desired result.

Step-By-Step Example
Calculation of Time Required for Electroplating
Calculate how long (in hours) a current of 5.0 amperes must be applied to molten calcium chloride (CaCl2) to electroplate 60.5 g of calcium.
Step 1
First, write the half-reactions.
$\begin{gathered}{\rm{Ca}}^{2+}+2{\rm{e}}^-\rightarrow{\rm{Ca}}\\2{\rm{Cl}}^-\rightarrow{\rm{Cl}}_2+2{\rm{e}}^-\end{gathered}$
Step 2
Next, use the molar mass of calcium to determine the number of moles in 60.5 g, carrying extra digits for the intermediate calculation.
$\frac{60.5\;{\rm{g}}}{40.08\;{\rm{g/mol}}}=1.509\;{\rm{mol\;Ca}}$
Step 3
Multiply the number of moles of calcium by the number of electrons released, as shown by the half-reaction equation.
$(1.509\;{\rm{mol\;Ca}})\left(\frac{2\;{\rm{mol\;e}}^-}{1\;{\rm{mol\;Ca}}}\right)=\;3.018\;{\rm{mol\;e}}^-$
Step 4
Use the moles of electrons to determine the electric charge in coulombs (C).
$(3.018\;{\rm{mol\;e^-}})\left(\frac{96\rm{,}485\;{\rm{C}}}{1\;{\rm{mol\;e}}^-}\right)=2.912\times10^5\;{\rm{C}}$
Step 5
Rearrange the equation relating coulombs and amperes to solve for seconds.
$\begin{gathered}{\text{coulombs }}{\rm{(C)}}={\text{amperes }}{\rm{(A)}}\times{\text{seconds }}{\rm{(s)}}\\{\text{seconds }}{\rm{(s)}}=\frac{{\text{coulombs }}{\rm{(C)}}}{{\text{amperes }}{\rm{(A)}}}\end{gathered}$
Solution
Use this equation to determine the time required for the electroplating, converting seconds to hours.
\begin{aligned}{\text{time for electroplating}}&=\left(\frac{\text{charge in coulombs}}{\text{current in amperes}}\right)\!\left(\frac{1\;{\rm{h}}}{3{\rm{,}}600\;{\rm{s}}}\right)\\&=\left(\frac{2.912\times10^5\;{\rm{C}}}{5.0\;{\rm{A}}}\right)\!\left(\frac{1\;{\rm{h}}}{3\rm{,}600\;{\rm{s}}}\right)\\&=16\;{\rm{h}}\end{aligned}
Importantly, electrolytic cells do not always function the way calculations describe they will. The difference between a half-cell's calculated reduction potential and its empirically observed reduction potential is its overpotential. Overpotential indicates the efficiency of the half-cell. Some energy put into the cell is lost as heat, resulting in half-cells that are less than 100 percent efficient. If the efficiency of the cell is known, the calculations for it are much more accurate.