Stoichiometry of Electrolytic Reactions — Faraday's Law

We've been studying voltaic and electrolytic cells. Let's refresh ourselves on where we've been. We started by defining a redox reaction. So, we know that losing electrons is oxidation and gaining electrons is reduction. Next, we learned to balance redox reactions. The most critical method was the half-reaction method. After balancing redox reactions, we studied the properties of electrochemical cells. So, we studied their voltage and current. We saw that we could manipulate these to do electrical work. As we moved into electrolysis, we learned that you could apply an external voltage. What was the result? We converted ions into their elemental states. For molten salts, we converted metal ions into solid or liquid metal. For water, we split it into hydrogen and oxygen gas. Eventually, we'll apply electrolysis to electroplating. We can deposit metals onto each other. But first, we need to revisit our dear friend, stoichiometry. With electrochemistry, stoichiometry takes a unique twist.

Stoichiometry Refresher

Do you need a quick stoichiometry refresher? I recommend reexamining the extended stoichiometry road map. But remember, the essence of stoichiometry is dimensional analysis. We are using constants to convert between units. So, the process is almost always the same. We always start by writing a balanced chemical equation. Next, we convert a compound's mass to moles with its molar mass. Then we use the mole ratios to convert between moles of reactant and moles of product. Finally, we can convert back to the mass of the reactant.

Faraday's Law

Electrochemistry provides a novel addition to stoichiometry problems. We often don't measure mass. Instead, we measure current and time. Do you remember the unit for the electric current? It's the ampere or amp. We calculate it as the charge per second. But what's generating the charge? In an electrochemical cell, it's the electrons! So, we can incorporate electrons into our stoichiometry problems.

What's the relationship between charge, current, and time? It's Faraday's law! We write Faraday's law mathematically as...

$I=\frac{Q}{t}$

$I$ is current in amperes.
$Q$ is the charge in Coulombs.
$t$ is the time in seconds.

But how do we relate Faraday's law to stoichiometry? We need to use Faraday's constant. Do you remember it? It's $F= 96,485\frac{C}{mol_{e^{-}}}$ . So, we have a way to use the moles of electrons!

Mass to Mass in Galvanic Cells

Before we explore using charge, let's do mass-to-mass stoichiometry with galvanic cells. How would it work with an experimental setup? You would measure the electrodes' masses before and after the reaction. But, let's study an example problem.

Let's say we made a galvanic cell with sodium and silver. If our sodium electrode lost 5.00 grams. How much silver metal precipitated from the solution?

Start by writing the half-reactions.

Anode: $Na_{(s)}\rightarrow Na^{+} + e^{-}$

Cathode: $Ag^{+} + e^{-} \rightarrow Ag_{(s)}$

Overall: $Na_{(s)} + Ag^{+} \rightarrow Ag_{(s)} + Na^{+}$

Convert the mass of sodium to moles of sodium. Remember, you find the molar mass on the periodic table.

$5.00g_{Na} \times \frac{1 mol_{Na}}{22.99g}=0.22mols$

Now, we use the mole ratio to convert moles of sodium to moles of silver. From our overall reaction, we see the mole ratio is 1:1.

$0.22mols_{Na}\times \frac{1 mol_{Ag}}{1mol_{Na}}=0.22mols_{Ag}$

Now, we can use silver's molar mass to convert moles to grams.

$0.22mols_{Ag}\times \frac{107.87g}{1mol_{Ag}}=23.46g$

Charge to Mass in Galvanic Cells

Now let's look at converting charge to mass. Let's say we made a galvanic cell from aluminum and copper. The cell generated a current of 5.00 amps for 1000.00 seconds. How much copper did we produce?

Our first step is to write our balanced reaction.

Anode: $Al_{(s)} \rightarrow Al_{(aq)}^{3+} + 3e^{-}$

Cathode: $Cu_{(aq)}^{2+} + 2e^{-} \rightarrow Cu_{(s)}$

Overall: $2Al_{(s)} + 3Cu_{(aq)} \rightarrow 2Al_{(aq)}^{3+} + 3Cu_{(s)}$

Now, we can calculate the charge using Faraday's law.

$Q=I\times t=5.00amp\times1000.00 seconds= 5000.00 C$

Then we can use Faraday's constant to convert to moles of electrons.

$5000.0C\times \frac{1 mol_{e^{-}}}{96,485C}=0.052 mols_{e^{-}}$

Now, we'll use the cathode half-reaction. With it, we can determine the moles of aluminum that oxidized.

$0.052 mols_{e^{-}}\times \frac{1mol_{Al}}{3mols_{e^{-}}}=0.017 mols_{Al}$

We can convert moles of aluminum to moles of copper using the mole ratio from our overall reaction.

$0.017 mols_{Al}\times\frac{2mols_{Cu}}{3mols_{Al}}=0.012mols_{Cu}$

Finally, we can convert moles of aluminum to mass.

$0.012mols_{Cu}\times26.98\frac{g}{mol}=0.31 g_{Cu}$

Charge to Mass in Electrolytic Cells

Finally, let's look at the stoichiometry of electrolytic cells. When we studied the electrolysis of molten salts, we looked at sodium chloride. So let's use it for our example. Let's say we run 10.00 amps through a solution of sodium chloride. The current is applied for 1 day. What mass of sodium metal did we produce.

We need to start by finding the charge. So, we'll use Faraday's law.

$Q=I\times t=10.00amps\times (1 day\times\frac{24hrs}{1day}\times\frac{60 min}{1hr}\times\frac{60s}{1 min})=865000.00C$

Now, we can convert our charge to moles of electrons with Faraday's constant.

$865000.00C\times\frac{1mol_{e^{-}}}{96,485C}=8.95mols_{e^{-}}$

Now, let's write our half-reaction.

$Na_{(aq)}^{+} + e^{-} \rightarrow Na_{(s)}$

From our half-reaction, we know the mole ratio is 1:1. Now we can convert moles of electrons to moles of sodium metal.

$8.95mols_{e^{-}}\times \frac{1 mol_{Na}}{1 mol_{e^{-}}}=8.95 mols_{Na}$

Finally, we can convert moles to grams.

$8.95 mols_{Na}\times\frac{22.99g}{1 mol_{Na}}=205.87g_{Na}$