# AP Chemistry/Electrochemistry

## Finding Oxidation Numbers

Oxidation numbers are simply the charges of individual elements in each compound. To find the oxidation numbers of something, there are a few rules to keep in mind:

Any pure, 0 charge species has an oxidation number of 0

## Reduction and Oxidation

Example equation:

${\displaystyle {\ce {Cu (aq)^2+ + Zn (s) -> Cu (s) + Zn (aq)^2+}}}$

Reduction: When the ion/element gains electrons (${\displaystyle {\ce {Cu^2+}}}$  is reduced to ${\displaystyle {\ce {Cu}}}$ )

Oxidation:When the ion/element loses electrons (${\displaystyle {\ce {Zn}}}$  is oxidized to ${\displaystyle {\ce {Zn^2+}}}$ )

Reducing Agent:The ion/element that gives the electron to the other one (${\displaystyle {\ce {Zn}}}$  is the reducing agent to ${\displaystyle {\ce {Cu^2+}}}$ )

Oxidation Agent: The ion/element that takes the electron from the other (${\displaystyle {\ce {Cu^2+}}}$  is the oxidation agent to ${\displaystyle {\ce {Zn}}}$ )

### Standard Cell Potentials

In redox reactions, electrons are transferred. These transfers allow for there to be a Standard Cell Potential (measured in volts) between the two metals. As AP chemistry students, you should be able to calculate the Standard Cell Potential by using something called a Standard Reduction potential table. This gives you the standard reduction potentials for each element/compound, which is the amount of volts of electricity needed to turn 1 mole of ions to a reduced state (it could become another ion, or the pure form of the metal). To find the Standard Cell potential, you need to find the reduction potential of the element reduced, and the oxidation potential of the element oxidized. The oxidation potential is determined by changing the sign of the reduction potential found on the table (the reduction potential of ${\displaystyle {\ce {Sr+}}}$  is -4.10, the oxidation potential of ${\displaystyle {\ce {Sr}}}$  is 4.10). When you find the reduction potential of the element reduced, and the oxidation potential of the element oxidized, simply add the two values together to get the standard cell potential.

When looking at the galvanic cell, you need to be able to calculated how much mass of a certain metal is plated out in a given time. There are a few new units you need to know to do this:

Coulombs: Coulombs are the charge of 6.241×1018 electrons.

Amperes: Amperes translate to Coulombs per second. It measures the current passing trough the wire.

Faraday's Constant: This basically says that there are 96,500 Coulombs per one mole of electrons

## Galvanic Cell

Here is the galvanic cell:

### Parts of a cell

There are a few different components to a galvanic cell:

• Anode: Where the metal is oxidized, and electrons are released into the wire, causing the solid metal anode to turn into ions
• Cathode: Where the incoming electrons are received by the ions in the ion bath, causing reduction, and turning the ions to solid metal.
• Salt Bridge: Where the salt is dissolved in water, and each ion moves to either the anode or the cathode, balancing whatever charges need to be balanced. The anion moves to the anode, the cation moves to the cathode. The salt in the salt bridge must be a salt that, if the anion is bonded to the ion of either metal, the compound will still remain dissolved and will not solidify.
• Metal ion bath: where the metals that are oxidized go when they become ions. It is also where the ions that are reduced solidify and become solid metal
• Wire: Where the electrons pass from the anode to the cathode

### How it works

According to the Standard Reduction Potential Table, the galvanic cell with the higher reduction potential (or lower oxidation potential) will be the cathode, and the other metal will be the anode. The electrons in the anode are oxidized, passing the electrons through the wire, all the way to the cathode. This is how voltage is determined, by measuring the current of the electrons in the wire.

### Determining the mass of the metal produced

Given this information, we need to figure out how much mass is plated out in a galvanic cell, given the amount of Amps and the time. Lets looks at an example problem:

How many grams of mercury could be produced by electrolyzing a ${\displaystyle 1.0{\text{ M}}}$  ${\displaystyle {\ce {Hg(NO3)2}}}$  solution with a current of ${\displaystyle 2.00{\text{ A}}}$  for ${\displaystyle 3.00{\text{ h}}}$ ?

Here, we are given three important pieces of information:

1. ${\displaystyle 1.0{\text{ M}}}$  ${\displaystyle {\ce {Hg(NO3)2}}}$
2. The current is ${\displaystyle 2.00{\text{ A}}}$
3. The time taken is ${\displaystyle 3.00{\text{ h}}}$

The first piece of information tells us how to st up our net ionic half reaction. In these types of reactions, there are two metals that are going to be reacting, one oxidizing, one reducing. We really only need to worry about the reduction reaction in this case, because we are looking for solid grams of mercury, and reduction is where the solid mercury is created.

(NOTE: You might have a problem that says "plated out". This means the same thing as "produced".)

To set up our reduction half reaction, we need to know the oxidation value of the metal that is being reduced (basically when the metal becomes and ion, what its charge is). Transition metals have two or more oxidation states, but it is usually safe to go with the most common one (It's usually 2+). For Mercury, the oxidation state is 2+ in this case. Therefore, our reaction is as such:

${\displaystyle {\ce {Hg^2+ (aq) + 2e- -> Hg (s)}}}$

With this equation, we now know all the mole ratios we need to know. Let's begin setting up the conversion factors, by using the other two pieces of information.

${\displaystyle {\frac {3{\text{ hr}}}{1}}\times {\frac {60{\text{ mins}}}{1{\text{ hr}}}}\times {\frac {60{\text{ s}}}{1{\text{ min}}}}\times {\frac {2{\text{ C}}}{1{\text{ s}}}}\times {\frac {1{\text{ mol e}}^{-}}{96,500{\text{ C}}}}\times {\frac {1{\text{ mol Hg}}}{2{\text{ mol e}}^{-}}}\times {\frac {200.59{\text{ g}}}{1{\text{ mol Hg}}}}\approx 22.4{\text{ g}}}$

Since Amperes are ${\displaystyle {\text{C mol}}^{-1},}$  the hours needed to be converted into seconds so the units are able to cancel out. Then from there, Faraday's Constant was used to get the moles of electrons, and then that was multiplied by the mole ratio, and then by molar mass.

Know how to do problems like this, and also know how to go from grams plated out to time, etc.

### Electrolytic Cell

This is the opposite of a galvanic cell. In this kind of cell, a certain voltage is applied to a body of liquid, either the molten version of the salt, or the salt dissolved in water. This time one is adding energy to the system, meaning the standard cell potential will be negative. This current goes to the ions/molten salt and basically separates them out, giving you the pure version of whatever substance you are using. The "anode" is the metal that has the lower oxidation potential, the cathode with the lower reduction potential. This is mainly used to purify metals, and to extract pure substances from the salts.