Electrolytic Cells

Voltaic cells are electrochemical cells that use spontaneous(thermodynamically favorable) redox reactions to drive cell reactions. Electrolytic cells use an outside voltage source to drive a non-spontaneous redox reaction. However, the same rules still otherwise apply: oxidation is at the anode and reduction is at the cathode. An intuitive example of an electrolytic cell is a phone battery. It uses voltage from your charger to power itself, but won't spontaneously do so.

With reduction potentials in voltaic cells, you find positive cell potentials but for electrolytic cells, you find negative ones.

Molten electrolytic cells like the one below have an anion at the anode and cation at the cathode and the cell operates with a positive external voltage that is higher in magnitude than the negative cell potential.

Water can be electrolyzed and is often added with a strong acid to improve its electrical conductivity.

If a compound is dissolved in water, the more potentials between the solute half-reactions and the water half-reactions are at the electrodes. For example, let's take an aqueous sodium chloride electrolytic cell.

Here are some potential anode reactions:

2Cl^- → Cl₂ + 2e^- (+1.36 V)

2H₂O → O₂ + 4H⁺ + 4e^-(+1.23 V)

Out of these two, the first one with chloride will happen at the anode because its potential is more positive.

Here are some potential cathode reactions:

2H₂O +2e^-→ H₂ + 2OH^-(-0.83 V)

Na⁺ + e^- → Na(-2.71 V)

The first one will occur at the cathode because it is more positive. We can use this and the first one above to find a total cell potential and reaction then.

Quantitative Electrolysis

It's important to remember what an electrochemical cell is at its heart. It's a cell with a bunch of electrons flowing between two half-cells. The rate at which these electrons flow is called current, measured in amperes(amps), or coulombs/second.

Let's take this innocent half-reaction below and assume it happens at the cathode of an electrolytic cell:

Ag⁺+e^- → Ag

This reaction stoichiometrically tells us that one mole of Ag is produced for every 1 mole of electrons that flows into the reduction half-cell. If we knew the current of the electrons in the cell and how long they flowed, we could use quantitative electrolysis to figure out the mass of Ag produced.

The equation for current is below. If we know the cell's current and how long it's been running, we can multiply those two quantities to get the charge flowing through in coulombs.

If we get the charge in coulombs, we can divide by Faraday's constant, equal to the charge in coulombs per mole of electrons, to get moles of electrons.

Then, from there, it's all stoichiometry. We can convert moles of electrons to moles of Ag through the half-reaction above. From here, we multiply the moles of Ag to grams of Ag through Ag's molar mass. If you do this, you'll see all the units cancel out nicely to give you your mass.

This can be done the opposite way, too, if you take the outlined process and trace through it backwards.